exec_outcome stringclasses 1 value | code_uid stringlengths 32 32 | file_name stringclasses 111 values | prob_desc_created_at stringlengths 10 10 | prob_desc_description stringlengths 63 3.8k | prob_desc_memory_limit stringclasses 18 values | source_code stringlengths 117 65.5k | lang_cluster stringclasses 1 value | prob_desc_sample_inputs stringlengths 2 802 | prob_desc_time_limit stringclasses 27 values | prob_desc_sample_outputs stringlengths 2 796 | prob_desc_notes stringlengths 4 3k ⌀ | lang stringclasses 5 values | prob_desc_input_from stringclasses 3 values | tags listlengths 0 11 | src_uid stringlengths 32 32 | prob_desc_input_spec stringlengths 28 2.37k ⌀ | difficulty int64 -1 3.5k ⌀ | prob_desc_output_spec stringlengths 17 1.47k ⌀ | prob_desc_output_to stringclasses 3 values | hidden_unit_tests stringclasses 1 value |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
PASSED | d15715cb6e8a2833974a0c47f07598df | train_108.jsonl | 1650722700 | You are given an array $$$a$$$ of length $$$n$$$. We define the equality of the array as the number of indices $$$1 \le i \le n - 1$$$ such that $$$a_i = a_{i + 1}$$$. We are allowed to do the following operation: Select two integers $$$i$$$ and $$$x$$$ such that $$$1 \le i \le n - 1$$$ and $$$1 \le x \le 10^9$$$. Then, set $$$a_i$$$ and $$$a_{i + 1}$$$ to be equal to $$$x$$$. Find the minimum number of operations needed such that the equality of the array is less than or equal to $$$1$$$. | 256 megabytes | //package Java3rdyearPracticals;
import java.util.*;
public class Main {
//static int mod=998244353;
public static void main(String[] args)
{
// write your code here
Scanner sc = new Scanner(System.in);
int t = sc.nextInt();
for(int k=0;k<t;k++)
{
// HashMap<Integer,Integer> m=new HashMap<>();
int n=sc.nextInt();
/* char[] a=Integer.toString(n).toCharArray();
if((a.length==2 && a[1]-a[0]>=0))
{
System.out.println(a[1]);
}
else if(a.length==2 && a[1]-a[0]<0) {
System.out.println(a[1]);
}
else
{
Arrays.sort(a);
System.out.println(a[0]);
}
*/
//int q=sc.nextInt();
/*First think about when array contains 3 inverion is there any benefit to take this array ---No*/
//Question states that largest number of subarrays which means only one inversion is more benificial
//Maximum inversion count is possible either 1 or 2 and if 2 that means even
int a[] =new int[n];
for(int j=0;j<n;j++)
{
a[j]=sc.nextInt();
}
int ans=0,min=Integer.MAX_VALUE,max=0;
for(int i=0;i<n-1;i++)
{
if(a[i]==a[i+1] && min==Integer.MAX_VALUE)
{
min=Math.min(min,i);
// i++;
}
else if(a[i]==a[i+1])
{
max=Math.max(i,max);
//i++;
}
}
//T.C:- [1,1,1,1]
if(max==0 || min==max)
{
System.out.println("0");
}
else if((max-min)==1)
{
System.out.println("1");
}
else
{
System.out.println(max-min-1);
}
}
/*
long sum=Arrays.stream(a).sum();
for(int j=0;j<q;j++)
{
int type=sc.nextInt();
if(type==1)
{
int p_i=sc.nextInt();
int r=sc.nextInt();
//System.out.println("r is" + r);
//a[p_i-1]=r;
// System.out.println(a[p_i-1]);
System.out.println(r+(sum-a[p_i-1]));
sum=r+sum-a[p_i-1];
a[p_i-1]=r;
}
else
{
int r=sc.nextInt();
sum=r*(n);
System.out.println(sum);
Arrays.fill(a,r);
// System.out.println(Arrays.toString(a));
}
}
*/
}
}
/*static long fact(int n)
{
long f=1;
for(int k=1;k<=n;k++){
f=((f)%mod*k);
}
// System.out.println(f);
return f;
}
long gcd(long a, long b)
{
if (a == 0)
return b;
return gcd(b % a, a);
}
*/
| Java | ["4\n\n5\n\n1 1 1 1 1\n\n5\n\n2 1 1 1 2\n\n6\n\n1 1 2 3 3 4\n\n6\n\n1 2 1 4 5 4"] | 1 second | ["2\n1\n2\n0"] | NoteIn the first test case, we can select $$$i=2$$$ and $$$x=2$$$ to form $$$[1, 2, 2, 1, 1]$$$. Then, we can select $$$i=3$$$ and $$$x=3$$$ to form $$$[1, 2, 3, 3, 1]$$$.In the second test case, we can select $$$i=3$$$ and $$$x=100$$$ to form $$$[2, 1, 100, 100, 2]$$$. | Java 11 | standard input | [
"constructive algorithms",
"greedy",
"implementation"
] | a91be662101762bcb2c58b8db9ff61e0 | Each test contains multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains an integer $$$n$$$ ($$$2 \le n \le 2 \cdot 10 ^ 5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10^9$$$) — elements of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10 ^ 5$$$ | 1,100 | For each test case, print the minimum number of operations needed. | standard output | |
PASSED | 45ac1cf09efaa38a619fd29d6fb1ddb4 | train_108.jsonl | 1650722700 | You are given an array $$$a$$$ of length $$$n$$$. We define the equality of the array as the number of indices $$$1 \le i \le n - 1$$$ such that $$$a_i = a_{i + 1}$$$. We are allowed to do the following operation: Select two integers $$$i$$$ and $$$x$$$ such that $$$1 \le i \le n - 1$$$ and $$$1 \le x \le 10^9$$$. Then, set $$$a_i$$$ and $$$a_{i + 1}$$$ to be equal to $$$x$$$. Find the minimum number of operations needed such that the equality of the array is less than or equal to $$$1$$$. | 256 megabytes | import java.io.BufferedInputStream;
import java.util.Scanner;
public class Main {
public static void main(String[] args) {
int a[] = new int[250000];
Scanner sc = new Scanner(new BufferedInputStream(System.in));
int t = sc.nextInt();
while (t --> 0) {
int n = sc.nextInt();
for (int i = 0; i < n; i++) {
a[i] = sc.nextInt();
}
int l = n,r = -1;
for (int i = 0; i < n - 1; i++) {
if(a[i] == a[i+1]){
l = Math.min(i,l);
r = Math.max(r,i);
}
}
if(l>=r) System.out.println(0);
else System.out.println(Math.max(1,r-l-1));
}
}
} | Java | ["4\n\n5\n\n1 1 1 1 1\n\n5\n\n2 1 1 1 2\n\n6\n\n1 1 2 3 3 4\n\n6\n\n1 2 1 4 5 4"] | 1 second | ["2\n1\n2\n0"] | NoteIn the first test case, we can select $$$i=2$$$ and $$$x=2$$$ to form $$$[1, 2, 2, 1, 1]$$$. Then, we can select $$$i=3$$$ and $$$x=3$$$ to form $$$[1, 2, 3, 3, 1]$$$.In the second test case, we can select $$$i=3$$$ and $$$x=100$$$ to form $$$[2, 1, 100, 100, 2]$$$. | Java 11 | standard input | [
"constructive algorithms",
"greedy",
"implementation"
] | a91be662101762bcb2c58b8db9ff61e0 | Each test contains multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains an integer $$$n$$$ ($$$2 \le n \le 2 \cdot 10 ^ 5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10^9$$$) — elements of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10 ^ 5$$$ | 1,100 | For each test case, print the minimum number of operations needed. | standard output | |
PASSED | 0997808c869592b94f384064a9aca491 | train_108.jsonl | 1650722700 | You are given an array $$$a$$$ of length $$$n$$$. We define the equality of the array as the number of indices $$$1 \le i \le n - 1$$$ such that $$$a_i = a_{i + 1}$$$. We are allowed to do the following operation: Select two integers $$$i$$$ and $$$x$$$ such that $$$1 \le i \le n - 1$$$ and $$$1 \le x \le 10^9$$$. Then, set $$$a_i$$$ and $$$a_{i + 1}$$$ to be equal to $$$x$$$. Find the minimum number of operations needed such that the equality of the array is less than or equal to $$$1$$$. | 256 megabytes | import java.util.*;
import java.lang.*;
import java.io.*;
public class Main
{
static long mod = (int)1e9+7;
static PrintWriter out=new PrintWriter(new BufferedOutputStream(System.out));
public static void main (String[] args) throws java.lang.Exception
{
FastReader sc =new FastReader();
int t=sc.nextInt();
// int t=1;
O : while(t-->0)
{
int n=sc.nextInt();
int a[]=sc.readArray(n);
int min=Integer.MAX_VALUE,max=Integer.MIN_VALUE;
for(int i=0;i<n-1;i++){
if(a[i]==a[i+1]){
min=Math.min(min,i);
max=Math.max(max,i);
}
}
if(min==Integer.MAX_VALUE || min==max){
out.println(0);
continue;
}
out.println((max-min)<3?1:max-min-1);
}
out.flush();
}
static void printN()
{
System.out.println("NO");
}
static void printY()
{
System.out.println("YES");
}
static int findfrequencies(int a[],int n)
{
int count=0;
for(int i=0;i<a.length;i++)
{
if(a[i]==n)
{
count++;
}
}
return count;
}
static class FastReader {
BufferedReader br;
StringTokenizer st;
public FastReader()
{
br = new BufferedReader(
new InputStreamReader(System.in));
}
String next()
{
while (st == null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(br.readLine());
}
catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt() { return Integer.parseInt(next()); }
long nextLong() { return Long.parseLong(next()); }
double nextDouble()
{
return Double.parseDouble(next());
}
float nextFloat()
{
return Float.parseFloat(next());
}
String nextLine()
{
String str = "";
try {
str = br.readLine();
}
catch (IOException e) {
e.printStackTrace();
}
return str;
}
int[] readArray(int n) {
int[] a=new int[n];
for (int i=0; i<n; i++) a[i]=nextInt();
return a;
}
long[] readArrayLong(int n) {
long[] a=new long[n];
for (int i=0; i<n; i++) a[i]=nextLong();
return a;
}
}
public static int[] radixSort2(int[] a)
{
int n = a.length;
int[] c0 = new int[0x101];
int[] c1 = new int[0x101];
int[] c2 = new int[0x101];
int[] c3 = new int[0x101];
for(int v : a) {
c0[(v&0xff)+1]++;
c1[(v>>>8&0xff)+1]++;
c2[(v>>>16&0xff)+1]++;
c3[(v>>>24^0x80)+1]++;
}
for(int i = 0;i < 0xff;i++) {
c0[i+1] += c0[i];
c1[i+1] += c1[i];
c2[i+1] += c2[i];
c3[i+1] += c3[i];
}
int[] t = new int[n];
for(int v : a)t[c0[v&0xff]++] = v;
for(int v : t)a[c1[v>>>8&0xff]++] = v;
for(int v : a)t[c2[v>>>16&0xff]++] = v;
for(int v : t)a[c3[v>>>24^0x80]++] = v;
return a;
}
static int[] EvenOddArragement(int a[])
{
ArrayList<Integer> list=new ArrayList<>();
for(int i=0;i<a.length;i++)
{
if(a[i]%2==0)
{
list.add(a[i]);
}
}
for(int i=0;i<a.length;i++)
{
if(a[i]%2!=0)
{
list.add(a[i]);
}
}
for(int i=0;i<a.length;i++)
{
a[i]=list.get(i);
}
return a;
}
static int gcd(int a, int b) {
while (b != 0) {
int t = a;
a = b;
b = t % b;
}
return a;
}
static int lcm(int a, int b)
{
return (a / gcd(a, b)) * b;
}
public static HashMap<Integer, Integer> sortByValue(HashMap<Integer, Integer> hm)
{
// Create a list from elements of HashMap
List<Map.Entry<Integer, Integer> > list =
new LinkedList<Map.Entry<Integer, Integer> >(hm.entrySet());
// Sort the list
Collections.sort(list, new Comparator<Map.Entry<Integer, Integer> >() {
public int compare(Map.Entry<Integer, Integer> o1,
Map.Entry<Integer, Integer> o2)
{
return (o1.getValue()).compareTo(o2.getValue());
}
});
// put data from sorted list to hashmap
HashMap<Integer, Integer> temp = new LinkedHashMap<Integer, Integer>();
for (Map.Entry<Integer, Integer> aa : list) {
temp.put(aa.getKey(), aa.getValue());
}
return temp;
}
static int DigitSum(int n)
{
int r=0,sum=0;
while(n>=0)
{
r=n%10;
sum=sum+r;
n=n/10;
}
return sum;
}
static boolean checkPerfectSquare(int number)
{
double sqrt=Math.sqrt(number);
return ((sqrt - Math.floor(sqrt)) == 0);
}
static boolean isPowerOfTwo(int n)
{
if(n==0)
return false;
return (int)(Math.ceil((Math.log(n) / Math.log(2)))) == (int)(Math.floor(((Math.log(n) / Math.log(2)))));
}
static boolean isPrime2(int n)
{
if (n <= 1)
{
return false;
}
if (n == 2)
{
return true;
}
if (n % 2 == 0)
{
return false;
}
for (int i = 3; i <= Math.sqrt(n) + 1; i = i + 2)
{
if (n % i == 0)
{
return false;
}
}
return true;
}
static String minLexRotation(String str)
{
int n = str.length();
String arr[] = new String[n];
String concat = str + str;
for(int i=0;i<n;i++)
{
arr[i] = concat.substring(i, i + n);
}
Arrays.sort(arr);
return arr[0];
}
static String maxLexRotation(String str)
{
int n = str.length();
String arr[] = new String[n];
String concat = str + str;
for (int i = 0; i < n; i++)
{
arr[i] = concat.substring(i, i + n);
}
Arrays.sort(arr);
return arr[arr.length-1];
}
static class P implements Comparable<P> {
int i, j;
public P(int i, int j) {
this.i=i;
this.j=j;
}
public int compareTo(P o) {
return Integer.compare(i, o.i);
}
}
static class pair{
int i,j;
pair(int x,int y){
i=x;
j=y;
}
}
static int binary_search(int a[],int value)
{
int start=0;
int end=a.length-1;
int mid=start+(end-start)/2;
while(start<=end)
{
if(a[mid]==value)
{
return mid;
}
if(a[mid]>value)
{
end=mid-1;
}
else
{
start=mid+1;
}
mid=start+(end-start)/2;
}
return -1;
}
} | Java | ["4\n\n5\n\n1 1 1 1 1\n\n5\n\n2 1 1 1 2\n\n6\n\n1 1 2 3 3 4\n\n6\n\n1 2 1 4 5 4"] | 1 second | ["2\n1\n2\n0"] | NoteIn the first test case, we can select $$$i=2$$$ and $$$x=2$$$ to form $$$[1, 2, 2, 1, 1]$$$. Then, we can select $$$i=3$$$ and $$$x=3$$$ to form $$$[1, 2, 3, 3, 1]$$$.In the second test case, we can select $$$i=3$$$ and $$$x=100$$$ to form $$$[2, 1, 100, 100, 2]$$$. | Java 11 | standard input | [
"constructive algorithms",
"greedy",
"implementation"
] | a91be662101762bcb2c58b8db9ff61e0 | Each test contains multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains an integer $$$n$$$ ($$$2 \le n \le 2 \cdot 10 ^ 5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10^9$$$) — elements of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10 ^ 5$$$ | 1,100 | For each test case, print the minimum number of operations needed. | standard output | |
PASSED | 480184c1f14a80ad52819c4140199108 | train_108.jsonl | 1650722700 | You are given an array $$$a$$$ of length $$$n$$$. We define the equality of the array as the number of indices $$$1 \le i \le n - 1$$$ such that $$$a_i = a_{i + 1}$$$. We are allowed to do the following operation: Select two integers $$$i$$$ and $$$x$$$ such that $$$1 \le i \le n - 1$$$ and $$$1 \le x \le 10^9$$$. Then, set $$$a_i$$$ and $$$a_{i + 1}$$$ to be equal to $$$x$$$. Find the minimum number of operations needed such that the equality of the array is less than or equal to $$$1$$$. | 256 megabytes | import java.io.*;
import java.util.*;
public class Codeforces
{
public static void main(String args[])throws Exception
{
BufferedReader bu=new BufferedReader(new InputStreamReader(System.in));
StringBuilder sb=new StringBuilder();
int t=Integer.parseInt(bu.readLine());
while(t-->0)
{
int n=Integer.parseInt(bu.readLine());
int i,a[]=new int[n],b[]=new int[n];
String s[]=bu.readLine().split(" ");
for(i=0;i<n;i++) b[i]=a[i]=Integer.parseInt(s[i]);
int ans,l[]=new int[n],r[]=new int[n];
for(i=1;i<n;i++)
{
l[i]=Math.max(l[i],l[i-1]);
if(i+1<n && a[i]==a[i-1])
{
a[i-1]=a[i]=a[i+1]=-1;
l[i+1]=l[i]+1;
}
}
for(i=n-2;i>=0;i--)
{
r[i]=Math.max(r[i],r[i+1]);
if(i-1>=0 && b[i]==b[i+1])
{
b[i+1]=b[i]=b[i-1]=-1;
r[i-1]=r[i]+1;
}
}
ans=l[n-1];
for(i=0;i+1<n;i++)
{
int cur=l[i]+r[i+1],add=0;
if(i-1>=0 && a[i]==a[i-1]) add++;
if(i+2<n && b[i+2]==b[i+1]) add++;
cur+=Math.min(add,1);
ans=Math.min(ans,cur);
}
sb.append(ans+"\n");
}
System.out.print(sb);
}
} | Java | ["4\n\n5\n\n1 1 1 1 1\n\n5\n\n2 1 1 1 2\n\n6\n\n1 1 2 3 3 4\n\n6\n\n1 2 1 4 5 4"] | 1 second | ["2\n1\n2\n0"] | NoteIn the first test case, we can select $$$i=2$$$ and $$$x=2$$$ to form $$$[1, 2, 2, 1, 1]$$$. Then, we can select $$$i=3$$$ and $$$x=3$$$ to form $$$[1, 2, 3, 3, 1]$$$.In the second test case, we can select $$$i=3$$$ and $$$x=100$$$ to form $$$[2, 1, 100, 100, 2]$$$. | Java 11 | standard input | [
"constructive algorithms",
"greedy",
"implementation"
] | a91be662101762bcb2c58b8db9ff61e0 | Each test contains multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains an integer $$$n$$$ ($$$2 \le n \le 2 \cdot 10 ^ 5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10^9$$$) — elements of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10 ^ 5$$$ | 1,100 | For each test case, print the minimum number of operations needed. | standard output | |
PASSED | 075589d514f37c5878f107517063c578 | train_108.jsonl | 1650722700 | You are given an array $$$a$$$ of length $$$n$$$. We define the equality of the array as the number of indices $$$1 \le i \le n - 1$$$ such that $$$a_i = a_{i + 1}$$$. We are allowed to do the following operation: Select two integers $$$i$$$ and $$$x$$$ such that $$$1 \le i \le n - 1$$$ and $$$1 \le x \le 10^9$$$. Then, set $$$a_i$$$ and $$$a_{i + 1}$$$ to be equal to $$$x$$$. Find the minimum number of operations needed such that the equality of the array is less than or equal to $$$1$$$. | 256 megabytes | import java.io.*;
import java.util.*;
/**
*
* @author eslam
*/
public class IceCave {
static class FastReader {
BufferedReader br;
StringTokenizer st;
public FastReader() {
br = new BufferedReader(new InputStreamReader(System.in));
}
String next() {
while (st == null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt() {
return Integer.parseInt(next());
}
long nextLong() {
return Long.parseLong(next());
}
double nextDouble() {
return Double.parseDouble(next());
}
String nextLine() {
String str = "";
try {
str = br.readLine();
} catch (IOException e) {
e.printStackTrace();
}
return str;
}
}
static FastReader input = new FastReader();
static BufferedWriter log = new BufferedWriter(new OutputStreamWriter(System.out));
public static void main(String[] args) throws IOException {
int t = input.nextInt();
loop:
while (t-- > 0) {
int n = input.nextInt();
int a[] = new int[n];
int ans = 0;
int ba = -1;
int l = 0;
read(a);
for (int i = 0; i < n - 1; i++) {
if (a[i] == a[i + 1]) {
l++;
}
}
if (l < 2) {
log.write("0\n");
continue;
}
for (int i = 0; i < n - 1; i++) {
if (a[i] == a[i + 1]) {
if (ba == -1) {
ba = i + 2;
i += 2;
ans++;
} else {
ans += (i - ba);
ba = i;
}
}
}
log.write(ans + "\n");
}
log.flush();
}
static long mod(long a, long b) {
long r = a % b;
return r < 0 ? r + b : r;
}
public static long binaryToDecimal(String w) {
long r = 0;
long l = 0;
for (int i = w.length() - 1; i > -1; i--) {
long x = (w.charAt(i) - '0') * (long) Math.pow(2, l);
r = r + x;
l++;
}
return r;
}
public static String decimalToBinary(long n) {
String w = "";
while (n > 0) {
w = n % 2 + w;
n /= 2;
}
return w;
}
public static boolean isSorted(int[] a) {
for (int i = 0; i < a.length - 1; i++) {
if (a[i] > a[i + 1]) {
return false;
}
}
return true;
}
public static void print(int a[]) throws IOException {
for (int i = 0; i < a.length; i++) {
log.write(a[i] + " ");
}
log.write("\n");
}
public static void read(int[] a) {
for (int i = 0; i < a.length; i++) {
a[i] = input.nextInt();
}
}
static class pair {
int x;
int y;
public pair(int x, int y) {
this.x = x;
this.y = y;
}
@Override
public String toString() {
return x + " " + y;
}
}
public static long LCM(long x, long y) {
return (x * y) / GCD(x, y);
}
public static long GCD(long x, long y) {
if (y == 0) {
return x;
}
return GCD(y, x % y);
}
}
| Java | ["4\n\n5\n\n1 1 1 1 1\n\n5\n\n2 1 1 1 2\n\n6\n\n1 1 2 3 3 4\n\n6\n\n1 2 1 4 5 4"] | 1 second | ["2\n1\n2\n0"] | NoteIn the first test case, we can select $$$i=2$$$ and $$$x=2$$$ to form $$$[1, 2, 2, 1, 1]$$$. Then, we can select $$$i=3$$$ and $$$x=3$$$ to form $$$[1, 2, 3, 3, 1]$$$.In the second test case, we can select $$$i=3$$$ and $$$x=100$$$ to form $$$[2, 1, 100, 100, 2]$$$. | Java 11 | standard input | [
"constructive algorithms",
"greedy",
"implementation"
] | a91be662101762bcb2c58b8db9ff61e0 | Each test contains multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains an integer $$$n$$$ ($$$2 \le n \le 2 \cdot 10 ^ 5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10^9$$$) — elements of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10 ^ 5$$$ | 1,100 | For each test case, print the minimum number of operations needed. | standard output | |
PASSED | 1855468aee7f3e4feb35b1dcc4eb47b5 | train_108.jsonl | 1650722700 | You are given an array $$$a$$$ of length $$$n$$$. We define the equality of the array as the number of indices $$$1 \le i \le n - 1$$$ such that $$$a_i = a_{i + 1}$$$. We are allowed to do the following operation: Select two integers $$$i$$$ and $$$x$$$ such that $$$1 \le i \le n - 1$$$ and $$$1 \le x \le 10^9$$$. Then, set $$$a_i$$$ and $$$a_{i + 1}$$$ to be equal to $$$x$$$. Find the minimum number of operations needed such that the equality of the array is less than or equal to $$$1$$$. | 256 megabytes | import java.io.*;
import java.util.*;
public class Mainn {
static long M = (long) (1e9 + 7);
static class FastReader {
BufferedReader br;
StringTokenizer st;
public FastReader() {
br = new BufferedReader(
new InputStreamReader(System.in));
}
String next() {
while (st == null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt() {
return Integer.parseInt(next());
}
long nextLong() {
return Long.parseLong(next());
}
double nextDouble() {
return Double.parseDouble(next());
}
String nextLine() {
String str = "";
try {
if (st.hasMoreTokens()) {
str = st.nextToken("\n");
} else {
str = br.readLine();
}
} catch (IOException e) {
e.printStackTrace();
}
return str;
}
}
public static void heapify(int arr[], int n, int i) {
int largest = i;
int l = 2 * i + 1;
int r = 2 * i + 2;
if (l < n && arr[l] > arr[largest])
largest = l;
if (r < n && arr[r] > arr[largest])
largest = r;
if (largest != i) {
int swap = arr[i];
arr[i] = arr[largest];
arr[largest] = swap;
heapify(arr, n, largest);
}
}
public static void swap(int[] a, long i, long j) {
long temp = a[(int) i];
a[(int) i] = a[(int) j];
a[(int) j] = (int) temp;
}
public static long gcd(long a, long b) {
if (b == 0)
return a;
return gcd(b, a % b);
}
public static int lcm(int a, int b) {
return (int) (a * b / gcd(a, b));
}
public static String sortString(String inputString) {
// Converting input string to character array
char[] tempArray = inputString.toCharArray();
// Sorting temp array using
Arrays.sort(tempArray);
// Returning new sorted string
return new String(tempArray);
}
static boolean isSquare(int n) {
int v = (int) Math.sqrt(n);
return v * v == n;
}
static boolean PowerOfTwo(int n) {
if (n == 0) return false;
return (int) (Math.ceil((Math.log(n) / Math.log(2)))) ==
(int) (Math.floor(((Math.log(n) / Math.log(2)))));
}
static int power(long a, long b) {
long res = 1;
while (b > 0) {
if (b % 2 == 1) {
res = (res * a) % M;
}
a = ((a * a) % M);
b = b / 2;
}
return (int) res;
}
public static boolean isPrime(int n) {
for (int i = 2; i * i <= n; i++)
if (n % i == 0) {
return false;
}
return true;
}
static int pown(long n) {
if (n == 0)
return 1;
if (n == 1)
return 1;
if ((n & (-n)) == n)
return 0;
return 1;
}
static long computeXOR(long n) {
// If n is a multiple of 4
if (n % 4 == 0)
return n;
// If n%4 gives remainder 1
if (n % 4 == 1)
return 1;
// If n%4 gives remainder 2
if (n % 4 == 2)
return n + 1;
// If n%4 gives remainder 3
return 0;
}
static long binaryToInteger(String binary) {
char[] numbers = binary.toCharArray();
long result = 0;
for (int i = numbers.length - 1; i >= 0; i--)
if (numbers[i] == '1')
result += Math.pow(2, (numbers.length - i - 1));
return result;
}
static String reverseString(String str) {
char ch[] = str.toCharArray();
String rev = "";
for (int i = ch.length - 1; i >= 0; i--) {
rev += ch[i];
}
return rev;
}
static int countFreq(int[] arr, int n) {
HashMap<Integer, Integer> map = new HashMap<>();
int x = 0;
for (int i = 0; i < n; i++) {
map.put(arr[i], map.getOrDefault(arr[i], 0) + 1);
}
for (int i = 0; i < n; i++) {
if (map.get(arr[i]) == 1)
x++;
}
return x;
}
static boolean symm(String str) {
if (str.length() == 1 || str.length() == 0)
return true;
if (str.substring(0, (str.length() / 2)).equals(str.substring(str.length() / 2)))
return symm(str.substring(0, (str.length() / 2)));
return false;
}
static void reverse(int[] arr, int l, int r) {
int d = (r - l + 1) / 2;
for (int i = 0; i < d; i++) {
int t = arr[l + i];
arr[l + i] = arr[r - i];
arr[r - i] = t;
}
// print array here
}
static int f(int x) {
x++;
while (x % 10 == 0) {
x /= 10;
}
return x;
}
public static boolean isFair(long x) {
long n = x;
while (n != 0) {
long r = (long) n % 10;
if (r != 0 && x % r != 0)
return false;
n = n / 10;
}
return true;
}
static void sort(String[] s, int n) {
for (int i = 1; i < n; i++) {
String temp = s[i];
// Insert s[j] at its correct position
int j = i - 1;
while (j >= 0 && temp.length() < s[j].length()) {
s[j + 1] = s[j];
j--;
}
s[j + 1] = temp;
}
}
static boolean sign(int x) {
if (x > 0)
return true;
else
return false;
}
static int sqr(int n) {
double x = Math.sqrt(n);
if ((int) x == x)
return (int) x;
else
return (int) (x + 1);
}
static class Pair implements Comparable<Pair> {
int start, end;
public Pair(int x, int y) {
this.start = x;
this.end = y;
}
@Override
public int compareTo(Pair p) {
if (start == p.start) {
return end - p.end;
}
return start - p.start;
}
}
static int set_bits_count(int num) {
int count = 0;
while (num > 0) {
num &= (num - 1);
count++;
}
return count;
}
static long factorial(long n) {
if (n == 0)
return 1;
else
return (n * factorial(n - 1));
}
public static void main(String[] args) throws Exception {
FastReader sc = new FastReader();
BufferedWriter output = new BufferedWriter(new OutputStreamWriter(System.out));
int t=sc.nextInt();
while(t--!=0){
int n = sc.nextInt();
int[] arr = new int[n];
for (int i = 0; i < n; i++) {
arr[i] = sc.nextInt();
}
int min = -1;
int max = -1;
for (int i = 1; i < n; i++) {
if(arr[i]==arr[i-1]) {
if(min==-1)
min=i;
max=i;
}
}
if(min==max)
System.out.println(0);
else
System.out.println(Math.max(1,max-min-1));
}
}
} | Java | ["4\n\n5\n\n1 1 1 1 1\n\n5\n\n2 1 1 1 2\n\n6\n\n1 1 2 3 3 4\n\n6\n\n1 2 1 4 5 4"] | 1 second | ["2\n1\n2\n0"] | NoteIn the first test case, we can select $$$i=2$$$ and $$$x=2$$$ to form $$$[1, 2, 2, 1, 1]$$$. Then, we can select $$$i=3$$$ and $$$x=3$$$ to form $$$[1, 2, 3, 3, 1]$$$.In the second test case, we can select $$$i=3$$$ and $$$x=100$$$ to form $$$[2, 1, 100, 100, 2]$$$. | Java 11 | standard input | [
"constructive algorithms",
"greedy",
"implementation"
] | a91be662101762bcb2c58b8db9ff61e0 | Each test contains multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains an integer $$$n$$$ ($$$2 \le n \le 2 \cdot 10 ^ 5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10^9$$$) — elements of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10 ^ 5$$$ | 1,100 | For each test case, print the minimum number of operations needed. | standard output | |
PASSED | f6cd2cc198fc2b1d303577ef70bddd64 | train_108.jsonl | 1650722700 | You are given an array $$$a$$$ of length $$$n$$$. We define the equality of the array as the number of indices $$$1 \le i \le n - 1$$$ such that $$$a_i = a_{i + 1}$$$. We are allowed to do the following operation: Select two integers $$$i$$$ and $$$x$$$ such that $$$1 \le i \le n - 1$$$ and $$$1 \le x \le 10^9$$$. Then, set $$$a_i$$$ and $$$a_{i + 1}$$$ to be equal to $$$x$$$. Find the minimum number of operations needed such that the equality of the array is less than or equal to $$$1$$$. | 256 megabytes | import java.io.*;
import java.math.BigInteger;
import java.util.*;
public class MainClass {
static FastScanner fs;
static FastWriter fw;
static boolean checkOnlineJudge = System.getProperty("ONLINE_JUDGE") == null;
private static final int[][] kdir = new int[][]{{-1, 2}, {-2, 1}, {-2, -1}, {-1, -2}, {1, -2}, {2, -1}, {2, 1}, {1, 2}};
private static final int[][] dir = new int[][]{{-1, 0}, {1, 0}, {0, 1}, {0, -1}};
private static final int iMax = (int) (1e9 + 100), iMin = (int) (-1e9 - 100);
private static final long lMax = (long) (1e18 + 100), lMin = (long) (-1e18 - 100);
private static final int mod1 = (int) (1e9 + 7);
private static final int mod2 = 998244353;
public static void main(String[] args) throws IOException {
fs = new FastScanner();
fw = new FastWriter();
int t = 1;
t = fs.nextInt();
while (t-- > 0) {
solve();
}
fw.out.close();
}
private static class myPair{
int a; int b;
myPair(int a, int b){
this.a = a;
this.b = b;
}
}
private static void solve() {
int n = fs.nextInt();
int[] arr = new int[n];
for(int i =0; i<n; i++){
arr[i] = fs.nextInt();
}
int fir = -1;
int last = -1;
for(int i = 1 ; i<n; i++){
if(arr[i] == arr[i-1]){
if(fir == -1)
fir = i;
last = i;
}
}
//System.out.println(fir + " " + last);
if(fir == last){
fw.out.println(0);
}else{
fw.out.println(Math.max(1, last- fir -1));
}
}
private static class UnionFind {
private final int[] parent;
private final int[] rank;
UnionFind(int n) {
parent = new int[n + 5];
rank = new int[n + 5];
for (int i = 0; i <= n; i++) {
parent[i] = i;
rank[i] = 0;
}
}
private int find(int i) {
if (parent[i] == i)
return i;
return parent[i] = find(parent[i]);
}
private void union(int a, int b) {
a = find(a);
b = find(b);
if (a != b) {
if (rank[a] < rank[b]) {
int temp = a;
a = b;
b = temp;
}
parent[b] = a;
if (rank[a] == rank[b])
rank[a]++;
}
}
}
private static class Calc_nCr {
private final long[] fact;
private final long[] invfact;
private final int p;
Calc_nCr(int n, int prime) {
fact = new long[n + 5];
invfact = new long[n + 5];
p = prime;
fact[0] = 1;
for (int i = 1; i <= n; i++) {
fact[i] = (i * fact[i - 1]) % p;
}
invfact[n] = pow(fact[n], p - 2, p);
for (int i = n - 1; i >= 0; i--) {
invfact[i] = (invfact[i + 1] * (i + 1)) % p;
}
}
private long nCr(int n, int r) {
if (r > n || n < 0 || r < 0) return 0;
return (((fact[n] * invfact[r]) % p) * invfact[n - r]) % p;
}
}
private static long gcd(long a, long b) {
return (b == 0 ? a : gcd(b, a % b));
}
private static long lcm(long a, long b) {
return ((a * b) / gcd(a, b));
}
private static long pow(long a, long b, int mod) {
long result = 1;
while (b > 0) {
if ((b & 1L) == 1) {
result = (result * a) % mod;
}
a = (a * a) % mod;
b >>= 1;
}
return result;
}
private static long ceilDiv(long a, long b) {
return ((a + b - 1) / b);
}
private static long getMin(long... args) {
long min = lMax;
for (long arg : args)
min = Math.min(min, arg);
return min;
}
private static long getMax(long... args) {
long max = lMin;
for (long arg : args)
max = Math.max(max, arg);
return max;
}
private static boolean isPalindrome(String s, int l, int r) {
int i = l, j = r;
while (j - i >= 1) {
if (s.charAt(i) != s.charAt(j))
return false;
i++;
j--;
}
return true;
}
private static List<Integer> primes(int n) {
boolean[] primeArr = new boolean[n + 5];
Arrays.fill(primeArr, true);
for (int i = 2; (i * i) <= n; i++) {
if (primeArr[i]) {
for (int j = i * i; j <= n; j += i) {
primeArr[j] = false;
}
}
}
List<Integer> primeList = new ArrayList<>();
for (int i = 2; i <= n; i++) {
if (primeArr[i])
primeList.add(i);
}
return primeList;
}
private static int noOfSetBits(long x) {
int cnt = 0;
while (x != 0) {
x = x & (x - 1);
cnt++;
}
return cnt;
}
private static boolean isPerfectSquare(long num) {
long sqrt = (long) Math.sqrt(num);
return ((sqrt * sqrt) == num);
}
private static class Pair<U, V> {
private final U first;
private final V second;
@Override
public boolean equals(Object o) {
if (this == o) return true;
if (o == null || getClass() != o.getClass()) return false;
Pair<?, ?> pair = (Pair<?, ?>) o;
return first.equals(pair.first) && second.equals(pair.second);
}
@Override
public int hashCode() {
return Objects.hash(first, second);
}
@Override
public String toString() {
return "(" + first + ", " + second + ")";
}
private Pair(U ff, V ss) {
this.first = ff;
this.second = ss;
}
}
private static void randomizeIntArr(int[] arr, int n) {
Random r = new Random();
for (int i = (n - 1); i > 0; i--) {
int j = r.nextInt(i + 1);
swapInIntArr(arr, i, j);
}
}
private static void randomizeLongArr(long[] arr, int n) {
Random r = new Random();
for (int i = (n - 1); i > 0; i--) {
int j = r.nextInt(i + 1);
swapInLongArr(arr, i, j);
}
}
private static void swapInIntArr(int[] arr, int a, int b) {
int temp = arr[a];
arr[a] = arr[b];
arr[b] = temp;
}
private static void swapInLongArr(long[] arr, int a, int b) {
long temp = arr[a];
arr[a] = arr[b];
arr[b] = temp;
}
private static int[] readIntArray(int n) {
int[] arr = new int[n];
for (int i = 0; i < n; i++)
arr[i] = fs.nextInt();
return arr;
}
private static long[] readLongArray(int n) {
long[] arr = new long[n];
for (int i = 0; i < n; i++)
arr[i] = fs.nextLong();
return arr;
}
private static List<Integer> readIntList(int n) {
List<Integer> list = new ArrayList<>();
for (int i = 0; i < n; i++)
list.add(fs.nextInt());
return list;
}
private static List<Long> readLongList(int n) {
List<Long> list = new ArrayList<>();
for (int i = 0; i < n; i++)
list.add(fs.nextLong());
return list;
}
private static class FastScanner {
BufferedReader br;
StringTokenizer st;
FastScanner() throws IOException {
if (checkOnlineJudge)
this.br = new BufferedReader(new FileReader("src/input.txt"));
else
this.br = new BufferedReader(new InputStreamReader(System.in));
this.st = new StringTokenizer("");
}
public String next() {
while (!st.hasMoreTokens()) {
try {
st = new StringTokenizer(br.readLine());
} catch (IOException err) {
err.printStackTrace();
}
}
return st.nextToken();
}
public String nextLine() {
if (st.hasMoreTokens()) {
return st.nextToken("").trim();
}
try {
return br.readLine().trim();
} catch (IOException err) {
err.printStackTrace();
}
return "";
}
public int nextInt() {
return Integer.parseInt(next());
}
public long nextLong() {
return Long.parseLong(next());
}
public double nextDouble() {
return Double.parseDouble(next());
}
}
private static class FastWriter {
PrintWriter out;
FastWriter() throws IOException {
if (checkOnlineJudge)
out = new PrintWriter(new FileWriter("src/output.txt"));
else
out = new PrintWriter(System.out);
}
}
}
| Java | ["4\n\n5\n\n1 1 1 1 1\n\n5\n\n2 1 1 1 2\n\n6\n\n1 1 2 3 3 4\n\n6\n\n1 2 1 4 5 4"] | 1 second | ["2\n1\n2\n0"] | NoteIn the first test case, we can select $$$i=2$$$ and $$$x=2$$$ to form $$$[1, 2, 2, 1, 1]$$$. Then, we can select $$$i=3$$$ and $$$x=3$$$ to form $$$[1, 2, 3, 3, 1]$$$.In the second test case, we can select $$$i=3$$$ and $$$x=100$$$ to form $$$[2, 1, 100, 100, 2]$$$. | Java 11 | standard input | [
"constructive algorithms",
"greedy",
"implementation"
] | a91be662101762bcb2c58b8db9ff61e0 | Each test contains multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains an integer $$$n$$$ ($$$2 \le n \le 2 \cdot 10 ^ 5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10^9$$$) — elements of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10 ^ 5$$$ | 1,100 | For each test case, print the minimum number of operations needed. | standard output | |
PASSED | 704013cc6c1a38f134c3db770cb84493 | train_108.jsonl | 1650722700 | You are given an array $$$a$$$ of length $$$n$$$. We define the equality of the array as the number of indices $$$1 \le i \le n - 1$$$ such that $$$a_i = a_{i + 1}$$$. We are allowed to do the following operation: Select two integers $$$i$$$ and $$$x$$$ such that $$$1 \le i \le n - 1$$$ and $$$1 \le x \le 10^9$$$. Then, set $$$a_i$$$ and $$$a_{i + 1}$$$ to be equal to $$$x$$$. Find the minimum number of operations needed such that the equality of the array is less than or equal to $$$1$$$. | 256 megabytes |
import java.io.BufferedReader;
import java.io.BufferedWriter;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.OutputStreamWriter;
import java.util.*;
public class weird_algrithm {
static BufferedWriter output = new BufferedWriter(
new OutputStreamWriter(System.out));
static BufferedReader sc = new BufferedReader(new InputStreamReader(System.in));
static int mod = 998244353;
static String toReturn = "";
static int steps = Integer.MAX_VALUE;
static int maxlen = 1000005;
/*MATHEMATICS FUNCTIONS START HERE
MATHS
MATHS
MATHS
MATHS*/
static long gcd(long a, long b) {
if(b == 0) return a;
else return gcd(b, a % b);
}
static long powerMod(long x, long y, int mod) {
if(y == 0) return 1;
long temp = powerMod(x, y / 2, mod);
temp = ((temp % mod) * (temp % mod)) % mod;
if(y % 2 == 0) return temp;
else return ((x % mod) * (temp % mod)) % mod;
}
static long modInverse(long n, int p) {
return powerMod(n, p - 2, p);
}
static long nCr(int n, int r, int mod, long [] fact, long [] ifact) {
return ((fact[n] % mod) * ((ifact[r] * ifact[n - r]) % mod)) % mod;
}
static boolean isPrime(long a) {
if(a == 1) return false;
else if(a == 2 || a == 3 || a== 5) return true;
else if(a % 2 == 0 || a % 3 == 0) return false;
for(int i = 5; i * i <= a; i = i + 6) {
if(a % i == 0 || a % (i + 2) == 0) return false;
}
return true;
}
static int [] seive(int a) {
int [] toReturn = new int [a + 1];
for(int i = 0; i < a; i++) toReturn[i] = 1;
toReturn[0] = 0;
toReturn[1] = 0;
toReturn[2] = 1;
for(int i = 2; i * i <= a; i++) {
if(toReturn[i] == 0) continue;
for(int j = 2 * i; j <= a; j += i) toReturn[j] = 0;
}
return toReturn;
}
static long [] fact(int a) {
long [] arr = new long[a + 1];
arr[0] = 1;
for(int i = 1; i < a + 1; i++) {
arr[i] = (arr[i - 1] * i) % mod;
}
return arr;
}
static ArrayList<Integer> divisors(int n) {
ArrayList<Integer> arr = new ArrayList<Integer>();
for(int i = 1; i * i <= n; i++) {
if(n % i == 0) {
arr.add(i);
if(n / i == i) continue;
arr.add(n / i);
}
}
return arr;
}
/*MATHS
MATHS
MATHS
MATHS
MATHEMATICS FUNCTIONS END HERE */
/*SWAP FUNCTION START HERE
SWAP
SWAP
SWAP
SWAP
*/
static void swap(int i, int j, long[] arr) {
long temp = arr[i];
arr[i] = arr[j];
arr[j] = temp;
}
static void swap(int i, int j, int[] arr) {
int temp = arr[i];
arr[i] = arr[j];
arr[j] = temp;
}
static void swap(int i, int j, String [] arr) {
String temp = arr[i];
arr[i] = arr[j];
arr[j] = temp;
}
static void swap(int i, int j, char [] arr) {
char temp = arr[i];
arr[i] = arr[j];
arr[j] = temp;
}
/*SWAP
SWAP
SWAP
SWAP
SWAP FUNCTION END HERE*/
/*BINARY SEARCH METHODS START HERE
* BINARY SEARCH
* BINARY SEARCH
* BINARY SEARCH
* BINARY SEARCH
*/
static boolean BinaryCheck(long test, long [] arr, long health) {
for(int i = 0; i <= arr.length - 1; i++) {
if(i == arr.length - 1) health -= test;
else if(arr[i + 1] - arr[i] > test) {
health = health - test;
}else {
health = health - (arr[i + 1] - arr[i]);
}
if(health <= 0) return true;
}
return false;
}
static long binarySearchModified(long start1, long n, ArrayList<Long> arr, int a, long r) {
long start = start1, end = n, ans = -1;
while(start < end) {
long mid = (start + end) / 2;
//System.out.println(mid);
if(arr.get((int)mid) + arr.get(a) <= r && mid != start1) {
ans = mid;
start = mid + 1;
}else{
end = mid;
}
}
//System.out.println();
return ans;
}
static int upper(int start, int end, ArrayList<Long> arr, long val) {
while(start < end) {
int mid = (start + end) / 2;
if(arr.get(mid) <= val) start = mid + 1;
else end = mid;
}
return start;
}
static int lower(int start, int end, ArrayList<Long> arr, long val) {
while(start < end) {
int mid = (start + end) / 2;
if(arr.get(mid) >= val) end = mid;
else start = mid + 1;
}
return start;
}
/*BINARY SEARCH
* BINARY SEARCH
* BINARY SEARCH
* BINARY SEARCH
* BINARY SEARCH
BINARY SEARCH METHODS END HERE*/
/*RECURSIVE FUNCTION START HERE
* RECURSIVE
* RECURSIVE
* RECURSIVE
* RECURSIVE
*/
static int recurse(int x, int y, int n, int steps1, Integer [][] dp) {
if(x > n || y > n) return 0;
if(dp[x][y] != null) {
return dp[x][y];
}
else if(x == n || y == n) {
return steps1;
}
return dp[x][y] = Math.max(recurse(x + y, y, n, steps1 + 1, dp), recurse(x, x + y, n, steps1 + 1, dp));
}
/*RECURSIVE
* RECURSIVE
* RECURSIVE
* RECURSIVE
* RECURSIVE
RECURSIVE FUNCTION END HERE*/
/*GRAPH FUNCTIONS START HERE
* GRAPH
* GRAPH
* GRAPH
* GRAPH
* */
static class edge{
int from, to;
long weight;
public edge(int x, int y, long weight2) {
this.from = x;
this.to = y;
this.weight = weight2;
}
}
static class sort implements Comparator<pair>{
@Override
public int compare(pair o1, pair o2) {
// TODO Auto-generated method stub
return (int)o1.a - (int)o2.a;
}
}
static void addEdge(ArrayList<ArrayList<edge>> graph, int from, int to, long weight) {
edge temp = new edge(from, to, weight);
edge temp1 = new edge(to, from, weight);
graph.get(from).add(temp);
//graph.get(to).add(temp1);
}
static int ans = 0;
static void topoSort(ArrayList<ArrayList<Integer>> graph, int vertex, boolean [] visited, ArrayList<Integer> toReturn) {
if(visited[vertex]) return;
visited[vertex] = true;
for(int i = 0; i < graph.get(vertex).size(); i++) {
if(!visited[graph.get(vertex).get(i)]) topoSort(graph, graph.get(vertex).get(i), visited, toReturn);
}
toReturn.add(vertex);
}
static boolean isCyclicDirected(ArrayList<ArrayList<Integer>> graph, int vertex, boolean [] visited, boolean [] reStack) {
if(reStack[vertex]) return true;
if(visited[vertex]) return false;
reStack[vertex] = true;
visited[vertex] = true;
for(int i = 0; i < graph.get(vertex).size(); i++) {
if(isCyclicDirected(graph, graph.get(vertex).get(i), visited, reStack)) return true;
}
reStack[vertex] = false;
return false;
}
static int e = 0;
static long mst(PriorityQueue<edge> pq, int nodes) {
long weight = 0;
int [] size = new int[nodes + 1];
Arrays.fill(size, 1);
while(!pq.isEmpty()) {
edge temp = pq.poll();
int x = parent(parent, temp.to);
int y = parent(parent, temp.from);
if(x != y) {
//System.out.println(temp.weight);
union(x, y, rank, parent, size);
weight += temp.weight;
e++;
}
}
return weight;
}
static void floyd(long [][] dist) { // to find min distance between two nodes
for(int k = 0; k < dist.length; k++) {
for(int i = 0; i < dist.length; i++) {
for(int j = 0; j < dist.length; j++) {
if(dist[i][j] > dist[i][k] + dist[k][j]) {
dist[i][j] = dist[i][k] + dist[k][j];
}
}
}
}
}
static void dijkstra(ArrayList<ArrayList<edge>> graph, long [] dist, int src) {
for(int i = 0; i < dist.length; i++) dist[i] = Long.MAX_VALUE / 2;
dist[src] = 0;
boolean visited[] = new boolean[dist.length];
PriorityQueue<pair> pq = new PriorityQueue<>(new sort());
pq.add(new pair(src, 0));
while(!pq.isEmpty()) {
pair temp = pq.poll();
int index = (int)temp.a;
for(int i = 0; i < graph.get(index).size(); i++) {
if(dist[graph.get(index).get(i).to] > dist[index] + graph.get(index).get(i).weight) {
dist[graph.get(index).get(i).to] = dist[index] + graph.get(index).get(i).weight;
pq.add(new pair(graph.get(index).get(i).to, graph.get(index).get(i).weight));
}
}
}
}
static int parent1 = -1;
static boolean ford(ArrayList<ArrayList<edge>> graph1, ArrayList<edge> graph, long [] dist, int src, int [] parent) {
for(int i = 0; i < dist.length; i++) dist[i] = Long.MIN_VALUE / 2;
dist[src] = 0;
boolean hasNeg = false;
for(int i = 0; i < dist.length - 1; i++) {
for(int j = 0; j < graph.size(); j++) {
int from = graph.get(j).from;
int to = graph.get(j).to;
long weight = graph.get(j).weight;
if(dist[to] < dist[from] + weight) {
dist[to] = dist[from] + weight;
parent[to] = from;
}
}
}
for(int i = 0; i < graph.size(); i++) {
int from = graph.get(i).from;
int to = graph.get(i).to;
long weight = graph.get(i).weight;
if(dist[to] < dist[from] + weight) {
parent1 = from;
hasNeg = true;
/*
* dfs(graph1, parent1, new boolean[dist.length], dist.length - 1);
* //System.out.println(ans); dfs(graph1, 0, new boolean[dist.length], parent1);
*/
//System.out.println(ans);
if(ans == 2) break;
else ans = 0;
}
}
return hasNeg;
}
/*GRAPH FUNCTIONS END HERE
* GRAPH
* GRAPH
* GRAPH
* GRAPH
*/
/*disjoint Set START HERE
* disjoint Set
* disjoint Set
* disjoint Set
* disjoint Set
*/
static int [] rank;
static int [] parent;
static int parent(int [] parent, int x) {
if(parent[x] == x) return x;
else return parent[x] = parent(parent, parent[x]);
}
static boolean union(int x, int y, int [] rank, int [] parent, int [] setSize) {
if(parent(parent, x) == parent(parent, y)) {
return true;
}
if (rank[x] > rank[y]) {
parent[y] = x;
setSize[x] += setSize[y];
} else {
parent[x] = y;
setSize[y] += setSize[x];
if (rank[x] == rank[y]) rank[y]++;
}
return false;
}
/*disjoint Set END HERE
* disjoint Set
* disjoint Set
* disjoint Set
* disjoint Set
*/
/*INPUT START HERE
* INPUT
* INPUT
* INPUT
* INPUT
* INPUT
*/
static int nextInt() throws NumberFormatException, IOException {
return Integer.parseInt(sc.readLine());
}
static long nextLong() throws NumberFormatException, IOException {
return Long.parseLong(sc.readLine());
}
static long [] inputLongArr() throws NumberFormatException, IOException{
String [] s = sc.readLine().split(" ");
long [] toReturn = new long[s.length];
for(int i = 0; i < s.length; i++) {
toReturn[i] = Long.parseLong(s[i]);
}
return toReturn;
}
static int max = 0;
static int [] inputIntArr() throws NumberFormatException, IOException{
String [] s = sc.readLine().split(" ");
//System.out.println(s.length);
int [] toReturn = new int[s.length];
for(int i = 0; i < s.length; i++) {
toReturn[i] = Integer.parseInt(s[i]);
}
return toReturn;
}
/*INPUT
* INPUT
* INPUT
* INPUT
* INPUT
* INPUT END HERE
*/
static long [] preCompute(int level) {
long [] toReturn = new long[level];
toReturn[0] = 1;
toReturn[1] = 16;
for(int i = 2; i < level; i++) {
toReturn[i] = ((toReturn[i - 1] % mod) * (toReturn[i - 1] % mod)) % mod;
}
return toReturn;
}
static class pair{
long a;
long b;
long d;
public pair(long in, long y) {
this.a = in;
this.b = y;
this.d = 0;
}
}
static int [] nextGreaterBack(char [] s) {
Stack<Integer> stack = new Stack<>();
int [] toReturn = new int[s.length];
for(int i = 0; i < s.length; i++) {
if(!stack.isEmpty() && s[stack.peek()] >= s[i]) {
stack.pop();
}
if(stack.isEmpty()) {
stack.push(i);
toReturn[i] = -1;
}else {
toReturn[i] = stack.peek();
stack.push(i);
}
}
return toReturn;
}
static int [] nextGreaterFront(char [] s) {
Stack<Integer> stack = new Stack<>();
int [] toReturn = new int[s.length];
for(int i = s.length - 1; i >= 0; i--) {
if(!stack.isEmpty() && s[stack.peek()] >= s[i]) {
stack.pop();
}
if(stack.isEmpty()) {
stack.push(i);
toReturn[i] = -1;
}else {
toReturn[i] = stack.peek();
stack.push(i);
}
}
return toReturn;
}
static int lps(String s) {
int max = 0;
int [] lps = new int[s.length()];
lps[0] = 0;
int j = 0;
for(int i = 1; i < lps.length; i++) {
j = lps[i - 1];
while(j > 0 && s.charAt(i) != s.charAt(j)) j = lps[j - 1];
if(s.charAt(i) == s.charAt(j)) {
lps[i] = j + 1;
max = Math.max(max, lps[i]);
}
}
return max;
}
static int [][] vectors = {{1, 0}, {0, 1}, {-1, 0}, {0, -1}};
static String dir = "DRUL";
static boolean check(int i, int j, boolean [][] visited) {
if(i >= visited.length || j >= visited[0].length) return false;
if(i < 0 || j < 0) return false;
return true;
}
static void selectionSort(long arr[], long [] arr1, ArrayList<ArrayList<Integer>> ans)
{
int n = arr.length;
for (int i = 0; i < n-1; i++)
{
int min_idx = i;
for (int j = i+1; j < n; j++)
if (arr[j] < arr[min_idx])
min_idx = j;
else if(arr[j] == arr[min_idx]) {
if(arr1[j] < arr1[min_idx]) min_idx = j;
}
if(i == min_idx) {
continue;
}
ArrayList<Integer> p = new ArrayList<Integer>();
p.add(min_idx + 1);
p.add(i + 1);
ans.add(new ArrayList<Integer>(p));
swap(i, min_idx, arr);
swap(i, min_idx, arr1);
}
}
static int saved = Integer.MAX_VALUE;
static String ans1 = "";
public static boolean isValid(int x, int y, String [] mat) {
if(x >= mat.length || x < 0) return false;
if(y >= mat[0].length() || y < 0) return false;
return true;
}
public static void recurse3(ArrayList<Character> arr, int index, String s, int max, ArrayList<String> toReturn) {
if(s.length() == max) {
toReturn.add(s);
return;
}
if(index == arr.size()) return;
recurse3(arr, index + 1, s + arr.get(index), max, toReturn);
recurse3(arr, index + 1, s, max, toReturn);
}
/*
if(arr[i] > q) return Math.max(f(i + 1, q - 1) + 1, f(i + 1, q);
else return f(i + 1, q) + 1
*/
static void dfsDP(ArrayList<ArrayList<Integer>> graph, int src, int [] dp1, int [] dp2, int parent) {
int sum1 = 0; int sum2 = 0;
for(int x : graph.get(src)) {
if(x == parent) continue;
dfsDP(graph, x, dp1, dp2, src);
sum1 += Math.min(dp1[x], dp2[x]);
sum2 += dp1[x];
}
dp1[src] = 1 + sum1;
dp2[src] = sum2;
System.out.println(src + " " + dp1[src] + " " + dp2[src]);
}
static int balanced = 0;
static void dfs(ArrayList<ArrayList<ArrayList<Long>>> graph, long src, int [] dist, long sum1, long sum2, long parent, ArrayList<Long> arr, int index) {
index = 0;//binarySearch(index, arr.size() - 1, arr, sum1);
if(index < arr.size() && arr.get(index) <= sum1) {
dist[(int)src] = index + 1;
}
else dist[(int)src] = index;
for(ArrayList<Long> x : graph.get((int)src)) {
if(x.get(0) == parent) continue;
if(arr.size() != 0) arr.add(arr.get(arr.size() - 1) + x.get(2));
else arr.add(x.get(2));
dfs(graph, x.get(0), dist, sum1 + x.get(1), sum2, src, arr, index);
arr.remove(arr.size() - 1);
}
}
static int compare(String s1, String s2) {
Queue<Character> q1 = new LinkedList<>();
Queue<Character> q2 = new LinkedList<Character>();
for(int i = 0; i < s1.length(); i++) {
q1.add(s1.charAt(i));
q2.add(s2.charAt(i));
}
int k = 0;
while(k < s1.length()) {
if(q1.equals(q2)) {
break;
}
q2.add(q2.poll());
k++;
}
return k;
}
static long pro = 0;
public static int len(ArrayList<ArrayList<Integer>> graph, int src, boolean [] visited
) {
visited[src] = true;
int max = 0;
for(int x : graph.get(src)) {
if(!visited[x]) {
visited[x] = true;
int len = len(graph, x, visited) + 1;
//System.out.println(len);
pro = Math.max(max * (len - 1), pro);
max = Math.max(len, max);
}
}
return max;
}
public static void recurse(int l, int [] ans) {
if(l < 0) return;
int r = (int)Math.sqrt(l * 2);
int s = r * r;
r = s - l;
recurse(r - 1, ans);
while(r <= l) {
ans[r] = l;
ans[l] = r;
r++;
l--;
}
}
static boolean isSmaller(String str1, String str2)
{
// Calculate lengths of both string
int n1 = str1.length(), n2 = str2.length();
if (n1 < n2)
return true;
if (n2 < n1)
return false;
for (int i = 0; i < n1; i++)
if (str1.charAt(i) < str2.charAt(i))
return true;
else if (str1.charAt(i) > str2.charAt(i))
return false;
return false;
}
// Function for find difference of larger numbers
static String findDiff(String str1, String str2)
{
// Before proceeding further, make sure str1
// is not smaller
if (isSmaller(str1, str2)) {
String t = str1;
str1 = str2;
str2 = t;
}
// Take an empty string for storing result
String str = "";
// Calculate length of both string
int n1 = str1.length(), n2 = str2.length();
// Reverse both of strings
str1 = new StringBuilder(str1).reverse().toString();
str2 = new StringBuilder(str2).reverse().toString();
int carry = 0;
// Run loop till small string length
// and subtract digit of str1 to str2
for (int i = 0; i < n2; i++) {
// Do school mathematics, compute difference of
// current digits
int sub
= ((int)(str1.charAt(i) - '0')
- (int)(str2.charAt(i) - '0') - carry);
// If subtraction is less than zero
// we add then we add 10 into sub and
// take carry as 1 for calculating next step
if (sub < 0) {
sub = sub + 10;
carry = 1;
}
else
carry = 0;
str += (char)(sub + '0');
}
// subtract remaining digits of larger number
for (int i = n2; i < n1; i++) {
int sub = ((int)(str1.charAt(i) - '0') - carry);
// if the sub value is -ve, then make it
// positive
if (sub < 0) {
sub = sub + 10;
carry = 1;
}
else
carry = 0;
str += (char)(sub + '0');
}
// reverse resultant string
return new StringBuilder(str).reverse().toString();
}
static void solve() throws IOException {
int n = nextInt();
int [] arr = inputIntArr();
int prev = -1;
long count = 0;
boolean c = false;
int c1 = -1;
for(int i = 0; i < arr.length - 1; i++) {
if(arr[i] == arr[i + 1]) {
if(prev == -1) prev = i;
c1 = i;
}
}
if(c1 == prev) {
output.write(0 + "\n");
return;
}
output.write(Math.max(1, c1 - prev -1) + "\n");
}
public static void main(String[] args) throws IOException {
// TODO Auto-generated method stub
int t = Integer.parseInt(sc.readLine()); for(int i = 0; i < t; i++)
solve();
output.flush();
}
}
class TreeNode {
int val; int start; int end;
public TreeNode(int val, int x, int y) {
this.val = val;
this.start = x;
this.end = y;
}
public String toString() {
return Integer.toString(this.val);
}
}
/*
1
10
6 10 7 9 11 99 45 20 88 31
*/
| Java | ["4\n\n5\n\n1 1 1 1 1\n\n5\n\n2 1 1 1 2\n\n6\n\n1 1 2 3 3 4\n\n6\n\n1 2 1 4 5 4"] | 1 second | ["2\n1\n2\n0"] | NoteIn the first test case, we can select $$$i=2$$$ and $$$x=2$$$ to form $$$[1, 2, 2, 1, 1]$$$. Then, we can select $$$i=3$$$ and $$$x=3$$$ to form $$$[1, 2, 3, 3, 1]$$$.In the second test case, we can select $$$i=3$$$ and $$$x=100$$$ to form $$$[2, 1, 100, 100, 2]$$$. | Java 11 | standard input | [
"constructive algorithms",
"greedy",
"implementation"
] | a91be662101762bcb2c58b8db9ff61e0 | Each test contains multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains an integer $$$n$$$ ($$$2 \le n \le 2 \cdot 10 ^ 5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10^9$$$) — elements of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10 ^ 5$$$ | 1,100 | For each test case, print the minimum number of operations needed. | standard output | |
PASSED | 7e6de660789ae4901e3f8f5f26296555 | train_108.jsonl | 1650722700 | You are given an array $$$a$$$ of length $$$n$$$. We define the equality of the array as the number of indices $$$1 \le i \le n - 1$$$ such that $$$a_i = a_{i + 1}$$$. We are allowed to do the following operation: Select two integers $$$i$$$ and $$$x$$$ such that $$$1 \le i \le n - 1$$$ and $$$1 \le x \le 10^9$$$. Then, set $$$a_i$$$ and $$$a_{i + 1}$$$ to be equal to $$$x$$$. Find the minimum number of operations needed such that the equality of the array is less than or equal to $$$1$$$. | 256 megabytes | import java.util.Scanner;
public class cd1672C {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
int tt= in.nextInt();
for(int w=0; w<tt; w++) {
int n=in.nextInt();
long[] niz = new long[n];
int ans=0;
int l=0;
int r=0;
int ok=0;
for(int i=0; i<n; i++) {
niz[i]=in.nextLong();
}
for(int i=0; i<n-1; i++) {
if(niz[i]==niz[i+1] && ok==0) {
l=i+1;
ok=1;
}
else if(niz[i]==niz[i+1] && ok==1) {
r=i+1;
}
}
if(l==0) {
System.out.println(0);
}
else if(r==0) {
System.out.println(0);
}
else {
if(r-l==1 || r-l==2) {
ans=1;
}
else {
ans=r-l-1;
}
System.out.println(ans);
}
}
}
} | Java | ["4\n\n5\n\n1 1 1 1 1\n\n5\n\n2 1 1 1 2\n\n6\n\n1 1 2 3 3 4\n\n6\n\n1 2 1 4 5 4"] | 1 second | ["2\n1\n2\n0"] | NoteIn the first test case, we can select $$$i=2$$$ and $$$x=2$$$ to form $$$[1, 2, 2, 1, 1]$$$. Then, we can select $$$i=3$$$ and $$$x=3$$$ to form $$$[1, 2, 3, 3, 1]$$$.In the second test case, we can select $$$i=3$$$ and $$$x=100$$$ to form $$$[2, 1, 100, 100, 2]$$$. | Java 11 | standard input | [
"constructive algorithms",
"greedy",
"implementation"
] | a91be662101762bcb2c58b8db9ff61e0 | Each test contains multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains an integer $$$n$$$ ($$$2 \le n \le 2 \cdot 10 ^ 5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10^9$$$) — elements of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10 ^ 5$$$ | 1,100 | For each test case, print the minimum number of operations needed. | standard output | |
PASSED | 931470ce124f24b77e71f61923f89017 | train_108.jsonl | 1650722700 | You are given an array $$$a$$$ of length $$$n$$$. We define the equality of the array as the number of indices $$$1 \le i \le n - 1$$$ such that $$$a_i = a_{i + 1}$$$. We are allowed to do the following operation: Select two integers $$$i$$$ and $$$x$$$ such that $$$1 \le i \le n - 1$$$ and $$$1 \le x \le 10^9$$$. Then, set $$$a_i$$$ and $$$a_{i + 1}$$$ to be equal to $$$x$$$. Find the minimum number of operations needed such that the equality of the array is less than or equal to $$$1$$$. | 256 megabytes | import java.io.*;
import java.util.*;
public class Main {
private boolean oj = System.getProperty("ONLINE_JUDGE") != null;
private FastWriter wr;
private Reader rd;
public final int MOD = 1000000007;
/************************************************** FAST INPUT IMPLEMENTATION *********************************************/
class Reader {
BufferedReader br;
StringTokenizer st;
public Reader() {
br = new BufferedReader(
new InputStreamReader(System.in));
}
public Reader(String path) throws FileNotFoundException {
br = new BufferedReader(
new InputStreamReader(new FileInputStream(path)));
}
String next() {
while (st == null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt() {
return Integer.parseInt(next());
}
long nextLong() {
return Long.parseLong(next());
}
double nextDouble() {
return Double.parseDouble(next());
}
String nextLine() {
String str = "";
try {
str = br.readLine();
} catch (IOException e) {
e.printStackTrace();
}
return str;
}
}
public int ni() throws IOException {
return rd.nextInt();
}
public long nl() throws IOException {
return rd.nextLong();
}
public void yOrn(boolean flag){
if(flag){
wr.println("YES");
}else {
wr.println("NO");
}
}
char nc() throws IOException {
return rd.next().charAt(0);
}
public String ns() throws IOException {
return rd.nextLine();
}
public Double nd() throws IOException {
return rd.nextDouble();
}
public int[] nai(int n) throws IOException {
int[] arr = new int[n];
for (int i = 0; i < n; i++) {
arr[i] = ni();
}
return arr;
}
public long[] nal(int n) throws IOException {
long[] arr = new long[n];
for (int i = 0; i < n; i++) {
arr[i] = nl();
}
return arr;
}
/************************************************** FAST OUTPUT IMPLEMENTATION *********************************************/
public static class FastWriter {
private static final int BUF_SIZE = 1 << 13;
private final byte[] buf = new byte[BUF_SIZE];
private final OutputStream out;
private int ptr = 0;
private FastWriter() {
out = null;
}
public FastWriter(OutputStream os) {
this.out = os;
}
public FastWriter(String path) {
try {
this.out = new FileOutputStream(path);
} catch (FileNotFoundException e) {
throw new RuntimeException("FastWriter");
}
}
public FastWriter write(byte b) {
buf[ptr++] = b;
if (ptr == BUF_SIZE) innerflush();
return this;
}
public FastWriter write(char c) {
return write((byte) c);
}
public FastWriter write(char[] s) {
for (char c : s) {
buf[ptr++] = (byte) c;
if (ptr == BUF_SIZE) innerflush();
}
return this;
}
public FastWriter write(String s) {
s.chars().forEach(c -> {
buf[ptr++] = (byte) c;
if (ptr == BUF_SIZE) innerflush();
});
return this;
}
private static int countDigits(int l) {
if (l >= 1000000000) return 10;
if (l >= 100000000) return 9;
if (l >= 10000000) return 8;
if (l >= 1000000) return 7;
if (l >= 100000) return 6;
if (l >= 10000) return 5;
if (l >= 1000) return 4;
if (l >= 100) return 3;
if (l >= 10) return 2;
return 1;
}
public FastWriter write(int x) {
if (x == Integer.MIN_VALUE) {
return write((long) x);
}
if (ptr + 12 >= BUF_SIZE) innerflush();
if (x < 0) {
write((byte) '-');
x = -x;
}
int d = countDigits(x);
for (int i = ptr + d - 1; i >= ptr; i--) {
buf[i] = (byte) ('0' + x % 10);
x /= 10;
}
ptr += d;
return this;
}
private static int countDigits(long l) {
if (l >= 1000000000000000000L) return 19;
if (l >= 100000000000000000L) return 18;
if (l >= 10000000000000000L) return 17;
if (l >= 1000000000000000L) return 16;
if (l >= 100000000000000L) return 15;
if (l >= 10000000000000L) return 14;
if (l >= 1000000000000L) return 13;
if (l >= 100000000000L) return 12;
if (l >= 10000000000L) return 11;
if (l >= 1000000000L) return 10;
if (l >= 100000000L) return 9;
if (l >= 10000000L) return 8;
if (l >= 1000000L) return 7;
if (l >= 100000L) return 6;
if (l >= 10000L) return 5;
if (l >= 1000L) return 4;
if (l >= 100L) return 3;
if (l >= 10L) return 2;
return 1;
}
public FastWriter write(long x) {
if (x == Long.MIN_VALUE) {
return write("" + x);
}
if (ptr + 21 >= BUF_SIZE) innerflush();
if (x < 0) {
write((byte) '-');
x = -x;
}
int d = countDigits(x);
for (int i = ptr + d - 1; i >= ptr; i--) {
buf[i] = (byte) ('0' + x % 10);
x /= 10;
}
ptr += d;
return this;
}
public FastWriter write(double x, int precision) {
if (x < 0) {
write('-');
x = -x;
}
x += Math.pow(10, -precision) / 2;
// if(x < 0){ x = 0; }
write((long) x).write(".");
x -= (long) x;
for (int i = 0; i < precision; i++) {
x *= 10;
write((char) ('0' + (int) x));
x -= (int) x;
}
return this;
}
public FastWriter writeln(char c) {
return write(c).writeln();
}
public FastWriter writeln(int x) {
return write(x).writeln();
}
public FastWriter writeln(long x) {
return write(x).writeln();
}
public FastWriter writeln(double x, int precision) {
return write(x, precision).writeln();
}
public FastWriter write(int... xs) {
boolean first = true;
for (int x : xs) {
if (!first) write(' ');
first = false;
write(x);
}
return this;
}
public FastWriter write(long... xs) {
boolean first = true;
for (long x : xs) {
if (!first) write(' ');
first = false;
write(x);
}
return this;
}
public FastWriter writeln() {
return write((byte) '\n');
}
public FastWriter writeln(int... xs) {
return write(xs).writeln();
}
public FastWriter writeln(long... xs) {
return write(xs).writeln();
}
public FastWriter writeln(char[] line) {
return write(line).writeln();
}
public FastWriter writeln(char[]... map) {
for (char[] line : map) write(line).writeln();
return this;
}
public FastWriter writeln(String s) {
return write(s).writeln();
}
private void innerflush() {
try {
out.write(buf, 0, ptr);
ptr = 0;
} catch (IOException e) {
throw new RuntimeException("innerflush");
}
}
public void flush() {
innerflush();
try {
out.flush();
} catch (IOException e) {
throw new RuntimeException("flush");
}
}
public FastWriter print(byte b) {
return write(b);
}
public FastWriter print(char c) {
return write(c);
}
public FastWriter print(char[] s) {
return write(s);
}
public FastWriter print(String s) {
return write(s);
}
public FastWriter print(int x) {
return write(x);
}
public FastWriter print(long x) {
return write(x);
}
public FastWriter print(double x, int precision) {
return write(x, precision);
}
public FastWriter println(char c) {
return writeln(c);
}
public FastWriter println(int x) {
return writeln(x);
}
public FastWriter println(long x) {
return writeln(x);
}
public FastWriter println(double x, int precision) {
return writeln(x, precision);
}
public FastWriter print(int... xs) {
return write(xs);
}
public FastWriter print(long... xs) {
return write(xs);
}
public FastWriter println(int... xs) {
return writeln(xs);
}
public FastWriter println(long... xs) {
return writeln(xs);
}
public FastWriter println(char[] line) {
return writeln(line);
}
public FastWriter println(char[]... map) {
return writeln(map);
}
public FastWriter println(String s) {
return writeln(s);
}
public FastWriter println() {
return writeln();
}
}
/********************************************************* USEFUL CODE **************************************************/
boolean[] SAPrimeGenerator(int n) {
// TC-N*LOG(LOG N)
//Create Prime Marking Array and fill it with true value
boolean[] primeMarker = new boolean[n + 1];
Arrays.fill(primeMarker, true);
primeMarker[0] = false;
primeMarker[1] = false;
for (int i = 2; i <= n; i++) {
if (primeMarker[i]) {
// we start from 2*i because i*1 must be prime
for (int j = 2 * i; j <= n; j += i) {
primeMarker[j] = false;
}
}
}
return primeMarker;
}
private void tr(Object... o) {
if (!oj) System.out.println(Arrays.deepToString(o));
}
class Pair<F, S> {
private F first;
private S second;
Pair(F first, S second) {
this.first = first;
this.second = second;
}
public F getFirst() {
return first;
}
public S getSecond() {
return second;
}
@Override
public String toString() {
return "Pair{" +
"first=" + first +
", second=" + second +
'}';
}
@Override
public boolean equals(Object o) {
if (this == o) return true;
if (o == null || getClass() != o.getClass()) return false;
Pair<F, S> pair = (Pair<F, S>) o;
return first == pair.first && second == pair.second;
}
@Override
public int hashCode() {
return Objects.hash(first, second);
}
}
class PairThree<F, S, X> {
private F first;
private S second;
private X third;
PairThree(F first, S second,X third) {
this.first = first;
this.second = second;
this.third=third;
}
public F getFirst() {
return first;
}
public void setFirst(F first) {
this.first = first;
}
public S getSecond() {
return second;
}
public void setSecond(S second) {
this.second = second;
}
public X getThird() {
return third;
}
public void setThird(X third) {
this.third = third;
}
@Override
public boolean equals(Object o) {
if (this == o) return true;
if (o == null || getClass() != o.getClass()) return false;
PairThree<?, ?, ?> pair = (PairThree<?, ?, ?>) o;
return first.equals(pair.first) && second.equals(pair.second) && third.equals(pair.third);
}
@Override
public int hashCode() {
return Objects.hash(first, second, third);
}
}
public static void main(String[] args) throws IOException {
new Main().run();
}
public void run() throws IOException {
if (oj) {
rd = new Reader();
wr = new FastWriter(System.out);
} else {
File input = new File("input.txt");
File output = new File("output.txt");
if (input.exists() && output.exists()) {
rd = new Reader(input.getPath());
wr = new FastWriter(output.getPath());
} else {
rd = new Reader();
wr = new FastWriter(System.out);
oj = true;
}
}
long s = System.currentTimeMillis();
solve();
wr.flush();
tr(System.currentTimeMillis() - s + "ms");
}
/***************************************************************************************************************************
*********************************************************** MAIN CODE ******************************************************
****************************************************************************************************************************/
boolean[] sieve;
public void solve() throws IOException {
int t = 1;
t = ni();
while (t-- > 0) {
go();
}
}
/********************************************************* MAIN LOGIC HERE ****************************************************/
long gcd(long a, long b)
{
if (a == 0)
return b;
return gcd(b % a, a);
}
static int lower_bound(long array[], long key)
{
int low = 0, high = array.length;
int mid;
while (low < high) {
mid = low + (high - low) / 2;
if (key <= array[mid]) {
high = mid;
}
else {
low = mid + 1;
}
}
if (low < array.length && array[low] < key) {
low++;
}
return low;
}
public boolean getHealth(long[] arr,long k,long h){
long health=0;
health+=k;
for(int i=1;i<arr.length;i++){
if(arr[i]-arr[i-1]<=k){
health+=arr[i]-arr[i-1];
}else {
health+=k;
}
}
return health>=h;
}
public int Substr(String s2, String s1){
int counter = 0; //pointing s2
int i = 0;
for(;i<s1.length();i++){
if(counter==s2.length())
break;
if(s2.charAt(counter)==s1.charAt(i)){
counter++;
}else{
//Special case where character preceding the i'th character is duplicate
if(counter>0){
i -= counter;
}
counter = 0;
}
}
return counter < s2.length()?-1:i-counter;
}
public void go() throws IOException {
int n=ni();
int[] arr=nai(n);
int first=-1,last=-1;
for(int i=0;i<n-1;i++){
if(arr[i]==arr[i+1]){
if(first==-1){
first=i;
}
last=i;
}
}
if(first==last){
wr.println(0);
}else if(last-first==1){
wr.println(1);
} else {
wr.println(last-(first+1));
}
}
} | Java | ["4\n\n5\n\n1 1 1 1 1\n\n5\n\n2 1 1 1 2\n\n6\n\n1 1 2 3 3 4\n\n6\n\n1 2 1 4 5 4"] | 1 second | ["2\n1\n2\n0"] | NoteIn the first test case, we can select $$$i=2$$$ and $$$x=2$$$ to form $$$[1, 2, 2, 1, 1]$$$. Then, we can select $$$i=3$$$ and $$$x=3$$$ to form $$$[1, 2, 3, 3, 1]$$$.In the second test case, we can select $$$i=3$$$ and $$$x=100$$$ to form $$$[2, 1, 100, 100, 2]$$$. | Java 11 | standard input | [
"constructive algorithms",
"greedy",
"implementation"
] | a91be662101762bcb2c58b8db9ff61e0 | Each test contains multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains an integer $$$n$$$ ($$$2 \le n \le 2 \cdot 10 ^ 5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10^9$$$) — elements of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10 ^ 5$$$ | 1,100 | For each test case, print the minimum number of operations needed. | standard output | |
PASSED | b51aa311773470127cc5c06234c89528 | train_108.jsonl | 1650722700 | You are given an array $$$a$$$ of length $$$n$$$. We define the equality of the array as the number of indices $$$1 \le i \le n - 1$$$ such that $$$a_i = a_{i + 1}$$$. We are allowed to do the following operation: Select two integers $$$i$$$ and $$$x$$$ such that $$$1 \le i \le n - 1$$$ and $$$1 \le x \le 10^9$$$. Then, set $$$a_i$$$ and $$$a_{i + 1}$$$ to be equal to $$$x$$$. Find the minimum number of operations needed such that the equality of the array is less than or equal to $$$1$$$. | 256 megabytes | import java.util.*;
public class UnequalArray {
public static void main(String[] args){
Scanner s = new Scanner(System.in);
int test =s.nextInt();
while(test --> 0){
int n = s.nextInt();
int arr[] = new int[n];
for(int i=0; i < n;i++){
arr[i] = s.nextInt();
}
int min = -1;
int max = -1;
for(int i = 1; i < n; i++){
if(arr[i-1] == arr[i]){
if(min == -1){
min = i;
}
max = i;
}
}
if(min == max){
System.out.println(0);
}else {
System.out.println(Math.max(1, max - min - 1));
}
}
}
}
| Java | ["4\n\n5\n\n1 1 1 1 1\n\n5\n\n2 1 1 1 2\n\n6\n\n1 1 2 3 3 4\n\n6\n\n1 2 1 4 5 4"] | 1 second | ["2\n1\n2\n0"] | NoteIn the first test case, we can select $$$i=2$$$ and $$$x=2$$$ to form $$$[1, 2, 2, 1, 1]$$$. Then, we can select $$$i=3$$$ and $$$x=3$$$ to form $$$[1, 2, 3, 3, 1]$$$.In the second test case, we can select $$$i=3$$$ and $$$x=100$$$ to form $$$[2, 1, 100, 100, 2]$$$. | Java 11 | standard input | [
"constructive algorithms",
"greedy",
"implementation"
] | a91be662101762bcb2c58b8db9ff61e0 | Each test contains multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains an integer $$$n$$$ ($$$2 \le n \le 2 \cdot 10 ^ 5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10^9$$$) — elements of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10 ^ 5$$$ | 1,100 | For each test case, print the minimum number of operations needed. | standard output | |
PASSED | 71ad994b619cd4b8e7c044aa3dd24489 | train_108.jsonl | 1650722700 | You are given an array $$$a$$$ of length $$$n$$$. We define the equality of the array as the number of indices $$$1 \le i \le n - 1$$$ such that $$$a_i = a_{i + 1}$$$. We are allowed to do the following operation: Select two integers $$$i$$$ and $$$x$$$ such that $$$1 \le i \le n - 1$$$ and $$$1 \le x \le 10^9$$$. Then, set $$$a_i$$$ and $$$a_{i + 1}$$$ to be equal to $$$x$$$. Find the minimum number of operations needed such that the equality of the array is less than or equal to $$$1$$$. | 256 megabytes | import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.*;
public class Main{
static void solve() {
int n=in.nextInt();
int a[]=new int[n];
for(int i=0;i<n;i++)a[i]=in.nextInt();
int st=0,ed=-1;
for(int i=1;i<n;i++){
if(st==0&&a[i]==a[i-1]) st=i;
if(a[i]==a[i-1]) ed=i-1;
}
if(ed<st){out.println(0);return;}
out.println(Math.max(1,ed-st));
}
public static void main(String[] args) throws Exception {
int cnt=in.nextInt();
while (cnt-->0){
solve();
}
out.flush();
}
static class FastReader{
StringTokenizer st;
BufferedReader br;
public FastReader(){
br=new BufferedReader(new InputStreamReader(System.in));
}
String next(){
while (st==null||!st.hasMoreElements()){
try {
st=new StringTokenizer(br.readLine());
}catch (IOException e){
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt() {
return Integer.parseInt(next());
}
long nextLong() {
return Long.parseLong(next());
}
double nextDouble() {
return Double.parseDouble(next());
}
String nextLine() {
String str = "";
try {
str = br.readLine();
} catch (IOException e) {
e.printStackTrace();
}
return str;
}
}
static PrintWriter out = new PrintWriter(System.out);
static FastReader in = new FastReader();
}
| Java | ["4\n\n5\n\n1 1 1 1 1\n\n5\n\n2 1 1 1 2\n\n6\n\n1 1 2 3 3 4\n\n6\n\n1 2 1 4 5 4"] | 1 second | ["2\n1\n2\n0"] | NoteIn the first test case, we can select $$$i=2$$$ and $$$x=2$$$ to form $$$[1, 2, 2, 1, 1]$$$. Then, we can select $$$i=3$$$ and $$$x=3$$$ to form $$$[1, 2, 3, 3, 1]$$$.In the second test case, we can select $$$i=3$$$ and $$$x=100$$$ to form $$$[2, 1, 100, 100, 2]$$$. | Java 11 | standard input | [
"constructive algorithms",
"greedy",
"implementation"
] | a91be662101762bcb2c58b8db9ff61e0 | Each test contains multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains an integer $$$n$$$ ($$$2 \le n \le 2 \cdot 10 ^ 5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10^9$$$) — elements of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10 ^ 5$$$ | 1,100 | For each test case, print the minimum number of operations needed. | standard output | |
PASSED | c16b5441acb545a385d738bb698ba125 | train_108.jsonl | 1650722700 | You are given an array $$$a$$$ of length $$$n$$$. We define the equality of the array as the number of indices $$$1 \le i \le n - 1$$$ such that $$$a_i = a_{i + 1}$$$. We are allowed to do the following operation: Select two integers $$$i$$$ and $$$x$$$ such that $$$1 \le i \le n - 1$$$ and $$$1 \le x \le 10^9$$$. Then, set $$$a_i$$$ and $$$a_{i + 1}$$$ to be equal to $$$x$$$. Find the minimum number of operations needed such that the equality of the array is less than or equal to $$$1$$$. | 256 megabytes | import java.util.*;
public class Sov1
{
public static void main(String args[])
{
long t;int n,i,c=0,k=0,j,m=0;
Scanner x=new Scanner(System.in);
t=x.nextLong();
while(t>0)
{
t--;
n=x.nextInt();
int[] a=new int[n];
for(i=0;i<n;i++)
{
a[i]=x.nextInt();
}
for(j=0;j<n-1;j++)
{
if(a[j]==a[j+1])
{
if(c==0)
k=j;
m=j;
c++;
}}
if(c>1 && m-k!=1)
System.out.println(m-k-1);
else if(m-k==1 && c>0)
System.out.println("1");
else
System.out.println("0");
c=0;
k=0;
m=0;
}}} | Java | ["4\n\n5\n\n1 1 1 1 1\n\n5\n\n2 1 1 1 2\n\n6\n\n1 1 2 3 3 4\n\n6\n\n1 2 1 4 5 4"] | 1 second | ["2\n1\n2\n0"] | NoteIn the first test case, we can select $$$i=2$$$ and $$$x=2$$$ to form $$$[1, 2, 2, 1, 1]$$$. Then, we can select $$$i=3$$$ and $$$x=3$$$ to form $$$[1, 2, 3, 3, 1]$$$.In the second test case, we can select $$$i=3$$$ and $$$x=100$$$ to form $$$[2, 1, 100, 100, 2]$$$. | Java 11 | standard input | [
"constructive algorithms",
"greedy",
"implementation"
] | a91be662101762bcb2c58b8db9ff61e0 | Each test contains multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains an integer $$$n$$$ ($$$2 \le n \le 2 \cdot 10 ^ 5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10^9$$$) — elements of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10 ^ 5$$$ | 1,100 | For each test case, print the minimum number of operations needed. | standard output | |
PASSED | da4265c19fa9bc46fd131239e3476764 | train_108.jsonl | 1650722700 | You are given an array $$$a$$$ of length $$$n$$$. We define the equality of the array as the number of indices $$$1 \le i \le n - 1$$$ such that $$$a_i = a_{i + 1}$$$. We are allowed to do the following operation: Select two integers $$$i$$$ and $$$x$$$ such that $$$1 \le i \le n - 1$$$ and $$$1 \le x \le 10^9$$$. Then, set $$$a_i$$$ and $$$a_{i + 1}$$$ to be equal to $$$x$$$. Find the minimum number of operations needed such that the equality of the array is less than or equal to $$$1$$$. | 256 megabytes | import java.io.*;
import java.lang.*;
import java.util.*;
public class fast {
static final Random random = new Random();
static void sort(int arr[]) {
int n = arr.length;
for(int i = 0; i < n; i++) {
int j = random.nextInt(n),temp = arr[j];
arr[j] = arr[i];
arr[i] = temp;
}
Arrays.sort(arr);
}
static class FastScanner {
BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
StringTokenizer st=new StringTokenizer("");
String next() {
while (!st.hasMoreTokens())
try {
st=new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
return st.nextToken();
}
int nextInt() {
return Integer.parseInt(next());
}
long[] readArrayL(int n) {
long a[]=new long[n];
for(int i=0;i<n;i++) a[i]=nextLong();
return a;
}
int[] readArray(int n) {
int[] a=new int[n];
for (int i=0; i<n; i++) a[i]=nextInt();
return a;
}
long nextLong() {
return Long.parseLong(next());
}
double nextDouble() {
return Double.parseDouble(next());
}
}
static long factorial( long n )
{
long res = 1 ;
while(n>1)
{
res = (res*n)%998244353;
n-- ;
}
return res ;
}
static int gcd(int a, int b)
{
if (b == 0)
return a;
return gcd(b, a % b);
}
static int equal(int arr[])
{
int n = arr.length ;
int res = 0 ;
for(int i = 1 ; i< n ;i++)
{
if(arr[i] == arr[i-1])
{
res++ ;
}
} return res ;
}
static PrintWriter out = new PrintWriter(new OutputStreamWriter(System.out));//PRINTLN METHOD
static FastScanner sc = new FastScanner();//FASTSCANNER OBJECT
public static void main(String[] args){
int t = sc.nextInt() ;
while(t-- != 0 )
{
int n = sc.nextInt() ;
int arr[] = new int[n] ;
for(int i = 0; i< n ;i++)
{
arr[i] = sc.nextInt() ;
}
int a = -1 , b =-1 ;
int count = 0 ;
for(int i = 1 ;i<n ;i++)
{
if(a == -1 )
{if(arr[i-1] == arr[i])
{
a = i ; count++ ;
}}
else if(arr[i-1] == arr[i])
{
b = i-1 ; count++ ;
}
}
if(b == -1 || count <2 )
{
System.out.println(0) ;
}
else {
System.out.println(Math.max(1,b-a)) ;
}
}
////////////////////////////////////////////////////////////////////
out.flush();out.close();
}//END OF MAIN METHOD
}//END OF MAIN CLASS | Java | ["4\n\n5\n\n1 1 1 1 1\n\n5\n\n2 1 1 1 2\n\n6\n\n1 1 2 3 3 4\n\n6\n\n1 2 1 4 5 4"] | 1 second | ["2\n1\n2\n0"] | NoteIn the first test case, we can select $$$i=2$$$ and $$$x=2$$$ to form $$$[1, 2, 2, 1, 1]$$$. Then, we can select $$$i=3$$$ and $$$x=3$$$ to form $$$[1, 2, 3, 3, 1]$$$.In the second test case, we can select $$$i=3$$$ and $$$x=100$$$ to form $$$[2, 1, 100, 100, 2]$$$. | Java 11 | standard input | [
"constructive algorithms",
"greedy",
"implementation"
] | a91be662101762bcb2c58b8db9ff61e0 | Each test contains multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains an integer $$$n$$$ ($$$2 \le n \le 2 \cdot 10 ^ 5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10^9$$$) — elements of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10 ^ 5$$$ | 1,100 | For each test case, print the minimum number of operations needed. | standard output | |
PASSED | 9a67b0e20e85919723f89c95a46d50fa | train_108.jsonl | 1650722700 | You are given an array $$$a$$$ of length $$$n$$$. We define the equality of the array as the number of indices $$$1 \le i \le n - 1$$$ such that $$$a_i = a_{i + 1}$$$. We are allowed to do the following operation: Select two integers $$$i$$$ and $$$x$$$ such that $$$1 \le i \le n - 1$$$ and $$$1 \le x \le 10^9$$$. Then, set $$$a_i$$$ and $$$a_{i + 1}$$$ to be equal to $$$x$$$. Find the minimum number of operations needed such that the equality of the array is less than or equal to $$$1$$$. | 256 megabytes | import java.io.*;
import java.util.InputMismatchException;
public class E1672C {
public static void main(String[] args) {
FastIO io = new FastIO();
int t = io.nextInt();
while (t-- > 0) {
int n = io.nextInt();
int[] arr = new int[n];
for (int i = 0; i < n; i++) arr[i] = io.nextInt();
int start = -1;
int end = -1;
for (int i = 0; i < n - 1; i++)
if (arr[i] == arr[i + 1]) {
if (start == -1) start = i;
end = i;
}
io.println(end == start ? 0 : Math.max(1, end - start - 1));
}
io.close();
}
private static class FastIO extends PrintWriter {
private final InputStream stream;
private final byte[] buf = new byte[1 << 16];
private int curChar, numChars;
// standard input
public FastIO() {
this(System.in, System.out);
}
public FastIO(InputStream i, OutputStream o) {
super(o);
stream = i;
}
// file input
public FastIO(String i, String o) throws IOException {
super(new FileWriter(o));
stream = new FileInputStream(i);
}
// throws InputMismatchException() if previously detected end of file
private int nextByte() {
if (numChars == -1) throw new InputMismatchException();
if (curChar >= numChars) {
curChar = 0;
try {
numChars = stream.read(buf);
} catch (IOException e) {
throw new InputMismatchException();
}
if (numChars == -1) return -1; // end of file
}
return buf[curChar++];
}
// to read in entire lines, replace c <= ' '
// with a function that checks whether c is a line break
public String next() {
int c;
do {
c = nextByte();
} while (c <= ' ');
StringBuilder res = new StringBuilder();
do {
res.appendCodePoint(c);
c = nextByte();
} while (c > ' ');
return res.toString();
}
public int nextInt() { // nextLong() would be implemented similarly
int c;
do {
c = nextByte();
} while (c <= ' ');
int sgn = 1;
if (c == '-') {
sgn = -1;
c = nextByte();
}
int res = 0;
do {
if (c < '0' || c > '9')
throw new InputMismatchException();
res = 10 * res + c - '0';
c = nextByte();
} while (c > ' ');
return res * sgn;
}
public long nextLong() { // nextLong() would be implemented similarly
int c;
do {
c = nextByte();
} while (c <= ' ');
int sgn = 1;
if (c == '-') {
sgn = -1;
c = nextByte();
}
long res = 0;
do {
if (c < '0' || c > '9')
throw new InputMismatchException();
res = 10 * res + c - '0';
c = nextByte();
} while (c > ' ');
return res * sgn;
}
public double nextDouble() {
return Double.parseDouble(next());
}
}
}
| Java | ["4\n\n5\n\n1 1 1 1 1\n\n5\n\n2 1 1 1 2\n\n6\n\n1 1 2 3 3 4\n\n6\n\n1 2 1 4 5 4"] | 1 second | ["2\n1\n2\n0"] | NoteIn the first test case, we can select $$$i=2$$$ and $$$x=2$$$ to form $$$[1, 2, 2, 1, 1]$$$. Then, we can select $$$i=3$$$ and $$$x=3$$$ to form $$$[1, 2, 3, 3, 1]$$$.In the second test case, we can select $$$i=3$$$ and $$$x=100$$$ to form $$$[2, 1, 100, 100, 2]$$$. | Java 11 | standard input | [
"constructive algorithms",
"greedy",
"implementation"
] | a91be662101762bcb2c58b8db9ff61e0 | Each test contains multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains an integer $$$n$$$ ($$$2 \le n \le 2 \cdot 10 ^ 5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10^9$$$) — elements of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10 ^ 5$$$ | 1,100 | For each test case, print the minimum number of operations needed. | standard output | |
PASSED | f3a5806f44b44d3d9551f67e4ba94748 | train_108.jsonl | 1650722700 | You are given an array $$$a$$$ of length $$$n$$$. We define the equality of the array as the number of indices $$$1 \le i \le n - 1$$$ such that $$$a_i = a_{i + 1}$$$. We are allowed to do the following operation: Select two integers $$$i$$$ and $$$x$$$ such that $$$1 \le i \le n - 1$$$ and $$$1 \le x \le 10^9$$$. Then, set $$$a_i$$$ and $$$a_{i + 1}$$$ to be equal to $$$x$$$. Find the minimum number of operations needed such that the equality of the array is less than or equal to $$$1$$$. | 256 megabytes | import java.util.*;
public class a11 {
public static void main(String [] args) {
Scanner s=new Scanner (System.in);
int t=s.nextInt();
while(t-->0) {
int n=s.nextInt();
int []f=new int [n];
for(int i=0;i<n;i++)
f[i]=s.nextInt();
int min=-1,max=-1;
for(int i=1;i<n;i++) {
if(f[i-1]==f[i]) {
if(min==-1)
min=i;
max=i;
}
}
if(min==max)System.out.println(0);
else System.out.println(Math.max(1,max-min-1));
}
}
}
| Java | ["4\n\n5\n\n1 1 1 1 1\n\n5\n\n2 1 1 1 2\n\n6\n\n1 1 2 3 3 4\n\n6\n\n1 2 1 4 5 4"] | 1 second | ["2\n1\n2\n0"] | NoteIn the first test case, we can select $$$i=2$$$ and $$$x=2$$$ to form $$$[1, 2, 2, 1, 1]$$$. Then, we can select $$$i=3$$$ and $$$x=3$$$ to form $$$[1, 2, 3, 3, 1]$$$.In the second test case, we can select $$$i=3$$$ and $$$x=100$$$ to form $$$[2, 1, 100, 100, 2]$$$. | Java 11 | standard input | [
"constructive algorithms",
"greedy",
"implementation"
] | a91be662101762bcb2c58b8db9ff61e0 | Each test contains multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains an integer $$$n$$$ ($$$2 \le n \le 2 \cdot 10 ^ 5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10^9$$$) — elements of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10 ^ 5$$$ | 1,100 | For each test case, print the minimum number of operations needed. | standard output | |
PASSED | c52846396934e4218e6a27f962f31c8d | train_108.jsonl | 1650722700 | You are given an array $$$a$$$ of length $$$n$$$. We define the equality of the array as the number of indices $$$1 \le i \le n - 1$$$ such that $$$a_i = a_{i + 1}$$$. We are allowed to do the following operation: Select two integers $$$i$$$ and $$$x$$$ such that $$$1 \le i \le n - 1$$$ and $$$1 \le x \le 10^9$$$. Then, set $$$a_i$$$ and $$$a_{i + 1}$$$ to be equal to $$$x$$$. Find the minimum number of operations needed such that the equality of the array is less than or equal to $$$1$$$. | 256 megabytes | // 1672C
import java.util.*;
public class UnequalArray {
public static void main (String[] args) {
Scanner in = new Scanner(System.in);
int t = in.nextInt();
for (int i = 0; i < t; i++) {
int n = in.nextInt();
int maxDistanceBetweenTwoPairs = 0;
int indexOfFirstPair = -1;
int indexOfSecondPair = -1;
int prev = in.nextInt();
for (int j = 1; j < n; j++) {
int next = in.nextInt();
if (indexOfFirstPair == -1 && prev == next) {
indexOfFirstPair = j;
} else if (prev == next) {
indexOfSecondPair = j-1;
}
prev = next;
}
// System.out.println("first: " + indexOfFirstPair);
// System.out.println("second: " + indexOfSecondPair);
if (indexOfSecondPair > 0) {
maxDistanceBetweenTwoPairs = indexOfSecondPair - indexOfFirstPair;
if (maxDistanceBetweenTwoPairs == 0) {
maxDistanceBetweenTwoPairs++;
}
}
System.out.println(maxDistanceBetweenTwoPairs);
}
in.close();
}
}
| Java | ["4\n\n5\n\n1 1 1 1 1\n\n5\n\n2 1 1 1 2\n\n6\n\n1 1 2 3 3 4\n\n6\n\n1 2 1 4 5 4"] | 1 second | ["2\n1\n2\n0"] | NoteIn the first test case, we can select $$$i=2$$$ and $$$x=2$$$ to form $$$[1, 2, 2, 1, 1]$$$. Then, we can select $$$i=3$$$ and $$$x=3$$$ to form $$$[1, 2, 3, 3, 1]$$$.In the second test case, we can select $$$i=3$$$ and $$$x=100$$$ to form $$$[2, 1, 100, 100, 2]$$$. | Java 11 | standard input | [
"constructive algorithms",
"greedy",
"implementation"
] | a91be662101762bcb2c58b8db9ff61e0 | Each test contains multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains an integer $$$n$$$ ($$$2 \le n \le 2 \cdot 10 ^ 5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10^9$$$) — elements of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10 ^ 5$$$ | 1,100 | For each test case, print the minimum number of operations needed. | standard output | |
PASSED | 5cdb49e5bd95adbdbe6dc5d2210386d4 | train_108.jsonl | 1650722700 | You are given an array $$$a$$$ of length $$$n$$$. We define the equality of the array as the number of indices $$$1 \le i \le n - 1$$$ such that $$$a_i = a_{i + 1}$$$. We are allowed to do the following operation: Select two integers $$$i$$$ and $$$x$$$ such that $$$1 \le i \le n - 1$$$ and $$$1 \le x \le 10^9$$$. Then, set $$$a_i$$$ and $$$a_{i + 1}$$$ to be equal to $$$x$$$. Find the minimum number of operations needed such that the equality of the array is less than or equal to $$$1$$$. | 256 megabytes | // 1672C
import java.util.*;
public class UnequalArray {
public static void main (String[] args) {
Scanner in = new Scanner(System.in);
int t = in.nextInt();
for (int i = 0; i < t; i++) {
int n = in.nextInt();
int maxDistanceBetweenTwoPairs = 0;
int lastVal = -1;
int indexOfFirstPair = -1;
int indexOfSecondPair = -1;
int prev = in.nextInt();
for (int j = 1; j < n; j++) {
int next = in.nextInt();
if (indexOfFirstPair == -1 && prev == next) {
indexOfFirstPair = j;
} else if (prev == next) {
indexOfSecondPair = j-1;
lastVal = indexOfSecondPair;
}
prev = next;
}
// System.out.println("first: " + indexOfFirstPair);
// System.out.println("second: " + indexOfSecondPair);
if (lastVal > 0) {
maxDistanceBetweenTwoPairs = indexOfSecondPair - indexOfFirstPair;
if (maxDistanceBetweenTwoPairs == 0) {
maxDistanceBetweenTwoPairs++;
}
}
System.out.println(maxDistanceBetweenTwoPairs);
}
in.close();
}
}
| Java | ["4\n\n5\n\n1 1 1 1 1\n\n5\n\n2 1 1 1 2\n\n6\n\n1 1 2 3 3 4\n\n6\n\n1 2 1 4 5 4"] | 1 second | ["2\n1\n2\n0"] | NoteIn the first test case, we can select $$$i=2$$$ and $$$x=2$$$ to form $$$[1, 2, 2, 1, 1]$$$. Then, we can select $$$i=3$$$ and $$$x=3$$$ to form $$$[1, 2, 3, 3, 1]$$$.In the second test case, we can select $$$i=3$$$ and $$$x=100$$$ to form $$$[2, 1, 100, 100, 2]$$$. | Java 11 | standard input | [
"constructive algorithms",
"greedy",
"implementation"
] | a91be662101762bcb2c58b8db9ff61e0 | Each test contains multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains an integer $$$n$$$ ($$$2 \le n \le 2 \cdot 10 ^ 5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10^9$$$) — elements of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10 ^ 5$$$ | 1,100 | For each test case, print the minimum number of operations needed. | standard output | |
PASSED | 73c0a5e985e3ade3d3a4d52784415182 | train_108.jsonl | 1650722700 | You are given an array $$$a$$$ of length $$$n$$$. We define the equality of the array as the number of indices $$$1 \le i \le n - 1$$$ such that $$$a_i = a_{i + 1}$$$. We are allowed to do the following operation: Select two integers $$$i$$$ and $$$x$$$ such that $$$1 \le i \le n - 1$$$ and $$$1 \le x \le 10^9$$$. Then, set $$$a_i$$$ and $$$a_{i + 1}$$$ to be equal to $$$x$$$. Find the minimum number of operations needed such that the equality of the array is less than or equal to $$$1$$$. | 256 megabytes | import java.io.*;
import java.util.*;
import java.math.BigInteger;
public class Main
{
InputStream is;
PrintWriter out = new PrintWriter(System.out); ;
String INPUT = "";
void run() throws Exception
{
is = System.in;
solve();
out.flush();
out.close();
}
public static void main(String[] args) throws Exception { new Main().run(); }
public byte[] inbuf = new byte[1024];
public int lenbuf = 0, ptrbuf = 0;
public int readByte()
{
if(lenbuf == -1)throw new InputMismatchException();
if(ptrbuf >= lenbuf){ ptrbuf = 0;
try { lenbuf = is.read(inbuf); } catch (IOException e) { throw new InputMismatchException(); }
if(lenbuf <= 0)return -1; } return inbuf[ptrbuf++]; }
public boolean isSpaceChar(int c)
{
return !(c >= 33 && c <= 126);
}
public int skip()
{
// continue; String; charAt; println(); ArrayList; Integer; Long;
// long; Queue; Deque; LinkedList; Pair; double; binarySearch;
// s.toCharArray; length(); length; getOrDefault; break;
// Map.Entry<Integer, Integer> e; HashMap; TreeMap;
int b;
while((b = readByte()) != -1 && isSpaceChar(b));
return b;
}
public double nd()
{
return Double.parseDouble(ns());
}
public char nc()
{
return (char)skip();
}
private String ns()
{
int b = skip();
StringBuilder sb = new StringBuilder();
while(!(isSpaceChar(b)))
{
sb.appendCodePoint(b); b = readByte();
}
return sb.toString();
}
private int ni()
{
return (int)nl();
}
private long nl()
{
long num = 0;
int b;
boolean minus = false;
while((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-'));
if(b == '-')
{
minus = true; b = readByte();
}
while(true)
{
if(b >= '0' && b <= '9')
{
num = num * 10 + (b - '0');
}
else { return minus ? -num : num; } b = readByte();
}
}
class Pair
{
int first;
int second;
Pair(int a, int b)
{
first = a;
second = b;
}
}
// KMP ALGORITHM
void KMPSearch(String pat, String txt) {
int M = pat.length(); int N = txt.length();
int lps[] = new int[M]; int j = 0;
computeLPSArray(pat, M, lps); int i = 0;
while (i < N) {
if (pat.charAt(j) == txt.charAt(i)) {
j++;
i++;
}
if (j == M) {
j = lps[j - 1];
}
else if (i < N && pat.charAt(j) != txt.charAt(i)) {
if (j != 0) j = lps[j - 1];
else i = i + 1;
}
}
}
void computeLPSArray(String pat, int M, int lps[]) {
int len = 0; int i = 1; lps[0] = 0;
while (i < M) {
if (pat.charAt(i) == pat.charAt(len)) {
len++;
lps[i] = len;
i++;
}
else {
if (len != 0) {
len = lps[len - 1];
}
else {
lps[i] = len;
i++;
}
}
}
}
int FirstAndLastOccurrenceOfAnElement(int[] arr, int target, boolean findStartIndex)
{
int ans = -1;
int n = arr.length;
int low = 0; int high = n-1;
while(low <= high)
{
int mid = low + (high - low)/2;
if(arr[mid] > target) high = mid-1;
else if(arr[mid] < target) low = mid+1;
else
{
ans = mid;
if(findStartIndex) high = mid-1;
else low = mid+1;
}
}
return ans;
}
// Print All Subsequence.
static ArrayList<Integer> qwerty = new ArrayList<>();
void printAllSubsequence(int[] arr, int index, int n)
{
if(index >= n) {
for(int i=0; i<qwerty.size(); i++) out.print(qwerty.get(i) + " ");
if(qwerty.size() == 0) out.print(" "); out.println();
return;
}
qwerty.add(arr[index]); printAllSubsequence(arr, index+1, n);
qwerty.remove(qwerty.size()-1); printAllSubsequence(arr, index+1, n);
}
// Print All Subsequence.
// Print All Subsequence. with sum k.
static ArrayList<Integer> qwerty2 = new ArrayList<>();
void printSubsequenceWithSumK(int[] arr, int index, int K, int sum, int n)
{
if(index >= n) {
if(sum == K)
{
for(int i=0; i<qwerty2.size(); i++)
out.print(qwerty2.get(i) + " ");
out.println();
}
return;
}
qwerty2.add(arr[index]);
sum += arr[index];
printSubsequenceWithSumK(arr, index+1, K, sum, n);
sum -= arr[index];
qwerty2.remove(qwerty2.size()-1);
printSubsequenceWithSumK(arr, index+1, K, sum, n);
}
// Print All Subsequence. with sum k.
int[] na(int n)
{
int[] arr = new int[n];
for(int i=0; i<n; i++) arr[i]=ni();
return arr;
}
long[] nal(int n)
{
long[] arr = new long[n];
for(int i=0; i<n; i++) arr[i]=nl();
return arr;
}
void solve()
{
int test_case = ni();
while(test_case-- > 0)
{
int n = ni(); int[] a = na(n);
int p = -1, q = -1;
for (int i = 0; i < n-1; i++) {
if (a[i] == a[i+1]) {
if (p == -1) {
p = i;
}
q = i;
}
}
out.println(q - p < 2 ? q - p : q - p - 1);
}
}
} | Java | ["4\n\n5\n\n1 1 1 1 1\n\n5\n\n2 1 1 1 2\n\n6\n\n1 1 2 3 3 4\n\n6\n\n1 2 1 4 5 4"] | 1 second | ["2\n1\n2\n0"] | NoteIn the first test case, we can select $$$i=2$$$ and $$$x=2$$$ to form $$$[1, 2, 2, 1, 1]$$$. Then, we can select $$$i=3$$$ and $$$x=3$$$ to form $$$[1, 2, 3, 3, 1]$$$.In the second test case, we can select $$$i=3$$$ and $$$x=100$$$ to form $$$[2, 1, 100, 100, 2]$$$. | Java 11 | standard input | [
"constructive algorithms",
"greedy",
"implementation"
] | a91be662101762bcb2c58b8db9ff61e0 | Each test contains multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains an integer $$$n$$$ ($$$2 \le n \le 2 \cdot 10 ^ 5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10^9$$$) — elements of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10 ^ 5$$$ | 1,100 | For each test case, print the minimum number of operations needed. | standard output | |
PASSED | 738df9b8243f398252ec2bc2b2e16d20 | train_108.jsonl | 1650722700 | You are given an array $$$a$$$ of length $$$n$$$. We define the equality of the array as the number of indices $$$1 \le i \le n - 1$$$ such that $$$a_i = a_{i + 1}$$$. We are allowed to do the following operation: Select two integers $$$i$$$ and $$$x$$$ such that $$$1 \le i \le n - 1$$$ and $$$1 \le x \le 10^9$$$. Then, set $$$a_i$$$ and $$$a_{i + 1}$$$ to be equal to $$$x$$$. Find the minimum number of operations needed such that the equality of the array is less than or equal to $$$1$$$. | 256 megabytes | import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.OutputStream;
import java.io.PrintWriter;
import java.util.ArrayDeque;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Comparator;
import java.util.HashMap;
import java.util.Objects;
import java.util.Queue;
import java.util.StringTokenizer;
import java.util.function.BiFunction;
import java.util.function.Function;
public class Main {
static BiFunction<Integer, Integer, Integer> ADD = (x, y) -> (x + y);
static BiFunction<ArrayList<Integer>, ArrayList<Integer>, ArrayList<Integer>> ADD_ARRAY_LIST = (x, y) -> {
x.addAll(y);
return x;
};
static Function<Pair<Integer, Integer>, Integer> GET_FIRST = (x) -> (x.first);
static Function<Pair<Integer, Integer>, Integer> GET_SECOND = (x) -> (x.second);
static Comparator<Pair<Integer, Integer>> C = Comparator.comparing(x -> x.first + x.second);
static long MOD = 1_000_000_000 + 7;
public static void main(String[] args) throws Exception {
long startTime = System.nanoTime();
int t = in.nextInt();
while (t-- > 0) {
solve();
}
long endTime = System.nanoTime();
err.println("Execution Time : +" + (endTime - startTime) / 1000000 + " ms");
exit(0);
}
static void solve() {
int n = in.nextInt();
int[] data = in.readAllInts(n);
if (check(data) <= 1) {
out.println(0);
return;
}
int start = -1, end = -1;
for (int i = 0; i + 1 < n; i++) {
if (data[i] == data[i + 1]) {
if (start == -1) {
start = i + 1;
}
end = i;
}
}
int ans = (end - start);
if (ans == 0) {
ans++;
}
out.println(ans);
}
static int check(int[] data) {
int count = 0;
for (int i = 0; i + 1 < data.length; i++) {
if (data[i] == data[i + 1]) {
count++;
}
}
return count;
}
static void debug(Object... args) {
for (Object a : args) {
out.println(a);
}
}
static int dist(Pair<Integer, Integer> a, Pair<Integer, Integer> b) {
return Math.abs(a.first - b.first) + Math.abs(a.second - b.second);
}
static void y() {
out.println("YES");
}
static void n() {
out.println("NO");
}
static int[] stringToArray(String s) {
return s.chars().map(x -> Character.getNumericValue(x)).toArray();
}
static <T> T min(T a, T b, Comparator<T> C) {
if (C.compare(a, b) <= 0) {
return a;
}
return b;
}
static <T> T max(T a, T b, Comparator<T> C) {
if (C.compare(a, b) >= 0) {
return a;
}
return b;
}
static void fail() {
out.println("-1");
}
static class Pair<T, R> {
public T first;
public R second;
public Pair(T first, R second) {
this.first = first;
this.second = second;
}
@Override
public boolean equals(final Object o) {
if (this == o) {
return true;
}
if (o == null || getClass() != o.getClass()) {
return false;
}
final Pair<?, ?> pair = (Pair<?, ?>) o;
return Objects.equals(first, pair.first) && Objects.equals(second, pair.second);
}
@Override
public int hashCode() {
return Objects.hash(first, second);
}
@Override
public String toString() {
return "Pair{" + "a=" + first + ", b=" + second + '}';
}
public T getFirst() {
return first;
}
public R getSecond() {
return second;
}
}
static <T, R> Pair<T, R> make_pair(T a, R b) {
return new Pair<>(a, b);
}
static long mod_inverse(long a, long m) {
Number x = new Number(0);
Number y = new Number(0);
extended_gcd(a, m, x, y);
return (m + x.v % m) % m;
}
static long extended_gcd(long a, long b, Number x, Number y) {
long d = a;
if (b != 0) {
d = extended_gcd(b, a % b, y, x);
y.v -= (a / b) * x.v;
} else {
x.v = 1;
y.v = 0;
}
return d;
}
static class Number {
long v = 0;
public Number(long v) {
this.v = v;
}
}
static long lcm(long a, long b, long c) {
return lcm(a, lcm(b, c));
}
static long lcm(long a, long b) {
long p = 1L * a * b;
return p / gcd(a, b);
}
static long gcd(long a, long b) {
while (b != 0) {
long t = b;
b = a % b;
a = t;
}
return a;
}
static long gcd(long a, long b, long c) {
return gcd(a, gcd(b, c));
}
static class ArrayUtils {
static void swap(int[] a, int i, int j) {
int temp = a[i];
a[i] = a[j];
a[j] = temp;
}
static void swap(char[] a, int i, int j) {
char temp = a[i];
a[i] = a[j];
a[j] = temp;
}
static void print(char[] a) {
for (char c : a) {
out.print(c);
}
out.println("");
}
static int[] reverse(int[] data) {
int[] p = new int[data.length];
for (int i = 0, j = data.length - 1; i < data.length; i++, j--) {
p[i] = data[j];
}
return p;
}
static void prefixSum(long[] data) {
for (int i = 1; i < data.length; i++) {
data[i] += data[i - 1];
}
}
static void prefixSum(int[] data) {
for (int i = 1; i < data.length; i++) {
data[i] += data[i - 1];
}
}
static long[] reverse(long[] data) {
long[] p = new long[data.length];
for (int i = 0, j = data.length - 1; i < data.length; i++, j--) {
p[i] = data[j];
}
return p;
}
static char[] reverse(char[] data) {
char[] p = new char[data.length];
for (int i = 0, j = data.length - 1; i < data.length; i++, j--) {
p[i] = data[j];
}
return p;
}
static int[] MergeSort(int[] A) {
if (A.length > 1) {
int q = A.length / 2;
int[] left = new int[q];
int[] right = new int[A.length - q];
System.arraycopy(A, 0, left, 0, q);
System.arraycopy(A, q, right, 0, A.length - q);
int[] left_sorted = MergeSort(left);
int[] right_sorted = MergeSort(right);
return Merge(left_sorted, right_sorted);
} else {
return A;
}
}
static int[] Merge(int[] left, int[] right) {
int[] A = new int[left.length + right.length];
int i = 0;
int j = 0;
for (int k = 0; k < A.length; k++) {
// To handle left becoming empty
if (i == left.length && j < right.length) {
A[k] = right[j];
j++;
continue;
}
// To handle right becoming empty
if (j == right.length && i < left.length) {
A[k] = left[i];
i++;
continue;
}
if (left[i] <= right[j]) {
A[k] = left[i];
i++;
} else {
A[k] = right[j];
j++;
}
}
return A;
}
static long[] MergeSort(long[] A) {
if (A.length > 1) {
int q = A.length / 2;
long[] left = new long[q];
long[] right = new long[A.length - q];
System.arraycopy(A, 0, left, 0, q);
System.arraycopy(A, q, right, 0, A.length - q);
long[] left_sorted = MergeSort(left);
long[] right_sorted = MergeSort(right);
return Merge(left_sorted, right_sorted);
} else {
return A;
}
}
static long[] Merge(long[] left, long[] right) {
long[] A = new long[left.length + right.length];
int i = 0;
int j = 0;
for (int k = 0; k < A.length; k++) {
// To handle left becoming empty
if (i == left.length && j < right.length) {
A[k] = right[j];
j++;
continue;
}
// To handle right becoming empty
if (j == right.length && i < left.length) {
A[k] = left[i];
i++;
continue;
}
if (left[i] <= right[j]) {
A[k] = left[i];
i++;
} else {
A[k] = right[j];
j++;
}
}
return A;
}
static int upper_bound(int[] data, int num, int start) {
int low = start;
int high = data.length - 1;
int mid = 0;
int ans = -1;
while (low <= high) {
mid = (low + high) / 2;
if (data[mid] < num) {
low = mid + 1;
} else if (data[mid] >= num) {
high = mid - 1;
ans = mid;
}
}
if (ans == -1) {
return 100000000;
}
return data[ans];
}
static int lower_bound(int[] data, int num, int start) {
int low = start;
int high = data.length - 1;
int mid = 0;
int ans = -1;
while (low <= high) {
mid = (low + high) / 2;
if (data[mid] <= num) {
low = mid + 1;
ans = mid;
} else if (data[mid] > num) {
high = mid - 1;
}
}
if (ans == -1) {
return 100000000;
}
return data[ans];
}
}
static boolean[] primeSieve(int n) {
boolean[] primes = new boolean[n + 1];
Arrays.fill(primes, true);
primes[0] = false;
primes[1] = false;
for (int i = 2; i <= Math.sqrt(n); i++) {
if (primes[i]) {
for (int j = i * i; j <= n; j += i) {
primes[j] = false;
}
}
}
return primes;
}
// Iterative Version
static HashMap<Integer, Boolean> subsets_sum_iter(int[] data) {
HashMap<Integer, Boolean> temp = new HashMap<Integer, Boolean>();
temp.put(data[0], true);
for (int i = 1; i < data.length; i++) {
HashMap<Integer, Boolean> t1 = new HashMap<Integer, Boolean>(temp);
t1.put(data[i], true);
for (int j : temp.keySet()) {
t1.put(j + data[i], true);
}
temp = t1;
}
return temp;
}
static HashMap<Integer, Integer> subsets_sum_count(int[] data) {
HashMap<Integer, Integer> temp = new HashMap<>();
temp.put(data[0], 1);
for (int i = 1; i < data.length; i++) {
HashMap<Integer, Integer> t1 = new HashMap<>(temp);
t1.merge(data[i], 1, ADD);
for (int j : temp.keySet()) {
t1.merge(j + data[i], temp.get(j) + 1, ADD);
}
temp = t1;
}
return temp;
}
static class Graph {
ArrayList<Integer>[] g;
boolean[] visited;
ArrayList<Integer>[] graph(int n) {
g = new ArrayList[n];
visited = new boolean[n];
for (int i = 0; i < n; i++) {
g[i] = new ArrayList<>();
}
return g;
}
void BFS(int s) {
Queue<Integer> Q = new ArrayDeque<>();
visited[s] = true;
Q.add(s);
while (!Q.isEmpty()) {
int v = Q.poll();
for (int a : g[v]) {
if (!visited[a]) {
visited[a] = true;
Q.add(a);
}
}
}
}
}
static class SparseTable {
int[] log;
int[][] st;
public SparseTable(int n, int k, int[] data, BiFunction<Integer, Integer, Integer> f) {
log = new int[n + 1];
st = new int[n][k + 1];
log[1] = 0;
for (int i = 2; i <= n; i++) {
log[i] = log[i / 2] + 1;
}
for (int i = 0; i < data.length; i++) {
st[i][0] = data[i];
}
for (int j = 1; j <= k; j++)
for (int i = 0; i + (1 << j) <= data.length; i++)
st[i][j] = f.apply(st[i][j - 1], st[i + (1 << (j - 1))][j - 1]);
}
public int query(int L, int R, BiFunction<Integer, Integer, Integer> f) {
int j = log[R - L + 1];
return f.apply(st[L][j], st[R - (1 << j) + 1][j]);
}
}
static class InputReader {
public BufferedReader reader;
public StringTokenizer tokenizer;
public InputReader(InputStream stream) {
reader = new BufferedReader(new InputStreamReader(stream), 2048);
tokenizer = null;
}
public String next() {
while (tokenizer == null || !tokenizer.hasMoreTokens()) {
try {
tokenizer = new StringTokenizer(reader.readLine());
} catch (IOException e) {
throw new RuntimeException(e);
}
}
return tokenizer.nextToken();
}
public long nextLong() {
return Long.parseLong(next());
}
public int nextInt() {
return Integer.parseInt(next());
}
public int[] readAllInts(int n) {
int[] p = new int[n];
for (int i = 0; i < n; i++) {
p[i] = in.nextInt();
}
return p;
}
public int[] readAllInts(int n, int s) {
int[] p = new int[n + s];
for (int i = s; i < n + s; i++) {
p[i] = in.nextInt();
}
return p;
}
public long[] readAllLongs(int n) {
long[] p = new long[n];
for (int i = 0; i < n; i++) {
p[i] = in.nextLong();
}
return p;
}
public long[] readAllLongs(int n, int s) {
long[] p = new long[n + s];
for (int i = s; i < n + s; i++) {
p[i] = in.nextLong();
}
return p;
}
public double nextDouble() {
return Double.parseDouble(next());
}
}
static void exit(int a) {
out.close();
err.close();
System.exit(a);
}
static InputStream inputStream = System.in;
static OutputStream outputStream = System.out;
static OutputStream errStream = System.err;
static InputReader in = new InputReader(inputStream);
static PrintWriter out = new PrintWriter(outputStream);
static PrintWriter err = new PrintWriter(errStream);
static BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
}
| Java | ["4\n\n5\n\n1 1 1 1 1\n\n5\n\n2 1 1 1 2\n\n6\n\n1 1 2 3 3 4\n\n6\n\n1 2 1 4 5 4"] | 1 second | ["2\n1\n2\n0"] | NoteIn the first test case, we can select $$$i=2$$$ and $$$x=2$$$ to form $$$[1, 2, 2, 1, 1]$$$. Then, we can select $$$i=3$$$ and $$$x=3$$$ to form $$$[1, 2, 3, 3, 1]$$$.In the second test case, we can select $$$i=3$$$ and $$$x=100$$$ to form $$$[2, 1, 100, 100, 2]$$$. | Java 11 | standard input | [
"constructive algorithms",
"greedy",
"implementation"
] | a91be662101762bcb2c58b8db9ff61e0 | Each test contains multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains an integer $$$n$$$ ($$$2 \le n \le 2 \cdot 10 ^ 5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10^9$$$) — elements of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10 ^ 5$$$ | 1,100 | For each test case, print the minimum number of operations needed. | standard output | |
PASSED | 3f9f4a55e2746991556ffee60faf64b8 | train_108.jsonl | 1650722700 | You are given an array $$$a$$$ of length $$$n$$$. We define the equality of the array as the number of indices $$$1 \le i \le n - 1$$$ such that $$$a_i = a_{i + 1}$$$. We are allowed to do the following operation: Select two integers $$$i$$$ and $$$x$$$ such that $$$1 \le i \le n - 1$$$ and $$$1 \le x \le 10^9$$$. Then, set $$$a_i$$$ and $$$a_{i + 1}$$$ to be equal to $$$x$$$. Find the minimum number of operations needed such that the equality of the array is less than or equal to $$$1$$$. | 256 megabytes | import java.util.*;
import java.io.*;
public class Main {
public static void main(String args[]) throws IOException{
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
// Scanner scn=new Scanner(System.in);
int t = Integer.parseInt(br.readLine());
while (t-- > 0) {
int n=Integer.parseInt(br.readLine());
String line=br.readLine();
String strs[]=line.split(" ");
int[] a=new int[n];
for(int i=0;i<n;i++){
a[i]=Integer.parseInt(strs[i]);
}
int si=-1;
int ei=-1;
// for(int i=0;i<n;i++){
// System.out.print(a[i]+" ");
// }
// System.out.println();
for(int i=0;i<n-1;i++){
if(a[i]==a[i+1]){
if(si==-1){
si=i;
}
ei=i+1;
}
}
int range=ei-si+1;
// System.out.println("Range-"+range);
if(range==1 || range==2){
System.out.println(0);
} else if(range==3){
System.out.println(range-2);
} else {
System.out.println(range-3);
}
}
}
} | Java | ["4\n\n5\n\n1 1 1 1 1\n\n5\n\n2 1 1 1 2\n\n6\n\n1 1 2 3 3 4\n\n6\n\n1 2 1 4 5 4"] | 1 second | ["2\n1\n2\n0"] | NoteIn the first test case, we can select $$$i=2$$$ and $$$x=2$$$ to form $$$[1, 2, 2, 1, 1]$$$. Then, we can select $$$i=3$$$ and $$$x=3$$$ to form $$$[1, 2, 3, 3, 1]$$$.In the second test case, we can select $$$i=3$$$ and $$$x=100$$$ to form $$$[2, 1, 100, 100, 2]$$$. | Java 11 | standard input | [
"constructive algorithms",
"greedy",
"implementation"
] | a91be662101762bcb2c58b8db9ff61e0 | Each test contains multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains an integer $$$n$$$ ($$$2 \le n \le 2 \cdot 10 ^ 5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10^9$$$) — elements of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10 ^ 5$$$ | 1,100 | For each test case, print the minimum number of operations needed. | standard output | |
PASSED | 005a02abbb92df40667584f4166c38fd | train_108.jsonl | 1650722700 | You are given an array $$$a$$$ of length $$$n$$$. We define the equality of the array as the number of indices $$$1 \le i \le n - 1$$$ such that $$$a_i = a_{i + 1}$$$. We are allowed to do the following operation: Select two integers $$$i$$$ and $$$x$$$ such that $$$1 \le i \le n - 1$$$ and $$$1 \le x \le 10^9$$$. Then, set $$$a_i$$$ and $$$a_{i + 1}$$$ to be equal to $$$x$$$. Find the minimum number of operations needed such that the equality of the array is less than or equal to $$$1$$$. | 256 megabytes | import java.io.*;
import java.util.*;
public class A {
static PrintWriter pw;
static Scanner sc;
public static void main(String[] args) throws IOException {
sc = new Scanner(System.in);
pw = new PrintWriter(System.out);
int t = sc.nextInt();
while (t-- > 0) {
int n = sc.nextInt();
int[] arr = sc.nextIntArray(n);
if (n == 2) {
pw.println(0);
} else {
int cnt = 0;
for (int i = 1; i < n; i++) {
if (arr[i] == arr[i - 1])
cnt++;
}
if (cnt <= 1) {
pw.println(0);
} else {
int first = -1;
int last = 0;
for (int i = 1; i < n; i++) {
if (arr[i] == arr[i - 1]) {
if (first == -1) {
first = i;
}
last = i - 1;
}
}
// pw.println(first+" "+last);
int ans = 0;
if (first == last)
ans++;
boolean f = true;
while (first != last) {
if (f)
first++;
else
last--;
ans++;
}
pw.println(ans);
}
}
}
pw.flush();
}
static class Scanner {
StringTokenizer st;
BufferedReader br;
public Scanner(InputStream s) {
br = new BufferedReader(new InputStreamReader(s));
}
public Scanner(FileReader r) {
br = new BufferedReader(r);
}
public String readAllLines(BufferedReader reader) throws IOException {
StringBuilder content = new StringBuilder();
String line;
while ((line = reader.readLine()) != null) {
content.append(line);
content.append(System.lineSeparator());
}
return content.toString();
}
public String next() throws IOException {
while (st == null || !st.hasMoreTokens())
st = new StringTokenizer(br.readLine());
return st.nextToken();
}
public int nextInt() throws IOException {
return Integer.parseInt(next());
}
public long nextLong() throws IOException {
return Long.parseLong(next());
}
public String nextLine() throws IOException {
return br.readLine();
}
public double nextDouble() throws IOException {
String x = next();
StringBuilder sb = new StringBuilder("0");
double res = 0, f = 1;
boolean dec = false, neg = false;
int start = 0;
if (x.charAt(0) == '-') {
neg = true;
start++;
}
for (int i = start; i < x.length(); i++)
if (x.charAt(i) == '.') {
res = Long.parseLong(sb.toString());
sb = new StringBuilder("0");
dec = true;
} else {
sb.append(x.charAt(i));
if (dec)
f *= 10;
}
res += Long.parseLong(sb.toString()) / f;
return res * (neg ? -1 : 1);
}
public long[] nextlongArray(int n) throws IOException {
long[] a = new long[n];
for (int i = 0; i < n; i++)
a[i] = nextLong();
return a;
}
public Long[] nextLongArray(int n) throws IOException {
Long[] a = new Long[n];
for (int i = 0; i < n; i++)
a[i] = nextLong();
return a;
}
public int[] nextIntArray(int n) throws IOException {
int[] a = new int[n];
for (int i = 0; i < n; i++)
a[i] = nextInt();
return a;
}
public Integer[] nextIntegerArray(int n) throws IOException {
Integer[] a = new Integer[n];
for (int i = 0; i < n; i++)
a[i] = nextInt();
return a;
}
public boolean ready() throws IOException {
return br.ready();
}
}
} | Java | ["4\n\n5\n\n1 1 1 1 1\n\n5\n\n2 1 1 1 2\n\n6\n\n1 1 2 3 3 4\n\n6\n\n1 2 1 4 5 4"] | 1 second | ["2\n1\n2\n0"] | NoteIn the first test case, we can select $$$i=2$$$ and $$$x=2$$$ to form $$$[1, 2, 2, 1, 1]$$$. Then, we can select $$$i=3$$$ and $$$x=3$$$ to form $$$[1, 2, 3, 3, 1]$$$.In the second test case, we can select $$$i=3$$$ and $$$x=100$$$ to form $$$[2, 1, 100, 100, 2]$$$. | Java 11 | standard input | [
"constructive algorithms",
"greedy",
"implementation"
] | a91be662101762bcb2c58b8db9ff61e0 | Each test contains multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains an integer $$$n$$$ ($$$2 \le n \le 2 \cdot 10 ^ 5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10^9$$$) — elements of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10 ^ 5$$$ | 1,100 | For each test case, print the minimum number of operations needed. | standard output | |
PASSED | 24527b1d377ff444aae1254e2f8b3186 | train_108.jsonl | 1650722700 | You are given an array $$$a$$$ of length $$$n$$$. We define the equality of the array as the number of indices $$$1 \le i \le n - 1$$$ such that $$$a_i = a_{i + 1}$$$. We are allowed to do the following operation: Select two integers $$$i$$$ and $$$x$$$ such that $$$1 \le i \le n - 1$$$ and $$$1 \le x \le 10^9$$$. Then, set $$$a_i$$$ and $$$a_{i + 1}$$$ to be equal to $$$x$$$. Find the minimum number of operations needed such that the equality of the array is less than or equal to $$$1$$$. | 256 megabytes | import java.io.*;
import java.util.*;
public class A {
static PrintWriter pw;
static Scanner sc;
public static void main(String[] args) throws IOException {
sc = new Scanner(System.in);
pw = new PrintWriter(System.out);
int t = sc.nextInt();
while (t-- > 0) {
int n = sc.nextInt();
int[] arr = sc.nextIntArray(n);
if (n == 2) {
pw.println(0);
} else {
int cnt = 0;
for (int i = 1; i < n; i++) {
if (arr[i] == arr[i - 1])
cnt++;
}
if (cnt <= 1) {
pw.println(0);
} else {
int first = -1;
int last = 0;
for (int i = 1; i < n; i++) {
if (arr[i] == arr[i - 1]) {
if (first == -1) {
first = i;
}
last = i - 1;
}
}
// pw.println(first+" "+last);
int ans = 0;
if (first == last)
ans++;
boolean f = true;
while (first != last) {
if (f)
first++;
else
last--;
f=!f;
ans++;
}
pw.println(ans);
}
}
}
pw.flush();
}
static class Scanner {
StringTokenizer st;
BufferedReader br;
public Scanner(InputStream s) {
br = new BufferedReader(new InputStreamReader(s));
}
public Scanner(FileReader r) {
br = new BufferedReader(r);
}
public String readAllLines(BufferedReader reader) throws IOException {
StringBuilder content = new StringBuilder();
String line;
while ((line = reader.readLine()) != null) {
content.append(line);
content.append(System.lineSeparator());
}
return content.toString();
}
public String next() throws IOException {
while (st == null || !st.hasMoreTokens())
st = new StringTokenizer(br.readLine());
return st.nextToken();
}
public int nextInt() throws IOException {
return Integer.parseInt(next());
}
public long nextLong() throws IOException {
return Long.parseLong(next());
}
public String nextLine() throws IOException {
return br.readLine();
}
public double nextDouble() throws IOException {
String x = next();
StringBuilder sb = new StringBuilder("0");
double res = 0, f = 1;
boolean dec = false, neg = false;
int start = 0;
if (x.charAt(0) == '-') {
neg = true;
start++;
}
for (int i = start; i < x.length(); i++)
if (x.charAt(i) == '.') {
res = Long.parseLong(sb.toString());
sb = new StringBuilder("0");
dec = true;
} else {
sb.append(x.charAt(i));
if (dec)
f *= 10;
}
res += Long.parseLong(sb.toString()) / f;
return res * (neg ? -1 : 1);
}
public long[] nextlongArray(int n) throws IOException {
long[] a = new long[n];
for (int i = 0; i < n; i++)
a[i] = nextLong();
return a;
}
public Long[] nextLongArray(int n) throws IOException {
Long[] a = new Long[n];
for (int i = 0; i < n; i++)
a[i] = nextLong();
return a;
}
public int[] nextIntArray(int n) throws IOException {
int[] a = new int[n];
for (int i = 0; i < n; i++)
a[i] = nextInt();
return a;
}
public Integer[] nextIntegerArray(int n) throws IOException {
Integer[] a = new Integer[n];
for (int i = 0; i < n; i++)
a[i] = nextInt();
return a;
}
public boolean ready() throws IOException {
return br.ready();
}
}
} | Java | ["4\n\n5\n\n1 1 1 1 1\n\n5\n\n2 1 1 1 2\n\n6\n\n1 1 2 3 3 4\n\n6\n\n1 2 1 4 5 4"] | 1 second | ["2\n1\n2\n0"] | NoteIn the first test case, we can select $$$i=2$$$ and $$$x=2$$$ to form $$$[1, 2, 2, 1, 1]$$$. Then, we can select $$$i=3$$$ and $$$x=3$$$ to form $$$[1, 2, 3, 3, 1]$$$.In the second test case, we can select $$$i=3$$$ and $$$x=100$$$ to form $$$[2, 1, 100, 100, 2]$$$. | Java 11 | standard input | [
"constructive algorithms",
"greedy",
"implementation"
] | a91be662101762bcb2c58b8db9ff61e0 | Each test contains multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains an integer $$$n$$$ ($$$2 \le n \le 2 \cdot 10 ^ 5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10^9$$$) — elements of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10 ^ 5$$$ | 1,100 | For each test case, print the minimum number of operations needed. | standard output | |
PASSED | c6dad25c04fce92afe6ef99ba67921d2 | train_108.jsonl | 1650722700 | You are given an array $$$a$$$ of length $$$n$$$. We define the equality of the array as the number of indices $$$1 \le i \le n - 1$$$ such that $$$a_i = a_{i + 1}$$$. We are allowed to do the following operation: Select two integers $$$i$$$ and $$$x$$$ such that $$$1 \le i \le n - 1$$$ and $$$1 \le x \le 10^9$$$. Then, set $$$a_i$$$ and $$$a_{i + 1}$$$ to be equal to $$$x$$$. Find the minimum number of operations needed such that the equality of the array is less than or equal to $$$1$$$. | 256 megabytes |
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
public class Main {
public static void main(String[] args) throws IOException {
BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
int t=Integer.parseInt(br.readLine());
while(t-->0){
br.readLine();
String[] s= br.readLine().split(" ");
int began=0;
for (int i = 0; i < s.length-1; i++) {
if (s[i].equals(s[i+1])){
began=i;
break;
}
}
int last=0;
for (int i = s.length-1; i >0 ; i--) {
// System.out.println(s);
if (s[i].equals(s[i-1])){
last=i;
break;
}
}
// System.out.println(began+" "+last);
if (last==0){
System.out.println(0);
}
else{
if (last-began-2==0)
System.out.println(1);
else
if (last-began-2<0)
System.out.println(0);
else
System.out.println(last-began-2);
}
}
}
}
| Java | ["4\n\n5\n\n1 1 1 1 1\n\n5\n\n2 1 1 1 2\n\n6\n\n1 1 2 3 3 4\n\n6\n\n1 2 1 4 5 4"] | 1 second | ["2\n1\n2\n0"] | NoteIn the first test case, we can select $$$i=2$$$ and $$$x=2$$$ to form $$$[1, 2, 2, 1, 1]$$$. Then, we can select $$$i=3$$$ and $$$x=3$$$ to form $$$[1, 2, 3, 3, 1]$$$.In the second test case, we can select $$$i=3$$$ and $$$x=100$$$ to form $$$[2, 1, 100, 100, 2]$$$. | Java 11 | standard input | [
"constructive algorithms",
"greedy",
"implementation"
] | a91be662101762bcb2c58b8db9ff61e0 | Each test contains multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains an integer $$$n$$$ ($$$2 \le n \le 2 \cdot 10 ^ 5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10^9$$$) — elements of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10 ^ 5$$$ | 1,100 | For each test case, print the minimum number of operations needed. | standard output | |
PASSED | 21723bb6cf39b65bbc2407b741eec664 | train_108.jsonl | 1650722700 | You are given an array $$$a$$$ of length $$$n$$$. We define the equality of the array as the number of indices $$$1 \le i \le n - 1$$$ such that $$$a_i = a_{i + 1}$$$. We are allowed to do the following operation: Select two integers $$$i$$$ and $$$x$$$ such that $$$1 \le i \le n - 1$$$ and $$$1 \le x \le 10^9$$$. Then, set $$$a_i$$$ and $$$a_{i + 1}$$$ to be equal to $$$x$$$. Find the minimum number of operations needed such that the equality of the array is less than or equal to $$$1$$$. | 256 megabytes | /*
"Everything in the universe is balanced. Every disappointment
you face in life will be balanced by something good for you!
Keep going, never give up."
Just have Patience + 1...
*/
import java.util.*;
import java.lang.*;
import java.io.*;
public class Solution {
public static void main(String[] args) throws java.lang.Exception {
out = new PrintWriter(new BufferedOutputStream(System.out));
sc = new FastReader();
int test = sc.nextInt();
for (int t = 1; t <= test; t++) {
solve();
}
out.close();
}
private static void solve() {
int n = sc.nextInt();
int[] arr = new int[n];
for (int i = 0; i < n; i++) {
arr[i] = sc.nextInt();
}
int start = 0, end = 0;
for (int i = 0; i < n - 1; i++) {
if (arr[i] == arr[i + 1]) {
start = i;
break;
}
}
for (int i = n - 1; i > 0; i--) {
if (arr[i] == arr[i - 1]) {
end = i;
break;
}
}
long remainingSize = end - start + 1;
if (remainingSize < 3) {
out.println(0);
return;
}
if (remainingSize == 3 || remainingSize == 4) {
out.println(1);
return;
}
out.println(remainingSize - 3);
}
public static FastReader sc;
public static PrintWriter out;
static class FastReader
{
BufferedReader br;
StringTokenizer str;
public FastReader()
{
br = new BufferedReader(new
InputStreamReader(System.in));
}
String next()
{
while (str == null || !str.hasMoreElements())
{
try
{
str = new StringTokenizer(br.readLine());
}
catch (IOException lastMonthOfVacation)
{
lastMonthOfVacation.printStackTrace();
}
}
return str.nextToken();
}
int nextInt()
{
return Integer.parseInt(next());
}
long nextLong()
{
return Long.parseLong(next());
}
double nextDouble()
{
return Double.parseDouble(next());
}
String nextLine()
{
String str = "";
try
{
str = br.readLine();
}
catch (IOException lastMonthOfVacation)
{
lastMonthOfVacation.printStackTrace();
}
return str;
}
}
} | Java | ["4\n\n5\n\n1 1 1 1 1\n\n5\n\n2 1 1 1 2\n\n6\n\n1 1 2 3 3 4\n\n6\n\n1 2 1 4 5 4"] | 1 second | ["2\n1\n2\n0"] | NoteIn the first test case, we can select $$$i=2$$$ and $$$x=2$$$ to form $$$[1, 2, 2, 1, 1]$$$. Then, we can select $$$i=3$$$ and $$$x=3$$$ to form $$$[1, 2, 3, 3, 1]$$$.In the second test case, we can select $$$i=3$$$ and $$$x=100$$$ to form $$$[2, 1, 100, 100, 2]$$$. | Java 11 | standard input | [
"constructive algorithms",
"greedy",
"implementation"
] | a91be662101762bcb2c58b8db9ff61e0 | Each test contains multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains an integer $$$n$$$ ($$$2 \le n \le 2 \cdot 10 ^ 5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10^9$$$) — elements of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10 ^ 5$$$ | 1,100 | For each test case, print the minimum number of operations needed. | standard output | |
PASSED | e949feb3552b3bcde42402306891c98a | train_108.jsonl | 1650722700 | You are given an array $$$a$$$ of length $$$n$$$. We define the equality of the array as the number of indices $$$1 \le i \le n - 1$$$ such that $$$a_i = a_{i + 1}$$$. We are allowed to do the following operation: Select two integers $$$i$$$ and $$$x$$$ such that $$$1 \le i \le n - 1$$$ and $$$1 \le x \le 10^9$$$. Then, set $$$a_i$$$ and $$$a_{i + 1}$$$ to be equal to $$$x$$$. Find the minimum number of operations needed such that the equality of the array is less than or equal to $$$1$$$. | 256 megabytes | import java.util.*;
import java.io.*;
public class Main {
public static Scanner obj = new Scanner(System.in);
public static PrintWriter out = new PrintWriter(System.out);
public static int i() {
return obj.nextInt();
}
public static void main(String[] args) {
int len = i();
while (len-- != 0) {
int n= obj.nextInt();
long[] a=new long[n];
long v=(long)1e9+1;
for(int i=0;i<n;i++)a[i]=obj.nextInt();
int ans=0;
int e=n-1;
for(int i=n-1;i>=1;i--)
{
if(a[i]==a[i-1]) {
e=i-1;
break;
}
}
for(int i=0;i<e-1;i++)
{
if(a[i]==a[i+1]) {
a[i+1]=v+ans;
a[i+2]=v+ans;
ans++;
}
}
//for(int i=0;i<n;i++)out.print(a[i]+" ");
if(ans==0)
{
for(int i=0;i<n-1;i++)if(a[i]==a[i+1])ans+=1;
if(ans>0)ans-=1;
}
out.println(ans);
}
out.flush();
}
} | Java | ["4\n\n5\n\n1 1 1 1 1\n\n5\n\n2 1 1 1 2\n\n6\n\n1 1 2 3 3 4\n\n6\n\n1 2 1 4 5 4"] | 1 second | ["2\n1\n2\n0"] | NoteIn the first test case, we can select $$$i=2$$$ and $$$x=2$$$ to form $$$[1, 2, 2, 1, 1]$$$. Then, we can select $$$i=3$$$ and $$$x=3$$$ to form $$$[1, 2, 3, 3, 1]$$$.In the second test case, we can select $$$i=3$$$ and $$$x=100$$$ to form $$$[2, 1, 100, 100, 2]$$$. | Java 11 | standard input | [
"constructive algorithms",
"greedy",
"implementation"
] | a91be662101762bcb2c58b8db9ff61e0 | Each test contains multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains an integer $$$n$$$ ($$$2 \le n \le 2 \cdot 10 ^ 5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10^9$$$) — elements of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10 ^ 5$$$ | 1,100 | For each test case, print the minimum number of operations needed. | standard output | |
PASSED | 98be5f2f5395d044c850ed4e01aec2d3 | train_108.jsonl | 1650722700 | You are given an array $$$a$$$ of length $$$n$$$. We define the equality of the array as the number of indices $$$1 \le i \le n - 1$$$ such that $$$a_i = a_{i + 1}$$$. We are allowed to do the following operation: Select two integers $$$i$$$ and $$$x$$$ such that $$$1 \le i \le n - 1$$$ and $$$1 \le x \le 10^9$$$. Then, set $$$a_i$$$ and $$$a_{i + 1}$$$ to be equal to $$$x$$$. Find the minimum number of operations needed such that the equality of the array is less than or equal to $$$1$$$. | 256 megabytes | import java.util.*;
import java.io.*;
public class C {
static class Scan {
private byte[] buf=new byte[1024];
private int index;
private InputStream in;
private int total;
public Scan()
{
in=System.in;
}
public int scan()throws IOException
{
if(total<0)
throw new InputMismatchException();
if(index>=total)
{
index=0;
total=in.read(buf);
if(total<=0)
return -1;
}
return buf[index++];
}
public int scanInt()throws IOException
{
int integer=0;
int n=scan();
while(isWhiteSpace(n))
n=scan();
int neg=1;
if(n=='-')
{
neg=-1;
n=scan();
}
while(!isWhiteSpace(n))
{
if(n>='0'&&n<='9')
{
integer*=10;
integer+=n-'0';
n=scan();
}
else throw new InputMismatchException();
}
return neg*integer;
}
public double scanDouble()throws IOException
{
double doub=0;
int n=scan();
while(isWhiteSpace(n))
n=scan();
int neg=1;
if(n=='-')
{
neg=-1;
n=scan();
}
while(!isWhiteSpace(n)&&n!='.')
{
if(n>='0'&&n<='9')
{
doub*=10;
doub+=n-'0';
n=scan();
}
else throw new InputMismatchException();
}
if(n=='.')
{
n=scan();
double temp=1;
while(!isWhiteSpace(n))
{
if(n>='0'&&n<='9')
{
temp/=10;
doub+=(n-'0')*temp;
n=scan();
}
else throw new InputMismatchException();
}
}
return doub*neg;
}
public String scanString()throws IOException
{
StringBuilder sb=new StringBuilder();
int n=scan();
while(isWhiteSpace(n))
n=scan();
while(!isWhiteSpace(n))
{
sb.append((char)n);
n=scan();
}
return sb.toString();
}
private boolean isWhiteSpace(int n)
{
if(n==' '||n=='\n'||n=='\r'||n=='\t'||n==-1)
return true;
return false;
}
}
public static void sort(int arr[],int l,int r) { //sort(arr,0,n-1);
if(l==r) {
return;
}
int mid=(l+r)/2;
sort(arr,l,mid);
sort(arr,mid+1,r);
merge(arr,l,mid,mid+1,r);
}
public static void merge(int arr[],int l1,int r1,int l2,int r2) {
int tmp[]=new int[r2-l1+1];
int indx1=l1,indx2=l2;
//sorting the two halves using a tmp array
for(int i=0;i<tmp.length;i++) {
if(indx1>r1) {
tmp[i]=arr[indx2];
indx2++;
continue;
}
if(indx2>r2) {
tmp[i]=arr[indx1];
indx1++;
continue;
}
if(arr[indx1]<arr[indx2]) {
tmp[i]=arr[indx1];
indx1++;
continue;
}
tmp[i]=arr[indx2];
indx2++;
}
//Copying the elements of tmp into the main array
for(int i=0,j=l1;i<tmp.length;i++,j++) {
arr[j]=tmp[i];
}
}
public static void sort(long arr[],int l,int r) { //sort(arr,0,n-1);
if(l==r) {
return;
}
int mid=(l+r)/2;
sort(arr,l,mid);
sort(arr,mid+1,r);
merge(arr,l,mid,mid+1,r);
}
public static void merge(long arr[],int l1,int r1,int l2,int r2) {
long tmp[]=new long[r2-l1+1];
int indx1=l1,indx2=l2;
//sorting the two halves using a tmp array
for(int i=0;i<tmp.length;i++) {
if(indx1>r1) {
tmp[i]=arr[indx2];
indx2++;
continue;
}
if(indx2>r2) {
tmp[i]=arr[indx1];
indx1++;
continue;
}
if(arr[indx1]<arr[indx2]) {
tmp[i]=arr[indx1];
indx1++;
continue;
}
tmp[i]=arr[indx2];
indx2++;
}
//Copying the elements of tmp into the main array
for(int i=0,j=l1;i<tmp.length;i++,j++) {
arr[j]=tmp[i];
}
}
public static void main(String args[]) throws IOException {
Scan input=new Scan();
StringBuilder ans=new StringBuilder("");
int test=input.scanInt();
for(int tt=1;tt<=test;tt++) {
int n=input.scanInt();
int arr[]=new int[n];
for(int i=0;i<n;i++) {
arr[i]=input.scanInt();
}
int min=Integer.MAX_VALUE;
min=Math.min(min,solve(n,arr));
for(int i=0,j=n-1;i<=j;i++,j--) {
int tmp=arr[i];
arr[i]=arr[j];
arr[j]=tmp;
}
min=Math.min(min,solve(n,arr));
ans.append(min+"\n");
}
System.out.println(ans);
}
public static int solve(int n,int arr[]) {
int l_indx=-1;
int cnt=0;
for(int i=0;i<n-1;i++) {
if(arr[i]==arr[i+1]) {
if(cnt==0) {
l_indx=i+1;
}
cnt++;
}
}
if(cnt==1 || l_indx==-1) {
return 0;
}
int r_indx=-1;
for(int i=n-1;i>=1;i--) {
if(arr[i]==arr[i-1]) {
r_indx=i-1;
break;
}
}
int val=r_indx-l_indx;
if(l_indx==r_indx) {
val=1;
}
return val;
}
}
| Java | ["4\n\n5\n\n1 1 1 1 1\n\n5\n\n2 1 1 1 2\n\n6\n\n1 1 2 3 3 4\n\n6\n\n1 2 1 4 5 4"] | 1 second | ["2\n1\n2\n0"] | NoteIn the first test case, we can select $$$i=2$$$ and $$$x=2$$$ to form $$$[1, 2, 2, 1, 1]$$$. Then, we can select $$$i=3$$$ and $$$x=3$$$ to form $$$[1, 2, 3, 3, 1]$$$.In the second test case, we can select $$$i=3$$$ and $$$x=100$$$ to form $$$[2, 1, 100, 100, 2]$$$. | Java 11 | standard input | [
"constructive algorithms",
"greedy",
"implementation"
] | a91be662101762bcb2c58b8db9ff61e0 | Each test contains multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains an integer $$$n$$$ ($$$2 \le n \le 2 \cdot 10 ^ 5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10^9$$$) — elements of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10 ^ 5$$$ | 1,100 | For each test case, print the minimum number of operations needed. | standard output | |
PASSED | a27c878f827e28141725091c0e8361dc | train_108.jsonl | 1650722700 | You are given an array $$$a$$$ of length $$$n$$$. We define the equality of the array as the number of indices $$$1 \le i \le n - 1$$$ such that $$$a_i = a_{i + 1}$$$. We are allowed to do the following operation: Select two integers $$$i$$$ and $$$x$$$ such that $$$1 \le i \le n - 1$$$ and $$$1 \le x \le 10^9$$$. Then, set $$$a_i$$$ and $$$a_{i + 1}$$$ to be equal to $$$x$$$. Find the minimum number of operations needed such that the equality of the array is less than or equal to $$$1$$$. | 256 megabytes | import java.io.*;
import java.math.*;
import java.util.*;
public class Main {
static double pi = 3.141592653589;
static int mod = (int) 1e9 + 7;
public static void main(String[] args) throws IOException {
int qwe = in.nextInt();
while (qwe-- > 0) {
solve();
}
out.close();
}
public static void solve() {
int n = in.nextInt();
int[] num = new int[n];
for (int i = 0; i < n; i++) {
num[i] = in.nextInt();
}
int idx1 = -1;
for (int i = 0; i < n-1; i++) {
if (num[i]==num[i+1]){
idx1 = i+1;
break;
}
}
int idx2 = -1;
for (int i = n-2; i >=0; i--) {
if (num[i]==num[i+1]){
idx2 = i;
break;
}
}
if (idx1==-1){
System.out.println(0);
}else if (idx1==idx2){
System.out.println(1);
}else if (idx2-idx1==-1){
System.out.println(0);
}else {
System.out.println(idx2-idx1);
}
}
/*
6
-1 0 1 2 -1 -4
*/
// static int N = 100010;
// static int idx = 0; //节点位置
// static char[] str = new char[N];
// static int[] cnt = new int[N];
// static int[][] son = new int[N][26];
// public static void insert(char[] str) {
// int p = 0; //从根节点出发
// for (char i : str) {
// int u = i - 'a';
// if (son[p][u] == 0) son[p][u] = ++idx;
// p = son[p][u];
// }
// cnt[p]++;
// }
// public static int query(char[] str) {
// int p = 0;
// for (char i : str) {
// int u = i - 'a';
// if (son[p][u] == 0) return 0;
// p = son[p][u];
// }
// return cnt[p];
// }
// static int bsearch_l(int l, int r, long target) {
// while (l < r) {
// int mid = l + r >> 1;
// if (num[mid] >= target) r = mid;
// else l = mid + 1;
// }
// return l;
// }
//
// static int bsearch_r(int l, int r, long target) {
// while (l < r) {
// int mid = l + r + 1 >> 1;
// if (num[mid] <= target) l = mid;
// else r = mid - 1;
// }
// return l;
// }
// static double bsearch_d(double l, double r) {
// double eps = 1e-8;
// while (Math.abs(r - l) > eps) {
// double mid = (l + r) / 2;
// if (check(mid)) r = mid;
// else l = mid;
// }
// return l;
// }
// static boolean check(int mid) {
// return true;
// }
public static long ksm(long n, long m) {//本质为n^m
long res = 1, base = n;
while (m != 0) {
if ((m & 1) == 1) { //等价于m%2==1，也就是m为奇数时
res = mul(res, base) % mod;
}
base = mul(base, base) % mod;//m为偶数时，base自乘
m = m >> 1;//等价于m/2
}
return res % mod;
}
static long mul(long a, long b) {//本质为a*b
long ans = 0;
while (b != 0) {
if ((b & 1) == 1) {
ans = (ans + a) % mod;
}
a = (a + a) % mod;
b = b >> 1;
}
return ans % mod;
}
public static long cnm(int a, int b) {
long sum = 1;
int i = a, j = 1;
while (j <= b) {
sum = sum * i / j;
i--;
j++;
}
return sum;
}
public static long gcd(long a, long b) {
return b == 0 ? a : gcd(b, a % b);
}
public static long lcm(long a, long b) {
return (a * b) / gcd(a, b);
}
public static void gbSort(int[] a, int l, int r) {
if (l < r) {
int m = (l + r) >> 1;
gbSort(a, l, m);
gbSort(a, m + 1, r);
int[] t = new int[r - l + 1];
int idx = 0, i = l, j = m + 1;
while (i <= m && j <= r)
if (a[i] <= a[j])
t[idx++] = a[i++];
else
t[idx++] = a[j++];
while (i <= m)
t[idx++] = a[i++];
while (j <= r)
t[idx++] = a[j++];
for (int z = 0; z < t.length; z++)
a[l + z] = t[z];
}
}
static InputReader in = new InputReader(System.in);
static PrintWriter out = new PrintWriter(System.out);
static class InputReader {
public BufferedReader reader;
public StringTokenizer tokenizer;
public InputReader(InputStream stream) {
reader = new BufferedReader(new InputStreamReader(stream), 32768);
tokenizer = null;
}
public String next() {
while (tokenizer == null || !tokenizer.hasMoreTokens()) {
try {
tokenizer = new StringTokenizer(reader.readLine());
} catch (IOException e) {
throw new RuntimeException(e);
}
}
return tokenizer.nextToken();
}
boolean hasNext() {
while (tokenizer == null || !tokenizer.hasMoreTokens()) {
try {
tokenizer = new StringTokenizer(reader.readLine());
} catch (Exception e) {
return false;
// TODO: handle exception
}
}
return true;
}
public String nextLine() {
String str = null;
try {
str = reader.readLine();
} catch (IOException e) {
e.printStackTrace();
}
return str;
}
public int nextInt() {
return Integer.parseInt(next());
}
public long nextLong() {
return Long.parseLong(next());
}
public Double nextDouble() {
return Double.parseDouble(next());
}
public BigInteger nextBigInteger() {
return new BigInteger(next());
}
public BigDecimal nextBigDecimal() {
return new BigDecimal(next());
}
}
}
| Java | ["4\n\n5\n\n1 1 1 1 1\n\n5\n\n2 1 1 1 2\n\n6\n\n1 1 2 3 3 4\n\n6\n\n1 2 1 4 5 4"] | 1 second | ["2\n1\n2\n0"] | NoteIn the first test case, we can select $$$i=2$$$ and $$$x=2$$$ to form $$$[1, 2, 2, 1, 1]$$$. Then, we can select $$$i=3$$$ and $$$x=3$$$ to form $$$[1, 2, 3, 3, 1]$$$.In the second test case, we can select $$$i=3$$$ and $$$x=100$$$ to form $$$[2, 1, 100, 100, 2]$$$. | Java 11 | standard input | [
"constructive algorithms",
"greedy",
"implementation"
] | a91be662101762bcb2c58b8db9ff61e0 | Each test contains multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains an integer $$$n$$$ ($$$2 \le n \le 2 \cdot 10 ^ 5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10^9$$$) — elements of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10 ^ 5$$$ | 1,100 | For each test case, print the minimum number of operations needed. | standard output | |
PASSED | 4706d4ef91f2aafe516497b22f705297 | train_108.jsonl | 1650722700 | You are given an array $$$a$$$ of length $$$n$$$. We define the equality of the array as the number of indices $$$1 \le i \le n - 1$$$ such that $$$a_i = a_{i + 1}$$$. We are allowed to do the following operation: Select two integers $$$i$$$ and $$$x$$$ such that $$$1 \le i \le n - 1$$$ and $$$1 \le x \le 10^9$$$. Then, set $$$a_i$$$ and $$$a_{i + 1}$$$ to be equal to $$$x$$$. Find the minimum number of operations needed such that the equality of the array is less than or equal to $$$1$$$. | 256 megabytes | /*
Rating: 1378
Date: 23-04-2022
Time: 19-55-11
Author: Kartik Papney
Linkedin: https://www.linkedin.com/in/kartik-papney-4951161a6/
Leetcode: https://leetcode.com/kartikpapney/
Codechef: https://www.codechef.com/users/kartikpapney
----------------------------Jai Shree Ram----------------------------
*/
import java.util.*;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
public class C_Unequal_Array {
public static void s() {
int n = sc.nextInt();
int[] arr = sc.readArray(n);
int count = 0;
ArrayList<Integer> rr = new ArrayList<>();
ArrayList<Integer> cc = new ArrayList<>();
for(int i=0; i<n-1; i++) {
if(arr[i] == arr[i+1]) {
cc.add(i);
count++;
} else {
if(count != 0) rr.add(count);
count = 0;
}
}
// p.writes(rr);
// p.writeln();
if(count != 0) rr.add(count);
if(rr.size() == 0) {
p.writeln(0);
return;
} else if(rr.size() == 1) {
int val = rr.get(0);
if(val == 1) {
p.writeln(0);
return;
}
p.writeln(Math.max(1, val-2));
} else {
int gv = cc.get(0);
int tv = cc.get(cc.size()-1);
gv+=1;
p.writeln(tv-gv);
}
}
public static void main(String[] args) {
int t = 1;
t = sc.nextInt();
while (t-- != 0) {
s();
}
p.print();
}
public static boolean debug = false;
static void debug(String st) {
if(debug) p.writeln(st);
}
static final Integer MOD = (int) 1e9 + 7;
static final FastReader sc = new FastReader();
static final Print p = new Print();
static class Functions {
static void sort(int... a) {
ArrayList<Integer> l = new ArrayList<>();
for (int i : a) l.add(i);
Collections.sort(l);
for (int i = 0; i < a.length; i++) a[i] = l.get(i);
}
static void sort(long... a) {
ArrayList<Long> l = new ArrayList<>();
for (long i : a) l.add(i);
Collections.sort(l);
for (int i = 0; i < a.length; i++) a[i] = l.get(i);
}
static int max(int... a) {
int max = Integer.MIN_VALUE;
for (int val : a) max = Math.max(val, max);
return max;
}
static int min(int... a) {
int min = Integer.MAX_VALUE;
for (int val : a) min = Math.min(val, min);
return min;
}
static long min(long... a) {
long min = Long.MAX_VALUE;
for (long val : a) min = Math.min(val, min);
return min;
}
static long max(long... a) {
long max = Long.MIN_VALUE;
for (long val : a) max = Math.max(val, max);
return max;
}
static long sum(long... a) {
long sum = 0;
for (long val : a) sum += val;
return sum;
}
static int sum(int... a) {
int sum = 0;
for (int val : a) sum += val;
return sum;
}
public static long mod_add(long a, long b) {
return (a % MOD + b % MOD + MOD) % MOD;
}
public static long pow(long a, long b) {
long res = 1;
while (b > 0) {
if ((b & 1) != 0)
res = mod_mul(res, a);
a = mod_mul(a, a);
b >>= 1;
}
return res;
}
public static long mod_mul(long a, long b) {
long res = 0;
a %= MOD;
while (b > 0) {
if ((b & 1) > 0) {
res = mod_add(res, a);
}
a = (2 * a) % MOD;
b >>= 1;
}
return res;
}
public static long gcd(long a, long b) {
if (a == 0) return b;
return gcd(b % a, a);
}
public static long factorial(long n) {
long res = 1;
for (int i = 1; i <= n; i++) {
res = (i % MOD * res % MOD) % MOD;
}
return res;
}
public static int count(int[] arr, int x) {
int count = 0;
for (int val : arr) if (val == x) count++;
return count;
}
public static ArrayList<Integer> generatePrimes(int n) {
boolean[] primes = new boolean[n];
for (int i = 2; i < primes.length; i++) primes[i] = true;
for (int i = 2; i < primes.length; i++) {
if (primes[i]) {
for (int j = i * i; j < primes.length; j += i) {
primes[j] = false;
}
}
}
ArrayList<Integer> arr = new ArrayList<>();
for (int i = 0; i < primes.length; i++) {
if (primes[i]) arr.add(i);
}
return arr;
}
}
static class Print {
StringBuffer strb = new StringBuffer();
public void write(Object str) {
strb.append(str);
}
public void writes(Object str) {
char c = ' ';
strb.append(str).append(c);
}
public void writeln(Object str) {
char c = '\n';
strb.append(str).append(c);
}
public void writeln() {
char c = '\n';
strb.append(c);
}
public void yes() {
char c = '\n';
writeln("YES");
}
public void no() {
writeln("NO");
}
public void writes(int... arr) {
for (int val : arr) {
write(val);
write(' ');
}
}
public void writes(long... arr) {
for (long val : arr) {
write(val);
write(' ');
}
}
public void writeln(int... arr) {
for (int val : arr) {
writeln(val);
}
}
public void print() {
System.out.print(strb);
}
public void println() {
System.out.println(strb);
}
}
static class FastReader {
BufferedReader br;
StringTokenizer st;
public FastReader() {
br = new BufferedReader(new InputStreamReader(System.in));
}
String next() {
while (st == null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt() {
return Integer.parseInt(next());
}
long nextLong() {
return Long.parseLong(next());
}
double nextDouble() {
return Double.parseDouble(next());
}
int[] readArray(int n) {
int[] a = new int[n];
for (int i = 0; i < n; i++) a[i] = nextInt();
return a;
}
long[] readLongArray(int n) {
long[] a = new long[n];
for (int i = 0; i < n; i++) a[i] = nextLong();
return a;
}
double[] readArrayDouble(int n) {
double[] a = new double[n];
for (int i = 0; i < n; i++) a[i] = nextInt();
return a;
}
String nextLine() {
String str = new String();
try {
str = br.readLine();
} catch (IOException e) {
e.printStackTrace();
}
return str;
}
}
} | Java | ["4\n\n5\n\n1 1 1 1 1\n\n5\n\n2 1 1 1 2\n\n6\n\n1 1 2 3 3 4\n\n6\n\n1 2 1 4 5 4"] | 1 second | ["2\n1\n2\n0"] | NoteIn the first test case, we can select $$$i=2$$$ and $$$x=2$$$ to form $$$[1, 2, 2, 1, 1]$$$. Then, we can select $$$i=3$$$ and $$$x=3$$$ to form $$$[1, 2, 3, 3, 1]$$$.In the second test case, we can select $$$i=3$$$ and $$$x=100$$$ to form $$$[2, 1, 100, 100, 2]$$$. | Java 11 | standard input | [
"constructive algorithms",
"greedy",
"implementation"
] | a91be662101762bcb2c58b8db9ff61e0 | Each test contains multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains an integer $$$n$$$ ($$$2 \le n \le 2 \cdot 10 ^ 5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10^9$$$) — elements of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10 ^ 5$$$ | 1,100 | For each test case, print the minimum number of operations needed. | standard output | |
PASSED | 7ba1f666469cc422373e491a602bdd4d | train_108.jsonl | 1650722700 | You are given an array $$$a$$$ of length $$$n$$$. We define the equality of the array as the number of indices $$$1 \le i \le n - 1$$$ such that $$$a_i = a_{i + 1}$$$. We are allowed to do the following operation: Select two integers $$$i$$$ and $$$x$$$ such that $$$1 \le i \le n - 1$$$ and $$$1 \le x \le 10^9$$$. Then, set $$$a_i$$$ and $$$a_{i + 1}$$$ to be equal to $$$x$$$. Find the minimum number of operations needed such that the equality of the array is less than or equal to $$$1$$$. | 256 megabytes |
import java.util.ArrayList;
import java.util.Scanner;
public class Cf {
public static void main(String[] args) {
// TODO Auto-generated method stub
Scanner s=new Scanner(System.in);
int t=s.nextInt();
int p=0;
while(p<t)
{
int j=0,n=s.nextInt(),k;
int[] v= new int[n+3];
int a,i;
int sum=0;
for(j=0;j<n;j++)
{
a=s.nextInt();
v[j+1]=a;
}
for(i=1;i<=n;i++)
{
if(i<n && v[i]==v[i+1])
{
k=i+1;
for(j=n;j>k;j--)
{
if(j-1>=k && v[j]==v[j-1])
{
k=j;
break;
}
}
if(k==i+1)sum=0;
else if(k==i+2)sum=1;
else sum=k-i-2;
break;
}
}
System.out.println(sum);
p++;
}
}
}
| Java | ["4\n\n5\n\n1 1 1 1 1\n\n5\n\n2 1 1 1 2\n\n6\n\n1 1 2 3 3 4\n\n6\n\n1 2 1 4 5 4"] | 1 second | ["2\n1\n2\n0"] | NoteIn the first test case, we can select $$$i=2$$$ and $$$x=2$$$ to form $$$[1, 2, 2, 1, 1]$$$. Then, we can select $$$i=3$$$ and $$$x=3$$$ to form $$$[1, 2, 3, 3, 1]$$$.In the second test case, we can select $$$i=3$$$ and $$$x=100$$$ to form $$$[2, 1, 100, 100, 2]$$$. | Java 11 | standard input | [
"constructive algorithms",
"greedy",
"implementation"
] | a91be662101762bcb2c58b8db9ff61e0 | Each test contains multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains an integer $$$n$$$ ($$$2 \le n \le 2 \cdot 10 ^ 5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10^9$$$) — elements of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10 ^ 5$$$ | 1,100 | For each test case, print the minimum number of operations needed. | standard output | |
PASSED | fa211d15cc5b82b7e1ea65c54846a95e | train_108.jsonl | 1650722700 | You are given an array $$$a$$$ of length $$$n$$$. We define the equality of the array as the number of indices $$$1 \le i \le n - 1$$$ such that $$$a_i = a_{i + 1}$$$. We are allowed to do the following operation: Select two integers $$$i$$$ and $$$x$$$ such that $$$1 \le i \le n - 1$$$ and $$$1 \le x \le 10^9$$$. Then, set $$$a_i$$$ and $$$a_{i + 1}$$$ to be equal to $$$x$$$. Find the minimum number of operations needed such that the equality of the array is less than or equal to $$$1$$$. | 256 megabytes | import java.util.*;
import java.lang.*;
import java.io.*;
public class Main
{
static final PrintWriter out =new PrintWriter(System.out);
static final FastReader sc = new FastReader();
public static boolean sorted(int a[])
{
int n=a.length,i;
int b[]=new int[n];
for(i=0;i<n;i++)
b[i]=a[i];
Arrays.sort(b);
for(i=0;i<n;i++)
{
if(a[i]!=b[i])
return false;
}
return true;
}
public static void main (String[] args) throws java.lang.Exception
{
int tes=sc.nextInt();
while(tes-->0)
{
int n=sc.nextInt();
int a[]=new int[n];
int i;
ArrayList<Integer> arr=new ArrayList<>();
for(i=0;i<n;i++)
a[i]=sc.nextInt();
for(i=1;i<n;i++)
{
if(a[i]==a[i-1])
arr.add(i);
}
if(arr.size()<=1)
{
System.out.println(0);
continue;
}
System.out.println(Math.max(1,arr.get(arr.size()-1)-arr.get(0)-1));
}
}
public static int first(ArrayList<Integer> arr, int low, int high, int x, int n)
{
if (high >= low) {
int mid = low + (high - low) / 2;
if ((mid == 0 || x > arr.get(mid-1)) && arr.get(mid) == x)
return mid;
else if (x > arr.get(mid))
return first(arr, (mid + 1), high, x, n);
else
return first(arr, low, (mid - 1), x, n);
}
return -1;
}
public static int last(ArrayList<Integer> arr, int low, int high, int x, int n)
{
if (high >= low) {
int mid = low + (high - low) / 2;
if ((mid == n - 1 || x < arr.get(mid+1)) && arr.get(mid) == x)
return mid;
else if (x < arr.get(mid))
return last(arr, low, (mid - 1), x, n);
else
return last(arr, (mid + 1), high, x, n);
}
return -1;
}
public static int lis(int[] arr) {
int n = arr.length;
ArrayList<Integer> al = new ArrayList<Integer>();
al.add(arr[0]);
for(int i = 1 ; i<n;i++) {
int x = al.get(al.size()-1);
if(arr[i]>=x) {
al.add(arr[i]);
}else {
int v = upper_bound(al, 0, al.size(), arr[i]);
al.set(v, arr[i]);
}
}
return al.size();
}
public static int lower_bound(ArrayList<Long> ar,int lo , int hi , long k)
{
Collections.sort(ar);
int s=lo;
int e=hi;
while (s !=e)
{
int mid = s+e>>1;
if (ar.get((int)mid) <k)
{
s=mid+1;
}
else
{
e=mid;
}
}
if(s==ar.size())
{
return -1;
}
return s;
}
public static int upper_bound(ArrayList<Integer> ar,int lo , int hi, int k)
{
Collections.sort(ar);
int s=lo;
int e=hi;
while (s !=e)
{
int mid = s+e>>1;
if (ar.get(mid) <=k)
{
s=mid+1;
}
else
{
e=mid;
}
}
if(s==ar.size())
{
return -1;
}
return s;
}
static boolean isPrime(long N)
{
if (N<=1) return false;
if (N<=3) return true;
if (N%2 == 0 || N%3 == 0) return false;
for (int i=5; i*i<=N; i=i+6)
if (N%i == 0 || N%(i+2) == 0)
return false;
return true;
}
static int countBits(long a)
{
return (int)(Math.log(a)/Math.log(2)+1);
}
static long fact(long N)
{
long mod=1000000007;
long n=2;
if(N<=1)return 1;
else
{
for(int i=3; i<=N; i++)n=(n*i)%mod;
}
return n;
}
private static boolean isInteger(String s) {
try {
Integer.parseInt(s);
} catch (NumberFormatException e) {
return false;
} catch (NullPointerException e) {
return false;
}
return true;
}
private static long gcd(long a, long b) {
if (b == 0)
return a;
return gcd(b, a % b);
}
private static long lcm(long a, long b) {
return (a / gcd(a, b)) * b;
}
private static boolean isPalindrome(String str) {
int i = 0, j = str.length() - 1;
while (i < j)
if (str.charAt(i++) != str.charAt(j--))
return false;
return true;
}
private static String reverseString(String str) {
StringBuilder sb = new StringBuilder(str);
return sb.reverse().toString();
}
static class FastReader {
BufferedReader br;
StringTokenizer st;
public FastReader()
{
br = new BufferedReader(
new InputStreamReader(System.in));
}
String next()
{
while (st == null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(br.readLine());
}
catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt() { return Integer.parseInt(next()); }
long nextLong() { return Long.parseLong(next()); }
double nextDouble()
{
return Double.parseDouble(next());
}
String nextLine()
{
String str = "";
try {
str = br.readLine();
}
catch (IOException e) {
e.printStackTrace();
}
return str;
}
}
} | Java | ["4\n\n5\n\n1 1 1 1 1\n\n5\n\n2 1 1 1 2\n\n6\n\n1 1 2 3 3 4\n\n6\n\n1 2 1 4 5 4"] | 1 second | ["2\n1\n2\n0"] | NoteIn the first test case, we can select $$$i=2$$$ and $$$x=2$$$ to form $$$[1, 2, 2, 1, 1]$$$. Then, we can select $$$i=3$$$ and $$$x=3$$$ to form $$$[1, 2, 3, 3, 1]$$$.In the second test case, we can select $$$i=3$$$ and $$$x=100$$$ to form $$$[2, 1, 100, 100, 2]$$$. | Java 11 | standard input | [
"constructive algorithms",
"greedy",
"implementation"
] | a91be662101762bcb2c58b8db9ff61e0 | Each test contains multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains an integer $$$n$$$ ($$$2 \le n \le 2 \cdot 10 ^ 5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10^9$$$) — elements of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10 ^ 5$$$ | 1,100 | For each test case, print the minimum number of operations needed. | standard output | |
PASSED | bc57a30de248313c625921949571084e | train_108.jsonl | 1650722700 | You are given an array $$$a$$$ of length $$$n$$$. We define the equality of the array as the number of indices $$$1 \le i \le n - 1$$$ such that $$$a_i = a_{i + 1}$$$. We are allowed to do the following operation: Select two integers $$$i$$$ and $$$x$$$ such that $$$1 \le i \le n - 1$$$ and $$$1 \le x \le 10^9$$$. Then, set $$$a_i$$$ and $$$a_{i + 1}$$$ to be equal to $$$x$$$. Find the minimum number of operations needed such that the equality of the array is less than or equal to $$$1$$$. | 256 megabytes | import java.io.*;
import java.util.*;
public class Unequal_Array {
static FastScanner fs;
static FastWriter fw;
static boolean checkOnlineJudge = System.getProperty("ONLINE_JUDGE") == null;
private static final int[][] kdir = new int[][]{{-1, 2}, {-2, 1}, {-2, -1}, {-1, -2}, {1, -2}, {2, -1}, {2, 1}, {1, 2}};
private static final int[][] dir = new int[][]{{-1, 0}, {1, 0}, {0, 1}, {0, -1}};
private static final int iMax = Integer.MAX_VALUE, iMin = Integer.MIN_VALUE;
private static final long lMax = Long.MAX_VALUE, lMin = Long.MIN_VALUE;
private static final int mod1 = (int) (1e9 + 7);
private static final int mod2 = 998244353;
public static void main(String[] args) throws IOException {
fs = new FastScanner();
fw = new FastWriter();
int t = 1;
t = fs.nextInt();
while (t-- > 0) {
solve();
}
fw.out.close();
}
private static void solve() {
int n = fs.nextInt();
int[] arr = readIntArray(n);
int[] temp = new int[]{iMax, iMin};
for (int i = 0; i < n - 1; i++) {
if (arr[i] == arr[i + 1]) {
temp[0] = Math.min(temp[0], i);
temp[1] = Math.max(temp[1], i);
}
}
if (temp[0] == iMax || temp[0] == temp[1]) {
fw.out.println(0);
return;
}
int ans = 0;
for (int ii = temp[0] + 1; ; ii++) {
ans++;
if (ii >= temp[1] - 1) break;
}
fw.out.println(ans);
}
private static class UnionFind {
private final int[] parent;
private final int[] rank;
UnionFind(int n) {
parent = new int[n + 5];
rank = new int[n + 5];
for (int i = 0; i <= n; i++) {
parent[i] = i;
rank[i] = 0;
}
}
private int find(int i) {
if (parent[i] == i)
return i;
return parent[i] = find(parent[i]);
}
private void union(int a, int b) {
a = find(a);
b = find(b);
if (a != b) {
if (rank[a] < rank[b]) {
int temp = a;
a = b;
b = temp;
}
parent[b] = a;
if (rank[a] == rank[b])
rank[a]++;
}
}
}
private static class SCC {
private final int n;
private final List<List<Integer>> adjList;
SCC(int _n, List<List<Integer>> _adjList) {
n = _n;
adjList = _adjList;
}
private List<List<Integer>> getSCC() {
List<List<Integer>> ans = new ArrayList<>();
Stack<Integer> stack = new Stack<>();
boolean[] vis = new boolean[n];
for (int i = 0; i < n; i++) {
if (!vis[i]) dfs(i, adjList, vis, stack);
}
vis = new boolean[n];
List<List<Integer>> rev_adjList = rev_graph(n, adjList);
while (!stack.isEmpty()) {
int curr = stack.pop();
if (!vis[curr]) {
List<Integer> scc_list = new ArrayList<>();
dfs2(curr, rev_adjList, vis, scc_list);
ans.add(scc_list);
}
}
return ans;
}
private void dfs(int curr, List<List<Integer>> adjList, boolean[] vis, Stack<Integer> stack) {
vis[curr] = true;
for (int x : adjList.get(curr)) {
if (!vis[x]) dfs(x, adjList, vis, stack);
}
stack.add(curr);
}
private void dfs2(int curr, List<List<Integer>> adjList, boolean[] vis, List<Integer> scc_list) {
vis[curr] = true;
scc_list.add(curr);
for (int x : adjList.get(curr)) {
if (!vis[x]) dfs2(x, adjList, vis, scc_list);
}
}
}
private static List<List<Integer>> rev_graph(int n, List<List<Integer>> adjList) {
List<List<Integer>> ans = new ArrayList<>();
for (int i = 0; i < n; i++) ans.add(new ArrayList<>());
for (int i = 0; i < n; i++) {
for (int x : adjList.get(i)) {
ans.get(x).add(i);
}
}
return ans;
}
private static class Calc_nCr {
private final long[] fact;
private final long[] invfact;
private final int p;
Calc_nCr(int n, int prime) {
fact = new long[n + 5];
invfact = new long[n + 5];
p = prime;
fact[0] = 1;
for (int i = 1; i <= n; i++) {
fact[i] = (i * fact[i - 1]) % p;
}
invfact[n] = pow_with_mod(fact[n], p - 2, p);
for (int i = n - 1; i >= 0; i--) {
invfact[i] = (invfact[i + 1] * (i + 1)) % p;
}
}
private long nCr(int n, int r) {
if (r > n || n < 0 || r < 0) return 0;
return (((fact[n] * invfact[r]) % p) * invfact[n - r]) % p;
}
}
private static int random_between_two_numbers(int l, int r) {
Random ran = new Random();
return ran.nextInt(r - l) + l;
}
private static long gcd(long a, long b) {
return (b == 0 ? a : gcd(b, a % b));
}
private static long lcm(long a, long b) {
return ((a * b) / gcd(a, b));
}
private static long pow(long a, long b) {
long result = 1;
while (b > 0) {
if ((b & 1L) == 1) {
result = (result * a);
}
a = (a * a);
b >>= 1;
}
return result;
}
private static long pow_with_mod(long a, long b, int mod) {
long result = 1;
while (b > 0) {
if ((b & 1L) == 1) {
result = (result * a) % mod;
}
a = (a * a) % mod;
b >>= 1;
}
return result;
}
private static long ceilDiv(long a, long b) {
return ((a + b - 1) / b);
}
private static long getMin(long... args) {
long min = lMax;
for (long arg : args)
min = Math.min(min, arg);
return min;
}
private static long getMax(long... args) {
long max = lMin;
for (long arg : args)
max = Math.max(max, arg);
return max;
}
private static boolean isPalindrome(String s, int l, int r) {
int i = l, j = r;
while (j - i >= 1) {
if (s.charAt(i) != s.charAt(j))
return false;
i++;
j--;
}
return true;
}
private static List<Integer> primes(int n) {
boolean[] primeArr = new boolean[n + 5];
Arrays.fill(primeArr, true);
for (int i = 2; (i * i) <= n; i++) {
if (primeArr[i]) {
for (int j = i * i; j <= n; j += i) {
primeArr[j] = false;
}
}
}
List<Integer> primeList = new ArrayList<>();
for (int i = 2; i <= n; i++) {
if (primeArr[i])
primeList.add(i);
}
return primeList;
}
private static int noOfSetBits(long x) {
int cnt = 0;
while (x != 0) {
x = x & (x - 1);
cnt++;
}
return cnt;
}
private static int sumOfDigits(long num) {
int cnt = 0;
while (num > 0) {
cnt += (num % 10);
num /= 10;
}
return cnt;
}
private static int noOfDigits(long num) {
int cnt = 0;
while (num > 0) {
cnt++;
num /= 10;
}
return cnt;
}
private static boolean isPerfectSquare(long num) {
long sqrt = (long) Math.sqrt(num);
return ((sqrt * sqrt) == num);
}
private static class Pair<U, V> {
private final U first;
private final V second;
@Override
public boolean equals(Object o) {
if (this == o) return true;
if (o == null || getClass() != o.getClass()) return false;
Pair<?, ?> pair = (Pair<?, ?>) o;
return first.equals(pair.first) && second.equals(pair.second);
}
@Override
public int hashCode() {
return Objects.hash(first, second);
}
@Override
public String toString() {
return "(" + first + ", " + second + ")";
}
private Pair(U ff, V ss) {
this.first = ff;
this.second = ss;
}
}
private static void randomizeIntArr(int[] arr, int n) {
Random r = new Random();
for (int i = (n - 1); i > 0; i--) {
int j = r.nextInt(i + 1);
swapInIntArr(arr, i, j);
}
}
private static void randomizeLongArr(long[] arr, int n) {
Random r = new Random();
for (int i = (n - 1); i > 0; i--) {
int j = r.nextInt(i + 1);
swapInLongArr(arr, i, j);
}
}
private static void swapInIntArr(int[] arr, int a, int b) {
int temp = arr[a];
arr[a] = arr[b];
arr[b] = temp;
}
private static void swapInLongArr(long[] arr, int a, int b) {
long temp = arr[a];
arr[a] = arr[b];
arr[b] = temp;
}
private static int[] readIntArray(int n) {
int[] arr = new int[n];
for (int i = 0; i < n; i++)
arr[i] = fs.nextInt();
return arr;
}
private static long[] readLongArray(int n) {
long[] arr = new long[n];
for (int i = 0; i < n; i++)
arr[i] = fs.nextLong();
return arr;
}
private static List<Integer> readIntList(int n) {
List<Integer> list = new ArrayList<>();
for (int i = 0; i < n; i++)
list.add(fs.nextInt());
return list;
}
private static List<Long> readLongList(int n) {
List<Long> list = new ArrayList<>();
for (int i = 0; i < n; i++)
list.add(fs.nextLong());
return list;
}
private static class FastScanner {
BufferedReader br;
StringTokenizer st;
FastScanner() throws IOException {
if (checkOnlineJudge)
this.br = new BufferedReader(new FileReader("src/input.txt"));
else
this.br = new BufferedReader(new InputStreamReader(System.in));
this.st = new StringTokenizer("");
}
public String next() {
while (!st.hasMoreTokens()) {
try {
st = new StringTokenizer(br.readLine());
} catch (IOException err) {
err.printStackTrace();
}
}
return st.nextToken();
}
public String nextLine() {
if (st.hasMoreTokens()) {
return st.nextToken("").trim();
}
try {
return br.readLine().trim();
} catch (IOException err) {
err.printStackTrace();
}
return "";
}
public int nextInt() {
return Integer.parseInt(next());
}
public long nextLong() {
return Long.parseLong(next());
}
public double nextDouble() {
return Double.parseDouble(next());
}
}
private static class FastWriter {
PrintWriter out;
FastWriter() throws IOException {
if (checkOnlineJudge)
out = new PrintWriter(new FileWriter("src/output.txt"));
else
out = new PrintWriter(System.out);
}
}
}
| Java | ["4\n\n5\n\n1 1 1 1 1\n\n5\n\n2 1 1 1 2\n\n6\n\n1 1 2 3 3 4\n\n6\n\n1 2 1 4 5 4"] | 1 second | ["2\n1\n2\n0"] | NoteIn the first test case, we can select $$$i=2$$$ and $$$x=2$$$ to form $$$[1, 2, 2, 1, 1]$$$. Then, we can select $$$i=3$$$ and $$$x=3$$$ to form $$$[1, 2, 3, 3, 1]$$$.In the second test case, we can select $$$i=3$$$ and $$$x=100$$$ to form $$$[2, 1, 100, 100, 2]$$$. | Java 11 | standard input | [
"constructive algorithms",
"greedy",
"implementation"
] | a91be662101762bcb2c58b8db9ff61e0 | Each test contains multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains an integer $$$n$$$ ($$$2 \le n \le 2 \cdot 10 ^ 5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10^9$$$) — elements of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10 ^ 5$$$ | 1,100 | For each test case, print the minimum number of operations needed. | standard output | |
PASSED | 027d680f4880133602164b2c037bd595 | train_108.jsonl | 1650722700 | You are given an array $$$a$$$ of length $$$n$$$. We define the equality of the array as the number of indices $$$1 \le i \le n - 1$$$ such that $$$a_i = a_{i + 1}$$$. We are allowed to do the following operation: Select two integers $$$i$$$ and $$$x$$$ such that $$$1 \le i \le n - 1$$$ and $$$1 \le x \le 10^9$$$. Then, set $$$a_i$$$ and $$$a_{i + 1}$$$ to be equal to $$$x$$$. Find the minimum number of operations needed such that the equality of the array is less than or equal to $$$1$$$. | 256 megabytes | import java.io.*;
import java.util.*;
public class TimePass {
public static int solve(int n,int arr[]) {
int count=0;
int i=0;
int prev=-1;
while(i<n) {
int j=i;
while(j<n&&arr[j]==arr[i]) j++;
int len=j-i;
if(len==1) {
i=j;
continue;
}
if(len>1&&prev!=-1) count+=j-prev-1;
else count+=len-2;
prev=j-1;
i=j;
}
if(count<=1) return count;
return count-1;
}
public static void main (String[] args) throws java.lang.Exception
{
BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
PrintWriter out=new PrintWriter(System.out);
int testCases=Integer.parseInt(br.readLine());
while(testCases-->0){
int n=Integer.parseInt(br.readLine());
String input[]=br.readLine().split(" ");
int arr[]=new int[n];
for(int i=0;i<n;i++) {
arr[i]=Integer.parseInt(input[i]);
}
int ans=solve(n,arr);
out.print(ans);
out.print('\n');
}
out.close();
}
}
| Java | ["4\n\n5\n\n1 1 1 1 1\n\n5\n\n2 1 1 1 2\n\n6\n\n1 1 2 3 3 4\n\n6\n\n1 2 1 4 5 4"] | 1 second | ["2\n1\n2\n0"] | NoteIn the first test case, we can select $$$i=2$$$ and $$$x=2$$$ to form $$$[1, 2, 2, 1, 1]$$$. Then, we can select $$$i=3$$$ and $$$x=3$$$ to form $$$[1, 2, 3, 3, 1]$$$.In the second test case, we can select $$$i=3$$$ and $$$x=100$$$ to form $$$[2, 1, 100, 100, 2]$$$. | Java 11 | standard input | [
"constructive algorithms",
"greedy",
"implementation"
] | a91be662101762bcb2c58b8db9ff61e0 | Each test contains multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains an integer $$$n$$$ ($$$2 \le n \le 2 \cdot 10 ^ 5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10^9$$$) — elements of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10 ^ 5$$$ | 1,100 | For each test case, print the minimum number of operations needed. | standard output | |
PASSED | 10009a77faf6a5c8afc89a34a76e1749 | train_108.jsonl | 1650722700 | You are given an array $$$a$$$ of length $$$n$$$. We define the equality of the array as the number of indices $$$1 \le i \le n - 1$$$ such that $$$a_i = a_{i + 1}$$$. We are allowed to do the following operation: Select two integers $$$i$$$ and $$$x$$$ such that $$$1 \le i \le n - 1$$$ and $$$1 \le x \le 10^9$$$. Then, set $$$a_i$$$ and $$$a_{i + 1}$$$ to be equal to $$$x$$$. Find the minimum number of operations needed such that the equality of the array is less than or equal to $$$1$$$. | 256 megabytes | import java.io.BufferedReader;
import java.io.FileInputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.InterruptedIOException;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.HashMap;
import java.util.HashSet;
import java.util.Iterator;
import java.util.Scanner;
import java.util.Set;
import java.util.*;
public class code {
public static BufferedReader in;
public static void main(String[] args) throws IOException {
// TODO Auto-generated method stub
// System.setIn(new FileInputStream("input.txt"));
BufferedReader in= new BufferedReader(new InputStreamReader(System.in));
// Scanner scanner = new Scanner(System.in);
String nstr = in.readLine();
int n_ = Integer.parseInt(nstr);
for(int jj=0; jj < n_ ;jj++)
{
String line2 = in.readLine();
String line1 = in.readLine();
// char[] line = line1.toCharArray();;
String[] lineStr1 = line1.split(" ");
int[] arr = new int[lineStr1.length];
for(int i= 0; i< arr.length; i++)
{
arr[i] = (Integer.parseInt(lineStr1[i]));
}
//
// String line2 = in.readLine();
// String[] lineStr2 = line2.split(" ");
// int[] arr2 = new int[lineStr2.length];
// for(int i= 0; i< arr2.length; i++)
// {
// arr2[i] = (Integer.parseInt(lineStr2[i]));
// }
int mm= 0;
int j=1;
int count =0;
int operations =0;
int start =-1;
int end =-1;
while(j<arr.length)
{
if(arr[j-1] == arr[j])
{
if(start == -1)
{
start = j;
}
else
{
end = j-1;
}
}
j++;
}
if(end ==-1)
{
System.out.println(0);
}
else
{
if(start == end)
{
System.out.println(end - start+1);
}
else
{
System.out.println(end - start);
}
}
}
}
}
| Java | ["4\n\n5\n\n1 1 1 1 1\n\n5\n\n2 1 1 1 2\n\n6\n\n1 1 2 3 3 4\n\n6\n\n1 2 1 4 5 4"] | 1 second | ["2\n1\n2\n0"] | NoteIn the first test case, we can select $$$i=2$$$ and $$$x=2$$$ to form $$$[1, 2, 2, 1, 1]$$$. Then, we can select $$$i=3$$$ and $$$x=3$$$ to form $$$[1, 2, 3, 3, 1]$$$.In the second test case, we can select $$$i=3$$$ and $$$x=100$$$ to form $$$[2, 1, 100, 100, 2]$$$. | Java 11 | standard input | [
"constructive algorithms",
"greedy",
"implementation"
] | a91be662101762bcb2c58b8db9ff61e0 | Each test contains multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains an integer $$$n$$$ ($$$2 \le n \le 2 \cdot 10 ^ 5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10^9$$$) — elements of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10 ^ 5$$$ | 1,100 | For each test case, print the minimum number of operations needed. | standard output | |
PASSED | 737274e308dcd54510c01e53bbe4bc63 | train_108.jsonl | 1650722700 | You are given an array $$$a$$$ of length $$$n$$$. We define the equality of the array as the number of indices $$$1 \le i \le n - 1$$$ such that $$$a_i = a_{i + 1}$$$. We are allowed to do the following operation: Select two integers $$$i$$$ and $$$x$$$ such that $$$1 \le i \le n - 1$$$ and $$$1 \le x \le 10^9$$$. Then, set $$$a_i$$$ and $$$a_{i + 1}$$$ to be equal to $$$x$$$. Find the minimum number of operations needed such that the equality of the array is less than or equal to $$$1$$$. | 256 megabytes | import java.io.BufferedReader;
import java.io.FileInputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.InterruptedIOException;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.HashMap;
import java.util.HashSet;
import java.util.Iterator;
import java.util.Scanner;
import java.util.Set;
import java.util.*;
public class code {
public static BufferedReader in;
public static void main(String[] args) throws IOException {
// TODO Auto-generated method stub
// System.setIn(new FileInputStream("input.txt"));
BufferedReader in= new BufferedReader(new InputStreamReader(System.in));
// Scanner scanner = new Scanner(System.in);
String nstr = in.readLine();
int n_ = Integer.parseInt(nstr);
for(int jj=0; jj < n_ ;jj++)
{
String line2 = in.readLine();
String line1 = in.readLine();
// char[] line = line1.toCharArray();;
String[] lineStr1 = line1.split(" ");
int[] arr = new int[lineStr1.length];
for(int i= 0; i< arr.length; i++)
{
arr[i] = (Integer.parseInt(lineStr1[i]));
}
//
// String line2 = in.readLine();
// String[] lineStr2 = line2.split(" ");
// int[] arr2 = new int[lineStr2.length];
// for(int i= 0; i< arr2.length; i++)
// {
// arr2[i] = (Integer.parseInt(lineStr2[i]));
// }
int j=1;
int count =0;
int operations =0;
int start =-1;
int end =-1;
while(j<arr.length)
{
if(arr[j-1] == arr[j])
{
if(start == -1)
{
start = j;
}
else
{
end = j-1;
}
}
j++;
}
if(end ==-1)
{
System.out.println(0);
}
else
{
if(start == end)
{
System.out.println(end - start+1);
}
else
{
System.out.println(end - start);
}
}
}
}
}
| Java | ["4\n\n5\n\n1 1 1 1 1\n\n5\n\n2 1 1 1 2\n\n6\n\n1 1 2 3 3 4\n\n6\n\n1 2 1 4 5 4"] | 1 second | ["2\n1\n2\n0"] | NoteIn the first test case, we can select $$$i=2$$$ and $$$x=2$$$ to form $$$[1, 2, 2, 1, 1]$$$. Then, we can select $$$i=3$$$ and $$$x=3$$$ to form $$$[1, 2, 3, 3, 1]$$$.In the second test case, we can select $$$i=3$$$ and $$$x=100$$$ to form $$$[2, 1, 100, 100, 2]$$$. | Java 11 | standard input | [
"constructive algorithms",
"greedy",
"implementation"
] | a91be662101762bcb2c58b8db9ff61e0 | Each test contains multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains an integer $$$n$$$ ($$$2 \le n \le 2 \cdot 10 ^ 5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10^9$$$) — elements of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10 ^ 5$$$ | 1,100 | For each test case, print the minimum number of operations needed. | standard output | |
PASSED | 4dff445554eb187ac7b1a2a522dc341b | train_108.jsonl | 1650722700 | You are given an array $$$a$$$ of length $$$n$$$. We define the equality of the array as the number of indices $$$1 \le i \le n - 1$$$ such that $$$a_i = a_{i + 1}$$$. We are allowed to do the following operation: Select two integers $$$i$$$ and $$$x$$$ such that $$$1 \le i \le n - 1$$$ and $$$1 \le x \le 10^9$$$. Then, set $$$a_i$$$ and $$$a_{i + 1}$$$ to be equal to $$$x$$$. Find the minimum number of operations needed such that the equality of the array is less than or equal to $$$1$$$. | 256 megabytes | import java.util.*;
public class Main {
public static void main(String[] args) throws Exception {
Scanner boom=new Scanner(System.in);
int t=boom.nextInt();
for(int j=0;j<t;j++){
int n=boom.nextInt();
int[] A=new int[n];
for(int i=0;i<n;i++){
A[i]=boom.nextInt();
}
int start=0;int end=0;boolean flag=false;
for(int i=0;i<n-1;i++){
if(A[i]==A[i+1]){
start=i+1;
break;
}
}
for(int i=n-1;i>0;i--){
if(A[i]==A[i-1]){
end=i-1;
flag=true;
break;
}
}
if(end-start>=0){
if(end==start && flag==true){
System.out.println(1);
}
else{
System.out.println(end-start);
}
}
else{
System.out.println(0);
}
}
}
} | Java | ["4\n\n5\n\n1 1 1 1 1\n\n5\n\n2 1 1 1 2\n\n6\n\n1 1 2 3 3 4\n\n6\n\n1 2 1 4 5 4"] | 1 second | ["2\n1\n2\n0"] | NoteIn the first test case, we can select $$$i=2$$$ and $$$x=2$$$ to form $$$[1, 2, 2, 1, 1]$$$. Then, we can select $$$i=3$$$ and $$$x=3$$$ to form $$$[1, 2, 3, 3, 1]$$$.In the second test case, we can select $$$i=3$$$ and $$$x=100$$$ to form $$$[2, 1, 100, 100, 2]$$$. | Java 11 | standard input | [
"constructive algorithms",
"greedy",
"implementation"
] | a91be662101762bcb2c58b8db9ff61e0 | Each test contains multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains an integer $$$n$$$ ($$$2 \le n \le 2 \cdot 10 ^ 5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10^9$$$) — elements of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10 ^ 5$$$ | 1,100 | For each test case, print the minimum number of operations needed. | standard output | |
PASSED | f31f97bc07c43325c54bea0f4b885824 | train_108.jsonl | 1650722700 | You are given an array $$$a$$$ of length $$$n$$$. We define the equality of the array as the number of indices $$$1 \le i \le n - 1$$$ such that $$$a_i = a_{i + 1}$$$. We are allowed to do the following operation: Select two integers $$$i$$$ and $$$x$$$ such that $$$1 \le i \le n - 1$$$ and $$$1 \le x \le 10^9$$$. Then, set $$$a_i$$$ and $$$a_{i + 1}$$$ to be equal to $$$x$$$. Find the minimum number of operations needed such that the equality of the array is less than or equal to $$$1$$$. | 256 megabytes | //Utilities
import java.io.*;
import java.util.*;
public class a {
static int t;
static int n;
static int[] a;
// static ArrayList<Integer> arr;
public static void main(String[] args) throws IOException {
t = in.iscan();
outer : while (t-- > 0) {
n = in.iscan(); a = new int[n];
// arr = new ArrayList<Integer>();
for (int i = 0; i < n; i++) {
a[i] = in.iscan();
}
int l = -1, r = -1, res = 0;
for (int i = 0; i < n; i++) {
if (i + 1 < n && a[i] == a[i+1]) {
if (l == -1) {
l = i+1;
r = i+1;
}
else {
r = i+1;
}
}
}
if (l < r) {
res = Math.max(1, r - l - 1);
}
else {
res = 0;
}
out.println(res);
}
out.close();
}
static INPUT in = new INPUT(System.in);
static PrintWriter out = new PrintWriter(System.out);
private static class INPUT {
private InputStream stream;
private byte[] buf = new byte[1024];
private int curChar, numChars;
public INPUT (InputStream stream) {
this.stream = stream;
}
public INPUT (String file) throws IOException {
this.stream = new FileInputStream (file);
}
public int cscan () throws IOException {
if (curChar >= numChars) {
curChar = 0;
numChars = stream.read (buf);
}
if (numChars == -1)
return numChars;
return buf[curChar++];
}
public int iscan () throws IOException {
int c = cscan (), sgn = 1;
while (space (c))
c = cscan ();
if (c == '-') {
sgn = -1;
c = cscan ();
}
int res = 0;
do {
res = (res << 1) + (res << 3);
res += c - '0';
c = cscan ();
}
while (!space (c));
return res * sgn;
}
public String sscan () throws IOException {
int c = cscan ();
while (space (c))
c = cscan ();
StringBuilder res = new StringBuilder ();
do {
res.appendCodePoint (c);
c = cscan ();
}
while (!space (c));
return res.toString ();
}
public double dscan () throws IOException {
int c = cscan (), sgn = 1;
while (space (c))
c = cscan ();
if (c == '-') {
sgn = -1;
c = cscan ();
}
double res = 0;
while (!space (c) && c != '.') {
if (c == 'e' || c == 'E')
return res * UTILITIES.fast_pow (10, iscan ());
res *= 10;
res += c - '0';
c = cscan ();
}
if (c == '.') {
c = cscan ();
double m = 1;
while (!space (c)) {
if (c == 'e' || c == 'E')
return res * UTILITIES.fast_pow (10, iscan ());
m /= 10;
res += (c - '0') * m;
c = cscan ();
}
}
return res * sgn;
}
public long lscan () throws IOException {
int c = cscan (), sgn = 1;
while (space (c))
c = cscan ();
if (c == '-') {
sgn = -1;
c = cscan ();
}
long res = 0;
do {
res = (res << 1) + (res << 3);
res += c - '0';
c = cscan ();
}
while (!space (c));
return res * sgn;
}
public boolean space (int c) {
return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
}
}
public static class UTILITIES {
static final double EPS = 10e-6;
public static void sort(int[] a, boolean increasing) {
ArrayList<Integer> arr = new ArrayList<Integer>();
int n = a.length;
for (int i = 0; i < n; i++) {
arr.add(a[i]);
}
Collections.sort(arr);
for (int i = 0; i < n; i++) {
if (increasing) {
a[i] = arr.get(i);
}
else {
a[i] = arr.get(n-1-i);
}
}
}
public static void sort(long[] a, boolean increasing) {
ArrayList<Long> arr = new ArrayList<Long>();
int n = a.length;
for (int i = 0; i < n; i++) {
arr.add(a[i]);
}
Collections.sort(arr);
for (int i = 0; i < n; i++) {
if (increasing) {
a[i] = arr.get(i);
}
else {
a[i] = arr.get(n-1-i);
}
}
}
public static void sort(double[] a, boolean increasing) {
ArrayList<Double> arr = new ArrayList<Double>();
int n = a.length;
for (int i = 0; i < n; i++) {
arr.add(a[i]);
}
Collections.sort(arr);
for (int i = 0; i < n; i++) {
if (increasing) {
a[i] = arr.get(i);
}
else {
a[i] = arr.get(n-1-i);
}
}
}
public static int lower_bound (int[] arr, int x) {
int low = 0, high = arr.length, mid = -1;
while (low < high) {
mid = (low + high) / 2;
if (arr[mid] >= x)
high = mid;
else
low = mid + 1;
}
return low;
}
public static int upper_bound (int[] arr, int x) {
int low = 0, high = arr.length, mid = -1;
while (low < high) {
mid = (low + high) / 2;
if (arr[mid] > x)
high = mid;
else
low = mid + 1;
}
return low;
}
public static void updateMap(HashMap<Integer, Integer> map, int key, int v) {
if (!map.containsKey(key)) {
map.put(key, v);
}
else {
map.put(key, map.get(key) + v);
}
if (map.get(key) == 0) {
map.remove(key);
}
}
public static long gcd (long a, long b) {
return b == 0 ? a : gcd (b, a % b);
}
public static long lcm (long a, long b) {
return a * b / gcd (a, b);
}
public static long fast_pow_mod (long b, long x, int mod) {
if (x == 0) return 1;
if (x == 1) return b;
if (x % 2 == 0) return fast_pow_mod (b * b % mod, x / 2, mod) % mod;
return b * fast_pow_mod (b * b % mod, x / 2, mod) % mod;
}
public static long fast_pow (long b, long x) {
if (x == 0) return 1;
if (x == 1) return b;
if (x % 2 == 0) return fast_pow (b * b, x / 2);
return b * fast_pow (b * b, x / 2);
}
public static long choose (long n, long k) {
k = Math.min (k, n - k);
long val = 1;
for (int i = 0; i < k; ++i)
val = val * (n - i) / (i + 1);
return val;
}
public static long permute (int n, int k) {
if (n < k) return 0;
long val = 1;
for (int i = 0; i < k; ++i)
val = (val * (n - i));
return val;
}
// start of permutation and lower/upper bound template
public static void nextPermutation(int[] nums) {
//find first decreasing digit
int mark = -1;
for (int i = nums.length - 1; i > 0; i--) {
if (nums[i] > nums[i - 1]) {
mark = i - 1;
break;
}
}
if (mark == -1) {
reverse(nums, 0, nums.length - 1);
return;
}
int idx = nums.length-1;
for (int i = nums.length-1; i >= mark+1; i--) {
if (nums[i] > nums[mark]) {
idx = i;
break;
}
}
swap(nums, mark, idx);
reverse(nums, mark + 1, nums.length - 1);
}
public static void swap(int[] nums, int i, int j) {
int t = nums[i];
nums[i] = nums[j];
nums[j] = t;
}
public static void reverse(int[] nums, int i, int j) {
while (i < j) {
swap(nums, i, j);
i++;
j--;
}
}
static int lower_bound (int[] arr, int hi, int cmp) {
int low = 0, high = hi, mid = -1;
while (low < high) {
mid = (low + high) / 2;
if (arr[mid] >= cmp) high = mid;
else low = mid + 1;
}
return low;
}
static int upper_bound (int[] arr, int hi, int cmp) {
int low = 0, high = hi, mid = -1;
while (low < high) {
mid = (low + high) / 2;
if (arr[mid] > cmp) high = mid;
else low = mid + 1;
}
return low;
}
// end of permutation and lower/upper bound template
}
} | Java | ["4\n\n5\n\n1 1 1 1 1\n\n5\n\n2 1 1 1 2\n\n6\n\n1 1 2 3 3 4\n\n6\n\n1 2 1 4 5 4"] | 1 second | ["2\n1\n2\n0"] | NoteIn the first test case, we can select $$$i=2$$$ and $$$x=2$$$ to form $$$[1, 2, 2, 1, 1]$$$. Then, we can select $$$i=3$$$ and $$$x=3$$$ to form $$$[1, 2, 3, 3, 1]$$$.In the second test case, we can select $$$i=3$$$ and $$$x=100$$$ to form $$$[2, 1, 100, 100, 2]$$$. | Java 11 | standard input | [
"constructive algorithms",
"greedy",
"implementation"
] | a91be662101762bcb2c58b8db9ff61e0 | Each test contains multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains an integer $$$n$$$ ($$$2 \le n \le 2 \cdot 10 ^ 5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10^9$$$) — elements of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10 ^ 5$$$ | 1,100 | For each test case, print the minimum number of operations needed. | standard output | |
PASSED | b8c399f11cfdd0421b2e1a90c7489570 | train_108.jsonl | 1650722700 | You are given an array $$$a$$$ of length $$$n$$$. We define the equality of the array as the number of indices $$$1 \le i \le n - 1$$$ such that $$$a_i = a_{i + 1}$$$. We are allowed to do the following operation: Select two integers $$$i$$$ and $$$x$$$ such that $$$1 \le i \le n - 1$$$ and $$$1 \le x \le 10^9$$$. Then, set $$$a_i$$$ and $$$a_{i + 1}$$$ to be equal to $$$x$$$. Find the minimum number of operations needed such that the equality of the array is less than or equal to $$$1$$$. | 256 megabytes | import java.util.Scanner;
import java.util.stream.IntStream;
public class Main {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int t = sc.nextInt();
for (int tc = 0; tc < t; ++tc) {
int n = sc.nextInt();
int[] a = new int[n];
for (int i = 0; i < a.length; ++i) {
a[i] = sc.nextInt();
}
System.out.println(solve(a));
}
sc.close();
}
static int solve(int[] a) {
int[] indices = IntStream.range(0, a.length - 1).filter(i -> a[i] == a[i + 1]).toArray();
return (indices.length <= 1) ? 0 : Math.max(1, indices[indices.length - 1] - indices[0] - 1);
}
} | Java | ["4\n\n5\n\n1 1 1 1 1\n\n5\n\n2 1 1 1 2\n\n6\n\n1 1 2 3 3 4\n\n6\n\n1 2 1 4 5 4"] | 1 second | ["2\n1\n2\n0"] | NoteIn the first test case, we can select $$$i=2$$$ and $$$x=2$$$ to form $$$[1, 2, 2, 1, 1]$$$. Then, we can select $$$i=3$$$ and $$$x=3$$$ to form $$$[1, 2, 3, 3, 1]$$$.In the second test case, we can select $$$i=3$$$ and $$$x=100$$$ to form $$$[2, 1, 100, 100, 2]$$$. | Java 11 | standard input | [
"constructive algorithms",
"greedy",
"implementation"
] | a91be662101762bcb2c58b8db9ff61e0 | Each test contains multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains an integer $$$n$$$ ($$$2 \le n \le 2 \cdot 10 ^ 5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10^9$$$) — elements of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10 ^ 5$$$ | 1,100 | For each test case, print the minimum number of operations needed. | standard output | |
PASSED | 2b85973804e198db0ab8e46da94bc9b5 | train_108.jsonl | 1650722700 | You are given an array $$$a$$$ of length $$$n$$$. We define the equality of the array as the number of indices $$$1 \le i \le n - 1$$$ such that $$$a_i = a_{i + 1}$$$. We are allowed to do the following operation: Select two integers $$$i$$$ and $$$x$$$ such that $$$1 \le i \le n - 1$$$ and $$$1 \le x \le 10^9$$$. Then, set $$$a_i$$$ and $$$a_{i + 1}$$$ to be equal to $$$x$$$. Find the minimum number of operations needed such that the equality of the array is less than or equal to $$$1$$$. | 256 megabytes | import java.io.*;
import java.util.*;
import java.math.*;
public class Solution{
public static void main(String[] args) {
TaskA solver = new TaskA();
int t = in.nextInt();
for (int i = 1; i <= t ; i++) {
solver.solve(i, in, out);
}
// solver.solve(1, in, out);
out.flush();
out.close();
}
static class TaskA {
public void solve(int testNumber, InputReader in, PrintWriter out) {
int n= in.nextInt();
int[]arr=input(n);int start=-1;int end=-1;int ct=0;int c=-1;int len=0;
for(int i=0;i<n-1;i++) {
if(arr[i]==arr[i+1]) {
len++;
if(c==-1) {
c=arr[i];
ct++;
}
if(start==-1) {
start=i;
}
else {
end=i;
}
}
else {
c=-1;
}
}
// println(ct);println(len);
if(start==-1||end==-1) {
println(0);return;
}
if(ct==1&&len<=2) {
println(len-1);return;
}
if(ct==1&&len>=3) {
println(len-2);return;
}
if(ct>1) {
println(end-start-1);return;
}
else {
println(end-start);return;
}
}
}
static int gcd(int a, int b)
{
if (b == 0)
return a;
return gcd(b, a % b);
}
static int len(long x) {
String s=String.valueOf(x);
return s.length();
}
static void sort(int[] a) {
ArrayList<Integer> q = new ArrayList<>();
for (int i : a) q.add(i);
Collections.sort(q);
for (int i = 0; i < a.length; i++) a[i] = q.get(i);
}
static void sort(long[] a) {
ArrayList<Long> q = new ArrayList<>();
for (long i : a) q.add(i);
Collections.sort(q);
for (int i = 0; i < a.length; i++) a[i] = q.get(i);
}
static void printArr(int[]arr){
for(int i=0;i<arr.length;i++) {
print(arr[i]+" ");
}
print("\n");
}
static void printArr(long[]arr){
for(int i=0;i<arr.length;i++) {
print(arr[i]+" ");
}
print("\n");
}
static int[]input(int n){
int[]arr=new int[n];
for(int i=0;i<n;i++) {
arr[i]=in.nextInt();
}
return arr;
}
static int[]input(){
int n= in.nextInt();
int[]arr=new int[n];
for(int i=0;i<n;i++) {
arr[i]=in.nextInt();
}
return arr;
}
////////////////////////////////////////////////////////
static class Pair {
int first;
int second;
Pair(int x, int y)
{
this.first = x;
this.second = y;
}
}
static void sortS(Pair arr[])
{
Arrays.sort(arr, new Comparator<Pair>() {
@Override public int compare(Pair p1, Pair p2)
{
return p1.second - p2.second;
}
});
}
static void sortF(Pair arr[])
{
Arrays.sort(arr, new Comparator<Pair>() {
@Override public int compare(Pair p1, Pair p2)
{
return p1.first - p2.first;
}
});
}
/////////////////////////////////////////////////////////////
static InputStream inputStream = System.in;
static OutputStream outputStream = System.out;
static InputReader in = new InputReader(inputStream);
static PrintWriter out = new PrintWriter(outputStream);
static void println(long c) {
out.println(c);
}
static void print(long c) {
out.print(c);
}
static void print(int c) {
out.print(c);
}
static void println(int x) {
out.println(x);
}
static void print(String s) {
out.print(s);
}
static void println(String s) {
out.println(s);
}
static class InputReader {
public BufferedReader reader;
public StringTokenizer tokenizer;
public InputReader(InputStream stream) {
reader = new BufferedReader(new InputStreamReader(stream), 32768);
tokenizer = null;
}
public String next() {
while (tokenizer == null || !tokenizer.hasMoreTokens()) {
try {
tokenizer = new StringTokenizer(reader.readLine());
} catch (IOException e) {
throw new RuntimeException(e);
}
}
return tokenizer.nextToken();
}
public int nextInt() {
return Integer.parseInt(next());
}
public long nextLong() {
return Long.parseLong(next());
}
}
} | Java | ["4\n\n5\n\n1 1 1 1 1\n\n5\n\n2 1 1 1 2\n\n6\n\n1 1 2 3 3 4\n\n6\n\n1 2 1 4 5 4"] | 1 second | ["2\n1\n2\n0"] | NoteIn the first test case, we can select $$$i=2$$$ and $$$x=2$$$ to form $$$[1, 2, 2, 1, 1]$$$. Then, we can select $$$i=3$$$ and $$$x=3$$$ to form $$$[1, 2, 3, 3, 1]$$$.In the second test case, we can select $$$i=3$$$ and $$$x=100$$$ to form $$$[2, 1, 100, 100, 2]$$$. | Java 11 | standard input | [
"constructive algorithms",
"greedy",
"implementation"
] | a91be662101762bcb2c58b8db9ff61e0 | Each test contains multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains an integer $$$n$$$ ($$$2 \le n \le 2 \cdot 10 ^ 5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10^9$$$) — elements of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10 ^ 5$$$ | 1,100 | For each test case, print the minimum number of operations needed. | standard output | |
PASSED | fbb14969678897de584e21a5cacf96e4 | train_108.jsonl | 1650722700 | You are given an array $$$a$$$ of length $$$n$$$. We define the equality of the array as the number of indices $$$1 \le i \le n - 1$$$ such that $$$a_i = a_{i + 1}$$$. We are allowed to do the following operation: Select two integers $$$i$$$ and $$$x$$$ such that $$$1 \le i \le n - 1$$$ and $$$1 \le x \le 10^9$$$. Then, set $$$a_i$$$ and $$$a_{i + 1}$$$ to be equal to $$$x$$$. Find the minimum number of operations needed such that the equality of the array is less than or equal to $$$1$$$. | 256 megabytes | import java.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.StringTokenizer;
import java.io.IOException;
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.InputStream;
import java.util.*;
/**
* Built using CHelper plug-in
* Actual solution is at the top
*/
public class Main {
public static void main(String[] args) {
InputStream inputStream = System.in;
OutputStream outputStream = System.out;
InputReader in = new InputReader(inputStream);
PrintWriter out = new PrintWriter(outputStream);
TaskA solver = new TaskA();
int testNum = in.nextInt();
solver.solve(testNum, in, out);
out.close();
}
static class TaskA {
public void solve(int testNumber, InputReader in, PrintWriter out) {
for (int z = 0; z < testNumber; z++) {
int n = in.nextInt();
int[] a = new int[n];
for (int i = 0; i <n ;i++) {
a[i] = in.nextInt();
}
int numDouble = 0;
for (int i = 1; i <n ;i++) {
if (a[i] == a[i-1]) {
numDouble++;
}
}
int indFirstDouble = -1;
int indLastDouble = -1;
for (int i = 1; i < n; i++) {
if (a[i] == a[i-1]) {
indFirstDouble = i;
break;
}
}
for (int i = n-1; i >= 1; i--) {
if (a[i] == a[i-1]) {
indLastDouble = i-1;
break;
}
}
if (numDouble <= 1) {
out.println(0);
} else {
out.println(Math.max(1, indLastDouble - indFirstDouble));
}
}
}
}
static class InputReader {
public BufferedReader reader;
public StringTokenizer tokenizer;
public InputReader(InputStream stream) {
reader = new BufferedReader(new InputStreamReader(stream), 32768);
tokenizer = null;
}
public String next() {
while (tokenizer == null || !tokenizer.hasMoreTokens()) {
try {
tokenizer = new StringTokenizer(reader.readLine());
} catch (IOException e) {
throw new RuntimeException(e);
}
}
return tokenizer.nextToken();
}
public int nextInt() {
return Integer.parseInt(next());
}
}
}
| Java | ["4\n\n5\n\n1 1 1 1 1\n\n5\n\n2 1 1 1 2\n\n6\n\n1 1 2 3 3 4\n\n6\n\n1 2 1 4 5 4"] | 1 second | ["2\n1\n2\n0"] | NoteIn the first test case, we can select $$$i=2$$$ and $$$x=2$$$ to form $$$[1, 2, 2, 1, 1]$$$. Then, we can select $$$i=3$$$ and $$$x=3$$$ to form $$$[1, 2, 3, 3, 1]$$$.In the second test case, we can select $$$i=3$$$ and $$$x=100$$$ to form $$$[2, 1, 100, 100, 2]$$$. | Java 11 | standard input | [
"constructive algorithms",
"greedy",
"implementation"
] | a91be662101762bcb2c58b8db9ff61e0 | Each test contains multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains an integer $$$n$$$ ($$$2 \le n \le 2 \cdot 10 ^ 5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10^9$$$) — elements of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10 ^ 5$$$ | 1,100 | For each test case, print the minimum number of operations needed. | standard output | |
PASSED | 1fce0c88d2b28617bed4408f9082360d | train_108.jsonl | 1650722700 | You are given an array $$$a$$$ of length $$$n$$$. We define the equality of the array as the number of indices $$$1 \le i \le n - 1$$$ such that $$$a_i = a_{i + 1}$$$. We are allowed to do the following operation: Select two integers $$$i$$$ and $$$x$$$ such that $$$1 \le i \le n - 1$$$ and $$$1 \le x \le 10^9$$$. Then, set $$$a_i$$$ and $$$a_{i + 1}$$$ to be equal to $$$x$$$. Find the minimum number of operations needed such that the equality of the array is less than or equal to $$$1$$$. | 256 megabytes |
import java.io.*;
import java.util.*;
import java.lang.*;
public class Codeforces {
static class Reader {
final private int BUFFER_SIZE = 1 << 16;
private DataInputStream din;
private byte[] buffer;
private int bufferPointer, bytesRead;
public Reader()
{
din = new DataInputStream(System.in);
buffer = new byte[BUFFER_SIZE];
bufferPointer = bytesRead = 0;
}
public Reader(String file_name) throws IOException
{
din = new DataInputStream(
new FileInputStream(file_name));
buffer = new byte[BUFFER_SIZE];
bufferPointer = bytesRead = 0;
}
public String readLine() throws IOException
{
byte[] buf = new byte[200001]; // line length
int cnt = 0, c;
while ((c = read()) != -1) {
if (c == '\n') {
if (cnt != 0) {
break;
}
else {
continue;
}
}
buf[cnt++] = (byte)c;
}
return new String(buf, 0, cnt);
}
public int nextInt() throws IOException
{
int ret = 0;
byte c = read();
while (c <= ' ') {
c = read();
}
boolean neg = (c == '-');
if (neg)
c = read();
do {
ret = ret * 10 + c - '0';
} while ((c = read()) >= '0' && c <= '9');
if (neg)
return -ret;
return ret;
}
public long nextLong() throws IOException
{
long ret = 0;
byte c = read();
while (c <= ' ')
c = read();
boolean neg = (c == '-');
if (neg)
c = read();
do {
ret = ret * 10 + c - '0';
} while ((c = read()) >= '0' && c <= '9');
if (neg)
return -ret;
return ret;
}
public double nextDouble() throws IOException
{
double ret = 0, div = 1;
byte c = read();
while (c <= ' ')
c = read();
boolean neg = (c == '-');
if (neg)
c = read();
do {
ret = ret * 10 + c - '0';
} while ((c = read()) >= '0' && c <= '9');
if (c == '.') {
while ((c = read()) >= '0' && c <= '9') {
ret += (c - '0') / (div *= 10);
}
}
if (neg)
return -ret;
return ret;
}
private void fillBuffer() throws IOException
{
bytesRead = din.read(buffer, bufferPointer = 0,
BUFFER_SIZE);
if (bytesRead == -1)
buffer[0] = -1;
}
private byte read() throws IOException
{
if (bufferPointer == bytesRead)
fillBuffer();
return buffer[bufferPointer++];
}
public void close() throws IOException
{
if (din == null)
return;
din.close();
}}
static void cout(Object line) {System.out.println(line);}
static void iop() {
try {
System.setIn(new FileInputStream("input.txt"));
System.setOut(new PrintStream(new FileOutputStream("output.txt")));
} catch (Exception e) {
System.err.println("Error");
}}
static int gcd(int a, int b) {
if(b == 0) return a;
else return gcd(b,a%b);}
static long gcd(long a, long b) {
if(b == 0) return a;
else return gcd(b,a%b);}
static boolean isPerfectSquare(long n){
if (Math.ceil((double)Math.sqrt(n)) ==Math.floor((double)Math.sqrt(n))) return true;
else return false;}
static int fact(int n){
return
(n == 1 || n == 0) ? 1 : n * fact(n - 1);}
static boolean isPowerOfTwo (long x) {return x!=0 && ((x&(x-1)) == 0);}
static void printArray(int arr[]) {
for(int i = 0; i < arr.length; i++) {
System.out.print(arr[i] + " ");
}
System.out.println();}
static int highestPowerof2lessthanx(int x){
// check for the set bits
x |= x >> 1;
x |= x >> 2;
x |= x >> 4;
x |= x >> 8;
x |= x >> 16;
// Then we remove all but the top bit by xor'ing the
// string of 1's with that string of 1's shifted one to
// the left, and we end up with just the one top bit
// followed by 0's.
return x ^ (x >> 1); }
static boolean isPalindrome(String str){
// Pointers pointing to the beginning
// and the end of the string
int i = 0, j = str.length() - 1;
// While there are characters to compare
while (i < j) {
// If there is a mismatch
if (str.charAt(i) != str.charAt(j))
return false;
// Increment first pointer and
// decrement the other
i++;
j--;
}
// Given string is a palindrome
return true;}
static void reverse(long[] array) {
// Length of the array
int n = array.length;
// Swaping the first half elements with last half
// elements
for (int i = 0; i < n / 2; i++) {
// Storing the first half elements temporarily
long temp = array[i];
// Assigning the first half to the last half
array[i] = array[n - i - 1];
// Assigning the last half to the first half
array[n - i - 1] = temp;
}}
static int sumPairs(int arr[], int n) {
// final result
int sum = 0;
for (int i = n - 1; i >= 0; i--)
sum += i * arr[i] - (n - 1 - i) * arr[i];
return sum;}
static int Right_most_setbit(int num) {
int pos = 1;
// counting the position of first set bit
for (int i = 0; i < Integer.MAX_VALUE; i++) {
if ((num & (1 << i))== 0)
pos++;
else
break;
}
return pos;}
static int getItself(int itself, int dummy) {return itself;}
public static void main(String[] args) throws IOException{
iop();
Reader s = new Reader();
// Scanner s = new Scanner(System.in);
int t = s.nextInt();
while(t-- > 0) {
int n = s.nextInt();
int arr[] = new int[n];
for(int i = 0; i < n; i++) {
arr[i] = s.nextInt();
}
int min = -1, max = -1,res = 0;
for(int i = 0; i < n-1; i++) {
if(arr[i] == arr[i+1]) {
if(min == -1) min = i+1;
max = i+1;
}
}
if(max == min) {res = 0;}
else res = Math.max(max-min-1,1);
cout(res);
}
}
}
| Java | ["4\n\n5\n\n1 1 1 1 1\n\n5\n\n2 1 1 1 2\n\n6\n\n1 1 2 3 3 4\n\n6\n\n1 2 1 4 5 4"] | 1 second | ["2\n1\n2\n0"] | NoteIn the first test case, we can select $$$i=2$$$ and $$$x=2$$$ to form $$$[1, 2, 2, 1, 1]$$$. Then, we can select $$$i=3$$$ and $$$x=3$$$ to form $$$[1, 2, 3, 3, 1]$$$.In the second test case, we can select $$$i=3$$$ and $$$x=100$$$ to form $$$[2, 1, 100, 100, 2]$$$. | Java 11 | standard input | [
"constructive algorithms",
"greedy",
"implementation"
] | a91be662101762bcb2c58b8db9ff61e0 | Each test contains multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains an integer $$$n$$$ ($$$2 \le n \le 2 \cdot 10 ^ 5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10^9$$$) — elements of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10 ^ 5$$$ | 1,100 | For each test case, print the minimum number of operations needed. | standard output | |
PASSED | baa05e71e90adfac81ec1605007a6502 | train_108.jsonl | 1650722700 | You are given an array $$$a$$$ of length $$$n$$$. We define the equality of the array as the number of indices $$$1 \le i \le n - 1$$$ such that $$$a_i = a_{i + 1}$$$. We are allowed to do the following operation: Select two integers $$$i$$$ and $$$x$$$ such that $$$1 \le i \le n - 1$$$ and $$$1 \le x \le 10^9$$$. Then, set $$$a_i$$$ and $$$a_{i + 1}$$$ to be equal to $$$x$$$. Find the minimum number of operations needed such that the equality of the array is less than or equal to $$$1$$$. | 256 megabytes | import java.util.*;
import java.io.*;
public class A1{
static BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
static BufferedWriter bw=new BufferedWriter(new OutputStreamWriter(System.out));
static StringTokenizer st;
static final long mod=1000000007;
public static void Solve() throws IOException{
st=new StringTokenizer(br.readLine());
int n=Integer.parseInt(st.nextToken());
int[] ar=getArrIn(n);
int ans=0,min=-1,max=-1;
for(int i=1;i<n;i++){
if(ar[i]==ar[i-1]){
if(min==-1) min=i;
max=i;
}
}
ans=(min==max?0:Math.max(1,max-min-1));
bw.write(ans+"\n");
}
/** Main Method**/
public static void main(String[] YDSV) throws IOException{
//int t=1;
int t=Integer.parseInt(br.readLine());
while(t-->0) Solve();
bw.flush();
}
/** Helpers**/
private static char[] getStr()throws IOException{
return br.readLine().toCharArray();
}
private static int Gcd(int a,int b){
if(b==0) return a;
return Gcd(b,a%b);
}
private static long Gcd(long a,long b){
if(b==0) return a;
return Gcd(b,a%b);
}
private static int[] getArrIn(int n) throws IOException{
st=new StringTokenizer(br.readLine());
int[] ar=new int[n];
for(int i=0;i<n;i++) ar[i]=Integer.parseInt(st.nextToken());
return ar;
}
private static Integer[] getArrInP(int n) throws IOException{
st=new StringTokenizer(br.readLine());
Integer[] ar=new Integer[n];
for(int i=0;i<n;i++) ar[i]=Integer.parseInt(st.nextToken());
return ar;
}
private static long[] getArrLo(int n) throws IOException{
st=new StringTokenizer(br.readLine());
long[] ar=new long[n];
for(int i=0;i<n;i++) ar[i]=Long.parseLong(st.nextToken());
return ar;
}
private static List<Integer> getListIn(int n) throws IOException{
st=new StringTokenizer(br.readLine());
List<Integer> al=new ArrayList<>();
for(int i=0;i<n;i++) al.add(Integer.parseInt(st.nextToken()));
return al;
}
private static List<Long> getListLo(int n) throws IOException{
st=new StringTokenizer(br.readLine());
List<Long> al=new ArrayList<>();
for(int i=0;i<n;i++) al.add(Long.parseLong(st.nextToken()));
return al;
}
private static long pow_mod(long a,long b) {
long result=1;
while(b!=0){
if((b&1)!=0) result=(result*a)%mod;
a=(a*a)%mod;
b>>=1;
}
return result;
}
private static int lower_bound(int a[],int x){
int l=-1,r=a.length;
while(l+1<r){
int m=(l+r)>>>1;
if(a[m]>=x) r=m;
else l=m;
}
return r+1;
}
private static long lower_bound(long a[],long x){
int l=-1,r=a.length;
while(l+1<r){
int m=(l+r)>>>1;
if(a[m]>=x) r=m;
else l=m;
}
return r+1;
}
private static int upper_bound(int a[],int x){
int l=-1,r=a.length;
while(l+1<r){
int m=(l+r)>>>1;
if(a[m]<=x) l=m;
else r=m;
}
return l+1;
}
private static long upper_bound(long a[],long x){
int l=-1,r=a.length;
while(l+1<r){
int m=(l+r)>>>1;
if(a[m]<=x) l=m;
else r=m;
}
return l+1;
}
private static boolean Sqrt(int x){
int a=(int)Math.sqrt(x);
return a*a==x;
}
private static boolean Sqrt(long x){
long a=(long)Math.sqrt(x);
return a*a==x;
}
} | Java | ["4\n\n5\n\n1 1 1 1 1\n\n5\n\n2 1 1 1 2\n\n6\n\n1 1 2 3 3 4\n\n6\n\n1 2 1 4 5 4"] | 1 second | ["2\n1\n2\n0"] | NoteIn the first test case, we can select $$$i=2$$$ and $$$x=2$$$ to form $$$[1, 2, 2, 1, 1]$$$. Then, we can select $$$i=3$$$ and $$$x=3$$$ to form $$$[1, 2, 3, 3, 1]$$$.In the second test case, we can select $$$i=3$$$ and $$$x=100$$$ to form $$$[2, 1, 100, 100, 2]$$$. | Java 11 | standard input | [
"constructive algorithms",
"greedy",
"implementation"
] | a91be662101762bcb2c58b8db9ff61e0 | Each test contains multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains an integer $$$n$$$ ($$$2 \le n \le 2 \cdot 10 ^ 5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10^9$$$) — elements of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10 ^ 5$$$ | 1,100 | For each test case, print the minimum number of operations needed. | standard output | |
PASSED | 886235f88d3a3935204387799b751215 | train_108.jsonl | 1650722700 | You are given an array $$$a$$$ of length $$$n$$$. We define the equality of the array as the number of indices $$$1 \le i \le n - 1$$$ such that $$$a_i = a_{i + 1}$$$. We are allowed to do the following operation: Select two integers $$$i$$$ and $$$x$$$ such that $$$1 \le i \le n - 1$$$ and $$$1 \le x \le 10^9$$$. Then, set $$$a_i$$$ and $$$a_{i + 1}$$$ to be equal to $$$x$$$. Find the minimum number of operations needed such that the equality of the array is less than or equal to $$$1$$$. | 256 megabytes | import java.io.*;
import java.util.*;
public class Q1672C{
static int mod = (int) (1e9 + 7);
static void solve() {
int n=i();
long[]arr=new long[n];
for(int i=0;i<arr.length;i++){
arr[i]=l();
}
int l=0;
int r=n;
for(int i=0;i<n-1;i++){
if(arr[i]==arr[i+1]){
l=i;
break;
}
}
for(int i=n-2;i>=0;i--){
if(arr[i]==arr[i+1]){
r=i;
break;
}
}
if(l==0 && r==n || l==r){
System.out.println(0);
return;
}
System.out.println(Math.max(1, r-l-1));
}
public static void main(String[] args) {
int test = i();
while (test-- > 0) {
solve();
}
}
// -----> POWER ---> long power(long x, long y) <---- power
// -----> LCM ---> long lcm(long x, long y) <---- lcm
// -----> GCD ---> long gcd(long x, long y) <---- gcd
// -----> SIEVE --> ArrayList<Integer> sieve(int N) <-----sieve
// -----> NCR ---> long ncr(int n, int r) <---- ncr
// -----> BINARY_SEARCH ---> int binary_Search(long[]arr,long x) <----binary_Search
// -----> (SORTING OF LONG, CHAR,INT) -->long[] sortLong(long[] a2)<--
// -----> (INPUT OF INT,LONG ,STRING) -> int i() long l() String s()<-
// tempstart
static class InputReader {
private InputStream stream;
private byte[] buf = new byte[1024];
private int curChar;
private int numChars;
private SpaceCharFilter filter;
public InputReader(InputStream stream) {
this.stream = stream;
}
public int read() {
if (numChars == -1)
throw new InputMismatchException();
if (curChar >= numChars) {
curChar = 0;
try {
numChars = stream.read(buf);
} catch (IOException e) {
throw new InputMismatchException();
}
if (numChars <= 0)
return -1;
}
return buf[curChar++];
}
public int Int() {
int c = read();
while (isSpaceChar(c))
c = read();
int sgn = 1;
if (c == '-') {
sgn = -1;
c = read();
}
int res = 0;
do {
if (c < '0' || c > '9')
throw new InputMismatchException();
res *= 10;
res += c - '0';
c = read();
} while (!isSpaceChar(c));
return res * sgn;
}
public String String() {
int c = read();
while (isSpaceChar(c))
c = read();
StringBuilder res = new StringBuilder();
do {
res.appendCodePoint(c);
c = read();
} while (!isSpaceChar(c));
return res.toString();
}
public boolean isSpaceChar(int c) {
if (filter != null)
return filter.isSpaceChar(c);
return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
}
public String next() {
return String();
}
public interface SpaceCharFilter {
public boolean isSpaceChar(int ch);
}
}
static InputReader in = new InputReader(System.in);
public static int i() {
return in.Int();
}
public static long l() {
String s = in.String();
return Long.parseLong(s);
}
public static String s() {
return in.String();
}
} | Java | ["4\n\n5\n\n1 1 1 1 1\n\n5\n\n2 1 1 1 2\n\n6\n\n1 1 2 3 3 4\n\n6\n\n1 2 1 4 5 4"] | 1 second | ["2\n1\n2\n0"] | NoteIn the first test case, we can select $$$i=2$$$ and $$$x=2$$$ to form $$$[1, 2, 2, 1, 1]$$$. Then, we can select $$$i=3$$$ and $$$x=3$$$ to form $$$[1, 2, 3, 3, 1]$$$.In the second test case, we can select $$$i=3$$$ and $$$x=100$$$ to form $$$[2, 1, 100, 100, 2]$$$. | Java 11 | standard input | [
"constructive algorithms",
"greedy",
"implementation"
] | a91be662101762bcb2c58b8db9ff61e0 | Each test contains multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains an integer $$$n$$$ ($$$2 \le n \le 2 \cdot 10 ^ 5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10^9$$$) — elements of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10 ^ 5$$$ | 1,100 | For each test case, print the minimum number of operations needed. | standard output | |
PASSED | 7ed9a2cff10bfaf4e48d6cabcdd2156a | train_108.jsonl | 1650722700 | You are given an array $$$a$$$ of length $$$n$$$. We define the equality of the array as the number of indices $$$1 \le i \le n - 1$$$ such that $$$a_i = a_{i + 1}$$$. We are allowed to do the following operation: Select two integers $$$i$$$ and $$$x$$$ such that $$$1 \le i \le n - 1$$$ and $$$1 \le x \le 10^9$$$. Then, set $$$a_i$$$ and $$$a_{i + 1}$$$ to be equal to $$$x$$$. Find the minimum number of operations needed such that the equality of the array is less than or equal to $$$1$$$. | 256 megabytes | import java.util.Scanner;
public class Main {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
int casesNumber = in.nextInt();
while (casesNumber-- != 0) {
int length = in.nextInt();
int[] array = new int[length];
int first = 0;
int end = 0;
int equal = 0;
boolean firstFound = false;
for (int i = 0; i < length; i++) {
array[i] = in.nextInt();
if (i >= 1 && array[i] == array[i - 1]) {
if (!firstFound) {
first = i;
firstFound = true;
} else end = i - 1;
equal++;
}
}
if (equal <= 1)
System.out.println(0);
else
System.out.println(end - first == 0 ? 1 : end - first);
}
}
} | Java | ["4\n\n5\n\n1 1 1 1 1\n\n5\n\n2 1 1 1 2\n\n6\n\n1 1 2 3 3 4\n\n6\n\n1 2 1 4 5 4"] | 1 second | ["2\n1\n2\n0"] | NoteIn the first test case, we can select $$$i=2$$$ and $$$x=2$$$ to form $$$[1, 2, 2, 1, 1]$$$. Then, we can select $$$i=3$$$ and $$$x=3$$$ to form $$$[1, 2, 3, 3, 1]$$$.In the second test case, we can select $$$i=3$$$ and $$$x=100$$$ to form $$$[2, 1, 100, 100, 2]$$$. | Java 11 | standard input | [
"constructive algorithms",
"greedy",
"implementation"
] | a91be662101762bcb2c58b8db9ff61e0 | Each test contains multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains an integer $$$n$$$ ($$$2 \le n \le 2 \cdot 10 ^ 5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10^9$$$) — elements of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10 ^ 5$$$ | 1,100 | For each test case, print the minimum number of operations needed. | standard output | |
PASSED | a8eed66f3fd3e883e3005398eca683ea | train_108.jsonl | 1650722700 | You are given an array $$$a$$$ of length $$$n$$$. We define the equality of the array as the number of indices $$$1 \le i \le n - 1$$$ such that $$$a_i = a_{i + 1}$$$. We are allowed to do the following operation: Select two integers $$$i$$$ and $$$x$$$ such that $$$1 \le i \le n - 1$$$ and $$$1 \le x \le 10^9$$$. Then, set $$$a_i$$$ and $$$a_{i + 1}$$$ to be equal to $$$x$$$. Find the minimum number of operations needed such that the equality of the array is less than or equal to $$$1$$$. | 256 megabytes | import java.io.ByteArrayInputStream;
import java.io.File;
import java.io.FileInputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.security.cert.X509CRL;
import java.util.*;
import java.lang.*;
import java.util.stream.Collector;
import java.util.stream.Collectors;
@SuppressWarnings("unused")
public class Main {
static InputStream is;
static PrintWriter out;
static String INPUT = "";
static String OUTPUT = "";
//global
//private final static long BASE = 998244353L;
private final static int BASE = 1000000007;
private final static int INF_I = 1001001001;
private final static long INF_L = 1001001001001001001L;
private final static int MAXN = 100100;
private static List<Integer> palin = new ArrayList<>();
static void solve() {
int ntest = readInt();
for (int test=0;test<ntest;test++) {
int N = readInt();
int[] A = readIntArray(N);
int L=INF_I,R=-INF_I;
for (int i=0;i<N-1;i++) {
if (A[i] != A[i+1]) continue;
L = Math.min(L,i);
R = Math.max(R,i);
}
if (L==INF_I || R==L) out.println(0);
else out.println(Math.max(1, R-L-1));
}
}
public static void main(String[] args) throws Exception
{
long S = System.currentTimeMillis();
if (INPUT=="") {
is = System.in;
} else {
File file = new File(INPUT);
is = new FileInputStream(file);
}
if (OUTPUT == "") out = new PrintWriter(System.out);
else out = new PrintWriter(OUTPUT);
solve();
out.flush();
long G = System.currentTimeMillis();
}
private static class Point<T extends Number & Comparable<T>> implements Comparable<Point<T>> {
private T x;
private T y;
public Point(T x, T y) {
this.x = x;
this.y = y;
}
public T getX() {return x;}
public T getY() {return y;}
@Override
public int compareTo(Point<T> o) {
int cmp = x.compareTo(o.getX());
if (cmp==0) return y.compareTo(o.getY());
return cmp;
}
}
private static class ClassComparator<T extends Comparable<T>> implements Comparator<T> {
public ClassComparator() {}
@Override
public int compare(T a, T b) {
return a.compareTo(b);
}
}
private static class ListComparator<T extends Comparable<T>> implements Comparator<List<T>> {
public ListComparator() {}
@Override
public int compare(List<T> o1, List<T> o2) {
for (int i = 0; i < Math.min(o1.size(), o2.size()); i++) {
int c = o1.get(i).compareTo(o2.get(i));
if (c != 0) {
return c;
}
}
return Integer.compare(o1.size(), o2.size());
}
}
private static boolean eof()
{
if(lenbuf == -1)return true;
int lptr = ptrbuf;
while(lptr < lenbuf)if(!isSpaceChar(inbuf[lptr++]))return false;
try {
is.mark(1000);
while(true){
int b = is.read();
if(b == -1){
is.reset();
return true;
}else if(!isSpaceChar(b)){
is.reset();
return false;
}
}
} catch (IOException e) {
return true;
}
}
private static byte[] inbuf = new byte[1024];
static int lenbuf = 0, ptrbuf = 0;
private static int readByte()
{
if(lenbuf == -1)throw new InputMismatchException();
if(ptrbuf >= lenbuf){
ptrbuf = 0;
try { lenbuf = is.read(inbuf); } catch (IOException e) { throw new InputMismatchException(); }
if(lenbuf <= 0)return -1;
}
return inbuf[ptrbuf++];
}
private static boolean isSpaceChar(int c) { return !(c >= 33 && c <= 126); }
// private static boolean isSpaceChar(int c) { return !(c >= 32 && c <= 126); }
private static int skip() { int b; while((b = readByte()) != -1 && isSpaceChar(b)); return b; }
private static double readDouble() { return Double.parseDouble(readString()); }
private static char readChar() { return (char)skip(); }
private static String readString()
{
int b = skip();
StringBuilder sb = new StringBuilder();
while(!(isSpaceChar(b))){
sb.appendCodePoint(b);
b = readByte();
}
return sb.toString();
}
private static char[] readChar(int n)
{
char[] buf = new char[n];
int b = skip(), p = 0;
while(p < n && !(isSpaceChar(b))){
buf[p++] = (char)b;
b = readByte();
}
return n == p ? buf : Arrays.copyOf(buf, p);
}
private static char[][] readTable(int n, int m)
{
char[][] map = new char[n][];
for(int i = 0;i < n;i++)map[i] = readChar(m);
return map;
}
private static int[] readIntArray(int n)
{
int[] a = new int[n];
for(int i = 0;i < n;i++)a[i] = readInt();
return a;
}
private static long[] readLongArray(int n) {
long[] a = new long[n];
for (int i=0;i<n;i++) a[i] = readLong();
return a;
}
private static int readInt()
{
int num = 0, b;
boolean minus = false;
while((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-'));
if(b == '-'){
minus = true;
b = readByte();
}
while(true){
if(b >= '0' && b <= '9'){
num = num * 10 + (b - '0');
}else{
return minus ? -num : num;
}
b = readByte();
}
}
private static long readLong()
{
long num = 0;
int b;
boolean minus = false;
while((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-'));
if(b == '-'){
minus = true;
b = readByte();
}
while(true){
if(b >= '0' && b <= '9'){
num = num * 10 + (b - '0');
}else{
return minus ? -num : num;
}
b = readByte();
}
}
private static void tr(Object... o) { if(INPUT.length() != 0)System.out.println(Arrays.deepToString(o)); }
} | Java | ["4\n\n5\n\n1 1 1 1 1\n\n5\n\n2 1 1 1 2\n\n6\n\n1 1 2 3 3 4\n\n6\n\n1 2 1 4 5 4"] | 1 second | ["2\n1\n2\n0"] | NoteIn the first test case, we can select $$$i=2$$$ and $$$x=2$$$ to form $$$[1, 2, 2, 1, 1]$$$. Then, we can select $$$i=3$$$ and $$$x=3$$$ to form $$$[1, 2, 3, 3, 1]$$$.In the second test case, we can select $$$i=3$$$ and $$$x=100$$$ to form $$$[2, 1, 100, 100, 2]$$$. | Java 11 | standard input | [
"constructive algorithms",
"greedy",
"implementation"
] | a91be662101762bcb2c58b8db9ff61e0 | Each test contains multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains an integer $$$n$$$ ($$$2 \le n \le 2 \cdot 10 ^ 5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10^9$$$) — elements of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10 ^ 5$$$ | 1,100 | For each test case, print the minimum number of operations needed. | standard output | |
PASSED | 0fbee8b19460dba5542defc438efded7 | train_108.jsonl | 1650722700 | You are given an array $$$a$$$ of length $$$n$$$. We define the equality of the array as the number of indices $$$1 \le i \le n - 1$$$ such that $$$a_i = a_{i + 1}$$$. We are allowed to do the following operation: Select two integers $$$i$$$ and $$$x$$$ such that $$$1 \le i \le n - 1$$$ and $$$1 \le x \le 10^9$$$. Then, set $$$a_i$$$ and $$$a_{i + 1}$$$ to be equal to $$$x$$$. Find the minimum number of operations needed such that the equality of the array is less than or equal to $$$1$$$. | 256 megabytes | import java.util.*;
import java.io.*;
import java.math.*;
public class Main {
public static void main(String[] args) {
Scanner sc=new Scanner(System.in);
int t=sc.nextInt();
while(t-->0){
int n=sc.nextInt();
long a[]=new long[n];
for(int i=0;i<n;i++) a[i]=sc.nextLong();
int start=-1;
int end=-1;
for(int i=0;i<n-1;i++){
if(a[i]==a[i+1])
{
start=i;
break;
}
}
for(int i=n-1;i>0;i--){
if(a[i]==a[i-1])
{
end=i;
break;
}
}
long len=end-start+1;
if(len<3)
System.out.println("0");
else if(len==3)
System.out.println("1");
else
System.out.println(len-3);
}
}
} | Java | ["4\n\n5\n\n1 1 1 1 1\n\n5\n\n2 1 1 1 2\n\n6\n\n1 1 2 3 3 4\n\n6\n\n1 2 1 4 5 4"] | 1 second | ["2\n1\n2\n0"] | NoteIn the first test case, we can select $$$i=2$$$ and $$$x=2$$$ to form $$$[1, 2, 2, 1, 1]$$$. Then, we can select $$$i=3$$$ and $$$x=3$$$ to form $$$[1, 2, 3, 3, 1]$$$.In the second test case, we can select $$$i=3$$$ and $$$x=100$$$ to form $$$[2, 1, 100, 100, 2]$$$. | Java 11 | standard input | [
"constructive algorithms",
"greedy",
"implementation"
] | a91be662101762bcb2c58b8db9ff61e0 | Each test contains multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains an integer $$$n$$$ ($$$2 \le n \le 2 \cdot 10 ^ 5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10^9$$$) — elements of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10 ^ 5$$$ | 1,100 | For each test case, print the minimum number of operations needed. | standard output | |
PASSED | 3d5a439aa311e2373fdbd6102094348f | train_108.jsonl | 1650722700 | You are given an array $$$a$$$ of length $$$n$$$. We define the equality of the array as the number of indices $$$1 \le i \le n - 1$$$ such that $$$a_i = a_{i + 1}$$$. We are allowed to do the following operation: Select two integers $$$i$$$ and $$$x$$$ such that $$$1 \le i \le n - 1$$$ and $$$1 \le x \le 10^9$$$. Then, set $$$a_i$$$ and $$$a_{i + 1}$$$ to be equal to $$$x$$$. Find the minimum number of operations needed such that the equality of the array is less than or equal to $$$1$$$. | 256 megabytes | import java.util.*;
public class QUESTION {
public static int compute(int[] a){
int n=a.length;
if(n<=2) return 0;
int res = 0;
int s=1,f=0;
while(f<n-2 && a[f]!=a[f+1]){
f++;
s=f+1;
}
while(f<n-2 && s<n-1){
while(s<n-1 && a[s]!=a[s+1]){
s++;
}
if(s>=n-1) {return res;}
if(s==f+1){
res+=1;
f=s;
s=f+2;
}
else{
res+=(s-f-1);
f=s-1;
s=f+2;
}
//System.out.println(res+" "+f+" "+s);
}
return res;
}
public static void main (String[] args) {
Scanner in = new Scanner(System.in);
int t = in.nextInt();
for(int i=0;i<t;i++){
int n = in.nextInt();
int[] a = new int[n];
for(int j=0;j<n;j++){
a[j] = in.nextInt();
}
System.out.println(compute(a));
}
}
} | Java | ["4\n\n5\n\n1 1 1 1 1\n\n5\n\n2 1 1 1 2\n\n6\n\n1 1 2 3 3 4\n\n6\n\n1 2 1 4 5 4"] | 1 second | ["2\n1\n2\n0"] | NoteIn the first test case, we can select $$$i=2$$$ and $$$x=2$$$ to form $$$[1, 2, 2, 1, 1]$$$. Then, we can select $$$i=3$$$ and $$$x=3$$$ to form $$$[1, 2, 3, 3, 1]$$$.In the second test case, we can select $$$i=3$$$ and $$$x=100$$$ to form $$$[2, 1, 100, 100, 2]$$$. | Java 11 | standard input | [
"constructive algorithms",
"greedy",
"implementation"
] | a91be662101762bcb2c58b8db9ff61e0 | Each test contains multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains an integer $$$n$$$ ($$$2 \le n \le 2 \cdot 10 ^ 5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10^9$$$) — elements of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10 ^ 5$$$ | 1,100 | For each test case, print the minimum number of operations needed. | standard output | |
PASSED | b5358acf699d0b9e229735ddbbfba449 | train_108.jsonl | 1650722700 | You are given an array $$$a$$$ of length $$$n$$$. We define the equality of the array as the number of indices $$$1 \le i \le n - 1$$$ such that $$$a_i = a_{i + 1}$$$. We are allowed to do the following operation: Select two integers $$$i$$$ and $$$x$$$ such that $$$1 \le i \le n - 1$$$ and $$$1 \le x \le 10^9$$$. Then, set $$$a_i$$$ and $$$a_{i + 1}$$$ to be equal to $$$x$$$. Find the minimum number of operations needed such that the equality of the array is less than or equal to $$$1$$$. | 256 megabytes | import java.util.*;
import java.io.*;
public class Practice {
static boolean multipleTC = true;
final static int mod2 = 1000000007;
final static int mod = 998244353;
final double E = 2.7182818284590452354;
final double PI = 3.14159265358979323846;
int MAX = 200001;
void pre() throws Exception {
}
// All the best
void solve(int TC) throws Exception {
int n = ni(), arr[] = readArr(n);
int first = -1, last = -1;
for (int i = 1; i < n; i++) {
if (arr[i] == arr[i - 1]) {
if (first == -1) {
first = i;
}
last = i;
}
}
if (first == -1 || first == last) {
pn("0");
}else {
int ans = last-first-1;
if(ans<=0) {
ans =1;
}
pn(ans);
}
}
public static long gcd(long a, long b) {
if (a > b)
a = (a + b) - (b = a);
if (a == 0L)
return b;
return gcd(b % a, a);
}
double dist(int x1, int y1, int x2, int y2) {
double a = x1 - x2, b = y1 - y2;
return Math.sqrt((a * a) + (b * b));
}
int[] readArr(int n) throws Exception {
int arr[] = new int[n];
for (int i = 0; i < n; i++) {
arr[i] = ni();
}
return arr;
}
void sort(int arr[], int left, int right) {
ArrayList<Integer> list = new ArrayList<>();
for (int i = left; i <= right; i++)
list.add(arr[i]);
Collections.sort(list);
for (int i = left; i <= right; i++)
arr[i] = list.get(i - left);
}
void sort(int arr[]) {
ArrayList<Integer> list = new ArrayList<>();
for (int i = 0; i < arr.length; i++)
list.add(arr[i]);
Collections.sort(list);
for (int i = 0; i < arr.length; i++)
arr[i] = list.get(i);
}
public long max(long... arr) {
long max = arr[0];
for (long itr : arr)
max = Math.max(max, itr);
return max;
}
public int max(int... arr) {
int max = arr[0];
for (int itr : arr)
max = Math.max(max, itr);
return max;
}
public long min(long... arr) {
long min = arr[0];
for (long itr : arr)
min = Math.min(min, itr);
return min;
}
public int min(int... arr) {
int min = arr[0];
for (int itr : arr)
min = Math.min(min, itr);
return min;
}
public long sum(long... arr) {
long sum = 0;
for (long itr : arr)
sum += itr;
return sum;
}
public long sum(int... arr) {
long sum = 0;
for (int itr : arr)
sum += itr;
return sum;
}
String bin(long n) {
return Long.toBinaryString(n);
}
String bin(int n) {
return Integer.toBinaryString(n);
}
static int bitCount(int x) {
return x == 0 ? 0 : (1 + bitCount(x & (x - 1)));
}
static void dbg(Object... o) {
System.err.println(Arrays.deepToString(o));
}
int bit(long n) {
return (n == 0) ? 0 : (1 + bit(n & (n - 1)));
}
int abs(int a) {
return (a < 0) ? -a : a;
}
long abs(long a) {
return (a < 0) ? -a : a;
}
void p(Object o) {
out.print(o);
}
void pn(Object o) {
out.println(o);
}
void pni(Object o) {
out.println(o);
out.flush();
}
void pn(int[] arr) {
int n = arr.length;
StringBuilder sb = new StringBuilder();
for (int i = 0; i < n; i++) {
sb.append(arr[i] + " ");
}
pn(sb);
}
void pn(long[] arr) {
int n = arr.length;
StringBuilder sb = new StringBuilder();
for (int i = 0; i < n; i++) {
sb.append(arr[i] + " ");
}
pn(sb);
}
String n() throws Exception {
return in.next();
}
String nln() throws Exception {
return in.nextLine();
}
int ni() throws Exception {
return Integer.parseInt(in.next());
}
long nl() throws Exception {
return Long.parseLong(in.next());
}
double nd() throws Exception {
return Double.parseDouble(in.next());
}
public static void main(String[] args) throws Exception {
new Practice().run();
}
FastReader in;
PrintWriter out;
void run() throws Exception {
in = new FastReader();
out = new PrintWriter(System.out);
int T = (multipleTC) ? ni() : 1;
pre();
for (int t = 1; t <= T; t++)
solve(t);
out.flush();
out.close();
}
class FastReader {
BufferedReader br;
StringTokenizer st;
public FastReader() {
br = new BufferedReader(new InputStreamReader(System.in));
}
public FastReader(String s) throws Exception {
br = new BufferedReader(new FileReader(s));
}
String next() throws Exception {
while (st == null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(br.readLine());
} catch (IOException e) {
throw new Exception(e.toString());
}
}
return st.nextToken();
}
String nextLine() throws Exception {
String str = "";
try {
str = br.readLine();
} catch (IOException e) {
throw new Exception(e.toString());
}
return str;
}
}
} | Java | ["4\n\n5\n\n1 1 1 1 1\n\n5\n\n2 1 1 1 2\n\n6\n\n1 1 2 3 3 4\n\n6\n\n1 2 1 4 5 4"] | 1 second | ["2\n1\n2\n0"] | NoteIn the first test case, we can select $$$i=2$$$ and $$$x=2$$$ to form $$$[1, 2, 2, 1, 1]$$$. Then, we can select $$$i=3$$$ and $$$x=3$$$ to form $$$[1, 2, 3, 3, 1]$$$.In the second test case, we can select $$$i=3$$$ and $$$x=100$$$ to form $$$[2, 1, 100, 100, 2]$$$. | Java 11 | standard input | [
"constructive algorithms",
"greedy",
"implementation"
] | a91be662101762bcb2c58b8db9ff61e0 | Each test contains multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains an integer $$$n$$$ ($$$2 \le n \le 2 \cdot 10 ^ 5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10^9$$$) — elements of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10 ^ 5$$$ | 1,100 | For each test case, print the minimum number of operations needed. | standard output | |
PASSED | 83a339ceba4735c8080cff4d280d4586 | train_108.jsonl | 1650722700 | You are given an array $$$a$$$ of length $$$n$$$. We define the equality of the array as the number of indices $$$1 \le i \le n - 1$$$ such that $$$a_i = a_{i + 1}$$$. We are allowed to do the following operation: Select two integers $$$i$$$ and $$$x$$$ such that $$$1 \le i \le n - 1$$$ and $$$1 \le x \le 10^9$$$. Then, set $$$a_i$$$ and $$$a_{i + 1}$$$ to be equal to $$$x$$$. Find the minimum number of operations needed such that the equality of the array is less than or equal to $$$1$$$. | 256 megabytes |
import java.util.*;
import java.io.*;
public class Solution{
public static void main(String[] args) throws IOException {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
StringTokenizer st = new StringTokenizer(br.readLine());
int t = Integer.valueOf(st.nextToken());
for(int tc = 1; tc <= t; tc++){
st = new StringTokenizer(br.readLine());
int n = Integer.valueOf(st.nextToken());
int[] arr = new int[n];
st = new StringTokenizer(br.readLine());
for(int i = 0; i < n; i++){
arr[i] = Integer.valueOf(st.nextToken());
}
int min = 20000000;
int max = -1;
for(int i = 0; i < n - 1; i++){
if(arr[i] == arr[i + 1]){
min = Math.min(min, i);
max = Math.max(max, i);
}
}
if(max == -1 || (min==max)){
System.out.println( 0);
}else{
if(max - min == 1)
System.out.println( 1);
else{
System.out.println((max - min - 1));
}
}
}
}
}
class IN{
BufferedReader br;
StringTokenizer st;
} | Java | ["4\n\n5\n\n1 1 1 1 1\n\n5\n\n2 1 1 1 2\n\n6\n\n1 1 2 3 3 4\n\n6\n\n1 2 1 4 5 4"] | 1 second | ["2\n1\n2\n0"] | NoteIn the first test case, we can select $$$i=2$$$ and $$$x=2$$$ to form $$$[1, 2, 2, 1, 1]$$$. Then, we can select $$$i=3$$$ and $$$x=3$$$ to form $$$[1, 2, 3, 3, 1]$$$.In the second test case, we can select $$$i=3$$$ and $$$x=100$$$ to form $$$[2, 1, 100, 100, 2]$$$. | Java 11 | standard input | [
"constructive algorithms",
"greedy",
"implementation"
] | a91be662101762bcb2c58b8db9ff61e0 | Each test contains multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains an integer $$$n$$$ ($$$2 \le n \le 2 \cdot 10 ^ 5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10^9$$$) — elements of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10 ^ 5$$$ | 1,100 | For each test case, print the minimum number of operations needed. | standard output | |
PASSED | 6802ec790ee5ffd4276c04a7724f8ada | train_108.jsonl | 1650722700 | You are given an array $$$a$$$ of length $$$n$$$. We define the equality of the array as the number of indices $$$1 \le i \le n - 1$$$ such that $$$a_i = a_{i + 1}$$$. We are allowed to do the following operation: Select two integers $$$i$$$ and $$$x$$$ such that $$$1 \le i \le n - 1$$$ and $$$1 \le x \le 10^9$$$. Then, set $$$a_i$$$ and $$$a_{i + 1}$$$ to be equal to $$$x$$$. Find the minimum number of operations needed such that the equality of the array is less than or equal to $$$1$$$. | 256 megabytes | import java.util.Scanner;
public class A {
public static void main(String[] args) {
Scanner s = new Scanner(System.in);
int t = s.nextInt();
for(int tc = 0; tc < t; tc++){
int moves = 0;
int[] a = new int[s.nextInt()];
for(int i = 0; i < a.length; i++){a[i] = s.nextInt();}
int x = 0; int y = 0;
for(int i = 0; i < a.length-1; i++){
if(a[i] == a[i+1]){
x = i;
y = i+1;
break;
}
}
for(int i = 0; i < a.length-1; i++){
if(a[i] == a[i+1]){
y=i+1;
}
}
if(y-x == 0 || y-x == 1) System.out.println(0);
else System.out.println(Math.max(1, y-x-2));
}
}
}
| Java | ["4\n\n5\n\n1 1 1 1 1\n\n5\n\n2 1 1 1 2\n\n6\n\n1 1 2 3 3 4\n\n6\n\n1 2 1 4 5 4"] | 1 second | ["2\n1\n2\n0"] | NoteIn the first test case, we can select $$$i=2$$$ and $$$x=2$$$ to form $$$[1, 2, 2, 1, 1]$$$. Then, we can select $$$i=3$$$ and $$$x=3$$$ to form $$$[1, 2, 3, 3, 1]$$$.In the second test case, we can select $$$i=3$$$ and $$$x=100$$$ to form $$$[2, 1, 100, 100, 2]$$$. | Java 11 | standard input | [
"constructive algorithms",
"greedy",
"implementation"
] | a91be662101762bcb2c58b8db9ff61e0 | Each test contains multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains an integer $$$n$$$ ($$$2 \le n \le 2 \cdot 10 ^ 5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10^9$$$) — elements of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10 ^ 5$$$ | 1,100 | For each test case, print the minimum number of operations needed. | standard output | |
PASSED | a45b8252f5488d3149ec2ea580260a65 | train_108.jsonl | 1650722700 | You are given an array $$$a$$$ of length $$$n$$$. We define the equality of the array as the number of indices $$$1 \le i \le n - 1$$$ such that $$$a_i = a_{i + 1}$$$. We are allowed to do the following operation: Select two integers $$$i$$$ and $$$x$$$ such that $$$1 \le i \le n - 1$$$ and $$$1 \le x \le 10^9$$$. Then, set $$$a_i$$$ and $$$a_{i + 1}$$$ to be equal to $$$x$$$. Find the minimum number of operations needed such that the equality of the array is less than or equal to $$$1$$$. | 256 megabytes | import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.List;
import java.util.StringTokenizer;
public class C {
static final FastReader sc = new FastReader();
static final PrintWriter out = new PrintWriter(System.out, true);
private static boolean debug = System.getProperty("ONLINE_JUDGE") == null;
private static void trace(Object... o) {
if (debug) {
System.err.println(Arrays.deepToString(o));
}
}
public static void main(String[] args) {
int t = sc.nextInt();
while (t-- > 0) {
int n = sc.nextInt();
int[] a = new int[n];
for (int i = 0; i < n; i++) {
a[i] = sc.nextInt();
}
int c = 0;
for (int i = 0; i < n - 1; i++) {
if (a[i] == a[i + 1])
c++;
}
if (c <= 1) {
out.println(0);
} else {
int l = Integer.MAX_VALUE, r = -1;
for (int i = 0; i < n - 1; i++) {
if (a[i] == a[i + 1]) {
l = Math.min(l, i);
r = Math.max(r, i);
}
}
out.println(Math.max(1, r - l - 1));
}
}
}
static class FastReader {
BufferedReader br;
StringTokenizer st;
public FastReader() {
br = new BufferedReader(new InputStreamReader(System.in));
}
String next() {
while (st == null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt() {
return Integer.parseInt(next());
}
long nextLong() {
return Long.parseLong(next());
}
double nextDouble() {
return Double.parseDouble(next());
}
String nextLine() {
String str = "";
try {
str = br.readLine();
} catch (IOException e) {
e.printStackTrace();
}
return str;
}
}
}
| Java | ["4\n\n5\n\n1 1 1 1 1\n\n5\n\n2 1 1 1 2\n\n6\n\n1 1 2 3 3 4\n\n6\n\n1 2 1 4 5 4"] | 1 second | ["2\n1\n2\n0"] | NoteIn the first test case, we can select $$$i=2$$$ and $$$x=2$$$ to form $$$[1, 2, 2, 1, 1]$$$. Then, we can select $$$i=3$$$ and $$$x=3$$$ to form $$$[1, 2, 3, 3, 1]$$$.In the second test case, we can select $$$i=3$$$ and $$$x=100$$$ to form $$$[2, 1, 100, 100, 2]$$$. | Java 11 | standard input | [
"constructive algorithms",
"greedy",
"implementation"
] | a91be662101762bcb2c58b8db9ff61e0 | Each test contains multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains an integer $$$n$$$ ($$$2 \le n \le 2 \cdot 10 ^ 5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10^9$$$) — elements of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10 ^ 5$$$ | 1,100 | For each test case, print the minimum number of operations needed. | standard output | |
PASSED | 134383886e79d468cbe1fbd494ea20b4 | train_108.jsonl | 1650722700 | You are given an array $$$a$$$ of length $$$n$$$. We define the equality of the array as the number of indices $$$1 \le i \le n - 1$$$ such that $$$a_i = a_{i + 1}$$$. We are allowed to do the following operation: Select two integers $$$i$$$ and $$$x$$$ such that $$$1 \le i \le n - 1$$$ and $$$1 \le x \le 10^9$$$. Then, set $$$a_i$$$ and $$$a_{i + 1}$$$ to be equal to $$$x$$$. Find the minimum number of operations needed such that the equality of the array is less than or equal to $$$1$$$. | 256 megabytes | /***** ---> :) Vijender Srivastava (: <--- *****/
import java.util.*;
import javax.sql.rowset.serial.SerialArray;
import java.lang.*;
// import java.lang.reflect.Array;
import java.io.*;
public class Main
{
static FastReader sc =new FastReader();
static PrintWriter out=new PrintWriter(System.out);
static long mod=(long)32768;
static StringBuilder sb = new StringBuilder();
/* start */
public static void main(String [] args)
{
// int testcases = 1;
int testcases = i();
// calc();
while(testcases-->0)
{
solve();
}
out.flush();
out.close();
}
static void solve()
{
int n = i();
int a[] = input(n);
int x = -1,y=-1;
for(int i=0;i<n-1;i++)
{
if(a[i]==a[i+1])
{
if(x==-1) x = i+1;
y = i+1;
}
}
if(x==y)
{
pl(0);
return;
}
if(y-x==1)pl(1);
else
pl((y-x-1));
}
/* end */
static class FastReader
{
BufferedReader br;
StringTokenizer st;
public FastReader()
{
br = new BufferedReader(new
InputStreamReader(System.in));
}
String next()
{
while (st == null || !st.hasMoreElements())
{
try
{
st = new StringTokenizer(br.readLine());
}
catch (IOException e)
{
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt()
{
return Integer.parseInt(next());
}
long nextLong()
{
return Long.parseLong(next());
}
double nextDouble()
{
return Double.parseDouble(next());
}
String nextLine()
{
String str = "";
try
{
str = br.readLine();
}
catch (IOException e)
{
e.printStackTrace();
}
return str;
}
}
static void p(Object o)
{
out.print(o);
}
static void pl(Object o)
{
out.println(o);
}
static int i() {
return sc.nextInt();
}
static String s() {
return sc.next();
}
static long l() {
return sc.nextLong();
}
static char[] inputC()
{
String s = sc.nextLine();
return s.toCharArray();
}
static int[] input(int n) {
int A[]=new int[n];
for(int i=0;i<n;i++) {
A[i]=sc.nextInt();
}
return A;
}
static long[] inputL(int n) {
long A[]=new long[n];
for(int i=0;i<n;i++) {
A[i]=sc.nextLong();
}
return A;
}
static long[] putL(long a[]) {
long A[]=new long[a.length];
for(int i=0;i<a.length;i++) {
A[i]=a[i];
}
return A;
}
static String[] inputS(int n) {
String A[]=new String[n];
for(int i=0;i<n;i++) {
A[i]=sc.next();
}
return A;
}
static int gcd(int a, int b)
{
if (a == 0)
return b;
return gcd(b % a, a);
}
static long gcd(long a, long b)
{
if (a == 0)
return b;
return gcd(b % a, a);
}
static int lcm(int a, int b)
{
return (a / gcd(a, b)) * b;
}
static String reverse(String s) {
StringBuffer p=new StringBuffer(s);
p.reverse();
return p.toString();
}
static int min(int a,int b) {
return Math.min(a, b);
}
static int min(int a,int b,int c) {
return Math.min(a, Math.min(b, c));
}
static int min(int a,int b,int c,int d) {
return Math.min(a, Math.min(b, Math.min(c, d)));
}
static int max(int a,int b) {
return Math.max(a, b);
}
static int max(int a,int b,int c) {
return Math.max(a, Math.max(b, c));
}
static int max(int a,int b,int c,int d) {
return Math.max(a, Math.max(b, Math.max(c, d)));
}
static long min(long a,long b) {
return Math.min(a, b);
}
static long min(long a,long b,long c) {
return Math.min(a, Math.min(b, c));
}
static long min(long a,long b,long c,long d) {
return Math.min(a, Math.min(b, Math.min(c, d)));
}
static long max(long a,long b) {
return Math.max(a, b);
}
static long max(long a,long b,long c) {
return Math.max(a, Math.max(b, c));
}
static long max(long a,long b,long c,long d) {
return Math.max(a, Math.max(b, Math.max(c, d)));
}
static long sum(int A[]) {
long sum=0;
for(int i : A) {
sum+=i;
}
return sum;
}
static long sum(long A[]) {
long sum=0;
for(long i : A) {
sum+=i;
}
return sum;
}
static void print(int A[]) {
for(int i : A) {
System.out.print(i+" ");
}
System.out.println();
}
static void print(long A[]) {
for(long i : A) {
System.out.print(i+" ");
}
System.out.println();
}
static long mod(long x) {
return ((x%mod + mod)%mod);
}
static long power(long x, long y)
{
if(y==0)
return 1;
if(x==0)
return 0;
long res = 1;
while (y > 0) {
if (y % 2 == 1)
res = (res * x) ;
y = y >> 1;
x = (x * x);
}
return res;
}
static boolean prime(int n)
{
if (n <= 1)
return false;
if (n <= 3)
return true;
if (n % 2 == 0 || n % 3 == 0)
return false;
double sq=Math.sqrt(n);
for (int i = 5; i <= sq; i = i + 6)
if (n % i == 0 || n % (i + 2) == 0)
return false;
return true;
}
static boolean prime(long n)
{
if (n <= 1)
return false;
if (n <= 3)
return true;
if (n % 2 == 0 || n % 3 == 0)
return false;
double sq=Math.sqrt(n);
for (int i = 5; i <= sq; i = i + 6)
if (n % i == 0 || n % (i + 2) == 0)
return false;
return true;
}
static long[] sort(long a[]) {
ArrayList<Long> arr = new ArrayList<>();
for(long i : a) {
arr.add(i);
}
Collections.sort(arr);
for(int i = 0; i < arr.size(); i++) {
a[i] = arr.get(i);
}
return a;
}
static int[] sort(int a[])
{
ArrayList<Integer> arr = new ArrayList<>();
for(Integer i : a) {
arr.add(i);
}
Collections.sort(arr);
for(int i = 0; i < arr.size(); i++) {
a[i] = arr.get(i);
}
return a;
}
//pair class
private static class Pair implements Comparable<Pair> {
long first, second;
public Pair(long f, long s) {
first = f;
second = s;
}
@Override
public int compareTo(Pair p) {
if (first < p.first)
return 1;
else if (first > p.first)
return -1;
else {
if (second > p.second)
return 1;
else if (second < p.second)
return -1;
else
return 0;
}
}
}
} | Java | ["4\n\n5\n\n1 1 1 1 1\n\n5\n\n2 1 1 1 2\n\n6\n\n1 1 2 3 3 4\n\n6\n\n1 2 1 4 5 4"] | 1 second | ["2\n1\n2\n0"] | NoteIn the first test case, we can select $$$i=2$$$ and $$$x=2$$$ to form $$$[1, 2, 2, 1, 1]$$$. Then, we can select $$$i=3$$$ and $$$x=3$$$ to form $$$[1, 2, 3, 3, 1]$$$.In the second test case, we can select $$$i=3$$$ and $$$x=100$$$ to form $$$[2, 1, 100, 100, 2]$$$. | Java 11 | standard input | [
"constructive algorithms",
"greedy",
"implementation"
] | a91be662101762bcb2c58b8db9ff61e0 | Each test contains multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains an integer $$$n$$$ ($$$2 \le n \le 2 \cdot 10 ^ 5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10^9$$$) — elements of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10 ^ 5$$$ | 1,100 | For each test case, print the minimum number of operations needed. | standard output | |
PASSED | 14d195a87ac6fb242819f65b9f510182 | train_108.jsonl | 1650722700 | You are given an array $$$a$$$ of length $$$n$$$. We define the equality of the array as the number of indices $$$1 \le i \le n - 1$$$ such that $$$a_i = a_{i + 1}$$$. We are allowed to do the following operation: Select two integers $$$i$$$ and $$$x$$$ such that $$$1 \le i \le n - 1$$$ and $$$1 \le x \le 10^9$$$. Then, set $$$a_i$$$ and $$$a_{i + 1}$$$ to be equal to $$$x$$$. Find the minimum number of operations needed such that the equality of the array is less than or equal to $$$1$$$. | 256 megabytes | import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
public class UnequalArray {
static int MAXN = 1123456;
static boolean[] prime = new boolean[MAXN];
static boolean[] used = new boolean[MAXN];
public static void seive() {
for (int i = 2; i < MAXN; ++i) if (!used[i]){
prime[i] = true;
for (int j = i; j < MAXN; j += i) {
used[j] = true;
}
}
prime[1] = false;
}
// pair
// class Pair implements Comparable<Pair>{
// int ind,val;
// Pair(int i,int v){ ind=i;val=v;}
// @Override
// public int compareTo(Pair o) {return o.val-val ;}
// }
public static void main (String[] args) throws IOException {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
int t = Integer.parseInt(br.readLine());
while (t-- > 0) {
String st[] = br.readLine().split(" ");
int n = Integer.parseInt(st[0]);
String str[] = br.readLine().split(" ");
int[] a = new int[n];
for(int i = 0; i < n; i++) {
a[i] = Integer.parseInt(str[i]);
}
boolean found = false;
int l = -1;
int r = -1;
for(int i = 0; i < n-1; i++){
if(a[i] == a[i+1] && !found) {
found = true;
l = i;
}
else if(found && a[i] == a[i+1]) r = i;
}
if(l == -1 || r == -1) System.out.println(0);
else
System.out.println(Math.max(r-l+2 - 3, 1));
}
}
}
| Java | ["4\n\n5\n\n1 1 1 1 1\n\n5\n\n2 1 1 1 2\n\n6\n\n1 1 2 3 3 4\n\n6\n\n1 2 1 4 5 4"] | 1 second | ["2\n1\n2\n0"] | NoteIn the first test case, we can select $$$i=2$$$ and $$$x=2$$$ to form $$$[1, 2, 2, 1, 1]$$$. Then, we can select $$$i=3$$$ and $$$x=3$$$ to form $$$[1, 2, 3, 3, 1]$$$.In the second test case, we can select $$$i=3$$$ and $$$x=100$$$ to form $$$[2, 1, 100, 100, 2]$$$. | Java 11 | standard input | [
"constructive algorithms",
"greedy",
"implementation"
] | a91be662101762bcb2c58b8db9ff61e0 | Each test contains multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains an integer $$$n$$$ ($$$2 \le n \le 2 \cdot 10 ^ 5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10^9$$$) — elements of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10 ^ 5$$$ | 1,100 | For each test case, print the minimum number of operations needed. | standard output | |
PASSED | e39fd8414c4e66bab67a71b4d8a62a19 | train_108.jsonl | 1650722700 | You are given an array $$$a$$$ of length $$$n$$$. We define the equality of the array as the number of indices $$$1 \le i \le n - 1$$$ such that $$$a_i = a_{i + 1}$$$. We are allowed to do the following operation: Select two integers $$$i$$$ and $$$x$$$ such that $$$1 \le i \le n - 1$$$ and $$$1 \le x \le 10^9$$$. Then, set $$$a_i$$$ and $$$a_{i + 1}$$$ to be equal to $$$x$$$. Find the minimum number of operations needed such that the equality of the array is less than or equal to $$$1$$$. | 256 megabytes |
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.util.List;
import java.util.Scanner;
import java.util.StringTokenizer;
public class C {
static class RealScanner {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
StringTokenizer st = new StringTokenizer("");
String next() {
while (!st.hasMoreTokens())
try {
st = new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
return st.nextToken();
}
int nextInt() {
return Integer.parseInt(next());
}
int[] readArray(int n) {
int[] a = new int[n];
for (int i = 0; i < n; i++) a[i] = nextInt();
return a;
}
long nextLong() {
return Long.parseLong(next());
}
public boolean isSorted(List<Long> list) {
for (int i = 0; i < list.size() - 1; i++) {
if (list.get(i) > list.get(i + 1))
return false;
}
return true;
}
}
public static void main(String[] args) {
RealScanner sc = new RealScanner();
int t = sc.nextInt();
while (t-- > 0) {
int n = sc.nextInt();
int[] arr = new int[n];
for (int i = 0; i < n; i++) {
arr[i] = sc.nextInt();
}
List<Integer> list = new ArrayList<>();
int k = 0, count = 0;
for (int i = 0; i < n - 1; i++) {
if (arr[i] == arr[i + 1]) {
list.add(i);
k = i + 1;
} else {
count++;
}
}
if (list.size() != 0) {
list.add(k);
}
// System.out.println(list);
// boolean check = true;
// for (int i = 0; i < list.size() - 1; i++) {
// if (list.get(i) + 1 == list.get(i + 1)) {
//
// } else {
// check = false;
// break;
// }
// }
// if (check && list.size() != 0) {
// System.out.println(1);
// continue;
// }
if (count == n - 1) {
System.out.println(0);
continue;
}
int idx1 = 0;
for (int i = 0; i < n - 1; i++) {
if (arr[i] == arr[i + 1]) {
idx1 = i + 1;
break;
}
}
int idx2 = 0;
for (int i = n - 1; i > 0; i--) {
if (arr[i] == arr[i - 1]) {
idx2 = i - 1;
break;
}
}
// System.out.println(idx1 + " " + (n - idx2 - 1));
if (idx1 == idx2) {
System.out.println(1);
continue;
}
if (idx2-idx1==-1){
System.out.println(0);
continue;
}
System.out.println(idx2 - idx1);
}
}
}
| Java | ["4\n\n5\n\n1 1 1 1 1\n\n5\n\n2 1 1 1 2\n\n6\n\n1 1 2 3 3 4\n\n6\n\n1 2 1 4 5 4"] | 1 second | ["2\n1\n2\n0"] | NoteIn the first test case, we can select $$$i=2$$$ and $$$x=2$$$ to form $$$[1, 2, 2, 1, 1]$$$. Then, we can select $$$i=3$$$ and $$$x=3$$$ to form $$$[1, 2, 3, 3, 1]$$$.In the second test case, we can select $$$i=3$$$ and $$$x=100$$$ to form $$$[2, 1, 100, 100, 2]$$$. | Java 11 | standard input | [
"constructive algorithms",
"greedy",
"implementation"
] | a91be662101762bcb2c58b8db9ff61e0 | Each test contains multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains an integer $$$n$$$ ($$$2 \le n \le 2 \cdot 10 ^ 5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10^9$$$) — elements of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10 ^ 5$$$ | 1,100 | For each test case, print the minimum number of operations needed. | standard output | |
PASSED | 58cbf9aecfe3c423112ee90e6737c6b2 | train_108.jsonl | 1650722700 | You are given an array $$$a$$$ of length $$$n$$$. We define the equality of the array as the number of indices $$$1 \le i \le n - 1$$$ such that $$$a_i = a_{i + 1}$$$. We are allowed to do the following operation: Select two integers $$$i$$$ and $$$x$$$ such that $$$1 \le i \le n - 1$$$ and $$$1 \le x \le 10^9$$$. Then, set $$$a_i$$$ and $$$a_{i + 1}$$$ to be equal to $$$x$$$. Find the minimum number of operations needed such that the equality of the array is less than or equal to $$$1$$$. | 256 megabytes | import java.util.*;
import java.lang.*;
import java.io.*;
/* Name of the class has to be "Main" only if the class is public. */
public class Codechef
{
public static void main (String[] args) throws java.lang.Exception
{
Scanner sc=new Scanner(System.in);
int t=sc.nextInt();
for(int i=0;i<t;i++){
int n=sc.nextInt();
int[] arr=new int[n];
for(int j=0;j<n;j++){
arr[j]=sc.nextInt();
}
int temp=0;
int champ=0,count=0;
for(int j=0;j<n-1;j++){
if((arr[j]==arr[j+1])&&temp==0){
temp=j+1;
count++;
}
else if(arr[j]==arr[j+1]){
champ=j;
count++;
}
}
if(count==0){
System.out.println(0);
}
else{
if(champ-temp>=1){
System.out.println(champ-temp);
}
else if(champ-temp==0){
System.out.println(1);
}
else if(champ-temp<0){
System.out.println(0);
}
}
}
// your code goes here
}
} | Java | ["4\n\n5\n\n1 1 1 1 1\n\n5\n\n2 1 1 1 2\n\n6\n\n1 1 2 3 3 4\n\n6\n\n1 2 1 4 5 4"] | 1 second | ["2\n1\n2\n0"] | NoteIn the first test case, we can select $$$i=2$$$ and $$$x=2$$$ to form $$$[1, 2, 2, 1, 1]$$$. Then, we can select $$$i=3$$$ and $$$x=3$$$ to form $$$[1, 2, 3, 3, 1]$$$.In the second test case, we can select $$$i=3$$$ and $$$x=100$$$ to form $$$[2, 1, 100, 100, 2]$$$. | Java 11 | standard input | [
"constructive algorithms",
"greedy",
"implementation"
] | a91be662101762bcb2c58b8db9ff61e0 | Each test contains multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains an integer $$$n$$$ ($$$2 \le n \le 2 \cdot 10 ^ 5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10^9$$$) — elements of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10 ^ 5$$$ | 1,100 | For each test case, print the minimum number of operations needed. | standard output | |
PASSED | 50a2f07c47821145717527a2360edb2d | train_108.jsonl | 1650722700 | You are given an array $$$a$$$ of length $$$n$$$. We define the equality of the array as the number of indices $$$1 \le i \le n - 1$$$ such that $$$a_i = a_{i + 1}$$$. We are allowed to do the following operation: Select two integers $$$i$$$ and $$$x$$$ such that $$$1 \le i \le n - 1$$$ and $$$1 \le x \le 10^9$$$. Then, set $$$a_i$$$ and $$$a_{i + 1}$$$ to be equal to $$$x$$$. Find the minimum number of operations needed such that the equality of the array is less than or equal to $$$1$$$. | 256 megabytes | import java.io.*;
import java.util.StringTokenizer;
public class s {
public static void main(String[] args) throws IOException {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
PrintWriter pr = new PrintWriter(new OutputStreamWriter(System.out));
int t = Integer.parseInt(br.readLine());
while (t-- != 0) {
StringTokenizer st = new StringTokenizer(br.readLine());
int n = Integer.parseInt(st.nextToken());
int[] arr = new int[n];
int start =-1;
int end = -1;
st = new StringTokenizer(br.readLine());
for (int i = 0; i < n; i++) {
arr[i] = Integer.parseInt(st.nextToken());
if(i>0 && arr[i]==arr[i-1]){
if(start == -1){
start = i-1;
}
end = i;
}
}
int length = end-start +1;
if(length<3){
pr.println(0);
}else if (length == 3){
pr.println(1);
}else{
pr.println(length-3);
}
}
pr.close();
}
} | Java | ["4\n\n5\n\n1 1 1 1 1\n\n5\n\n2 1 1 1 2\n\n6\n\n1 1 2 3 3 4\n\n6\n\n1 2 1 4 5 4"] | 1 second | ["2\n1\n2\n0"] | NoteIn the first test case, we can select $$$i=2$$$ and $$$x=2$$$ to form $$$[1, 2, 2, 1, 1]$$$. Then, we can select $$$i=3$$$ and $$$x=3$$$ to form $$$[1, 2, 3, 3, 1]$$$.In the second test case, we can select $$$i=3$$$ and $$$x=100$$$ to form $$$[2, 1, 100, 100, 2]$$$. | Java 11 | standard input | [
"constructive algorithms",
"greedy",
"implementation"
] | a91be662101762bcb2c58b8db9ff61e0 | Each test contains multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains an integer $$$n$$$ ($$$2 \le n \le 2 \cdot 10 ^ 5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10^9$$$) — elements of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10 ^ 5$$$ | 1,100 | For each test case, print the minimum number of operations needed. | standard output | |
PASSED | 6f4cf970d7ff735edd5fcb12ef1a79fc | train_108.jsonl | 1650722700 | You are given an array $$$a$$$ of length $$$n$$$. We define the equality of the array as the number of indices $$$1 \le i \le n - 1$$$ such that $$$a_i = a_{i + 1}$$$. We are allowed to do the following operation: Select two integers $$$i$$$ and $$$x$$$ such that $$$1 \le i \le n - 1$$$ and $$$1 \le x \le 10^9$$$. Then, set $$$a_i$$$ and $$$a_{i + 1}$$$ to be equal to $$$x$$$. Find the minimum number of operations needed such that the equality of the array is less than or equal to $$$1$$$. | 256 megabytes |
import java.io.*;
import java.util.Iterator;
import java.util.StringTokenizer;
public class Unequal {
void solveCase(FScanner scanner) throws IOException {
int n = scanner.nextInt();
int[] input = scanner.nextIntArr(n);
int result = 0;
Iterator<Run> it = new RunIterator(input);
Run first = null;
Run last = null;
if (it.hasNext())
first = it.next();
while (it.hasNext()) {
last = it.next();
}
if (first != null) {
int len = first.length;
if (last != null) {
int a = first.start;
int b = last.start + last.length - 1;
len = b - a + 1;
}
if (len == 2)
result = 0;
else if (len == 3)
result = 1;
else
result = len - 3;
}
out.println(result);
out.flush();
}
static class Run {
int start;
int length;
public Run(int start, int length) {
this.start = start;
this.length = length;
}
}
static class RunIterator implements Iterator<Run> {
private final int[] arr;
private int curr;
private Run found;
public RunIterator(int[] arr) {
this.arr = arr;
curr = 0;
}
@Override
public boolean hasNext() {
found = findNext();
return found != null;
}
@Override
public Run next() {
if (found == null) throw new IllegalStateException();
Run tmp = found;
found = null;
return tmp;
}
private Run findNext() {
if (curr == arr.length) return null;
int next = curr + 1;
while (next < arr.length && arr[next] != arr[next - 1]) {
next++;
}
if (next == arr.length) {
curr = next;
return null;
}
int start = next - 1;
int len = 2;
next++;
while (next < arr.length && arr[next - 1] == arr[next]) {
len++;
next++;
}
curr = next;
return new Run(start, len);
}
}
//-----------------------------------------
InputStream sin;
PrintWriter out;
Unequal() {
this(System.in, System.out);
}
Unequal(InputStream sin, OutputStream sout) {
this.sin = sin;
out = new PrintWriter(new BufferedOutputStream(sout));
}
public FScanner createScanner() {
return new FScanner(sin);
}
void solve(FScanner scanner) throws IOException {
int t = scanner.nextInt();
for (int i = 0; i < t; i++) {
solveCase(scanner);
}
}
public static void main(String[] args) {
Unequal task = new Unequal();
FScanner scanner = task.createScanner();
try {
task.solve(scanner);
} catch (IOException e) {
throw new RuntimeException(e);
}
}
static class FScanner {
private BufferedReader bufferedReader;
public FScanner(InputStream inputStream) {
this.bufferedReader = new BufferedReader(new InputStreamReader(inputStream));
}
public int nextInt() throws IOException {
String line = bufferedReader.readLine();
return Integer.parseInt(line);
}
public String next() throws IOException {
String line = bufferedReader.readLine();
return line;
}
public int[] nextIntArr(int n) throws IOException {
String line = bufferedReader.readLine();
StringTokenizer tokenizer = new StringTokenizer(line);
int[] result = new int[n];
for (int i = 0; i < n; i++) {
result[i] = Integer.parseInt(tokenizer.nextToken());
}
return result;
}
public long[] nextLongArr(int n) throws IOException {
String line = bufferedReader.readLine();
StringTokenizer tokenizer = new StringTokenizer(line);
long[] result = new long[n];
for (int i = 0; i < n; i++) {
result[i] = Long.parseLong(tokenizer.nextToken());
}
return result;
}
public double[] nextDoubleArr(int n) throws IOException {
String line = bufferedReader.readLine();
StringTokenizer tokenizer = new StringTokenizer(line);
double[] result = new double[n];
for (int i = 0; i < n; i++) {
result[i] = Double.parseDouble(tokenizer.nextToken());
}
return result;
}
public String[] nextStringArr(int n) throws IOException {
String line = bufferedReader.readLine();
StringTokenizer tokenizer = new StringTokenizer(line);
String[] result = new String[n];
for (int i = 0; i < n; i++) {
result[i] = tokenizer.nextToken();
}
return result;
}
// --------------
long gcd(long a, long b) {
if (b == 0) return a;
return gcd(b, a % b);
}
long gcdAll(long[] arr, int start) {
long g = arr[start];
if (arr.length > 2) {
for (int j = start + 2; j < arr.length; j += 2) {
g = gcd(arr[j], g);
}
}
return g;
}
void reverse(int[] arr) {
int n = arr.length;
for (int i = 0; i < n / 2; i++) {
int tmp = arr[i];
arr[i] = arr[n - 1 - i];
arr[n - 1 - i] = tmp;
}
}
}
}
| Java | ["4\n\n5\n\n1 1 1 1 1\n\n5\n\n2 1 1 1 2\n\n6\n\n1 1 2 3 3 4\n\n6\n\n1 2 1 4 5 4"] | 1 second | ["2\n1\n2\n0"] | NoteIn the first test case, we can select $$$i=2$$$ and $$$x=2$$$ to form $$$[1, 2, 2, 1, 1]$$$. Then, we can select $$$i=3$$$ and $$$x=3$$$ to form $$$[1, 2, 3, 3, 1]$$$.In the second test case, we can select $$$i=3$$$ and $$$x=100$$$ to form $$$[2, 1, 100, 100, 2]$$$. | Java 11 | standard input | [
"constructive algorithms",
"greedy",
"implementation"
] | a91be662101762bcb2c58b8db9ff61e0 | Each test contains multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains an integer $$$n$$$ ($$$2 \le n \le 2 \cdot 10 ^ 5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10^9$$$) — elements of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10 ^ 5$$$ | 1,100 | For each test case, print the minimum number of operations needed. | standard output | |
PASSED | 016ebdedcaac36f020425df3ea7bee7d | train_108.jsonl | 1650722700 | You are given an array $$$a$$$ of length $$$n$$$. We define the equality of the array as the number of indices $$$1 \le i \le n - 1$$$ such that $$$a_i = a_{i + 1}$$$. We are allowed to do the following operation: Select two integers $$$i$$$ and $$$x$$$ such that $$$1 \le i \le n - 1$$$ and $$$1 \le x \le 10^9$$$. Then, set $$$a_i$$$ and $$$a_{i + 1}$$$ to be equal to $$$x$$$. Find the minimum number of operations needed such that the equality of the array is less than or equal to $$$1$$$. | 256 megabytes | import java.util.*;
import org.w3c.dom.css.CSSCharsetRule;
import java.io.*;
import java.math.*;
public class C {
// -- static variables --- //
static FastReader sc = new FastReader();
static PrintWriter out = new PrintWriter(System.out);
static int mod = (int) 1000000007;
public static void main(String[] args) throws Exception {
sieve();
int t = sc.nextInt();
while (t-- > 0)
C.go();
// out.println();
out.flush();
}
// >>>>>>>>>>>>>>>>>>> Code Starts <<<<<<<<<<<<<<<<<<<< //
static class pair {
int x, y;
pair(int x, int y) {
this.x = x;
this.y = y;
}
}
static void go() throws Exception {
int n=sc.nextInt();
int a[]=sc.intArray(n);
int lt=-1,rt=-1;
for(int i=1;i<n;i++) {
if(a[i]==a[i-1]) {
if(lt==-1)lt=i;
rt=i;
}
}
if(rt==lt) {
out.println(0);
return;
}
out.println(Math.max(1, rt-lt-1));
}
static long lcm(long a, long b) {
return a * b / gcd(a, b);
}
// >>>>>>>>>>> Code Ends <<<<<<<<< //
// --For Rounding--//
static double round(double value, int places) {
if (places < 0)
throw new IllegalArgumentException();
BigDecimal bd = new BigDecimal(Double.toString(value));
bd = bd.setScale(places, RoundingMode.HALF_UP);
return bd.doubleValue();
}
// ----Greatest Common Divisor-----//
static long gcd(long a, long b) {
if (b == 0) {
return a;
}
return gcd(b, a % b);
}
// --- permutations and Combinations ---//
static long fact[];
static long invfact[];
static long ncr(int n, int k) {
if (k < 0 || k > n) {
return 0;
}
long x = fact[n];
long y = fact[k];
long yy = fact[n - k];
long ans = (x / y);
ans = (ans / yy);
return ans;
}
// ---sieve---//
static int prime[] = new int[1000006];
static void sieve() {
Arrays.fill(prime, 1);
prime[0] = 0;
prime[1] = 0;
for (int i = 2; i * i <= 1000005; i++) {
if (prime[i] == 1)
for (int j = i * i; j <= 1000005; j += i) {
prime[j] = 0;
}
}
}
// ---- Manual sort ------//
static void sort(long[] a) {
ArrayList<Long> aa = new ArrayList<>();
for (long i : a) {
aa.add(i);
}
Collections.sort(aa);
for (int i = 0; i < a.length; i++)
a[i] = aa.get(i);
}
static void sort(int[] a) {
ArrayList<Integer> aa = new ArrayList<>();
for (int i : a) {
aa.add(i);
}
Collections.sort(aa);
for (int i = 0; i < a.length; i++)
a[i] = aa.get(i);
}
// --- Fast exponentiation ---//
static long pow(long x, long y) {
long res = 1l;
while (y != 0) {
if (y % 2 == 1) {
res = (x * res);
}
y /= 2;
x = (x * x);
}
return res;
}
// >>>>>>>>>>>>>>> Fast IO <<<<<<<<<<<<<< //
static class FastReader {
BufferedReader br;
StringTokenizer st;
public FastReader() {
br = new BufferedReader(new InputStreamReader(System.in));
}
String next() {
while (st == null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt() {
return Integer.parseInt(next());
}
long nextLong() {
return Long.parseLong(next());
}
double nextDouble() {
return Double.parseDouble(next());
}
int[] intArray(int n) {
int a[] = new int[n];
for (int i = 0; i < n; i++)
a[i] = sc.nextInt();
return a;
}
long[] longArray(int n) {
long a[] = new long[n];
for (int i = 0; i < n; i++)
a[i] = sc.nextLong();
return a;
}
String nextLine() {
String str = "";
try {
str = br.readLine();
} catch (IOException e) {
e.printStackTrace();
}
return str;
}
}
}
| Java | ["4\n\n5\n\n1 1 1 1 1\n\n5\n\n2 1 1 1 2\n\n6\n\n1 1 2 3 3 4\n\n6\n\n1 2 1 4 5 4"] | 1 second | ["2\n1\n2\n0"] | NoteIn the first test case, we can select $$$i=2$$$ and $$$x=2$$$ to form $$$[1, 2, 2, 1, 1]$$$. Then, we can select $$$i=3$$$ and $$$x=3$$$ to form $$$[1, 2, 3, 3, 1]$$$.In the second test case, we can select $$$i=3$$$ and $$$x=100$$$ to form $$$[2, 1, 100, 100, 2]$$$. | Java 11 | standard input | [
"constructive algorithms",
"greedy",
"implementation"
] | a91be662101762bcb2c58b8db9ff61e0 | Each test contains multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains an integer $$$n$$$ ($$$2 \le n \le 2 \cdot 10 ^ 5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10^9$$$) — elements of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10 ^ 5$$$ | 1,100 | For each test case, print the minimum number of operations needed. | standard output | |
PASSED | 895b0ff5899405dc8b2dc0166321effb | train_108.jsonl | 1650722700 | You are given an array $$$a$$$ of length $$$n$$$. We define the equality of the array as the number of indices $$$1 \le i \le n - 1$$$ such that $$$a_i = a_{i + 1}$$$. We are allowed to do the following operation: Select two integers $$$i$$$ and $$$x$$$ such that $$$1 \le i \le n - 1$$$ and $$$1 \le x \le 10^9$$$. Then, set $$$a_i$$$ and $$$a_{i + 1}$$$ to be equal to $$$x$$$. Find the minimum number of operations needed such that the equality of the array is less than or equal to $$$1$$$. | 256 megabytes |
import java.util.*;
import java.io.*;
public class CF{
public static void main(String[] args)throws IOException{
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
int test = Integer.parseInt(br.readLine());
for(int tst = 0; tst < test; tst++){
int n = Integer.parseInt(br.readLine());
//HashMap<Integer, Integer> hm = new HashMap<Integer,Integer>();
// TreeSet<Integer> ts= new TreeSet<Integer>();
// PriorityQueue<Integer> pq = new PriorityQueue<Integer>();
String[] inp1 = br.readLine().split(" ");
//String[] inp2 = br.readLine().split(" ");
//int n = Integer.parseInt(inp1[0]);
//long x = Long.parseLong(inp1[1]);
//String[] inp2 = br.readLine().split(" ");
//<Long> arr = new ArrayList<Long>();
int arr[] = new int[n];
for(int i =0; i<n; i++){
arr[i] = Integer.parseInt(inp1[i]);
}
int i=-1, j=0, f=0,ans=0;
while(j<n-1 && i<n){
if(arr[j] == arr[j+1]){
if(i == -1){
i=j;
j++;
}
else{
if(j-i == 1){
ans++;
i=j;
j+=2;
}else{
ans+=(j-i-1);
i = i+(j-i-1);
j+=1;
}
}
}else j++;
}
System.out.println(ans);
//System.out.println("standard output");
}
}
} | Java | ["4\n\n5\n\n1 1 1 1 1\n\n5\n\n2 1 1 1 2\n\n6\n\n1 1 2 3 3 4\n\n6\n\n1 2 1 4 5 4"] | 1 second | ["2\n1\n2\n0"] | NoteIn the first test case, we can select $$$i=2$$$ and $$$x=2$$$ to form $$$[1, 2, 2, 1, 1]$$$. Then, we can select $$$i=3$$$ and $$$x=3$$$ to form $$$[1, 2, 3, 3, 1]$$$.In the second test case, we can select $$$i=3$$$ and $$$x=100$$$ to form $$$[2, 1, 100, 100, 2]$$$. | Java 11 | standard input | [
"constructive algorithms",
"greedy",
"implementation"
] | a91be662101762bcb2c58b8db9ff61e0 | Each test contains multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains an integer $$$n$$$ ($$$2 \le n \le 2 \cdot 10 ^ 5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10^9$$$) — elements of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10 ^ 5$$$ | 1,100 | For each test case, print the minimum number of operations needed. | standard output | |
PASSED | 1619eb3f7f5f5bdeff4848634d3d2a3b | train_108.jsonl | 1650722700 | You are given an array $$$a$$$ of length $$$n$$$. We define the equality of the array as the number of indices $$$1 \le i \le n - 1$$$ such that $$$a_i = a_{i + 1}$$$. We are allowed to do the following operation: Select two integers $$$i$$$ and $$$x$$$ such that $$$1 \le i \le n - 1$$$ and $$$1 \le x \le 10^9$$$. Then, set $$$a_i$$$ and $$$a_{i + 1}$$$ to be equal to $$$x$$$. Find the minimum number of operations needed such that the equality of the array is less than or equal to $$$1$$$. | 256 megabytes |
import java.util.*;
import java.io.*;
import java.math.BigInteger;
public class Main {
private static FS sc = new FS();
private static class FS {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
StringTokenizer st = new StringTokenizer("");
String next() {
while (!st.hasMoreTokens())
try {
st=new StringTokenizer(br.readLine());
} catch (IOException e) {}
return st.nextToken();
}
int nextInt() {
return Integer.parseInt(next());
}
long nextLong() {
return Long.parseLong(next());
}
}
private static class extra {
static int[] intArr(int size) {
int[] a = new int[size];
for(int i = 0; i < size; i++) a[i] = sc.nextInt();
return a;
}
static long[] longArr(int size) {
Scanner scc = new Scanner(System.in);
long[] a = new long[size];
for(int i = 0; i < size; i++) a[i] = sc.nextLong();
return a;
}
static long intSum(int[] a) {
long sum = 0;
for(int i = 0; i < a.length; i++) {
sum += a[i];
}
return sum;
}
static long longSum(long[] a) {
long sum = 0;
for(int i = 0; i < a.length; i++) {
sum += a[i];
}
return sum;
}
static LinkedList[] graphD(int vertices, int edges) {
LinkedList<Integer>[] temp = new LinkedList[vertices+1];
for(int i = 0; i <= vertices; i++) temp[i] = new LinkedList<>();
for(int i = 0; i < edges; i++) {
int x = sc.nextInt();
int y = sc.nextInt();
temp[x].add(y);
}
return temp;
}
static LinkedList[] graphUD(int vertices, int edges) {
LinkedList<Integer>[] temp = new LinkedList[vertices+1];
for(int i = 0; i <= vertices; i++) temp[i] = new LinkedList<>();
for(int i = 0; i < edges; i++) {
int x = sc.nextInt();
int y = sc.nextInt();
temp[x].add(y);
temp[y].add(x);
}
return temp;
}
static void printG(LinkedList[] temp) { for(LinkedList<Integer> aa:temp) System.out.println(aa); }
static long cal(long val, long pow, long mod) {
if(pow == 0) return 1;
long ans = cal(val, pow / 2, mod);
long ret = (ans * ans) % mod;
if(pow % 2 == 0) return ret;
return (ret * val) % mod;
}
static long gcd(long a, long b) { return b == 0 ? a:gcd(b, a%b); }
}
static int mod = 998244353;
// static int mod = (int) 1e9 + 7;
//
static long inf = (long) Long.MAX_VALUE - 1;
static LinkedList<Integer>[] temp;
public static void main(String[] args) {
int t = sc.nextInt();
// int t = 1;
StringBuilder ret = new StringBuilder();
while(t-- > 0) {
int n = sc.nextInt();
int[] a = extra.intArr(n);
List<int[]> list = new ArrayList<>();
for(int i = 0; i < n; i++) {
int temp = i, count = 0;
while(temp < n && a[temp] == a[i]) {
temp++;
count++;
}
if(count > 1) {
list.add(new int[] {i, count});
i = temp - 1;
}
}
if(list.size() == 0) ret.append(0 + "\n");
else if(list.size() == 1) {
int now = list.get(0)[1];
if(now <= 2) ret.append(0 + "\n");
else if(now == 3 || now == 4) ret.append(1 + "\n");
else ret.append((now - 3) + "\n");
} else {
int first = list.get(0)[0] + 1, last = list.get(list.size() - 1)[0] + list.get(list.size() - 1)[1] - 3;
// System.out.println(first + " " + last);
ret.append((last - first + 1) + "\n");
}
}
System.out.println(ret);
}
}
| Java | ["4\n\n5\n\n1 1 1 1 1\n\n5\n\n2 1 1 1 2\n\n6\n\n1 1 2 3 3 4\n\n6\n\n1 2 1 4 5 4"] | 1 second | ["2\n1\n2\n0"] | NoteIn the first test case, we can select $$$i=2$$$ and $$$x=2$$$ to form $$$[1, 2, 2, 1, 1]$$$. Then, we can select $$$i=3$$$ and $$$x=3$$$ to form $$$[1, 2, 3, 3, 1]$$$.In the second test case, we can select $$$i=3$$$ and $$$x=100$$$ to form $$$[2, 1, 100, 100, 2]$$$. | Java 11 | standard input | [
"constructive algorithms",
"greedy",
"implementation"
] | a91be662101762bcb2c58b8db9ff61e0 | Each test contains multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains an integer $$$n$$$ ($$$2 \le n \le 2 \cdot 10 ^ 5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10^9$$$) — elements of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10 ^ 5$$$ | 1,100 | For each test case, print the minimum number of operations needed. | standard output | |
PASSED | b885fa8d05c7be8c6c666b23830c43c5 | train_108.jsonl | 1650722700 | You are given an array $$$a$$$ of length $$$n$$$. We define the equality of the array as the number of indices $$$1 \le i \le n - 1$$$ such that $$$a_i = a_{i + 1}$$$. We are allowed to do the following operation: Select two integers $$$i$$$ and $$$x$$$ such that $$$1 \le i \le n - 1$$$ and $$$1 \le x \le 10^9$$$. Then, set $$$a_i$$$ and $$$a_{i + 1}$$$ to be equal to $$$x$$$. Find the minimum number of operations needed such that the equality of the array is less than or equal to $$$1$$$. | 256 megabytes | import java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;
public class JaiShreeRam{
static Scanner in=new Scanner();
static long systemTime;
static long mod = 1000000007;
static ArrayList<ArrayList<Integer>> adj;
static int seive[]=new int[1000001];
static long C[][];
static ArrayList<Integer> ans=new ArrayList<>();
public static void main(String[] args) throws Exception{
int z=in.readInt();
for(int test=1;test<=z;test++) {
//setTime();
solve();
//printTime();
//printMemory();
}
}
static void solve() {
int n=in.readInt();
int a[]=nia(n);
int cnt=0;
int last=0,first=0;
int f=0;
for(int i=1;i<n;i++) {
if(a[i]==a[i-1]) {
if(f==0) {
first=i-1;
f=1;
}
cnt++;
last=i;
}
}
if(cnt<=1) {
print(0);
return;
}
int diff=last-first;
print(last-first-1-(diff==2?0:1));
}
static long pow(long n, long m) {
if(m==0)
return 1;
else if(m==1)
return n;
else {
long r=pow(n,m/2);
if(m%2==0)
return (r*r)%mod;
else
return (r*r*n)%mod;
}
}
static long maxsumsub(ArrayList<Long> al) {
long max=0;
long sum=0;
for(int i=0;i<al.size();i++) {
sum+=al.get(i);
if(sum<0) {
sum=0;
}
max=Math.max(max,sum);
}
return max;
}
static long abs(long a) {
return Math.abs(a);
}
static void ncr(int n, int k){
C= new long[n + 1][k + 1];
int i, j;
for (i = 0; i <= n; i++) {
for (j = 0; j <= Math.min(i, k); j++) {
if (j == 0 || j == i)
C[i][j] = 1;
else
C[i][j] = C[i - 1][j - 1] + C[i - 1][j];
}
}
}
static boolean isPalin(String s) {
int i=0,j=s.length()-1;
while(i<=j) {
if(s.charAt(i)!=s.charAt(j)) {
return false;
}
i++;
j--;
}
return true;
}
static int knapsack(int W, int wt[],int val[], int n){
int []dp = new int[W + 1];
for (int i = 1; i < n + 1; i++) {
for (int w = W; w >= 0; w--) {
if (wt[i - 1] <= w) {
dp[w] = Math.max(dp[w],dp[w - wt[i - 1]] + val[i - 1]);
}
}
}
return dp[W];
}
static void seive() {
Arrays.fill(seive, 1);
seive[0]=0;
seive[1]=0;
for(int i=2;i*i<1000001;i++) {
if(seive[i]==1) {
for(int j=i*i;j<1000001;j+=i) {
if(seive[j]==1) {
seive[j]=0;
}
}
}
}
}
static void sort(int[] a) {
ArrayList<Integer> l=new ArrayList<>();
for (int i:a)
l.add(i);
Collections.sort(l);
for (int i=0; i<a.length; i++)
a[i]=l.get(i);
}
static void sort(long[] a) {
ArrayList<Long> l=new ArrayList<>();
for (long i:a)
l.add(i);
Collections.sort(l);
for (int i=0; i<a.length; i++)
a[i]=l.get(i);
}
static int[] nia(int n){
int[] arr= new int[n];
int i=0;
while(i<n){
arr[i++]=in.readInt();
}
return arr;
}
static long[] nla(int n){
long[] arr= new long[n];
int i=0;
while(i<n){
arr[i++]=in.readLong();
}
return arr;
}
static int[] nia1(int n){
int[] arr= new int[n+1];
int i=1;
while(i<=n){
arr[i++]=in.readInt();
}
return arr;
}
static Integer[] nIa(int n){
Integer[] arr= new Integer[n];
int i=0;
while(i<n){
arr[i++]=in.readInt();
}
return arr;
}
static Long[] nLa(int n){
Long[] arr= new Long[n];
int i=0;
while(i<n){
arr[i++]=in.readLong();
}
return arr;
}
static long gcd(long a, long b) {
if (b==0) return a;
return gcd(b, a%b);
}
static void print(long i) {
System.out.println(i);
}
static void print(Object o) {
System.out.println(o);
}
static void print(int a[]) {
for(int i:a) {
System.out.print(i+" ");
}
System.out.println();
}
static void print(long a[]) {
for(long i:a) {
System.out.print(i+" ");
}
System.out.println();
}
static void print(ArrayList<Long> a) {
for(long i:a) {
System.out.print(i+" ");
}
System.out.println();
}
static void print(Object a[]) {
for(Object i:a) {
System.out.print(i+" ");
}
System.out.println();
}
static void setTime() {
systemTime = System.currentTimeMillis();
}
static void printTime() {
System.err.println("Time consumed: " + (System.currentTimeMillis() - systemTime));
}
static void printMemory() {
System.err.println("Memory consumed: " + (Runtime.getRuntime().totalMemory() - Runtime.getRuntime().freeMemory()) / 1000 + "kb");
}
static class Scanner{
BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
StringTokenizer st=new StringTokenizer("");
String readString() {
while (!st.hasMoreTokens())
try {
st=new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
return st.nextToken();
}
double readDouble() {
return Double.parseDouble(readString());
}
int readInt() {
return Integer.parseInt(readString());
}
long readLong() {
return Long.parseLong(readString());
}
}
} | Java | ["4\n\n5\n\n1 1 1 1 1\n\n5\n\n2 1 1 1 2\n\n6\n\n1 1 2 3 3 4\n\n6\n\n1 2 1 4 5 4"] | 1 second | ["2\n1\n2\n0"] | NoteIn the first test case, we can select $$$i=2$$$ and $$$x=2$$$ to form $$$[1, 2, 2, 1, 1]$$$. Then, we can select $$$i=3$$$ and $$$x=3$$$ to form $$$[1, 2, 3, 3, 1]$$$.In the second test case, we can select $$$i=3$$$ and $$$x=100$$$ to form $$$[2, 1, 100, 100, 2]$$$. | Java 11 | standard input | [
"constructive algorithms",
"greedy",
"implementation"
] | a91be662101762bcb2c58b8db9ff61e0 | Each test contains multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains an integer $$$n$$$ ($$$2 \le n \le 2 \cdot 10 ^ 5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10^9$$$) — elements of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10 ^ 5$$$ | 1,100 | For each test case, print the minimum number of operations needed. | standard output | |
PASSED | ff883c8c9cc94fb6b53d2b8e3d3765ca | train_108.jsonl | 1650722700 | You are given an array $$$a$$$ of length $$$n$$$. We define the equality of the array as the number of indices $$$1 \le i \le n - 1$$$ such that $$$a_i = a_{i + 1}$$$. We are allowed to do the following operation: Select two integers $$$i$$$ and $$$x$$$ such that $$$1 \le i \le n - 1$$$ and $$$1 \le x \le 10^9$$$. Then, set $$$a_i$$$ and $$$a_{i + 1}$$$ to be equal to $$$x$$$. Find the minimum number of operations needed such that the equality of the array is less than or equal to $$$1$$$. | 256 megabytes | import java.util.*;
import java.io.*;
import java.text.*;
public class CF_1672_C{
//SOLUTION BEGIN
void pre() throws Exception{}
void solve(int TC) throws Exception{
int N = ni();
int[] A = new int[N];
for(int i = 0; i< N; i++)A[i] = ni();
int[][] G = new int[N][2];
int v = A[0], c = 1;
int cnt = 0;
for(int i = 1; i< N; i++){
if(A[i] != v){
G[cnt++] = new int[]{v, c};
v = A[i];
c = 0;
}
if(v == A[i])c++;
}
G[cnt++] = new int[]{v, c};
int fi = -1, la = -1;
for(int i = 0; i< cnt; i++){
if(fi == -1 && G[i][1] > 1)fi = i;
if(G[i][1] > 1)la = i;
}
if(fi == -1){
pn(0);
return;
}
if(fi == la){
pn(f(G[fi][1]));
return;
}
int len = 0;
for(int i = fi; i <= la; i++)len += G[i][1];
pn(f(len));
}
int f(int len){
if(len <= 2)return 0;
return Math.max(1, len-3);
}
//SOLUTION END
void hold(boolean b)throws Exception{if(!b)throw new Exception("Hold right there, Sparky!");}
void exit(boolean b){if(!b)System.exit(0);}
static void dbg(Object... o){System.err.println(Arrays.deepToString(o));}
DecimalFormat df = new DecimalFormat("0.00000000000");
double PI = 3.141592653589793238462643383279502884197169399;
static boolean multipleTC = true, memory = true, fileIO = false;
FastReader in;PrintWriter out;
void run() throws Exception{
long ct = System.currentTimeMillis();
if (fileIO) {
in = new FastReader("");
out = new PrintWriter("");
} else {
in = new FastReader();
out = new PrintWriter(System.out);
}
//Solution Credits: Taranpreet Singh
int T = multipleTC? ni():1;
pre();
for (int t = 1; t <= T; t++) solve(t);
out.flush();
out.close();
System.err.println("Runtime: "+(System.currentTimeMillis() - ct));
}
public static void main(String[] args) throws Exception{
if(memory)new Thread(null, new Runnable() {public void run(){try{new CF_1672_C().run();}catch(Exception e){e.printStackTrace();System.exit(1);}}}, "1", 1 << 28).start();
else new CF_1672_C().run();
}
int[][] make(int n, int e, int[] from, int[] to, boolean f){
int[][] g = new int[n][];int[]cnt = new int[n];
for(int i = 0; i< e; i++){
cnt[from[i]]++;
if(f)cnt[to[i]]++;
}
for(int i = 0; i< n; i++)g[i] = new int[cnt[i]];
for(int i = 0; i< e; i++){
g[from[i]][--cnt[from[i]]] = to[i];
if(f)g[to[i]][--cnt[to[i]]] = from[i];
}
return g;
}
int[][][] makeS(int n, int e, int[] from, int[] to, boolean f){
int[][][] g = new int[n][][];int[]cnt = new int[n];
for(int i = 0; i< e; i++){
cnt[from[i]]++;
if(f)cnt[to[i]]++;
}
for(int i = 0; i< n; i++)g[i] = new int[cnt[i]][];
for(int i = 0; i< e; i++){
g[from[i]][--cnt[from[i]]] = new int[]{to[i], i, 0};
if(f)g[to[i]][--cnt[to[i]]] = new int[]{from[i], i, 1};
}
return g;
}
int find(int[] set, int u){return set[u] = (set[u] == u?u:find(set, set[u]));}
int digit(long s){int ans = 0;while(s>0){s/=10;ans++;}return ans;}
long gcd(long a, long b){return (b==0)?a:gcd(b,a%b);}
int gcd(int a, int b){return (b==0)?a:gcd(b,a%b);}
int bit(long n){return (n==0)?0:(1+bit(n&(n-1)));}
void p(Object... o){for(Object oo:o)out.print(oo+" ");}
void pn(Object... o){for(int i = 0; i< o.length; i++)out.print(o[i]+(i+1 < o.length?" ":"\n"));}
void pni(Object... o){for(Object oo:o)out.print(oo+" ");out.println();out.flush();}
String n()throws Exception{return in.next();}
String nln()throws Exception{return in.nextLine();}
int ni()throws Exception{return Integer.parseInt(in.next());}
long nl()throws Exception{return Long.parseLong(in.next());}
double nd()throws Exception{return Double.parseDouble(in.next());}
class FastReader{
BufferedReader br;
StringTokenizer st;
public FastReader(){
br = new BufferedReader(new InputStreamReader(System.in));
}
public FastReader(String s) throws Exception{
br = new BufferedReader(new FileReader(s));
}
String next() throws Exception{
while (st == null || !st.hasMoreElements()){
try{
st = new StringTokenizer(br.readLine());
}catch (IOException e){
throw new Exception(e.toString());
}
}
return st.nextToken();
}
String nextLine() throws Exception{
String str;
try{
str = br.readLine();
}catch (IOException e){
throw new Exception(e.toString());
}
return str;
}
}
} | Java | ["4\n\n5\n\n1 1 1 1 1\n\n5\n\n2 1 1 1 2\n\n6\n\n1 1 2 3 3 4\n\n6\n\n1 2 1 4 5 4"] | 1 second | ["2\n1\n2\n0"] | NoteIn the first test case, we can select $$$i=2$$$ and $$$x=2$$$ to form $$$[1, 2, 2, 1, 1]$$$. Then, we can select $$$i=3$$$ and $$$x=3$$$ to form $$$[1, 2, 3, 3, 1]$$$.In the second test case, we can select $$$i=3$$$ and $$$x=100$$$ to form $$$[2, 1, 100, 100, 2]$$$. | Java 11 | standard input | [
"constructive algorithms",
"greedy",
"implementation"
] | a91be662101762bcb2c58b8db9ff61e0 | Each test contains multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains an integer $$$n$$$ ($$$2 \le n \le 2 \cdot 10 ^ 5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10^9$$$) — elements of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10 ^ 5$$$ | 1,100 | For each test case, print the minimum number of operations needed. | standard output | |
PASSED | d727c61011bd072fccfb25ac16062190 | train_108.jsonl | 1650722700 | You are given an array $$$a$$$ of length $$$n$$$. We define the equality of the array as the number of indices $$$1 \le i \le n - 1$$$ such that $$$a_i = a_{i + 1}$$$. We are allowed to do the following operation: Select two integers $$$i$$$ and $$$x$$$ such that $$$1 \le i \le n - 1$$$ and $$$1 \le x \le 10^9$$$. Then, set $$$a_i$$$ and $$$a_{i + 1}$$$ to be equal to $$$x$$$. Find the minimum number of operations needed such that the equality of the array is less than or equal to $$$1$$$. | 256 megabytes | // package global_20;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.*;
public class C{
public static void main(String[] args){
FastReader sc = new FastReader();
int t=sc.nextInt();
while(t-->0){
int n=sc.nextInt();
int a[]=sc.fastArray(n);
int r=-2;
for(int i=n-1;i>=1;i--) {
if(a[i]==a[i-1]) {
r=i-1;
break;
}
}
int l=-1;
for(int i=0;i<n-1;i++) {
if(a[i]==a[i+1]) {
l=i+1;
break;
}
}
int ans=0;
if(l==r) ans=1;
else if(l>=0)ans=r-l;
System.out.println(Math.max(0, ans));
}
}
static class pair {
int x;int y;
pair(int x,int y){
this.x=x;
this.y=y;
}
}
static void sort(int a[]) {
ArrayList<Integer> aa= new ArrayList<>();
for(int i:a)aa.add(i);
Collections.sort(aa);
int j=0;
for(int i:aa)a[j++]=i;
}
static ArrayList<Integer> primeFac(int n){
ArrayList<Integer>ans = new ArrayList<Integer>();
int lp[]=new int [n+1];
Arrays.fill(lp, 0); //0-prime
for(int i=2;i<=n;i++) {
if(lp[i]==0) {
for(int j=i;j<=n;j+=i) {
if(lp[j]==0) lp[j]=i;
}
}
}
int fac=n;
while(fac>1) {
ans.add(lp[fac]);
fac=fac/lp[fac];
}
print(ans);
return ans;
}
static ArrayList<Long> prime_in_given_range(long l,long r){
ArrayList<Long> ans= new ArrayList<>();
int n=(int)Math.sqrt(r)+1;
int prime[]=sieve_of_Eratosthenes(n);
long res[]=new long [(int)(r-l)+1];
for(int i=0;i<=r-l;i++) {
res[i]=i+l;
}
for(int i=0;i<prime.length;i++) {
if(prime[i]==1) {
System.out.println(2);
for(int j=Math.max((int)i*i, (int)(l+i-1)/i*i);j<=r;j+=i) {
res[j-(int)l]=0;
}
}
}
for(long i:res) if(i!=0)ans.add(i);
return ans;
}
static int [] sieve_of_Eratosthenes(int n) {
int prime[]=new int [n];
Arrays.fill(prime, 1); // 1-prime | 0-not prime
prime[0]=prime[1]=0;
for(int i=2;i<n;i++) {
if(prime[i]==1) {
for(int j=i*i;j<n;j+=i) {
prime[j]=0;
}
}
}
return prime;
}
static long binpow(long a,long b) {
long res=1;
if(b==0)return a;
if(a==0)return 1;
while(b>0) {
if((b&1)==1) {
res*=a;
}
a*=a;
b>>=1;
}
return res;
}
static void print(int a[]) {
System.out.println(a.length);
for(int i:a) {
System.out.print(i+" ");
}
System.out.println();
}
static void print(long a[]) {
System.out.println(a.length);
for(long i:a) {
System.out.print(i+" ");
}
System.out.println();
}
static void print(ArrayList<Integer> a) {
System.out.println(a.size());
for(long i:a) {
System.out.print(i+" ");
}
System.out.println();
}
static class FastReader {
BufferedReader br;
StringTokenizer st;
public FastReader()
{
br = new BufferedReader(
new InputStreamReader(System.in));
}
int [] fastArray(int n) {
int a[]=new int [n];
for(int i=0;i<n;i++) {
a[i]=nextInt();
}
return a;
}
String next()
{
while (st == null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(br.readLine());
}
catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt() { return Integer.parseInt(next()); }
long nextLong() { return Long.parseLong(next()); }
double nextDouble()
{
return Double.parseDouble(next());
}
String nextLine()
{
String str = "";
try {
str = br.readLine();
}
catch (IOException e) {
e.printStackTrace();
}
return str;
}
}
} | Java | ["4\n\n5\n\n1 1 1 1 1\n\n5\n\n2 1 1 1 2\n\n6\n\n1 1 2 3 3 4\n\n6\n\n1 2 1 4 5 4"] | 1 second | ["2\n1\n2\n0"] | NoteIn the first test case, we can select $$$i=2$$$ and $$$x=2$$$ to form $$$[1, 2, 2, 1, 1]$$$. Then, we can select $$$i=3$$$ and $$$x=3$$$ to form $$$[1, 2, 3, 3, 1]$$$.In the second test case, we can select $$$i=3$$$ and $$$x=100$$$ to form $$$[2, 1, 100, 100, 2]$$$. | Java 11 | standard input | [
"constructive algorithms",
"greedy",
"implementation"
] | a91be662101762bcb2c58b8db9ff61e0 | Each test contains multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains an integer $$$n$$$ ($$$2 \le n \le 2 \cdot 10 ^ 5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10^9$$$) — elements of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10 ^ 5$$$ | 1,100 | For each test case, print the minimum number of operations needed. | standard output | |
PASSED | 6147daed2ac799287f87e85a00f58166 | train_108.jsonl | 1650722700 | You are given an array $$$a$$$ of length $$$n$$$. We define the equality of the array as the number of indices $$$1 \le i \le n - 1$$$ such that $$$a_i = a_{i + 1}$$$. We are allowed to do the following operation: Select two integers $$$i$$$ and $$$x$$$ such that $$$1 \le i \le n - 1$$$ and $$$1 \le x \le 10^9$$$. Then, set $$$a_i$$$ and $$$a_{i + 1}$$$ to be equal to $$$x$$$. Find the minimum number of operations needed such that the equality of the array is less than or equal to $$$1$$$. | 256 megabytes | import java.io.*;
import java.util.*;
public class CF {
public static void main(String[] args) throws IOException {
Scanner sc=new Scanner(System.in);
int T=sc.nextInt();
while(T-->0)
{
int n=sc.nextInt();
int[] arr=new int[n];
for(int i=0;i<n;i++)
{
arr[i]=sc.nextInt();
}
int startIndex=-1,endIndex=-1;
for(int i=0;i<n-1;i++)
{
if(arr[i]==arr[i+1])
{
if(startIndex==-1){
startIndex=i;
endIndex=i;
}
else
{
endIndex=i;
}
}
}
if(startIndex==-1 || startIndex==endIndex)
{
System.out.println(0);
continue;
}
else
{
int operations=Integer.MAX_VALUE;
for(int i=0;i<n-1;i++)
{
int left=Math.max(i-(startIndex+1),0);
int right=Math.max(endIndex-(i+1),0);
operations=Math.min(left+right+1, operations);
}
System.out.println(operations);
}
}
sc.close();
}
}
// Aman and garden
// Medium
// There is a garden and Aman is the gardener and he wants to arrange trees in a row
// of length n such that every Kth plant is non-decreasing in height. For example the plant
// at position 1 should be smaller than or equal to the tree planted at position 1+K and plant
// at position 1+K should be smaller than or equal to 1+2*K and so on, this
// should be true for every position(for eg. 2 <= 2+K <= 2+2*K ...).
// Now Aman can change a plant at any position with a plant of any height in order to create
// the required arrangment. He wants to know minimum number of trees he will have to change to get the required arrangement.
// We"ll be given a plants array which represents the height of plants at every position
// and a number K and we need to output the minimum number of plants we will have to change to get the required arrangement.
// Example:
// plants = [5,3,4,1,6,5];
// K=2;
// here plants at index (0,2,4) and (1,3,5) should be non decreasing.
// plants[0]=5;
// plants[2]=4;
// plants[4]=6;
// We can change the plant at index 2 with a plant of height 5(one change).
// Similarly
// plants[1]=3;
// plants[3]=1;
// plants[5]=5;
// We can change the plant at index 3 with a plant of height 4(one change).
// So minimum 2 changes are required.
// Constraints
// 1<=plants.length<=100000
// 1<=plants[i],K<=plants.length
// Format
// Input
// First line contains an integer N representing length of the plant array.
// Next line contains N space separated integers representing height of trees.
// Last line contains an integer K
// Output
// Output the minimum number of changes required.
// Example
// Sample Input
// 6
// 5
// 3
// 4
// 1
// 6
// 5
// 2
// Sample Output
// 2
// Tokyo Drift
// Easy
// There is a racing competition which will be held in your city next weekend. There are N racers who are going to take part in the competition. Each racer is at a given distance from the finish line, which is given in an integer array.
// Due to some safety reasons, total sum of speeds with which racers can drive is restricted with a given limit. Since, cost of organizing such an event depends on how long the event will last, you need to find out the minimum time in which all racers will cross the finishing line, but sum of speeds of all racers should be less than or equal to the given limit.
// If it is impossible to complete the race for all racers within given limit, we have to return -1.
// Note: Speed is defined as Ceil (distance/time), where ceil represents smaller than or equal to. For example if distance is 20 km and time taken to travel is 3 hrs then speed equals ceil(20/3) i.e. 7 km/hr.
// Example: Let us take 4 Racers with Distances from finishing line as [15 km, 25 km, 5 km, 20 km], and the maximum sum of speeds allowed is 12 km/hr. The minimum time in which all racers will reach the finishing line will be 7 hours.
// Racer 1 will have 15 km / 7 hr = 3 km/hr speed
// Racer 2 will have 25 km / 7 hr = 4 km/hr speed
// Racer 3 will have 05 km / 7 hr = 1 km/hr speed
// Racer 4 will have 20 km / 7 hr = 3 km/hr speed
// Hence, the total sum of speeds will be (3 + 4 + 1 + 3) = 11 km/hr which is less than or equal to 12 km/hr.
// Constraints
// 1 <= N <= 100000
// 1 <= racers[i] <= 10000000
// 1 <= Limit <= 10000000
// Format
// Input
// First line contains an integer N representing length of array.
// Next line contains N space seprated integers representing distance from finishing line.
// Last line contains an integer representing limit of sum of speeds with which racers can drive
// Output
// A single line integer representing minimum time in which all racers will cross the finishing line, If not possible print -1
// Example
// Sample Input
// 4
// 15 25 5 20
// 12
// Sample Output
// 7
| Java | ["4\n\n5\n\n1 1 1 1 1\n\n5\n\n2 1 1 1 2\n\n6\n\n1 1 2 3 3 4\n\n6\n\n1 2 1 4 5 4"] | 1 second | ["2\n1\n2\n0"] | NoteIn the first test case, we can select $$$i=2$$$ and $$$x=2$$$ to form $$$[1, 2, 2, 1, 1]$$$. Then, we can select $$$i=3$$$ and $$$x=3$$$ to form $$$[1, 2, 3, 3, 1]$$$.In the second test case, we can select $$$i=3$$$ and $$$x=100$$$ to form $$$[2, 1, 100, 100, 2]$$$. | Java 11 | standard input | [
"constructive algorithms",
"greedy",
"implementation"
] | a91be662101762bcb2c58b8db9ff61e0 | Each test contains multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains an integer $$$n$$$ ($$$2 \le n \le 2 \cdot 10 ^ 5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10^9$$$) — elements of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10 ^ 5$$$ | 1,100 | For each test case, print the minimum number of operations needed. | standard output | |
PASSED | 59a8654d9bb2d85a2943a62bc91d71aa | train_108.jsonl | 1650722700 | You are given an array $$$a$$$ of length $$$n$$$. We define the equality of the array as the number of indices $$$1 \le i \le n - 1$$$ such that $$$a_i = a_{i + 1}$$$. We are allowed to do the following operation: Select two integers $$$i$$$ and $$$x$$$ such that $$$1 \le i \le n - 1$$$ and $$$1 \le x \le 10^9$$$. Then, set $$$a_i$$$ and $$$a_{i + 1}$$$ to be equal to $$$x$$$. Find the minimum number of operations needed such that the equality of the array is less than or equal to $$$1$$$. | 256 megabytes | import java.util.*;
import java.io.*;
public class C {
private static BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
private static BufferedWriter bw = new BufferedWriter(new OutputStreamWriter(System.out));
public static void main(String[] args) throws IOException {
int t = Integer.parseInt(br.readLine());
while (t-- > 0) {
int n = Integer.parseInt(br.readLine());
String s[] = br.readLine().split(" ");
int a[] = new int[n];
for (int i = 0; i < n; i++) {
a[i] = Integer.parseInt(s[i]);
}
if (n == 2) {
bw.write("0\n");
continue;
}
int cnt = 0;
int ps[] = new int[n];
ps[0] = 0;
for (int i = 1; i < n; i++) {
ps[i] = ps[i - 1];
if (a[i] == a[i - 1]) {
ps[i]++;
cnt++;
}
}
if (cnt <= 1) {
bw.write("0\n");
continue;
}
int ans = 0;
for (int i = 1; i < n; i++) {
if (ps[i] != 0) {
if (ps[i] == cnt) {
break;
} else {
if (i == n - 1) {
if (ans != 0)
break;
}
if (ps[i] + 1 == ps[i + 1]) {
if (ps[i + 1] == cnt && ans != 0) {
break;
}
}
ans++;
}
}
}
bw.write(ans + " \n");
}
bw.flush();
}
} | Java | ["4\n\n5\n\n1 1 1 1 1\n\n5\n\n2 1 1 1 2\n\n6\n\n1 1 2 3 3 4\n\n6\n\n1 2 1 4 5 4"] | 1 second | ["2\n1\n2\n0"] | NoteIn the first test case, we can select $$$i=2$$$ and $$$x=2$$$ to form $$$[1, 2, 2, 1, 1]$$$. Then, we can select $$$i=3$$$ and $$$x=3$$$ to form $$$[1, 2, 3, 3, 1]$$$.In the second test case, we can select $$$i=3$$$ and $$$x=100$$$ to form $$$[2, 1, 100, 100, 2]$$$. | Java 11 | standard input | [
"constructive algorithms",
"greedy",
"implementation"
] | a91be662101762bcb2c58b8db9ff61e0 | Each test contains multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains an integer $$$n$$$ ($$$2 \le n \le 2 \cdot 10 ^ 5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10^9$$$) — elements of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10 ^ 5$$$ | 1,100 | For each test case, print the minimum number of operations needed. | standard output | |
PASSED | 2383424a15433883913bab2a1193179d | train_108.jsonl | 1650722700 | You are given an array $$$a$$$ of length $$$n$$$. We define the equality of the array as the number of indices $$$1 \le i \le n - 1$$$ such that $$$a_i = a_{i + 1}$$$. We are allowed to do the following operation: Select two integers $$$i$$$ and $$$x$$$ such that $$$1 \le i \le n - 1$$$ and $$$1 \le x \le 10^9$$$. Then, set $$$a_i$$$ and $$$a_{i + 1}$$$ to be equal to $$$x$$$. Find the minimum number of operations needed such that the equality of the array is less than or equal to $$$1$$$. | 256 megabytes | import java.io.IOException;
import java.io.*;
import java.util.*;
public class codeforces {
static int max = Integer.MAX_VALUE, min = Integer.MIN_VALUE;
long maxl = Long.MAX_VALUE, minl = Long.MIN_VALUE;
static PrintWriter pw = new PrintWriter(System.out);
static StringBuilder sb = new StringBuilder();
static int parent[] = new int[100001];
static int mod = 1000000007;
static boolean b1 = true;
public static void main(String[] args) throws IOException {
FastScanner sc = new FastScanner();
int ttt = sc.nextInt();
while (ttt-- > 0) {
int n = sc.nextInt();
int a[] = new int[n];
for (int i = 0; i < n; i++)
a[i] = sc.nextInt();
int count = 0, index = -1;
for (int i = 0; i < n - 1; i++) {
if (a[i] == a[i + 1]) {
count++;
index = i;
}
}
if (count == 0 || count == 1)
System.out.println("0");
else {
count = 0;
for (int i = 0; i < n - 1; i++) {
if (a[i] == a[i + 1]) {
count++;
i += 2;
count += Math.max(0, index - i);
break;
}
}
System.out.println(count);
}
}
}
public static int lis(int arr[], int n) {
int lis[] = new int[n];
int i, j, max = 0;
/* Initialize LIS values for all indexes */
for (i = 0; i < n; i++)
lis[i] = 1;
/*
* Compute optimized LIS values in
* bottom up manner
*/
for (i = 1; i < n; i++)
for (j = 0; j < i; j++)
if (arr[i] > arr[j] && lis[i] < lis[j] + 1)
lis[i] = lis[j] + 1;
/* Pick maximum of all LIS values */
for (i = 0; i < n; i++)
if (max < lis[i])
max = lis[i];
return max;
}
public static int binarySearch(int arr[], int x) {
int l = 0, r = arr.length - 1;
int result1 = -1;
while (l <= r) {
int mid = l + (r - l) / 2;
if (arr[mid] == x) {
return mid;
} else if (arr[mid] > x) {
r = mid - 1;
} else {
l = mid + 1;
}
}
return result1;
}
public static class Pair {
public int x;
public int y;
public Pair(int x, int y) {
this.x = x;
this.y = y;
}
}
public static long gcd(long a, long b) {
if (a == 0)
return b;
return gcd(b % a, a);
}
public static class FastScanner {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
StringTokenizer st = new StringTokenizer("");
String next() {
while (!st.hasMoreTokens())
try {
st = new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
return st.nextToken();
}
int nextInt() {
return Integer.parseInt(next());
}
int[] readArray(int n) {
int[] a = new int[n];
for (int i = 0; i < n; i++)
a[i] = nextInt();
return a;
}
long nextLong() {
return Long.parseLong(next());
}
}
}
| Java | ["4\n\n5\n\n1 1 1 1 1\n\n5\n\n2 1 1 1 2\n\n6\n\n1 1 2 3 3 4\n\n6\n\n1 2 1 4 5 4"] | 1 second | ["2\n1\n2\n0"] | NoteIn the first test case, we can select $$$i=2$$$ and $$$x=2$$$ to form $$$[1, 2, 2, 1, 1]$$$. Then, we can select $$$i=3$$$ and $$$x=3$$$ to form $$$[1, 2, 3, 3, 1]$$$.In the second test case, we can select $$$i=3$$$ and $$$x=100$$$ to form $$$[2, 1, 100, 100, 2]$$$. | Java 11 | standard input | [
"constructive algorithms",
"greedy",
"implementation"
] | a91be662101762bcb2c58b8db9ff61e0 | Each test contains multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains an integer $$$n$$$ ($$$2 \le n \le 2 \cdot 10 ^ 5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10^9$$$) — elements of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10 ^ 5$$$ | 1,100 | For each test case, print the minimum number of operations needed. | standard output | |
PASSED | 99f259e359ba22ddc338597b9e7d539b | train_108.jsonl | 1650722700 | You are given an array $$$a$$$ of length $$$n$$$. We define the equality of the array as the number of indices $$$1 \le i \le n - 1$$$ such that $$$a_i = a_{i + 1}$$$. We are allowed to do the following operation: Select two integers $$$i$$$ and $$$x$$$ such that $$$1 \le i \le n - 1$$$ and $$$1 \le x \le 10^9$$$. Then, set $$$a_i$$$ and $$$a_{i + 1}$$$ to be equal to $$$x$$$. Find the minimum number of operations needed such that the equality of the array is less than or equal to $$$1$$$. | 256 megabytes | import java.io.*;
import java.util.*;
public class Main2 {
public static void main(String[] args) {
InputStream inputStream = System.in;
OutputStream outputStream = System.out;
FastScanner in = new FastScanner(inputStream);
FastPrinter out = new FastPrinter(outputStream);
TaskC solver = new TaskC();
solver.solve(1, in, out);
out.close();
}
static class TaskC {
public void solve(int testNumber, FastScanner in, FastPrinter out) {
int t = in.nextInt();
while (t-- > 0) {
int n = in.nextInt();
int[] nums = new int[n];
int ans=0,pre = 0, preNum = 1;
Deque<Integer> l = new ArrayDeque<>();
for (int i = 0; i < n; i++) {
nums[i] = in.nextInt();
if (nums[i] == pre) {
preNum++;
}else{
l.add(preNum);
pre = nums[i];
preNum = 1;
}
}
l.add(preNum);
// System.out.println(l);
while (l.size() > 0 && l.peekFirst() == 1) {
l.pollFirst();
}
while (l.size() > 0 && l.peekLast() == 1) {
l.pollLast();
}
if (l.size() == 1) {
System.out.println(l.peek()-2>1?l.peek()-3:l.peek()-2 );
continue;
} else if (l.isEmpty()) {
System.out.println(0);
continue;
}
while (!l.isEmpty()) {
ans += l.poll();
}
System.out.println(ans-3);
}
}
String solution(String s) {
if (s.length()<2||s.charAt(s.length()-1)=='A') {
return "NO";
}
Deque<Character> dq = new ArrayDeque<>();
for (int i = 0; i < s.length(); i++) {
char c = s.charAt(i);
if (c == 'B') {
if (dq.isEmpty()) {
return "NO";
}
dq.pollLast();
} else if (c == 'A') {
dq.offer('A');
}
}
return "YES";
}
}
static class ru_ifmo_niyaz_arrayutils_ArrayUtils {
static final long seed = System.nanoTime();
static final Random rand = new Random(seed);
public static void sort(int[] a) {
shuffle(a);
Arrays.sort(a);
}
public static void shuffle(int[] a) {
for (int i = 0; i < a.length; i++) {
int j = rand.nextInt(i + 1);
int t = a[i];
a[i] = a[j];
a[j] = t;
}
}
}
static class FastPrinter extends PrintWriter {
public FastPrinter(OutputStream out) {
super(out);
}
public FastPrinter(Writer out) {
super(out);
}
}
static class net_egork_misc_ArrayUtils {
public static long sumArray(int[] array) {
long result = 0;
for (int element : array)
result += element;
return result;
}
}
static class FastScanner extends BufferedReader {
public FastScanner(InputStream is) {
super(new InputStreamReader(is));
}
public int read() {
try {
int ret = super.read();
// if (isEOF && ret < 0) {
// throw new InputMismatchException();
// }
// isEOF = ret == -1;
return ret;
} catch (IOException e) {
throw new InputMismatchException();
}
}
public String next() {
StringBuilder sb = new StringBuilder();
int c = read();
while (isWhiteSpace(c)) {
c = read();
}
if (c < 0) {
return null;
}
while (c >= 0 && !isWhiteSpace(c)) {
sb.appendCodePoint(c);
c = read();
}
return sb.toString();
}
static boolean isWhiteSpace(int c) {
return c >= 0 && c <= 32;
}
public int nextInt() {
int c = read();
while (isWhiteSpace(c)) {
c = read();
}
int sgn = 1;
if (c == '-') {
sgn = -1;
c = read();
}
int ret = 0;
while (c >= 0 && !isWhiteSpace(c)) {
if (c < '0' || c > '9') {
throw new NumberFormatException("digit expected " + (char) c + " found");
}
ret = ret * 10 + c - '0';
c = read();
}
return ret * sgn;
}
public long nextLong() {
return Long.parseLong(next());
}
public String readLine() {
try {
return super.readLine();
} catch (IOException e) {
return null;
}
}
public int[] readIntArray(int n) {
int[] ret = new int[n];
for (int i = 0; i < n; i++) {
ret[i] = nextInt();
}
return ret;
}
}
} | Java | ["4\n\n5\n\n1 1 1 1 1\n\n5\n\n2 1 1 1 2\n\n6\n\n1 1 2 3 3 4\n\n6\n\n1 2 1 4 5 4"] | 1 second | ["2\n1\n2\n0"] | NoteIn the first test case, we can select $$$i=2$$$ and $$$x=2$$$ to form $$$[1, 2, 2, 1, 1]$$$. Then, we can select $$$i=3$$$ and $$$x=3$$$ to form $$$[1, 2, 3, 3, 1]$$$.In the second test case, we can select $$$i=3$$$ and $$$x=100$$$ to form $$$[2, 1, 100, 100, 2]$$$. | Java 11 | standard input | [
"constructive algorithms",
"greedy",
"implementation"
] | a91be662101762bcb2c58b8db9ff61e0 | Each test contains multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains an integer $$$n$$$ ($$$2 \le n \le 2 \cdot 10 ^ 5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10^9$$$) — elements of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10 ^ 5$$$ | 1,100 | For each test case, print the minimum number of operations needed. | standard output | |
PASSED | 01302f3d78a3ea35b4f9a685c5eaff40 | train_108.jsonl | 1650722700 | You are given an array $$$a$$$ of length $$$n$$$. We define the equality of the array as the number of indices $$$1 \le i \le n - 1$$$ such that $$$a_i = a_{i + 1}$$$. We are allowed to do the following operation: Select two integers $$$i$$$ and $$$x$$$ such that $$$1 \le i \le n - 1$$$ and $$$1 \le x \le 10^9$$$. Then, set $$$a_i$$$ and $$$a_{i + 1}$$$ to be equal to $$$x$$$. Find the minimum number of operations needed such that the equality of the array is less than or equal to $$$1$$$. | 256 megabytes | import java.io.*;
import java.util.*;
public class AC {
public static void main(String args[]) throws IOException {
FastReader sc = new FastReader();
PrintWriter out = new PrintWriter(System.out);
int tc, i, j;
String s;
char p;
tc = sc.nextInt();
while (tc-- > 0) {
int n = sc.nextInt();
int arr[] = sc.readArray(n);
int ans=Integer.MAX_VALUE;
boolean flag=true;
int temp=0;
for (i = 0; i < n-1; i++) {
if(arr[i]==arr[i+1]){
ans=Math.min(ans,i);
temp=Math.max(temp,i);
}
}
if (temp==ans || ans==Integer.MAX_VALUE)
out.println(0);
else {
int temp2=temp-ans;
if(temp2<3)
out.println(1);
else out.println(temp-ans-1);
}
}
out.close();
}
/*--------------------------------------------------------
----------------------------------------------------------*/
//Pair Class
public static class Pair implements Comparable<Pair> {
int x, y;
Pair(int x, int y) {
this.x = x;
this.y = y;
}
@Override
public int compareTo(Pair o) {
return this.x - o.x;
}
}
/*
FASTREADER
*/
static class FastReader {
BufferedReader br;
StringTokenizer st;
public FastReader() {
br = new BufferedReader(new
InputStreamReader(System.in));
}
String next() {
while (st == null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt() {
return Integer.parseInt(next());
}
long nextLong() {
return Long.parseLong(next());
}
double nextDouble() {
return Double.parseDouble(next());
}
String nextLine() {
String str = "";
try {
str = br.readLine();
} catch (IOException e) {
e.printStackTrace();
}
return str;
}
/*DEFINED BY ME*/
//READING ARRAY
int[] readArray(int n) {
int arr[] = new int[n];
for (int i = 0; i < n; i++)
arr[i] = nextInt();
return arr;
}
//COLLECTIONS SORT
void sort(int arr[]) {
ArrayList<Integer> l = new ArrayList<>();
for (int i : arr)
l.add(i);
Collections.sort(l);
for (int i = 0; i < arr.length; i++)
arr[i] = l.get(i);
}
//EUCLID'S GCD
int gcd(int a, int b) {
if (b != 0)
return gcd(b, a % b);
else
return a;
}
void swap(int arr[], int i, int j) {
int temp = arr[i];
arr[i] = arr[j];
arr[j] = temp;
}
}
} | Java | ["4\n\n5\n\n1 1 1 1 1\n\n5\n\n2 1 1 1 2\n\n6\n\n1 1 2 3 3 4\n\n6\n\n1 2 1 4 5 4"] | 1 second | ["2\n1\n2\n0"] | NoteIn the first test case, we can select $$$i=2$$$ and $$$x=2$$$ to form $$$[1, 2, 2, 1, 1]$$$. Then, we can select $$$i=3$$$ and $$$x=3$$$ to form $$$[1, 2, 3, 3, 1]$$$.In the second test case, we can select $$$i=3$$$ and $$$x=100$$$ to form $$$[2, 1, 100, 100, 2]$$$. | Java 11 | standard input | [
"constructive algorithms",
"greedy",
"implementation"
] | a91be662101762bcb2c58b8db9ff61e0 | Each test contains multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains an integer $$$n$$$ ($$$2 \le n \le 2 \cdot 10 ^ 5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10^9$$$) — elements of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10 ^ 5$$$ | 1,100 | For each test case, print the minimum number of operations needed. | standard output | |
PASSED | 2bb46d9d7c128c751359869cc466d0f4 | train_108.jsonl | 1650722700 | You are given an array $$$a$$$ of length $$$n$$$. We define the equality of the array as the number of indices $$$1 \le i \le n - 1$$$ such that $$$a_i = a_{i + 1}$$$. We are allowed to do the following operation: Select two integers $$$i$$$ and $$$x$$$ such that $$$1 \le i \le n - 1$$$ and $$$1 \le x \le 10^9$$$. Then, set $$$a_i$$$ and $$$a_{i + 1}$$$ to be equal to $$$x$$$. Find the minimum number of operations needed such that the equality of the array is less than or equal to $$$1$$$. | 256 megabytes |
import java.io.ByteArrayInputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.Arrays;
import java.util.InputMismatchException;
public class C {
private static PrintWriter out = new PrintWriter(System.out);
public static void solve() {
In in = new In();
int tests = in.ni();
while (tests-- > 0) {
int n = in.ni();
Integer[] nums = in.nia(n);
int[][] counter = counter(nums);
int ans = 0;
int eq = 0;
for (int[] elem : counter) {
eq += elem[1] - 1;
}
if (eq <= 1) {
out.println(0);
continue;
}
int left = 0;
int right = counter.length - 1;
while (left < counter.length && counter[left][1] == 1)
left++;
while (right >= 0 && counter[right][1] == 1)
right--;
if (left == right) {
ans = counter[left][1] == 3 ? 1 : (counter[left][1] - 3);
} else {
int total = 0;
for (int i = left; i <= right; i++)
total += counter[i][1];
ans = total - 3;
}
out.println(ans);
}
}
// Consequtive element count of an int array; each count is denoted as {element,
// count}
public static int[][] counter(Integer[] a) {
int n = a.length, p = 0;
int[][] b = new int[n][];
for (int i = 0; i < n; i++) {
if (i == 0 || !a[i].equals(a[i - 1]))
b[p++] = new int[] { a[i], 0 };
b[p - 1][1]++;
}
return Arrays.copyOf(b, p);
}
public static void main(String[] args) throws IOException {
// long start = System.nanoTime();
solve();
// System.out.println("Elapsed: " + (System.nanoTime() - start) / 1000000 +
// "ns");
out.flush();
}
@SuppressWarnings("unused")
private static class In {
final private static int BUFFER_SIZE = 1024;
private byte[] buf;
private InputStream is;
private int buflen;
private int bufptr;
public In() {
is = System.in;
buf = new byte[BUFFER_SIZE];
buflen = bufptr = 0;
}
public In(String input) {
is = new ByteArrayInputStream(input.getBytes());
buf = new byte[BUFFER_SIZE];
buflen = bufptr = 0;
}
public int readByte() {
if (bufptr >= buflen) {
try {
buflen = is.read(buf);
} catch (IOException ioe) {
throw new InputMismatchException();
}
bufptr = 0;
}
if (buflen <= 0)
return -1;
return buf[bufptr++];
}
public boolean isSpaceChar(int c) {
return !(c >= 33 && c <= 126);
}
public int skip() {
int b;
while ((b = readByte()) != -1 && isSpaceChar(b))
;
return b;
}
/* Next character */
public char nc() {
return (char) skip();
}
/* Next double */
public double nd() {
return Double.parseDouble(ns());
}
/* Next string */
public String ns() {
final StringBuilder sb = new StringBuilder();
int b = skip();
while (!isSpaceChar(b)) {
sb.appendCodePoint(b);
b = readByte();
}
return sb.toString();
}
/* Next integer */
public int ni() {
boolean minus = false;
int num = 0;
int b;
while ((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-'))
;
if (b == '-') {
minus = true;
b = readByte();
}
while (b >= '0' && b <= '9') {
num = num * 10 + (b - '0');
b = readByte();
}
return minus ? -num : num;
}
/* Next long */
public long nl() {
boolean minus = false;
long num = 0;
int b;
while ((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-'))
;
if (b == '-') {
minus = true;
b = readByte();
}
while (b >= '0' && b <= '9') {
num = num * 10 + (b - '0');
b = readByte();
}
return minus ? -num : num;
}
/* Next integer 1D array */
public Integer[] nia(int n) {
final Integer[] arr = new Integer[n];
for (int i = 0; i < n; i++)
arr[i] = ni();
return arr;
}
/* Next long 1D array */
public long[] nla(int n) {
final long[] arr = new long[n];
for (int i = 0; i < n; i++)
arr[i] = nl();
return arr;
}
/* Next string 1D array */
public String[] nsa(int n) {
final String[] arr = new String[n];
for (int i = 0; i < n; i++)
arr[i] = ns();
return arr;
}
/* Next char 1D array */
public char[] nca(int n) {
final char[] arr = new char[n];
for (int i = 0; i < n; i++)
arr[i] = nc();
return arr;
}
/* Next double 1D array */
public double[] nda(int n) {
final double[] arr = new double[n];
for (int i = 0; i < n; i++)
arr[i] = nc();
return arr;
}
/* Next integer matrix */
public int[][] nim(int n, int m) {
final int[][] arr = new int[n][m];
for (int i = 0; i < n; i++)
for (int j = 0; j < m; j++)
arr[i][j] = ni();
return arr;
}
/* Next long matrix */
public long[][] nlm(int n, int m) {
final long[][] arr = new long[n][m];
for (int i = 0; i < n; i++)
for (int j = 0; j < m; j++)
arr[i][j] = nl();
return arr;
}
/* Next string matrix */
public String[][] nsm(int n, int m) {
final String[][] arr = new String[n][m];
for (int i = 0; i < n; i++)
for (int j = 0; j < m; j++)
arr[i][j] = ns();
return arr;
}
/* Next char matrix */
public char[][] ncm(int n, int m) {
final char[][] arr = new char[n][m];
for (int i = 0; i < n; i++)
for (int j = 0; j < m; j++)
arr[i][j] = nc();
return arr;
}
/* Next double matrix */
public double[][] ndm(int n, int m) {
final double[][] arr = new double[n][m];
for (int i = 0; i < n; i++)
for (int j = 0; j < m; j++)
arr[i][j] = nd();
return arr;
}
public static void log(Object... o) {
System.out.println(Arrays.deepToString(o));
}
}
}
| Java | ["4\n\n5\n\n1 1 1 1 1\n\n5\n\n2 1 1 1 2\n\n6\n\n1 1 2 3 3 4\n\n6\n\n1 2 1 4 5 4"] | 1 second | ["2\n1\n2\n0"] | NoteIn the first test case, we can select $$$i=2$$$ and $$$x=2$$$ to form $$$[1, 2, 2, 1, 1]$$$. Then, we can select $$$i=3$$$ and $$$x=3$$$ to form $$$[1, 2, 3, 3, 1]$$$.In the second test case, we can select $$$i=3$$$ and $$$x=100$$$ to form $$$[2, 1, 100, 100, 2]$$$. | Java 11 | standard input | [
"constructive algorithms",
"greedy",
"implementation"
] | a91be662101762bcb2c58b8db9ff61e0 | Each test contains multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains an integer $$$n$$$ ($$$2 \le n \le 2 \cdot 10 ^ 5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10^9$$$) — elements of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10 ^ 5$$$ | 1,100 | For each test case, print the minimum number of operations needed. | standard output | |
PASSED | 89237293538b4b0b978e54ec52f42ce8 | train_108.jsonl | 1650722700 | You are given an array $$$a$$$ of length $$$n$$$. We define the equality of the array as the number of indices $$$1 \le i \le n - 1$$$ such that $$$a_i = a_{i + 1}$$$. We are allowed to do the following operation: Select two integers $$$i$$$ and $$$x$$$ such that $$$1 \le i \le n - 1$$$ and $$$1 \le x \le 10^9$$$. Then, set $$$a_i$$$ and $$$a_{i + 1}$$$ to be equal to $$$x$$$. Find the minimum number of operations needed such that the equality of the array is less than or equal to $$$1$$$. | 256 megabytes | import java.io.*;
import java.util.Arrays;
import java.util.Random;
import java.util.StringTokenizer;
public class codeforces_G20_C {
private static void solve(FastIOAdapter in, PrintWriter out) {
int n = in.nextInt();
int[] a = in.readArray(n);
// -++-
int ans = 0;
var b = new boolean[n - 1];
int c = 0;
for (int i = 0; i < n - 1; i++) {
if (a[i] == a[i + 1]) {
b[i] = true;
c++;
}
}
if (c > 1) {
for (int i = 0; i < n - 1; i++) {
if (b[i]) {
for (int j = n - 2; j >= 0; j--) {
if (b[j]) {
ans = Math.max(1, j - i - 1);
break;
}
}
break;
}
}
}
out.println(ans);
}
public static void main(String[] args) throws Exception {
try (FastIOAdapter ioAdapter = new FastIOAdapter()) {
int count = 1;
count = ioAdapter.nextInt();
while (count-- > 0) {
solve(ioAdapter, ioAdapter.out);
}
}
}
static void ruffleSort(int[] arr) {
int n = arr.length;
Random rnd = new Random();
for (int i = 0; i < n; ++i) {
int tmp = arr[i];
int randomPos = i + rnd.nextInt(n - i);
arr[i] = arr[randomPos];
arr[randomPos] = tmp;
}
Arrays.sort(arr);
}
static class FastIOAdapter implements AutoCloseable {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
public PrintWriter out = new PrintWriter(new BufferedWriter(new OutputStreamWriter((System.out))));
StringTokenizer st = new StringTokenizer("");
String next() {
while (!st.hasMoreTokens())
try {
st = new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
return st.nextToken();
}
String nextLine() {
try {
return br.readLine();
} catch (IOException e) {
e.printStackTrace();
return null;
}
}
int nextInt() {
return Integer.parseInt(next());
}
int[] readArray(int n) {
int[] a = new int[n];
for (int i = 0; i < n; i++) a[i] = nextInt();
return a;
}
long[] readArrayLong(int n) {
long[] a = new long[n];
for (int i = 0; i < n; i++) a[i] = nextLong();
return a;
}
long nextLong() {
return Long.parseLong(next());
}
@Override
public void close() throws Exception {
out.flush();
out.close();
br.close();
}
}
}
| Java | ["4\n\n5\n\n1 1 1 1 1\n\n5\n\n2 1 1 1 2\n\n6\n\n1 1 2 3 3 4\n\n6\n\n1 2 1 4 5 4"] | 1 second | ["2\n1\n2\n0"] | NoteIn the first test case, we can select $$$i=2$$$ and $$$x=2$$$ to form $$$[1, 2, 2, 1, 1]$$$. Then, we can select $$$i=3$$$ and $$$x=3$$$ to form $$$[1, 2, 3, 3, 1]$$$.In the second test case, we can select $$$i=3$$$ and $$$x=100$$$ to form $$$[2, 1, 100, 100, 2]$$$. | Java 11 | standard input | [
"constructive algorithms",
"greedy",
"implementation"
] | a91be662101762bcb2c58b8db9ff61e0 | Each test contains multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains an integer $$$n$$$ ($$$2 \le n \le 2 \cdot 10 ^ 5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10^9$$$) — elements of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10 ^ 5$$$ | 1,100 | For each test case, print the minimum number of operations needed. | standard output | |
PASSED | 24d0a3153c474088ebdbe3ffba229f37 | train_108.jsonl | 1650722700 | You are given an array $$$a$$$ of length $$$n$$$. We define the equality of the array as the number of indices $$$1 \le i \le n - 1$$$ such that $$$a_i = a_{i + 1}$$$. We are allowed to do the following operation: Select two integers $$$i$$$ and $$$x$$$ such that $$$1 \le i \le n - 1$$$ and $$$1 \le x \le 10^9$$$. Then, set $$$a_i$$$ and $$$a_{i + 1}$$$ to be equal to $$$x$$$. Find the minimum number of operations needed such that the equality of the array is less than or equal to $$$1$$$. | 256 megabytes | /* package codechef; // don't place package name! */
import java.util.*;
import java.lang.*;
import java.io.*;
/* Name of the class has to be "Main" only if the class is public. */
public class Codechef
{static class FastReader
{
BufferedReader br;
StringTokenizer st;
public FastReader()
{
br = new BufferedReader(new
InputStreamReader(System.in));
}
String next()
{
while (st == null || !st.hasMoreElements())
{
try
{
st = new StringTokenizer(br.readLine());
}
catch (IOException e)
{
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt()
{
return Integer.parseInt(next());
}
long nextLong()
{
return Long.parseLong(next());
}
double nextDouble()
{
return Double.parseDouble(next());
}
String nextLine()
{
String str = "";
try
{
str = br.readLine();
}
catch (IOException e)
{
e.printStackTrace();
}
return str;
}
}
static int mod = 998244353 ;
static int N = 200005;
static long factorial_num_inv[] = new long[N+1];
static long natual_num_inv[] = new long[N+1];
static long fact[] = new long[N+1];
static void InverseofNumber()
{
natual_num_inv[0] = 1;
natual_num_inv[1] = 1;
for (int i = 2; i <= N; i++)
natual_num_inv[i] = natual_num_inv[mod % i] * (mod - mod / i) % mod;
}
static void InverseofFactorial()
{
factorial_num_inv[0] = factorial_num_inv[1] = 1;
for (int i = 2; i <= N; i++)
factorial_num_inv[i] = (natual_num_inv[i] * factorial_num_inv[i - 1]) % mod;
}
static long nCrModP(long N, long R)
{
long ans = ((fact[(int)N] * factorial_num_inv[(int)R]) % mod * factorial_num_inv[(int)(N - R)]) % mod;
return ans%mod;
}
static boolean prime[];
static void sieveOfEratosthenes(int n)
{
prime = new boolean[n+1];
for (int i = 0; i <= n; i++)
prime[i] = true;
for (int p = 2; p * p <= n; p++)
{
// If prime[p] is not changed, then it is a
// prime
if (prime[p] == true)
{
// Update all multiples of p
for (int i = p * p; i <= n; i += p)
prime[i] = false;
}
}
}
public static void main (String[] args) throws java.lang.Exception
{
// InverseofNumber();
// InverseofFactorial();
// fact[0] = 1;
// for (long i = 1; i <= 2*100000; i++)
// {
// fact[(int)i] = (fact[(int)i - 1] * i) % mod;
// }
FastReader scan = new FastReader();
PrintWriter pw = new PrintWriter(System.out);
int t = scan.nextInt();
while(t-->0){
int n = scan.nextInt();
long a[] = new long[n];
for(int i=0;i<n;i++){
a[i] = scan.nextLong();
}
int start = 0;
while(start<n-1){
if(a[start]==a[start+1]){
break;
}
start++;
}
int end = n-1;
while(end>0){
if(a[end]==a[end-1]){
break;
}
end--;
}
if(start==n-1)
pw.println(0);
else{
int len = end-start+1;
if(len==2)
pw.println(0);
else if(len==3)
pw.println(1);
else
pw.println(len-3);
}
pw.flush();
}
}
//static long bin_exp_mod(long a,long n){
// long res = 1;
// while(n!=0){
// if(n%2==1){
// res = ((res)*(a));
// }
// n = n/2;
// a = ((a)*(a));
// }
// return res;
//}
//static long bin_exp_mod(long a,long n){
// long mod = 998244353;
// long res = 1;
// while(n!=0){
// if(n%2==1){
// res = ((res%mod)*(a%mod))%mod;
// }
// n = n/2;
// a = ((a%mod)*(a%mod))%mod;
// }
// res = res%mod;
// return res;
//}
static long gcd(long a,long b){
if(a==0)
return b;
return gcd(b%a,a);
}
// static long lcm(long a,long b){
// return (a/gcd(a,b))*b;
// }
}
class Pair{
int x,y;
Pair(int x,int y){
this.x = x;
this.y = y;
}
//public boolean equals(Object obj) {
// // TODO Auto-generated method stub
// if(obj instanceof Pair)
// {
// Pair temp = (Pair) obj;
// if(this.x.equals(temp.x) && this.y.equals(temp.y))
// return true;
// }
// return false;
//}
//@Override
//public int hashCode() {
// // TODO Auto-generated method stub
// return (this.x.hashCode() + this.y.hashCode());
//}
}
//class Compar implements Comparator<Pair>{
// public int compare(Pair p1,Pair p2){
// if(p1.y==p2.y)
// return 0;
// else if(p1.y<p2.y)
// return -1;
// else
// return 1;
// }
//}
//class DisjointSet{
// Map<Long,Node> map = new HashMap<>();
// class Node{
// long data;
// Node parent;
// int rank;
// }
// public void makeSet(long data){
// Node node = new Node();
// node.data = data;
// node.parent = node;
// node.rank = 0;
// map.put(data,node);
// }
// //here we just need the rank of parent to be exact
// public void union(long data1,long data2){
// Node node1 = map.get(data1);
// Node node2 = map.get(data2);
// Node parent1 = findSet(node1);
// Node parent2 = findSet(node2);
// if(parent1.data==parent2.data){
// return;
// }
// if(parent1.rank>=parent2.rank){
// parent1.rank = (parent1.rank==parent2.rank)?parent1.rank+1:parent1.rank;
// parent2.parent = parent1;
// }
// else{
// parent1.parent = parent2;
// }
// }
// public long findSet(long data){
// return findSet(map.get(data)).data;
// }
// private Node findSet(Node node){
// Node parent = node.parent;
// if(parent==node){
// return parent;
// }
// node.parent = findSet(node.parent);
// return node.parent;
// }
// } | Java | ["4\n\n5\n\n1 1 1 1 1\n\n5\n\n2 1 1 1 2\n\n6\n\n1 1 2 3 3 4\n\n6\n\n1 2 1 4 5 4"] | 1 second | ["2\n1\n2\n0"] | NoteIn the first test case, we can select $$$i=2$$$ and $$$x=2$$$ to form $$$[1, 2, 2, 1, 1]$$$. Then, we can select $$$i=3$$$ and $$$x=3$$$ to form $$$[1, 2, 3, 3, 1]$$$.In the second test case, we can select $$$i=3$$$ and $$$x=100$$$ to form $$$[2, 1, 100, 100, 2]$$$. | Java 11 | standard input | [
"constructive algorithms",
"greedy",
"implementation"
] | a91be662101762bcb2c58b8db9ff61e0 | Each test contains multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains an integer $$$n$$$ ($$$2 \le n \le 2 \cdot 10 ^ 5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10^9$$$) — elements of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10 ^ 5$$$ | 1,100 | For each test case, print the minimum number of operations needed. | standard output | |
PASSED | 1203d7dadd3dc6d22fa354c1b7dd2e39 | train_108.jsonl | 1650722700 | You are given an array $$$a$$$ of length $$$n$$$. We define the equality of the array as the number of indices $$$1 \le i \le n - 1$$$ such that $$$a_i = a_{i + 1}$$$. We are allowed to do the following operation: Select two integers $$$i$$$ and $$$x$$$ such that $$$1 \le i \le n - 1$$$ and $$$1 \le x \le 10^9$$$. Then, set $$$a_i$$$ and $$$a_{i + 1}$$$ to be equal to $$$x$$$. Find the minimum number of operations needed such that the equality of the array is less than or equal to $$$1$$$. | 256 megabytes | import java.io.PrintWriter;
import java.util.Scanner;
public class Main {
public static void main(String[] args) {
var sc = new Scanner(System.in);
var pw = new PrintWriter(System.out);
int T = Integer.parseInt(sc.next());
for(int t = 0; t < T; t++){
int n = Integer.parseInt(sc.next());
var a = new int[n];
for(int i = 0; i < n; i++){
a[i] = Integer.parseInt(sc.next());
}
int equality = 0;
for(int i = 0; i < n-1; i++){
if(a[i] == a[i+1]){
equality++;
}
}
int ans = 0;
if(equality >= 2){
for(int i = 0; i < n-1; i++){
if(a[i] == a[i+1]){
ans++;
if(i+3 < n && a[i+2] == a[i+3]){
equality--;
}
if(i+2 < n && a[i+1] == a[i+2]){
equality--;
}
if(equality == 1){
break;
}
a[i+1] = -i;
if(i+2 < n){
a[i+2] = -i;
}
}
}
}
pw.println(ans);
}
pw.flush();
}
} | Java | ["4\n\n5\n\n1 1 1 1 1\n\n5\n\n2 1 1 1 2\n\n6\n\n1 1 2 3 3 4\n\n6\n\n1 2 1 4 5 4"] | 1 second | ["2\n1\n2\n0"] | NoteIn the first test case, we can select $$$i=2$$$ and $$$x=2$$$ to form $$$[1, 2, 2, 1, 1]$$$. Then, we can select $$$i=3$$$ and $$$x=3$$$ to form $$$[1, 2, 3, 3, 1]$$$.In the second test case, we can select $$$i=3$$$ and $$$x=100$$$ to form $$$[2, 1, 100, 100, 2]$$$. | Java 11 | standard input | [
"constructive algorithms",
"greedy",
"implementation"
] | a91be662101762bcb2c58b8db9ff61e0 | Each test contains multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains an integer $$$n$$$ ($$$2 \le n \le 2 \cdot 10 ^ 5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10^9$$$) — elements of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10 ^ 5$$$ | 1,100 | For each test case, print the minimum number of operations needed. | standard output | |
PASSED | 86d443b69f58099195b35a7bbcdf48f1 | train_108.jsonl | 1650722700 | You are given an array $$$a$$$ of length $$$n$$$. We define the equality of the array as the number of indices $$$1 \le i \le n - 1$$$ such that $$$a_i = a_{i + 1}$$$. We are allowed to do the following operation: Select two integers $$$i$$$ and $$$x$$$ such that $$$1 \le i \le n - 1$$$ and $$$1 \le x \le 10^9$$$. Then, set $$$a_i$$$ and $$$a_{i + 1}$$$ to be equal to $$$x$$$. Find the minimum number of operations needed such that the equality of the array is less than or equal to $$$1$$$. | 256 megabytes | import java.io.*;
import java.util.*;
public class Main {
static long nod(long a, long b) {
while (Math.min(a, b) != 0) {
if (a > b) {
a = a % b;
} else b = b % a;
}
return a + b;
}
public static void main(String[] args) throws IOException {
// br = new BufferedReader(new FileReader("pencils.in"));
// out = new PrintWriter("pencils.out");
int t = nextInt();
for (int i = 0; i < t; i++) {
int n=nextInt();
int last=0;
int l=-1;
int r=0;
for (int j = 0; j < n; j++) {
int a=nextInt();
if(l==-1&&last==a)l=j-1;
if(last==a)r=j;
last=a;
}
if(l==-1||r-l+1==2) out.println(0);
else if(r-l+1==3||r-l+1==4) out.println(1);
else out.println(r-l-2);
}
out.close();
}
static BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
static PrintWriter out = new PrintWriter(System.out);
static StringTokenizer in = new StringTokenizer("");
public static boolean hasNext() throws IOException {
if (in.hasMoreTokens()) return true;
String s;
while ((s = br.readLine()) != null) {
in = new StringTokenizer(s);
if (in.hasMoreTokens()) return true;
}
return false;
}
public static String nextToken() throws IOException {
while (!in.hasMoreTokens()) {
in = new StringTokenizer(br.readLine());
}
return in.nextToken();
}
public static int nextInt() throws IOException {
return Integer.parseInt(nextToken());
}
public static long nextLong() throws IOException {
return Long.parseLong(nextToken());
}
public static double nextDouble() throws IOException {
return Double.parseDouble(nextToken());
}
static class Maincraft implements Comparable<Maincraft> {
int x;
int y;
int z;
int type;
public Maincraft(int x, int y, int z, int type) {
this.x = x;
this.y = y;
this.z = z;
this.type = type;
}
@Override
public int compareTo(Maincraft o) {
if (x != o.x) return Integer.compare(x, o.x);
else if (y != o.y) return -Integer.compare(y, o.y);
else return -Integer.compare(z, o.z);
}
}
} | Java | ["4\n\n5\n\n1 1 1 1 1\n\n5\n\n2 1 1 1 2\n\n6\n\n1 1 2 3 3 4\n\n6\n\n1 2 1 4 5 4"] | 1 second | ["2\n1\n2\n0"] | NoteIn the first test case, we can select $$$i=2$$$ and $$$x=2$$$ to form $$$[1, 2, 2, 1, 1]$$$. Then, we can select $$$i=3$$$ and $$$x=3$$$ to form $$$[1, 2, 3, 3, 1]$$$.In the second test case, we can select $$$i=3$$$ and $$$x=100$$$ to form $$$[2, 1, 100, 100, 2]$$$. | Java 11 | standard input | [
"constructive algorithms",
"greedy",
"implementation"
] | a91be662101762bcb2c58b8db9ff61e0 | Each test contains multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains an integer $$$n$$$ ($$$2 \le n \le 2 \cdot 10 ^ 5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10^9$$$) — elements of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10 ^ 5$$$ | 1,100 | For each test case, print the minimum number of operations needed. | standard output | |
PASSED | ca55d1cba1910f4de50a007d01a989a1 | train_108.jsonl | 1650722700 | You are given an array $$$a$$$ of length $$$n$$$. We define the equality of the array as the number of indices $$$1 \le i \le n - 1$$$ such that $$$a_i = a_{i + 1}$$$. We are allowed to do the following operation: Select two integers $$$i$$$ and $$$x$$$ such that $$$1 \le i \le n - 1$$$ and $$$1 \le x \le 10^9$$$. Then, set $$$a_i$$$ and $$$a_{i + 1}$$$ to be equal to $$$x$$$. Find the minimum number of operations needed such that the equality of the array is less than or equal to $$$1$$$. | 256 megabytes | import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.StringTokenizer;
public class C {
static class FastReader {
BufferedReader br;
StringTokenizer st;
public FastReader() {
br = new BufferedReader(
new InputStreamReader(System.in));
}
String next() {
while (st == null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt() {
return Integer.parseInt(next());
}
long nextLong() {
return Long.parseLong(next());
}
double nextDouble() {
return Double.parseDouble(next());
}
String nextLine() {
String str = "";
try {
str = br.readLine();
} catch (IOException e) {
e.printStackTrace();
}
return str;
}
}
public static void main(String[] args) {
FastReader sc = new FastReader();
int t = sc.nextInt();
StringBuilder s = new StringBuilder();
while (t-- > 0) {
int n = sc.nextInt();
int[] a = new int[n];
for (int i=0;i<n;i++){
a[i] = sc.nextInt();
}
int eq = 0;
boolean hasEq = false;
int lastEq = 0;
int firstEq = 0;
for (int i=0;i<n-1;i++){
if (a[i]==a[i+1]){
if (!hasEq){
hasEq = true;
firstEq = i;
}
eq++;
lastEq = i;
}
}
int ans = 0;
if (eq<=1){
}else {
ans = Math.max(lastEq-firstEq-1,1);
}
s.append(ans+"\n");
}
System.out.println(s);
}
}
// 1 2 1 2
// 1 1 1 1 1 1 1 1 3 3 3 3 3 | Java | ["4\n\n5\n\n1 1 1 1 1\n\n5\n\n2 1 1 1 2\n\n6\n\n1 1 2 3 3 4\n\n6\n\n1 2 1 4 5 4"] | 1 second | ["2\n1\n2\n0"] | NoteIn the first test case, we can select $$$i=2$$$ and $$$x=2$$$ to form $$$[1, 2, 2, 1, 1]$$$. Then, we can select $$$i=3$$$ and $$$x=3$$$ to form $$$[1, 2, 3, 3, 1]$$$.In the second test case, we can select $$$i=3$$$ and $$$x=100$$$ to form $$$[2, 1, 100, 100, 2]$$$. | Java 11 | standard input | [
"constructive algorithms",
"greedy",
"implementation"
] | a91be662101762bcb2c58b8db9ff61e0 | Each test contains multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains an integer $$$n$$$ ($$$2 \le n \le 2 \cdot 10 ^ 5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10^9$$$) — elements of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10 ^ 5$$$ | 1,100 | For each test case, print the minimum number of operations needed. | standard output | |
PASSED | af4415c09f78d1e8b4ff33a4c780b849 | train_108.jsonl | 1650722700 | You are given an array $$$a$$$ of length $$$n$$$. We define the equality of the array as the number of indices $$$1 \le i \le n - 1$$$ such that $$$a_i = a_{i + 1}$$$. We are allowed to do the following operation: Select two integers $$$i$$$ and $$$x$$$ such that $$$1 \le i \le n - 1$$$ and $$$1 \le x \le 10^9$$$. Then, set $$$a_i$$$ and $$$a_{i + 1}$$$ to be equal to $$$x$$$. Find the minimum number of operations needed such that the equality of the array is less than or equal to $$$1$$$. | 256 megabytes | import java.util.*;
import java.lang.*;
import java.io.*;
// Created by @thesupremeone on 4/23/22
public class C {
void solve() {
int ts = getInt();
for (int t = 0; t < ts; t++) {
int n = getInt();
int[] arr = new int[n];
for (int i = 0; i < n; i++) {
arr[i] = getInt();
}
int last = arr[0];
int count = 0;
int min = -1;
int max = -1;
for (int i = 0; i < n; i++) {
int ai = arr[i];
if(last==ai){
count++;
}else {
if(count>1){
if(min==-1){
min = i-count;
}
max = i-1;
}
count = 1;
}
last = ai;
if(i==n-1){
if(count>1){
if(min==-1){
min = i-count+1;
}
max = i;
}
}
}
if(max==-1){
println(0);
}else {
int elements = max-min+1;
if(elements<3){
println(0);
}else if(elements==3){
println(1);
}else {
println(elements-3);
}
}
}
}
public static void main(String[] args) throws Exception {
if (isOnlineJudge()) {
in = new BufferedReader(new InputStreamReader(System.in));
out = new BufferedWriter(new OutputStreamWriter(System.out));
new C().solve();
out.flush();
} else {
localJudge = new Thread();
in = new BufferedReader(new FileReader("input.txt"));
out = new BufferedWriter(new FileWriter("output.txt"));
localJudge.start();
new C().solve();
out.flush();
localJudge.suspend();
}
}
static boolean isOnlineJudge() {
try {
return System.getProperty("ONLINE_JUDGE") != null
|| System.getProperty("LOCAL") == null;
} catch (Exception e) {
return true;
}
}
// Fast Input & Output
static Thread localJudge = null;
static BufferedReader in;
static StringTokenizer st;
static BufferedWriter out;
static String getLine() {
try {
return in.readLine();
} catch (Exception ignored) {
return "";
}
}
static String getToken() {
if (st == null || !st.hasMoreTokens())
st = new StringTokenizer(getLine());
return st.nextToken();
}
static int getInt() {
return Integer.parseInt(getToken());
}
static long getLong() {
return Long.parseLong(getToken());
}
static void print(Object s) {
try {
out.write(String.valueOf(s));
} catch (Exception ignored) {
}
}
static void println(Object s) {
try {
out.write(String.valueOf(s));
out.newLine();
} catch (Exception ignored) {
}
}
} | Java | ["4\n\n5\n\n1 1 1 1 1\n\n5\n\n2 1 1 1 2\n\n6\n\n1 1 2 3 3 4\n\n6\n\n1 2 1 4 5 4"] | 1 second | ["2\n1\n2\n0"] | NoteIn the first test case, we can select $$$i=2$$$ and $$$x=2$$$ to form $$$[1, 2, 2, 1, 1]$$$. Then, we can select $$$i=3$$$ and $$$x=3$$$ to form $$$[1, 2, 3, 3, 1]$$$.In the second test case, we can select $$$i=3$$$ and $$$x=100$$$ to form $$$[2, 1, 100, 100, 2]$$$. | Java 11 | standard input | [
"constructive algorithms",
"greedy",
"implementation"
] | a91be662101762bcb2c58b8db9ff61e0 | Each test contains multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains an integer $$$n$$$ ($$$2 \le n \le 2 \cdot 10 ^ 5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10^9$$$) — elements of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10 ^ 5$$$ | 1,100 | For each test case, print the minimum number of operations needed. | standard output | |
PASSED | 4d89bd1addf42e114cde08f1fd25abdb | train_108.jsonl | 1650722700 | You are given an array $$$a$$$ of length $$$n$$$. We define the equality of the array as the number of indices $$$1 \le i \le n - 1$$$ such that $$$a_i = a_{i + 1}$$$. We are allowed to do the following operation: Select two integers $$$i$$$ and $$$x$$$ such that $$$1 \le i \le n - 1$$$ and $$$1 \le x \le 10^9$$$. Then, set $$$a_i$$$ and $$$a_{i + 1}$$$ to be equal to $$$x$$$. Find the minimum number of operations needed such that the equality of the array is less than or equal to $$$1$$$. | 256 megabytes | //some updates in import stuff
import static java.lang.Math.max;
import static java.lang.Math.min;
import static java.lang.Math.abs;
import java.util.*;
import java.io.*;
import java.math.*;
//key points learned
//max space ever that could be alloted in a program to pass in cf
//int[][] prefixSum = new int[201][200_005]; -> not a single array more!!!
//never allocate memory again again to such bigg array, it will give memory exceeded for sure
//believe in your fucking solution and keep improving it!!! (sometimes)
//few things to figure around
//getting better and faster at taking input/output with normal method (buffered reader and printwriter)
//memorise all the key algos! a
public class C{
static int mod = (int) (Math.pow(10, 9)+7);
static final int dx[] = { -1, 0, 1, 0 }, dy[] = { 0, -1, 0, 1 };
static final int[] dx8 = { -1, -1, -1, 0, 0, 1, 1, 1 }, dy8 = { -1, 0, 1, -1, 1, -1, 0, 1 };
static final int[] dx9 = { -1, -1, -1, 0, 0, 0, 1, 1, 1 }, dy9 = { -1, 0, 1, -1, 0, 1, -1, 0, 1 };
static final double eps = 1e-10;
static List<Integer> primeNumbers = new ArrayList<>();
public static void main(String[] args) {
MyScanner sc = new MyScanner(); //pretty important for sure -
out = new PrintWriter(new BufferedOutputStream(System.out)); //dope shit output for sure
//code here
int test = sc.nextInt();
while(test --> 0){
int n = sc.nextInt();
int[] arr = new int[n];
int count = 0;
for(int i= 0; i< n; i++){
arr[i] = sc.nextInt();
}
for(int i= 0; i < n-1; i++){
if(arr[i] == arr[i+1]) count++;
}
if(count <= 1){
out.println(0);
continue;
}
int f = -1;
int s = -1;
for(int i= 0; i < n-1; i++){
if(arr[i] == arr[i + 1]){
if(f == -1) f = i;
s = i;
}
}
int diff = s - f;
// out.println(diff);
if(diff <= 2){
out.println(1);
}else{
out.println(diff - 1);
}
// out.println(ans);
}
out.close();
}
//new stuff to learn (whenever this is need for them, then only)
//Lazy Segment Trees
//Persistent Segment Trees
//Square Root Decomposition
//Geometry & Convex Hull
//High Level DP -- yk yk
//String Matching Algorithms
//Heavy light Decomposition
//Updation Required
//Fenwick Tree - both are done (sum)
//Segment Tree - both are done (min, max, sum)
//-----CURRENTLY PRESENT-------//
//Graph
//DSU
//powerMODe
//power
//Segment Tree (work on this one)
//Prime Sieve
//Count Divisors
//Next Permutation
//Get NCR
//isVowel
//Sort (int)
//Sort (long)
//Binomial Coefficient
//Pair
//Triplet
//lcm (int & long)
//gcd (int & long)
//gcd (for binomial coefficient)
//swap (int & char)
//reverse
//primeExponentCounts
//Fast input and output
//-------------------------------------------------------------------
//-------------------------------------------------------------------
//-------------------------------------------------------------------
//-------------------------------------------------------------------
//-------------------------------------------------------------------
//GRAPH (basic structure)
public static class Graph{
public int V;
public ArrayList<ArrayList<Integer>> edges;
//2 -> [0,1,2] (current)
Graph(int V){
this.V = V;
edges = new ArrayList<>(V+1);
for(int i= 0; i <= V; i++){
edges.add(new ArrayList<>());
}
}
public void addEdge(int from , int to){
edges.get(from).add(to);
edges.get(to).add(from);
}
}
//DSU (path and rank optimised)
public static class DisjointUnionSets {
int[] rank, parent;
int n;
public DisjointUnionSets(int n)
{
rank = new int[n];
parent = new int[n];
Arrays.fill(rank, 1);
Arrays.fill(parent,-1);
this.n = n;
}
public int find(int curr){
if(parent[curr] == -1)
return curr;
//path compression optimisation
return parent[curr] = find(parent[curr]);
}
public void union(int a, int b){
int s1 = find(a);
int s2 = find(b);
if(s1 != s2){
//union by size
if(rank[s1] < rank[s2]){
parent[s1] = s2;
rank[s2] += rank[s1];
}else{
parent[s2] = s1;
rank[s1] += rank[s2];
}
}
}
}
//with mod
public static long powerMOD(long x, long y)
{
long res = 1L;
while (y > 0)
{
// If y is odd, multiply x with result
if ((y & 1) != 0){
x %= mod;
res %= mod;
res = (res * x)%mod;
}
// y must be even now
y = y >> 1; // y = y/2
x%= mod;
x = (x * x)%mod; // Change x to x^2
}
return res%mod;
}
//without mod
public static long power(long x, long y)
{
long res = 1L;
while (y > 0)
{
// If y is odd, multiply x with result
if ((y & 1) != 0){
res = (res * x);
}
// y must be even now
y = y >> 1; // y = y/
x = (x * x);
}
return res;
}
public static class segmentTree{
//so let's make a constructor function for this bad boi for sure!!!
public long[] arr;
public long[] tree;
//COMPLEXITY (normal segment tree, stuff)
//build -> O(n)
//query -> O(logn)
//update -> O(logn)
//update-range -> O(n) (worst case)
//simple iteration and stuff for sure
public segmentTree(long[] arr){
int n = arr.length;
this.arr = new long[n];
for(int i= 0; i < n; i++){
this.arr[i] = arr[i];
}
tree = new long[4*n + 1];
}
//pretty basic idea if you read the code once
//first make child node once
//then form the parent node using them
public void buildTree(int s, int e, int index){
if(s == e){
tree[index] = arr[s];
return;
}
//recursive case
int mid = (s + e)/2;
buildTree(s, mid, 2 * index);
buildTree(mid + 1, e, 2*index + 1);
//the condition we want from children be like this
tree[index] = min(tree[2 * index], tree[2 * index + 1]);
return;
}
//definitely index based 0 query!!!
//only int index = 1!!
//baaki everything is simple as fuck
public long query(int s, int e, int qs , int qe, int index){
//complete overlap
if(s >= qs && e <= qe){
return tree[index];
}
//no overlap
if(qe < s || qs > e){
return Long.MAX_VALUE;
}
//partial overlap
int mid = (s + e)/2;
long left = query( s, mid , qs, qe, 2*index);
long right = query( mid + 1, e, qs, qe, 2*index + 1);
return min(left, right);
}
//gonna do range updates for sure now!!
//let's do this bois!!! (solve this problem for sure)
public void updateRange(int s, int e, int l, int r, long increment, int index){
//out of bounds
if(l > e || r < s){
return;
}
//leaf node
if(s == e){
tree[index] += increment;
return; //behnchoda return tera baap krvayege?
}
//recursive case
int mid = (s + e)/2;
updateRange(s, mid, l, r, increment, 2 * index);
updateRange(mid + 1, e, l, r, increment, 2 * index + 1);
tree[index] = min(tree[2 * index], tree[2 * index + 1]);
}
}
public static class segmentTreeLazy{
//so let's make a constructor function for this bad boi for sure!!!
public long[] arr;
public long[] tree;
public long[] lazy;
//COMPLEXITY (normal segment tree, stuff)
//build -> O(n)
//query-range -> O(logn)
//lazy update-range -> O(logn) (imp)
//simple iteration and stuff for sure
public segmentTreeLazy(long[] arr){
int n = arr.length;
this.arr = new long[n];
for(int i= 0; i < n; i++){
this.arr[i] = arr[i];
}
tree = new long[4*n + 1];
lazy = new long[1000000]; //pretty big for no inconvenience (no?) NONONONOONON! NO fucker NO!
}
//pretty basic idea if you read the code once
//first make child node once
//then form the parent node using them
public void buildTree(int s, int e, int index){
if(s == e){
tree[index] = arr[s];
return;
}
//recursive case
int mid = (s + e)/2;
buildTree(s, mid, 2 * index);
buildTree(mid + 1, e, 2*index + 1);
//the condition we want from children be like this
tree[index] = min(tree[2 * index], tree[2 * index + 1]);
return;
}
//definitely index based 0 query!!!
//only int index = 1!!
//baaki everything is simple as fuck
public long queryLazy(int s, int e, int qs, int qe, int index){
//before going down resolve if it exist
if(lazy[index] != 0){
tree[index] += lazy[index];
//non leaf node
if(s != e){
lazy[2*index] += lazy[index];
lazy[2*index + 1] += lazy[index];
}
lazy[index] = 0; //clear the lazy value at current node for sure
}
//no overlap
if(s > qe || e < qs){
return Long.MAX_VALUE;
}
//complete overlap
if(s >= qs && e <= qe){
return tree[index];
}
//partial overlap
int mid = (s + e)/2;
long left = queryLazy(s, mid, qs, qe, 2 * index);
long right = queryLazy(mid + 1, e, qs, qe, 2 * index + 1);
return Math.min(left, right);
}
//update range in O(logn) -- using lazy array
public void updateRangeLazy(int s, int e, int l, int r, int inc, int index){
//before going down resolve if it exist
if(lazy[index] != 0){
tree[index] += lazy[index];
//non leaf node
if(s != e){
lazy[2*index] += lazy[index];
lazy[2*index + 1] += lazy[index];
}
lazy[index] = 0; //clear the lazy value at current node for sure
}
//no overlap
if(s > r || l > e){
return;
}
//another case
if(l <= s && e <= r){
tree[index] += inc;
//create a new lazy value for children node
if(s != e){
lazy[2*index] += inc;
lazy[2*index + 1] += inc;
}
return;
}
//recursive case
int mid = (s + e)/2;
updateRangeLazy(s, mid, l, r, inc, 2*index);
updateRangeLazy(mid + 1, e, l, r, inc, 2*index + 1);
//update the tree index
tree[index] = Math.min(tree[2*index], tree[2*index + 1]);
return;
}
}
//prime sieve
public static void primeSieve(int n){
BitSet bitset = new BitSet(n+1);
for(long i = 0; i < n ; i++){
if (i == 0 || i == 1) {
bitset.set((int) i);
continue;
}
if(bitset.get((int) i)) continue;
primeNumbers.add((int)i);
for(long j = i; j <= n ; j+= i)
bitset.set((int)j);
}
}
//number of divisors
public static int countDivisors(long number){
if(number == 1) return 1;
List<Integer> primeFactors = new ArrayList<>();
int index = 0;
long curr = primeNumbers.get(index);
while(curr * curr <= number){
while(number % curr == 0){
number = number/curr;
primeFactors.add((int) curr);
}
index++;
curr = primeNumbers.get(index);
}
if(number != 1) primeFactors.add((int) number);
int current = primeFactors.get(0);
int totalDivisors = 1;
int currentCount = 2;
for (int i = 1; i < primeFactors.size(); i++) {
if (primeFactors.get(i) == current) {
currentCount++;
} else {
totalDivisors *= currentCount;
currentCount = 2;
current = primeFactors.get(i);
}
}
totalDivisors *= currentCount;
return totalDivisors;
}
//primeExponentCounts
public static int primeExponentsCount(int n) {
if (n <= 1)
return 0;
int sqrt = (int) Math.sqrt(n);
int remainingNumber = n;
int result = 0;
for (int i = 2; i <= sqrt; i++) {
while (remainingNumber % i == 0) {
result++;
remainingNumber /= i;
}
}
//in case of prime numbers this would happen
if (remainingNumber > 1) {
result++;
}
return result;
}
//now adding next permutation function to java hehe
public static boolean next_permutation(int[] p) {
for (int a = p.length - 2; a >= 0; --a)
if (p[a] < p[a + 1])
for (int b = p.length - 1;; --b)
if (p[b] > p[a]) {
int t = p[a];
p[a] = p[b];
p[b] = t;
for (++a, b = p.length - 1; a < b; ++a, --b) {
t = p[a];
p[a] = p[b];
p[b] = t;
}
return true;
}
return false;
}
//finding the value of NCR in O(RlogN) time and O(1) space
public static long getNcR(int n, int r)
{
long p = 1, k = 1;
if (n - r < r) r = n - r;
if (r != 0) {
while (r > 0) {
p *= n;
k *= r;
long m = __gcd(p, k);
p /= m;
k /= m;
n--;
r--;
}
}
else {
p = 1;
}
return p;
}
//is vowel function
public static boolean isVowel(char c)
{
return (c=='a' || c=='A' || c=='e' || c=='E' || c=='i' || c=='I' || c=='o' || c=='O' || c=='u' || c=='U');
}
//to sort the array with better method
public static void sort(int[] a) {
ArrayList<Integer> l=new ArrayList<>();
for (int i:a) l.add(i);
Collections.sort(l);
for (int i=0; i<a.length; i++) a[i]=l.get(i);
}
//sort long
public static void sort(long[] a) {
ArrayList<Long> l=new ArrayList<>();
for (long i:a) l.add(i);
Collections.sort(l);
for (int i=0; i<a.length; i++) a[i]=l.get(i);
}
//for calculating binomialCoeff
public static int binomialCoeff(int n, int k)
{
int C[] = new int[k + 1];
// nC0 is 1
C[0] = 1;
for (int i = 1; i <= n; i++) {
// Compute next row of pascal
// triangle using the previous row
for (int j = Math.min(i, k); j > 0; j--)
C[j] = C[j] + C[j - 1];
}
return C[k];
}
//Pair with int int
public static class Pair{
public int a;
public int b;
public int hashCode;
Pair(int a , int b){
this.a = a;
this.b = b;
this.hashCode = Objects.hash(a, b);
}
@Override
public String toString(){
return a + " -> " + b;
}
@Override
public boolean equals(Object o) {
if (this == o)
return true;
if (o == null || getClass() != o.getClass())
return false;
Pair that = (Pair) o;
return a == that.a && b == that.b;
}
@Override
public int hashCode() {
return this.hashCode;
}
}
//Triplet with int int int
public static class Triplet{
public int a;
public int b;
public int c;
Triplet(int a , int b, int c){
this.a = a;
this.b = b;
this.c = c;
}
@Override
public String toString(){
return a + " -> " + b;
}
}
//Shortcut function
public static long lcm(long a , long b){
return a * (b/gcd(a,b));
}
//let's make one for calculating lcm basically
public static int lcm(int a , int b){
return (a * b)/gcd(a,b);
}
//int version for gcd
public static int gcd(int a, int b){
if(b == 0)
return a;
return gcd(b , a%b);
}
//long version for gcd
public static long gcd(long a, long b){
if(b == 0)
return a;
return gcd(b , a%b);
}
//for ncr calculator(ignore this code)
public static long __gcd(long n1, long n2)
{
long gcd = 1;
for (int i = 1; i <= n1 && i <= n2; ++i) {
// Checks if i is factor of both integers
if (n1 % i == 0 && n2 % i == 0) {
gcd = i;
}
}
return gcd;
}
//swapping two elements in an array
public static void swap(int[] arr, int left , int right){
int temp = arr[left];
arr[left] = arr[right];
arr[right] = temp;
}
//for char array
public static void swap(char[] arr, int left , int right){
char temp = arr[left];
arr[left] = arr[right];
arr[right] = temp;
}
//reversing an array
public static void reverse(int[] arr){
int left = 0;
int right = arr.length-1;
while(left <= right){
swap(arr, left,right);
left++;
right--;
}
}
public static long expo(long a, long b, long mod) {
long res = 1;
while (b > 0) {
if ((b & 1) == 1L) res = (res * a) % mod; //think about this one for a second
a = (a * a) % mod;
b = b >> 1;
}
return res;
}
//SOME EXTRA DOPE FUNCTIONS
public static long mminvprime(long a, long b) {
return expo(a, b - 2, b);
}
public static long mod_add(long a, long b, long m) {
a = a % m;
b = b % m;
return (((a + b) % m) + m) % m;
}
public static long mod_sub(long a, long b, long m) {
a = a % m;
b = b % m;
return (((a - b) % m) + m) % m;
}
public static long mod_mul(long a, long b, long m) {
a = a % m;
b = b % m;
return (((a * b) % m) + m) % m;
}
public static long mod_div(long a, long b, long m) {
a = a % m;
b = b % m;
return (mod_mul(a, mminvprime(b, m), m) + m) % m;
}
//O(n) every single time remember that
public static long nCr(long N, long K , long mod){
long upper = 1L;
long lower = 1L;
long lowerr = 1L;
for(long i = 1; i <= N; i++){
upper = mod_mul(upper, i, mod);
}
for(long i = 1; i <= K; i++){
lower = mod_mul(lower, i, mod);
}
for(long i = 1; i <= (N - K); i++){
lowerr = mod_mul(lowerr, i, mod);
}
// out.println(upper + " " + lower + " " + lowerr);
long answer = mod_mul(lower, lowerr, mod);
answer = mod_div(upper, answer, mod);
return answer;
}
// long[] fact = new long[2 * n + 1];
// long[] ifact = new long[2 * n + 1];
// fact[0] = 1;
// ifact[0] = 1;
// for (long i = 1; i <= 2 * n; i++)
// {
// fact[(int)i] = mod_mul(fact[(int)i - 1], i, mod);
// ifact[(int)i] = mminvprime(fact[(int)i], mod);
// }
//ifact is basically inverse factorial in here!!!!!(imp)
public static long combination(long n, long r, long m, long[] fact, long[] ifact) {
long val1 = fact[(int)n];
long val2 = ifact[(int)(n - r)];
long val3 = ifact[(int)r];
return (((val1 * val2) % m) * val3) % m;
}
//-----------PrintWriter for faster output---------------------------------
public static PrintWriter out;
//-----------MyScanner class for faster input----------
public static class MyScanner {
BufferedReader br;
StringTokenizer st;
public MyScanner() {
br = new BufferedReader(new InputStreamReader(System.in));
}
String next() {
while (st == null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt() {
return Integer.parseInt(next());
}
long nextLong() {
return Long.parseLong(next());
}
double nextDouble() {
return Double.parseDouble(next());
}
String nextLine(){
String str = "";
try {
str = br.readLine();
} catch (IOException e) {
e.printStackTrace();
}
return str;
}
}
//--------------------------------------------------------
} | Java | ["4\n\n5\n\n1 1 1 1 1\n\n5\n\n2 1 1 1 2\n\n6\n\n1 1 2 3 3 4\n\n6\n\n1 2 1 4 5 4"] | 1 second | ["2\n1\n2\n0"] | NoteIn the first test case, we can select $$$i=2$$$ and $$$x=2$$$ to form $$$[1, 2, 2, 1, 1]$$$. Then, we can select $$$i=3$$$ and $$$x=3$$$ to form $$$[1, 2, 3, 3, 1]$$$.In the second test case, we can select $$$i=3$$$ and $$$x=100$$$ to form $$$[2, 1, 100, 100, 2]$$$. | Java 11 | standard input | [
"constructive algorithms",
"greedy",
"implementation"
] | a91be662101762bcb2c58b8db9ff61e0 | Each test contains multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains an integer $$$n$$$ ($$$2 \le n \le 2 \cdot 10 ^ 5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10^9$$$) — elements of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10 ^ 5$$$ | 1,100 | For each test case, print the minimum number of operations needed. | standard output | |
PASSED | 93315a65d7a556ef3491ba65d718ffeb | train_108.jsonl | 1650722700 | You are given an array $$$a$$$ of length $$$n$$$. We define the equality of the array as the number of indices $$$1 \le i \le n - 1$$$ such that $$$a_i = a_{i + 1}$$$. We are allowed to do the following operation: Select two integers $$$i$$$ and $$$x$$$ such that $$$1 \le i \le n - 1$$$ and $$$1 \le x \le 10^9$$$. Then, set $$$a_i$$$ and $$$a_{i + 1}$$$ to be equal to $$$x$$$. Find the minimum number of operations needed such that the equality of the array is less than or equal to $$$1$$$. | 256 megabytes | import java.util.*;
public class Main {
public static void main(String[] args){
Scanner sc = new Scanner(System.in);
// long mod = 1_000_000_007L;
// long mod = 998_244_353L;
int t = Integer.parseInt(sc.next());
for ( int zzz=0; zzz<t; zzz++ ) {
int n = Integer.parseInt(sc.next());
int[] a = new int[n];
int cl = 0;
for ( int i=0; i<n; i++ ) {
a[i] = Integer.parseInt(sc.next());
if ( i==0 ) continue;
if ( cl>0 ) continue;
if ( a[i]==a[i-1] ) {
cl = i;
}
}
int cr = n;
for ( int i=n-2; i>=0; i-- ) {
if ( a[i]==a[i+1] ) {
cr = i;
break;
}
}
int ans = 0;
if ( cl>0 ) {
if ( cl==cr ) {
ans = 1;
} else if ( cl<cr ) {
ans = cr-cl;
}
}
System.out.println(ans);
}
}
private static long gcd(long a, long b) {
if ( a<b ) return gcd(b,a);
if ( b==0L ) return a;
return gcd(b,a%b);
}
}
| Java | ["4\n\n5\n\n1 1 1 1 1\n\n5\n\n2 1 1 1 2\n\n6\n\n1 1 2 3 3 4\n\n6\n\n1 2 1 4 5 4"] | 1 second | ["2\n1\n2\n0"] | NoteIn the first test case, we can select $$$i=2$$$ and $$$x=2$$$ to form $$$[1, 2, 2, 1, 1]$$$. Then, we can select $$$i=3$$$ and $$$x=3$$$ to form $$$[1, 2, 3, 3, 1]$$$.In the second test case, we can select $$$i=3$$$ and $$$x=100$$$ to form $$$[2, 1, 100, 100, 2]$$$. | Java 11 | standard input | [
"constructive algorithms",
"greedy",
"implementation"
] | a91be662101762bcb2c58b8db9ff61e0 | Each test contains multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains an integer $$$n$$$ ($$$2 \le n \le 2 \cdot 10 ^ 5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10^9$$$) — elements of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10 ^ 5$$$ | 1,100 | For each test case, print the minimum number of operations needed. | standard output | |
PASSED | 4d42ce6d1b241c8a074b3ec17e92ca32 | train_108.jsonl | 1650722700 | You are given an array $$$a$$$ of length $$$n$$$. We define the equality of the array as the number of indices $$$1 \le i \le n - 1$$$ such that $$$a_i = a_{i + 1}$$$. We are allowed to do the following operation: Select two integers $$$i$$$ and $$$x$$$ such that $$$1 \le i \le n - 1$$$ and $$$1 \le x \le 10^9$$$. Then, set $$$a_i$$$ and $$$a_{i + 1}$$$ to be equal to $$$x$$$. Find the minimum number of operations needed such that the equality of the array is less than or equal to $$$1$$$. | 256 megabytes | import java.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.InputMismatchException;
import java.io.IOException;
import java.io.InputStream;
/**
* Built using CHelper plug-in
* Actual solution is at the top
*
* @author dauom
*/
public class Main {
public static void main(String[] args) {
InputStream inputStream = System.in;
OutputStream outputStream = System.out;
InputReader in = new InputReader(inputStream);
PrintWriter out = new PrintWriter(outputStream);
CUnequalArray solver = new CUnequalArray();
int testCount = Integer.parseInt(in.next());
for (int i = 1; i <= testCount; i++)
solver.solve(i, in, out);
out.close();
}
static class CUnequalArray {
public final void solve(int testNumber, InputReader in, PrintWriter out) {
int n = in.nextInt();
int[] a = in.nextIntArray(n);
int start = -1;
int end = -1;
for (int i = 1; i < n; i++) {
if (a[i] == a[i - 1]) {
if (start == -1) start = i;
end = i - 1;
}
}
if (start == end && start >= 0) {
start--;
}
out.println(Integer.max(0, end - start));
}
}
static final class InputReader {
private final InputStream stream;
private final byte[] buf = new byte[1 << 18];
private int curChar;
private int numChars;
public InputReader() {
this.stream = System.in;
}
public InputReader(final InputStream stream) {
this.stream = stream;
}
private int read() {
if (this.numChars == -1) {
throw new UnknownError();
} else {
if (this.curChar >= this.numChars) {
this.curChar = 0;
try {
this.numChars = this.stream.read(this.buf);
} catch (IOException ex) {
throw new InputMismatchException();
}
if (this.numChars <= 0) {
return -1;
}
}
return this.buf[this.curChar++];
}
}
public final int nextInt() {
int c;
for (c = this.read(); isSpaceChar(c); c = this.read()) {
}
byte sgn = 1;
if (c == 45) { // 45 == '-'
sgn = -1;
c = this.read();
}
int res = 0;
while (c >= 48 && c <= 57) { // 48 == '0', 57 == '9'
res *= 10;
res += c - 48; // 48 == '0'
c = this.read();
if (isSpaceChar(c)) {
return res * sgn;
}
}
throw new InputMismatchException();
}
public final String next() {
int c;
while (isSpaceChar(c = this.read())) {
}
StringBuilder result = new StringBuilder();
result.appendCodePoint(c);
while (!isSpaceChar(c = this.read())) {
result.appendCodePoint(c);
}
return result.toString();
}
private static boolean isSpaceChar(final int c) {
return c == 32 || c == 10 || c == 13 || c == 9
|| c == -1; // 32 == ' ', 10 == '\n', 13 == '\r', 9 == '\t'
}
public final int[] nextIntArray(final int n) {
int[] arr = new int[n];
for (int i = 0; i < n; i++) {
arr[i] = nextInt();
}
return arr;
}
}
}
| Java | ["4\n\n5\n\n1 1 1 1 1\n\n5\n\n2 1 1 1 2\n\n6\n\n1 1 2 3 3 4\n\n6\n\n1 2 1 4 5 4"] | 1 second | ["2\n1\n2\n0"] | NoteIn the first test case, we can select $$$i=2$$$ and $$$x=2$$$ to form $$$[1, 2, 2, 1, 1]$$$. Then, we can select $$$i=3$$$ and $$$x=3$$$ to form $$$[1, 2, 3, 3, 1]$$$.In the second test case, we can select $$$i=3$$$ and $$$x=100$$$ to form $$$[2, 1, 100, 100, 2]$$$. | Java 11 | standard input | [
"constructive algorithms",
"greedy",
"implementation"
] | a91be662101762bcb2c58b8db9ff61e0 | Each test contains multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains an integer $$$n$$$ ($$$2 \le n \le 2 \cdot 10 ^ 5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10^9$$$) — elements of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10 ^ 5$$$ | 1,100 | For each test case, print the minimum number of operations needed. | standard output | |
PASSED | d3da43b5ea36ddde0982d42eb262a6fb | train_108.jsonl | 1650722700 | You are given an array $$$a$$$ of length $$$n$$$. We define the equality of the array as the number of indices $$$1 \le i \le n - 1$$$ such that $$$a_i = a_{i + 1}$$$. We are allowed to do the following operation: Select two integers $$$i$$$ and $$$x$$$ such that $$$1 \le i \le n - 1$$$ and $$$1 \le x \le 10^9$$$. Then, set $$$a_i$$$ and $$$a_{i + 1}$$$ to be equal to $$$x$$$. Find the minimum number of operations needed such that the equality of the array is less than or equal to $$$1$$$. | 256 megabytes | import java.util.*;
import java.io.*;
public class Main {
static ContestScanner sc = new ContestScanner(System.in);
static PrintWriter pw = new PrintWriter(System.out);
static StringBuilder sb = new StringBuilder();
static long mod = (long) 1e9 + 7;
public static void main(String[] args) throws Exception {
int n = sc.nextInt();
for(int i= 0; i < n; i++) solve();
pw.flush();
}
public static void solve() {
int N = sc.nextInt();
int ans = 0;
int[] a = sc.nextIntArray(N);
int min = N;
int max = 0;
for(int i = 1; i < N; i++){
if(a[i-1] == a[i]){
min = Math.min(min,i);
max = Math.max(max,i);
}
}
if(max <= min){
pw.println(0);
}else{
pw.println(Math.max(1,max-min-1));
}
}
static class GeekInteger {
public static void save_sort(int[] array) {
shuffle(array);
Arrays.sort(array);
}
public static int[] shuffle(int[] array) {
int n = array.length;
Random random = new Random();
for (int i = 0, j; i < n; i++) {
j = i + random.nextInt(n - i);
int randomElement = array[j];
array[j] = array[i];
array[i] = randomElement;
}
return array;
}
public static void save_sort(long[] array) {
shuffle(array);
Arrays.sort(array);
}
public static long[] shuffle(long[] array) {
int n = array.length;
Random random = new Random();
for (int i = 0, j; i < n; i++) {
j = i + random.nextInt(n - i);
long randomElement = array[j];
array[j] = array[i];
array[i] = randomElement;
}
return array;
}
}
}
/**
* refercence : https://github.com/NASU41/AtCoderLibraryForJava/blob/master/ContestIO/ContestScanner.java
*/
class ContestScanner {
private final java.io.InputStream in;
private final byte[] buffer = new byte[1024];
private int ptr = 0;
private int buflen = 0;
private static final long LONG_MAX_TENTHS = 922337203685477580L;
private static final int LONG_MAX_LAST_DIGIT = 7;
private static final int LONG_MIN_LAST_DIGIT = 8;
public ContestScanner(java.io.InputStream in){
this.in = in;
}
public ContestScanner(java.io.File file) throws java.io.FileNotFoundException {
this(new java.io.BufferedInputStream(new java.io.FileInputStream(file)));
}
public ContestScanner(){
this(System.in);
}
private boolean hasNextByte() {
if (ptr < buflen) {
return true;
}else{
ptr = 0;
try {
buflen = in.read(buffer);
} catch (java.io.IOException e) {
e.printStackTrace();
}
if (buflen <= 0) {
return false;
}
}
return true;
}
private int readByte() {
if (hasNextByte()) return buffer[ptr++]; else return -1;
}
private static boolean isPrintableChar(int c) {
return 33 <= c && c <= 126;
}
public boolean hasNext() {
while(hasNextByte() && !isPrintableChar(buffer[ptr])) ptr++;
return hasNextByte();
}
public String next() {
if (!hasNext()) throw new java.util.NoSuchElementException();
StringBuilder sb = new StringBuilder();
int b = readByte();
while(isPrintableChar(b)) {
sb.appendCodePoint(b);
b = readByte();
}
return sb.toString();
}
public long nextLong() {
if (!hasNext()) throw new java.util.NoSuchElementException();
long n = 0;
boolean minus = false;
int b = readByte();
if (b == '-') {
minus = true;
b = readByte();
}
if (b < '0' || '9' < b) {
throw new NumberFormatException();
}
while (true) {
if ('0' <= b && b <= '9') {
int digit = b - '0';
if (n >= LONG_MAX_TENTHS) {
if (n == LONG_MAX_TENTHS) {
if (minus) {
if (digit <= LONG_MIN_LAST_DIGIT) {
n = -n * 10 - digit;
b = readByte();
if (!isPrintableChar(b)) {
return n;
} else if (b < '0' || '9' < b) {
throw new NumberFormatException(
String.format("%d%s... is not number", n, Character.toString(b))
);
}
}
} else {
if (digit <= LONG_MAX_LAST_DIGIT) {
n = n * 10 + digit;
b = readByte();
if (!isPrintableChar(b)) {
return n;
} else if (b < '0' || '9' < b) {
throw new NumberFormatException(
String.format("%d%s... is not number", n, Character.toString(b))
);
}
}
}
}
throw new ArithmeticException(
String.format("%s%d%d... overflows long.", minus ? "-" : "", n, digit)
);
}
n = n * 10 + digit;
}else if(b == -1 || !isPrintableChar(b)){
return minus ? -n : n;
}else{
throw new NumberFormatException();
}
b = readByte();
}
}
public int nextInt() {
long nl = nextLong();
if (nl < Integer.MIN_VALUE || nl > Integer.MAX_VALUE) throw new NumberFormatException();
return (int) nl;
}
public double nextDouble() {
return Double.parseDouble(next());
}
public long[] nextLongArray(int length){
long[] array = new long[length];
for(int i=0; i<length; i++) array[i] = this.nextLong();
return array;
}
public long[] nextLongArray(int length, java.util.function.LongUnaryOperator map){
long[] array = new long[length];
for(int i=0; i<length; i++) array[i] = map.applyAsLong(this.nextLong());
return array;
}
public int[] nextIntArray(int length){
int[] array = new int[length];
for(int i=0; i<length; i++) array[i] = this.nextInt();
return array;
}
public int[] nextIntArray(int length, java.util.function.IntUnaryOperator map){
int[] array = new int[length];
for(int i=0; i<length; i++) array[i] = map.applyAsInt(this.nextInt());
return array;
}
public double[] nextDoubleArray(int length){
double[] array = new double[length];
for(int i=0; i<length; i++) array[i] = this.nextDouble();
return array;
}
public double[] nextDoubleArray(int length, java.util.function.DoubleUnaryOperator map){
double[] array = new double[length];
for(int i=0; i<length; i++) array[i] = map.applyAsDouble(this.nextDouble());
return array;
}
public long[][] nextLongMatrix(int height, int width){
long[][] mat = new long[height][width];
for(int h=0; h<height; h++) for(int w=0; w<width; w++){
mat[h][w] = this.nextLong();
}
return mat;
}
public int[][] nextIntMatrix(int height, int width){
int[][] mat = new int[height][width];
for(int h=0; h<height; h++) for(int w=0; w<width; w++){
mat[h][w] = this.nextInt();
}
return mat;
}
public double[][] nextDoubleMatrix(int height, int width){
double[][] mat = new double[height][width];
for(int h=0; h<height; h++) for(int w=0; w<width; w++){
mat[h][w] = this.nextDouble();
}
return mat;
}
public char[][] nextCharMatrix(int height, int width){
char[][] mat = new char[height][width];
for(int h=0; h<height; h++){
String s = this.next();
for(int w=0; w<width; w++){
mat[h][w] = s.charAt(w);
}
}
return mat;
}
}
| Java | ["4\n\n5\n\n1 1 1 1 1\n\n5\n\n2 1 1 1 2\n\n6\n\n1 1 2 3 3 4\n\n6\n\n1 2 1 4 5 4"] | 1 second | ["2\n1\n2\n0"] | NoteIn the first test case, we can select $$$i=2$$$ and $$$x=2$$$ to form $$$[1, 2, 2, 1, 1]$$$. Then, we can select $$$i=3$$$ and $$$x=3$$$ to form $$$[1, 2, 3, 3, 1]$$$.In the second test case, we can select $$$i=3$$$ and $$$x=100$$$ to form $$$[2, 1, 100, 100, 2]$$$. | Java 11 | standard input | [
"constructive algorithms",
"greedy",
"implementation"
] | a91be662101762bcb2c58b8db9ff61e0 | Each test contains multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains an integer $$$n$$$ ($$$2 \le n \le 2 \cdot 10 ^ 5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10^9$$$) — elements of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10 ^ 5$$$ | 1,100 | For each test case, print the minimum number of operations needed. | standard output | |
PASSED | 0b59c43d2f0da3ac435ecf582afe772e | train_108.jsonl | 1650722700 | You are given an array $$$a$$$ of length $$$n$$$. We define the equality of the array as the number of indices $$$1 \le i \le n - 1$$$ such that $$$a_i = a_{i + 1}$$$. We are allowed to do the following operation: Select two integers $$$i$$$ and $$$x$$$ such that $$$1 \le i \le n - 1$$$ and $$$1 \le x \le 10^9$$$. Then, set $$$a_i$$$ and $$$a_{i + 1}$$$ to be equal to $$$x$$$. Find the minimum number of operations needed such that the equality of the array is less than or equal to $$$1$$$. | 256 megabytes | import java.io.*;
import java.util.*;
public class Codeforces {
static int mod = (int) (1e9 + 7);
static void solve() {
int n = i();
long[] arr = new long[n];
for (int i = 0; i < n; i++) {
arr[i] = l();
}
int count = 0;
int idx1 = 0;
int idx2 = 0;
int idx3 = 0;
int idx4 = 0;
for (int i = 0; i < n - 1; i++) {
// System.out.println(arr[i]+" "+arr[i+1]);
if (arr[i] == arr[i + 1]) {
count++;
idx1 = i;
idx2 = i + 1;
break;
}
}
// System.out.println(idx1+" "+idx2+" "+idx3+" "+idx4);
if (count == 0) {
System.out.println(0);
return;
}
for (int i = n - 1; i >= idx2 + 1; i--) {
if (arr[i] == arr[i - 1]) {
idx3 = i - 1;
count++;
idx4 = i;
break;
}
}
// System.out.println(count);
// System.out.println(idx1+" "+idx2+" "+idx3+" "+idx4);
if (count == 1) {
System.out.println(0);
return;
}
if(idx2==idx3){
System.out.println(1);
return;
}
System.out.println(idx3-idx2);
}
public static void main(String[] args) {
int test = i();
while (test-- > 0) {
solve();
}
}
// ------> swap(long[]arr,int idx1,int idx2)<---- swap int
// ----->segmentTree--> segTree as class
// ----->lazy_Seg_tree --> lazy_Seg_tree as class
// -----> Trie --->Trie as class
// ----->fenwick_Tree ---> fenwick_Tree
// -----> POWER ---> long power(long x, long y) <----
// -----> LCM ---> long lcm(long x, long y) <----
// -----> GCD ---> long gcd(long x, long y) <----
// -----> SIEVE --> ArrayList<Integer> sieve(int N) <-----
// -----> NCR ---> long ncr(int n, int r) <----
// -----> BINARY_SEARCH ---> int binary_Search(long[]arr,long x) <----
// -----> (SORTING OF LONG, CHAR,INT) -->long[] sortLong(long[] a2)<--
// ----> DFS ---> void dfs(ArrayList<ArrayList<Integer>>edges,int child,int
// parent)<---
// ---> NODETOROOT --> ArrayList<Integer>
// node2Root(ArrayList<ArrayList<Integer>>edges,int child,int parent,int tofind)
// <--
// ---> LEVELS_TREE -->levels_Trees(ArrayList<HashSet<Integer>> edges, int
// child, int parent,int[]level,int currLevel) <--
// ---> K_THPARENT --> int k_thparent(int node,int[][]parent,int k) <---
// ---> TWOPOWERITHPARENTFILL --> void twoPowerIthParentFill(int[][]parent) <---
// -----> (INPUT OF INT,LONG ,STRING) -> int i() long l() String s()<-
// tempstart
static class InputReader {
private InputStream stream;
private byte[] buf = new byte[1024];
private int curChar;
private int numChars;
private SpaceCharFilter filter;
public InputReader(InputStream stream) {
this.stream = stream;
}
public int read() {
if (numChars == -1)
throw new InputMismatchException();
if (curChar >= numChars) {
curChar = 0;
try {
numChars = stream.read(buf);
} catch (IOException e) {
throw new InputMismatchException();
}
if (numChars <= 0)
return -1;
}
return buf[curChar++];
}
public int Int() {
int c = read();
while (isSpaceChar(c))
c = read();
int sgn = 1;
if (c == '-') {
sgn = -1;
c = read();
}
int res = 0;
do {
if (c < '0' || c > '9')
throw new InputMismatchException();
res *= 10;
res += c - '0';
c = read();
} while (!isSpaceChar(c));
return res * sgn;
}
public String String() {
int c = read();
while (isSpaceChar(c))
c = read();
StringBuilder res = new StringBuilder();
do {
res.appendCodePoint(c);
c = read();
} while (!isSpaceChar(c));
return res.toString();
}
public boolean isSpaceChar(int c) {
if (filter != null)
return filter.isSpaceChar(c);
return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
}
public String next() {
return String();
}
public interface SpaceCharFilter {
public boolean isSpaceChar(int ch);
}
}
static InputReader in = new InputReader(System.in);
public static int i() {
return in.Int();
}
public static long l() {
String s = in.String();
return Long.parseLong(s);
}
public static String s() {
return in.String();
}
public static int compareChar(char a, char b) {
return Character.compare(a, b);
}
public static int compareInt(int a, int b) {
return Integer.compare(a, b);
}
public static int compareLong(long a, long b) {
return Long.compare(a, b);
}
public static boolean compareString(String a, String b) {
return a.equals(b);
}
public static boolean compareStringBuilder(StringBuilder a, StringBuilder b) {
return a.equals(b);
}
} | Java | ["4\n\n5\n\n1 1 1 1 1\n\n5\n\n2 1 1 1 2\n\n6\n\n1 1 2 3 3 4\n\n6\n\n1 2 1 4 5 4"] | 1 second | ["2\n1\n2\n0"] | NoteIn the first test case, we can select $$$i=2$$$ and $$$x=2$$$ to form $$$[1, 2, 2, 1, 1]$$$. Then, we can select $$$i=3$$$ and $$$x=3$$$ to form $$$[1, 2, 3, 3, 1]$$$.In the second test case, we can select $$$i=3$$$ and $$$x=100$$$ to form $$$[2, 1, 100, 100, 2]$$$. | Java 11 | standard input | [
"constructive algorithms",
"greedy",
"implementation"
] | a91be662101762bcb2c58b8db9ff61e0 | Each test contains multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains an integer $$$n$$$ ($$$2 \le n \le 2 \cdot 10 ^ 5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10^9$$$) — elements of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10 ^ 5$$$ | 1,100 | For each test case, print the minimum number of operations needed. | standard output | |
PASSED | 0fe32bfa622dcb0b02e09a05274950ea | train_108.jsonl | 1650722700 | You are given an array $$$a$$$ of length $$$n$$$. We define the equality of the array as the number of indices $$$1 \le i \le n - 1$$$ such that $$$a_i = a_{i + 1}$$$. We are allowed to do the following operation: Select two integers $$$i$$$ and $$$x$$$ such that $$$1 \le i \le n - 1$$$ and $$$1 \le x \le 10^9$$$. Then, set $$$a_i$$$ and $$$a_{i + 1}$$$ to be equal to $$$x$$$. Find the minimum number of operations needed such that the equality of the array is less than or equal to $$$1$$$. | 256 megabytes |
import java.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintStream;
import java.util.*;
import java.io.UncheckedIOException;
import java.io.Closeable;
import java.io.Writer;
import java.io.OutputStreamWriter;
/*
@author : sanskarXrawat
@date : 4/18/2022
@time : 4:49 PM
*/
@SuppressWarnings("ALL")
public class Demo {
public static void main(String[] args) throws Throwable {
Thread thread = new Thread (null, new TaskAdapter (), "", 1 << 29);
thread.start ();
thread.join ();
}
static class TaskAdapter implements Runnable {
@Override
public void run() {
InputStream inputStream = System.in;
OutputStream outputStream = System.out;
FastInput in = new FastInput (inputStream);
FastOutput out = new FastOutput (outputStream);
Solution solver = new Solution ();
try {
solver.solve (1, in, out);
} catch (Exception e) {
e.printStackTrace ();
}
in.close ();
out.close ();
}
}
@SuppressWarnings("unused")
static class Solution {
static final Debug debug = new Debug (true);
static int[][] dirs = new int[][]{{-1, 0}, {0, 1}, {1, 0}, {0, -1}};
public void solve(int testNumber, FastInput in, FastOutput out) throws Exception {
int test=in.ri ();
outer: while (test-->0){
int n=in.ri ();
int[] arr=new int[n];
for(int i=0;i<n;i++){
arr[i]=in.ri ();
}
int mx=-1,mn=-1;
for(int i=1;i<n;i++){
if(arr[i]==arr[i-1]){
if(mn==-1){
mn=i;
}
mx=i;
}
}
if(mx==mn){
out.prtl (0);
}else{
out.prtl (Math.max (1,mx-mn-1));
}
}
}
}
static class FastOutput implements AutoCloseable, Closeable, Appendable {
private final StringBuilder cache = new StringBuilder (THRESHOLD * 2);
private static final int THRESHOLD = 32 << 10;
private final Writer os;
public FastOutput append(CharSequence csq) {
cache.append (csq);
return this;
}
public FastOutput append(CharSequence csq, int start, int end) {
cache.append (csq, start, end);
return this;
}
private void afterWrite() {
if (cache.length () < THRESHOLD) {
return;
}
flush ();
}
public FastOutput(Writer os) {
this.os = os;
}
public FastOutput(OutputStream os) {
this (new OutputStreamWriter (os));
}
public FastOutput append(char c) {
cache.append (c);
afterWrite ();
return this;
}
public FastOutput append(String c) {
cache.append (c);
afterWrite ();
return this;
}
public FastOutput println(String c) {
return append (c).println ();
}
public FastOutput println() {
return append ('\n');
}
final <T> void prt(T a) {
append (a + " ");
}
final <T> void prtl(T a) {
append (a + "\n");
}
public FastOutput flush() {
try {
os.append (cache);
os.flush ();
cache.setLength (0);
} catch (IOException e) {
throw new UncheckedIOException (e);
}
return this;
}
public void close() {
flush ();
try {
os.close ();
} catch (IOException e) {
throw new UncheckedIOException (e);
}
}
public String toString() {
return cache.toString ();
}
public FastOutput printf(String format, Object... args) {
return append (String.format (format, args));
}
private static int countDigits(long l) {
if (l >= 1000000000000000000L) return 19;
if (l >= 100000000000000000L) return 18;
if (l >= 10000000000000000L) return 17;
if (l >= 1000000000000000L) return 16;
if (l >= 100000000000000L) return 15;
if (l >= 10000000000000L) return 14;
if (l >= 1000000000000L) return 13;
if (l >= 100000000000L) return 12;
if (l >= 10000000000L) return 11;
if (l >= 1000000000L) return 10;
if (l >= 100000000L) return 9;
if (l >= 10000000L) return 8;
if (l >= 1000000L) return 7;
if (l >= 100000L) return 6;
if (l >= 10000L) return 5;
if (l >= 1000L) return 4;
if (l >= 100L) return 3;
if (l >= 10L) return 2;
return 1;
}
private static int countDigits(int l) {
if (l >= 1000000000) return 10;
if (l >= 100000000) return 9;
if (l >= 10000000) return 8;
if (l >= 1000000) return 7;
if (l >= 100000) return 6;
if (l >= 10000) return 5;
if (l >= 1000) return 4;
if (l >= 100) return 3;
if (l >= 10) return 2;
return 1;
}
}
static class FastInput {
private final StringBuilder defaultStringBuf = new StringBuilder (1 << 13);
private final ByteBuffer tokenBuf = new ByteBuffer ();
private final byte[] buf = new byte[1 << 13];
private SpaceCharFilter filter;
private final InputStream is;
private int bufOffset;
private int bufLen;
private int next;
private int ptr;
public FastInput(InputStream is) {
this.is = is;
}
private int read() {
while (bufLen == bufOffset) {
bufOffset = 0;
try {
bufLen = is.read (buf);
} catch (IOException e) {
bufLen = -1;
}
if (bufLen == -1) {
return -1;
}
}
return buf[bufOffset++];
}
public void skipBlank() {
while (next >= 0 && next <= 32) {
next = read ();
}
}
public String next() {
return readString ();
}
public int ri() {
return readInt ();
}
public int readInt() {
boolean rev = false;
skipBlank ();
if (next == '+' || next == '-') {
rev = next == '-';
next = read ();
}
int val = 0;
while (next >= '0' && next <= '9') {
val = val * 10 - next + '0';
next = read ();
}
return rev ? val : -val;
}
public long readLong() {
boolean rev = false;
skipBlank ();
if (next == '+' || next == '-') {
rev = next == '-';
next = read ();
}
long val = 0L;
while (next >= '0' && next <= '9') {
val = val * 10 - next + '0';
next = read ();
}
return rev ? val : -val;
}
public long rl() {
return readLong ();
}
public String readString(StringBuilder builder) {
skipBlank ();
while (next > 32) {
builder.append ((char) next);
next = read ();
}
return builder.toString ();
}
public String readString() {
defaultStringBuf.setLength (0);
return readString (defaultStringBuf);
}
public int rs(char[] data, int offset) {
return readString (data, offset);
}
public char[] rsc() {
return readString ().toCharArray ();
}
public int rs(char[] data) {
return rs (data, 0);
}
public int readString(char[] data, int offset) {
skipBlank ();
int originalOffset = offset;
while (next > 32) {
data[offset++] = (char) next;
next = read ();
}
return offset - originalOffset;
}
public char rc() {
return readChar ();
}
public char readChar() {
skipBlank ();
char c = (char) next;
next = read ();
return c;
}
public double rd() {
return nextDouble ();
}
public double nextDouble() {
int c = read ();
while (isSpaceChar (c))
c = read ();
int sgn = 1;
if (c == '-') {
sgn = -1;
c = read ();
}
double res = 0;
while (!isSpaceChar (c) && c != '.') {
if (c == 'e' || c == 'E')
return res * Math.pow (10, readInt ());
if (c < '0' || c > '9')
throw new InputMismatchException ();
res *= 10;
res += c - '0';
c = read ();
}
if (c == '.') {
c = read ();
double m = 1;
while (!isSpaceChar (c)) {
if (c == 'e' || c == 'E')
return res * Math.pow (10, readInt ());
if (c < '0' || c > '9')
throw new InputMismatchException ();
m /= 10;
res += (c - '0') * m;
c = read ();
}
}
return res * sgn;
}
public boolean isSpaceChar(int c) {
if (filter != null)
return filter.isSpaceChar (c);
return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
}
public interface SpaceCharFilter {
public boolean isSpaceChar(int ch);
}
public final int readByteUnsafe() {
if (ptr < bufLen) return buf[ptr++];
ptr = 0;
try {
bufLen = is.read (buf);
if (bufLen > 0) {
return buf[ptr++];
} else {
return -1;
}
} catch (IOException e) {
throw new UncheckedIOException (e);
}
}
public final int readByte() {
if (ptr < bufLen) return buf[ptr++];
ptr = 0;
try {
bufLen = is.read (buf);
if (bufLen > 0) {
return buf[ptr++];
} else {
throw new java.io.EOFException ();
}
} catch (IOException e) {
throw new UncheckedIOException (e);
}
}
public final String nextLine() {
tokenBuf.clear ();
for (int b = readByte (); b != '\n'; b = readByteUnsafe ()) {
if (b == -1) break;
tokenBuf.append (b);
}
return new String (tokenBuf.getRawBuf (), 0, tokenBuf.size ());
}
public final String nl() {
return nextLine ();
}
public final boolean hasNext() {
for (int b = readByteUnsafe (); b <= 32 || b >= 127; b = readByteUnsafe ()) {
if (b == -1) return false;
}
--ptr;
return true;
}
public void readArray(Object T) {
if (T instanceof int[]) {
int[] arr = (int[]) T;
for (int i = 0; i < arr.length; i++) {
arr[i] = ri ();
}
}
if (T instanceof long[]) {
long[] arr = (long[]) T;
for (int i = 0; i < arr.length; i++) {
arr[i] = rl ();
}
}
if (T instanceof double[]) {
double[] arr = (double[]) T;
for (int i = 0; i < arr.length; i++) {
arr[i] = rd ();
}
}
if (T instanceof char[]) {
char[] arr = (char[]) T;
for (int i = 0; i < arr.length; i++) {
arr[i] = readChar ();
}
}
if (T instanceof String[]) {
String[] arr = (String[]) T;
for (int i = 0; i < arr.length; i++) {
arr[i] = next ();
}
}
if (T instanceof int[][]) {
int[][] arr = (int[][]) T;
for (int i = 0; i < arr.length; i++) {
for (int j = 0; j < arr[0].length; j++) {
arr[i][j] = ri ();
}
}
}
if (T instanceof char[][]) {
char[][] arr = (char[][]) T;
for (int i = 0; i < arr.length; i++) {
for (int j = 0; j < arr[0].length; j++) {
arr[i][j] = readChar ();
}
}
}
if (T instanceof long[][]) {
long[][] arr = (long[][]) T;
for (int i = 0; i < arr.length; i++) {
for (int j = 0; j < arr[0].length; j++) {
arr[i][j] = rl ();
}
}
}
}
public final void close() {
try {
is.close ();
} catch (IOException e) {
throw new UncheckedIOException (e);
}
}
private static final class ByteBuffer {
private static final int DEFAULT_BUF_SIZE = 1 << 12;
private byte[] buf;
private int ptr = 0;
private ByteBuffer(int capacity) {
this.buf = new byte[capacity];
}
private ByteBuffer() {
this (DEFAULT_BUF_SIZE);
}
private ByteBuffer append(int b) {
if (ptr == buf.length) {
int newLength = buf.length << 1;
byte[] newBuf = new byte[newLength];
System.arraycopy (buf, 0, newBuf, 0, buf.length);
buf = newBuf;
}
buf[ptr++] = (byte) b;
return this;
}
private char[] toCharArray() {
char[] chs = new char[ptr];
for (int i = 0; i < ptr; i++) {
chs[i] = (char) buf[i];
}
return chs;
}
private byte[] getRawBuf() {
return buf;
}
private int size() {
return ptr;
}
private void clear() {
ptr = 0;
}
}
}
static class Debug {
private final boolean offline;
private final PrintStream out = System.err;
static int[] empty = new int[0];
public Debug(boolean enable) {
offline = enable && System.getSecurityManager () == null;
}
public Debug debug(String name, Object x) {
return debug (name, x, empty);
}
public Debug debug(String name, long x) {
if (offline) {
debug (name, "" + x);
}
return this;
}
public Debug debug(String name, String x) {
if (offline) {
out.printf ("%s=%s", name, x);
out.println ();
}
return this;
}
public Debug debug(String name, Object x, int... indexes) {
if (offline) {
if (x == null || !x.getClass ().isArray ()) {
out.append (name);
for (int i : indexes) {
out.printf ("[%d]", i);
}
out.append ("=").append ("" + x);
out.println ();
} else {
indexes = Arrays.copyOf (indexes, indexes.length + 1);
if (x instanceof byte[]) {
byte[] arr = (byte[]) x;
for (int i = 0; i < arr.length; i++) {
indexes[indexes.length - 1] = i;
debug (name, arr[i], indexes);
}
} else if (x instanceof short[]) {
short[] arr = (short[]) x;
for (int i = 0; i < arr.length; i++) {
indexes[indexes.length - 1] = i;
debug (name, arr[i], indexes);
}
} else if (x instanceof boolean[]) {
boolean[] arr = (boolean[]) x;
for (int i = 0; i < arr.length; i++) {
indexes[indexes.length - 1] = i;
debug (name, arr[i], indexes);
}
} else if (x instanceof char[]) {
char[] arr = (char[]) x;
for (int i = 0; i < arr.length; i++) {
indexes[indexes.length - 1] = i;
debug (name, arr[i], indexes);
}
} else if (x instanceof int[]) {
int[] arr = (int[]) x;
for (int i = 0; i < arr.length; i++) {
indexes[indexes.length - 1] = i;
debug (name, arr[i], indexes);
}
} else if (x instanceof float[]) {
float[] arr = (float[]) x;
for (int i = 0; i < arr.length; i++) {
indexes[indexes.length - 1] = i;
debug (name, arr[i], indexes);
}
} else if (x instanceof double[]) {
double[] arr = (double[]) x;
for (int i = 0; i < arr.length; i++) {
indexes[indexes.length - 1] = i;
debug (name, arr[i], indexes);
}
} else if (x instanceof long[]) {
long[] arr = (long[]) x;
for (int i = 0; i < arr.length; i++) {
indexes[indexes.length - 1] = i;
debug (name, arr[i], indexes);
}
} else {
Object[] arr = (Object[]) x;
for (int i = 0; i < arr.length; i++) {
indexes[indexes.length - 1] = i;
debug (name, arr[i], indexes);
}
}
}
}
return this;
}
}
} | Java | ["4\n\n5\n\n1 1 1 1 1\n\n5\n\n2 1 1 1 2\n\n6\n\n1 1 2 3 3 4\n\n6\n\n1 2 1 4 5 4"] | 1 second | ["2\n1\n2\n0"] | NoteIn the first test case, we can select $$$i=2$$$ and $$$x=2$$$ to form $$$[1, 2, 2, 1, 1]$$$. Then, we can select $$$i=3$$$ and $$$x=3$$$ to form $$$[1, 2, 3, 3, 1]$$$.In the second test case, we can select $$$i=3$$$ and $$$x=100$$$ to form $$$[2, 1, 100, 100, 2]$$$. | Java 11 | standard input | [
"constructive algorithms",
"greedy",
"implementation"
] | a91be662101762bcb2c58b8db9ff61e0 | Each test contains multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains an integer $$$n$$$ ($$$2 \le n \le 2 \cdot 10 ^ 5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10^9$$$) — elements of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10 ^ 5$$$ | 1,100 | For each test case, print the minimum number of operations needed. | standard output | |
PASSED | 29013595057794fdea53d5c65b962eea | train_108.jsonl | 1650722700 | You are given an array $$$a$$$ of length $$$n$$$. We define the equality of the array as the number of indices $$$1 \le i \le n - 1$$$ such that $$$a_i = a_{i + 1}$$$. We are allowed to do the following operation: Select two integers $$$i$$$ and $$$x$$$ such that $$$1 \le i \le n - 1$$$ and $$$1 \le x \le 10^9$$$. Then, set $$$a_i$$$ and $$$a_{i + 1}$$$ to be equal to $$$x$$$. Find the minimum number of operations needed such that the equality of the array is less than or equal to $$$1$$$. | 256 megabytes | import static java.lang.Math.max;
import static java.lang.Math.min;
import static java.lang.Math.abs;
import java.util.*;
import java.io.*;
import java.math.*;
public class C_Unequal_Array {
public static void main(String[] args) {
OutputStream outputStream = System.out;
PrintWriter out = new PrintWriter(outputStream);
FastReader f = new FastReader();
int t = f.nextInt();
while(t-- > 0){
solve(f, out);
}
out.close();
}
public static void solve(FastReader f, PrintWriter out) {
int n = f.nextInt();
int arr[] = f.nextArray(n);
boolean flag = true;
for(int i = 1; i < n; i++) {
if(arr[i] == arr[i-1]) {
flag = false;
}
}
if(flag) {
out.println(0);
return;
}
int i = 1;
while(arr[i] != arr[i-1]) {
i++;
}
int j = n-2;
while(arr[j] != arr[j+1]) {
j--;
}
if(j < i) {
out.println(0);
return;
}
int ans = j-i;
if(j == i) ans++;
out.println(ans);
}
public static void sort(int arr[]) {
ArrayList<Integer> al = new ArrayList<>();
for(int i: arr) {
al.add(i);
}
Collections.sort(al);
for(int i = 0; i < arr.length; i++) {
arr[i] = al.get(i);
}
}
public static void allDivisors(int n) {
for(int i = 1; i*i <= n; i++) {
if(n%i == 0) {
System.out.println(i + " ");
if(i != n/i) {
System.out.println(n/i + " ");
}
}
}
}
public static boolean isPrime(int n) {
if(n < 1) return false;
if(n == 2 || n == 3) return true;
if(n % 2 == 0 || n % 3 == 0) return false;
for(int i = 5; i*i <= n; i += 6) {
if(n % i == 0 || n % (i+2) == 0) {
return false;
}
}
return true;
}
public static int gcd(int a, int b) {
int dividend = a > b ? a : b;
int divisor = a < b ? a : b;
while(divisor > 0) {
int reminder = dividend % divisor;
dividend = divisor;
divisor = reminder;
}
return dividend;
}
public static int lcm(int a, int b) {
int lcm = gcd(a, b);
int hcf = (a * b) / lcm;
return hcf;
}
public static String sortString(String inputString) {
char tempArray[] = inputString.toCharArray();
Arrays.sort(tempArray);
return new String(tempArray);
}
static class FastReader {
BufferedReader br;
StringTokenizer st;
public FastReader() {
br = new BufferedReader(new InputStreamReader(System.in));
}
String next() {
while (st == null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(br.readLine());
}
catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt() {
return Integer.parseInt(next());
}
long nextLong() {
return Long.parseLong(next());
}
double nextDouble() {
return Double.parseDouble(next());
}
float nextFloat() {
return Float.parseFloat(next());
}
boolean nextBoolean() {
return Boolean.parseBoolean(next());
}
String nextLine() {
String str = "";
try {
str = br.readLine();
}
catch (IOException e) {
e.printStackTrace();
}
return str;
}
int[] nextArray(int n) {
int[] a = new int[n];
for(int i=0; i<n; i++) {
a[i] = nextInt();
}
return a;
}
}
}
/**
Dec Char Dec Char Dec Char Dec Char
--------- --------- --------- ----------
0 NUL (null) 32 SPACE 64 @ 96 `
1 SOH (start of heading) 33 ! 65 A 97 a
2 STX (start of text) 34 " 66 B 98 b
3 ETX (end of text) 35 # 67 C 99 c
4 EOT (end of transmission) 36 $ 68 D 100 d
5 ENQ (enquiry) 37 % 69 E 101 e
6 ACK (acknowledge) 38 & 70 F 102 f
7 BEL (bell) 39 ' 71 G 103 g
8 BS (backspace) 40 ( 72 H 104 h
9 TAB (horizontal tab) 41 ) 73 I 105 i
10 LF (NL line feed, new line) 42 * 74 J 106 j
11 VT (vertical tab) 43 + 75 K 107 k
12 FF (NP form feed, new page) 44 , 76 L 108 l
13 CR (carriage return) 45 - 77 M 109 m
14 SO (shift out) 46 . 78 N 110 n
15 SI (shift in) 47 / 79 O 111 o
16 DLE (data link escape) 48 0 80 P 112 p
17 DC1 (device control 1) 49 1 81 Q 113 q
18 DC2 (device control 2) 50 2 82 R 114 r
19 DC3 (device control 3) 51 3 83 S 115 s
20 DC4 (device control 4) 52 4 84 T 116 t
21 NAK (negative acknowledge) 53 5 85 U 117 u
22 SYN (synchronous idle) 54 6 86 V 118 v
23 ETB (end of trans. block) 55 7 87 W 119 w
24 CAN (cancel) 56 8 88 X 120 x
25 EM (end of medium) 57 9 89 Y 121 y
26 SUB (substitute) 58 : 90 Z 122 z
27 ESC (escape) 59 ; 91 [ 123 {
28 FS (file separator) 60 < 92 \ 124 |
29 GS (group separator) 61 = 93 ] 125 }
30 RS (record separator) 62 > 94 ^ 126 ~
31 US (unit separator) 63 ? 95 _ 127 DEL
*/
| Java | ["4\n\n5\n\n1 1 1 1 1\n\n5\n\n2 1 1 1 2\n\n6\n\n1 1 2 3 3 4\n\n6\n\n1 2 1 4 5 4"] | 1 second | ["2\n1\n2\n0"] | NoteIn the first test case, we can select $$$i=2$$$ and $$$x=2$$$ to form $$$[1, 2, 2, 1, 1]$$$. Then, we can select $$$i=3$$$ and $$$x=3$$$ to form $$$[1, 2, 3, 3, 1]$$$.In the second test case, we can select $$$i=3$$$ and $$$x=100$$$ to form $$$[2, 1, 100, 100, 2]$$$. | Java 11 | standard input | [
"constructive algorithms",
"greedy",
"implementation"
] | a91be662101762bcb2c58b8db9ff61e0 | Each test contains multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains an integer $$$n$$$ ($$$2 \le n \le 2 \cdot 10 ^ 5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10^9$$$) — elements of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10 ^ 5$$$ | 1,100 | For each test case, print the minimum number of operations needed. | standard output | |
PASSED | 78607c76ef1185f233df1c815789822e | train_108.jsonl | 1650722700 | You are given an array $$$a$$$ of length $$$n$$$. We define the equality of the array as the number of indices $$$1 \le i \le n - 1$$$ such that $$$a_i = a_{i + 1}$$$. We are allowed to do the following operation: Select two integers $$$i$$$ and $$$x$$$ such that $$$1 \le i \le n - 1$$$ and $$$1 \le x \le 10^9$$$. Then, set $$$a_i$$$ and $$$a_{i + 1}$$$ to be equal to $$$x$$$. Find the minimum number of operations needed such that the equality of the array is less than or equal to $$$1$$$. | 256 megabytes | import java.util.*;
public class a11 {
public static void main(String [] args) {
Scanner s=new Scanner (System.in);
int t=s.nextInt();
while(t-->0) {
int n=s.nextInt();
int []f=new int [n];
for(int i=0;i<n;i++)
f[i]=s.nextInt();
int min=-1,max=-1;
for(int i=1;i<n;i++) {
if(f[i-1]==f[i]) {
if(min==-1)
min=i;
max=i;
}
}
if(min==max)System.out.println(0);
else System.out.println(Math.max(1,max-min-1));
}
}
}
| Java | ["4\n\n5\n\n1 1 1 1 1\n\n5\n\n2 1 1 1 2\n\n6\n\n1 1 2 3 3 4\n\n6\n\n1 2 1 4 5 4"] | 1 second | ["2\n1\n2\n0"] | NoteIn the first test case, we can select $$$i=2$$$ and $$$x=2$$$ to form $$$[1, 2, 2, 1, 1]$$$. Then, we can select $$$i=3$$$ and $$$x=3$$$ to form $$$[1, 2, 3, 3, 1]$$$.In the second test case, we can select $$$i=3$$$ and $$$x=100$$$ to form $$$[2, 1, 100, 100, 2]$$$. | Java 11 | standard input | [
"constructive algorithms",
"greedy",
"implementation"
] | a91be662101762bcb2c58b8db9ff61e0 | Each test contains multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains an integer $$$n$$$ ($$$2 \le n \le 2 \cdot 10 ^ 5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10^9$$$) — elements of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10 ^ 5$$$ | 1,100 | For each test case, print the minimum number of operations needed. | standard output | |
PASSED | f0de444805643f108ad7da3b5b3f6945 | train_108.jsonl | 1650722700 | You are given an array $$$a$$$ of length $$$n$$$. We define the equality of the array as the number of indices $$$1 \le i \le n - 1$$$ such that $$$a_i = a_{i + 1}$$$. We are allowed to do the following operation: Select two integers $$$i$$$ and $$$x$$$ such that $$$1 \le i \le n - 1$$$ and $$$1 \le x \le 10^9$$$. Then, set $$$a_i$$$ and $$$a_{i + 1}$$$ to be equal to $$$x$$$. Find the minimum number of operations needed such that the equality of the array is less than or equal to $$$1$$$. | 256 megabytes | import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.StringTokenizer;
public class UnequalArray_1672_C {
static class FastScanner {
BufferedReader bf = new BufferedReader(new InputStreamReader(System.in));
StringTokenizer st = new StringTokenizer("");
String next() {
while(!st.hasMoreTokens()) {
try {
st = new StringTokenizer(bf.readLine());
} catch(IOException e) { }
}
return st.nextToken();
}
int nextInt() {
return Integer.parseInt(next());
}
long nextLong() {
return Long.parseLong(next());
}
char nextChar() {
return next().charAt(0);
}
String nextLine() throws IOException{
return bf.readLine().trim();
}
}
public static void main(String[] args) {
FastScanner sc = new FastScanner();
PrintWriter out = new PrintWriter(System.out);
int t = sc.nextInt();
while(t-- > 0) {
int n = sc.nextInt();
int[] arr = new int[n];
for(int i=0;i<n;i++) arr[i] = sc.nextInt();
int first_outer = -1;
int last_inner = -1;
for(int i=0;i<n-1;i++) {
if(arr[i] == arr[i+1]) {
first_outer = i+1;
break;
}
}
for(int i=n-1;i>0;i--) {
if(arr[i] == arr[i-1]) {
last_inner = i-1;
break;
}
}
if(first_outer > last_inner || first_outer == -1) {
out.println(0);
} else if(first_outer == last_inner) {
out.println(1);
} else {
out.println(last_inner - first_outer);
}
}
out.close();
}
}
| Java | ["4\n\n5\n\n1 1 1 1 1\n\n5\n\n2 1 1 1 2\n\n6\n\n1 1 2 3 3 4\n\n6\n\n1 2 1 4 5 4"] | 1 second | ["2\n1\n2\n0"] | NoteIn the first test case, we can select $$$i=2$$$ and $$$x=2$$$ to form $$$[1, 2, 2, 1, 1]$$$. Then, we can select $$$i=3$$$ and $$$x=3$$$ to form $$$[1, 2, 3, 3, 1]$$$.In the second test case, we can select $$$i=3$$$ and $$$x=100$$$ to form $$$[2, 1, 100, 100, 2]$$$. | Java 11 | standard input | [
"constructive algorithms",
"greedy",
"implementation"
] | a91be662101762bcb2c58b8db9ff61e0 | Each test contains multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains an integer $$$n$$$ ($$$2 \le n \le 2 \cdot 10 ^ 5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10^9$$$) — elements of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10 ^ 5$$$ | 1,100 | For each test case, print the minimum number of operations needed. | standard output | |
PASSED | 2f077121652b754750a488f760628a32 | train_108.jsonl | 1650722700 | You are given an array $$$a$$$ of length $$$n$$$. We define the equality of the array as the number of indices $$$1 \le i \le n - 1$$$ such that $$$a_i = a_{i + 1}$$$. We are allowed to do the following operation: Select two integers $$$i$$$ and $$$x$$$ such that $$$1 \le i \le n - 1$$$ and $$$1 \le x \le 10^9$$$. Then, set $$$a_i$$$ and $$$a_{i + 1}$$$ to be equal to $$$x$$$. Find the minimum number of operations needed such that the equality of the array is less than or equal to $$$1$$$. | 256 megabytes |
import java.io.*;
import java.util.*;
public class ProblemA{
static long mod = 1000000007L;
// map.put(a[i],map.getOrDefault(a[i],0)+1);
// map.putIfAbsent;
static MyScanner sc = new MyScanner();
//<----------------------------------------------WRITE HERE------------------------------------------->
static void solve(){
int n = sc.nextInt();
int[] a= sc.readIntArray(n);
int fi = -1;
int li = -1;
int ans =0;
for(int i=0;i<n-1;i++) {
if(a[i] == a[i+1]) {
if(li == -1) {
fi = i;
li = i;
}
else fi = i;
}
}
if(fi == -1) {
out.println(0);
return;
}
// if(fi == li) {
// out.println(n-fi-1);
// return;
// }
// for(int i=fi;i<n-2;i++) {
// if(i+2>li) break;
// int v = Math.max(a[i+1], a[i+2])+1;
// a[i+1] = v;
// a[i+2] = v;
// ans++;
//
// }
if(fi-li>1)out.println((fi-li-1));
else out.println(fi-li);
}
//<----------------------------------------------WRITE HERE------------------------------------------->
static void swap(char[] a,int i,int j) {
char temp = a[i];
a[i] = a[j];
a[j] = temp;
}
static int LowerBound(int a[], int x) {
int l = -1, r = a.length;
while (l + 1 < r) {
int m = (l + r) >>> 1;
if (a[m] >= x)
r = m;
else
l = m;
}
return r;
}
static int UpperBound(int a[], int x) {
int l = -1, r = a.length;
while (l + 1 < r) {
int m = (l + r) >>> 1;
if (a[m] <= x)
l = m;
else
r = m;
}
return l + 1;
}
static String sortString(String inputString) {
char tempArray[] = inputString.toCharArray();
Arrays.sort(tempArray);
return new String(tempArray);
}
static boolean isPali(String str){
int i = 0;
int j = str.length()-1;
while(i<j){
if(str.charAt(i)!=str.charAt(j)){
return false;
}
i++;
j--;
}
return true;
}
static void priArr(int[] a) {
for(int i=0;i<a.length;i++) {
out.print(a[i] + " ");
}
out.println();
}
static long gcd(long a,long b){
if(b==0) return a;
return gcd(b,a%b);
}
static String reverse(String str){
char arr[] = str.toCharArray();
int i = 0;
int j = arr.length-1;
while(i<j){
char temp = arr[i];
arr[i] = arr[j];
arr[j] = temp;
i++;
j--;
}
String st = new String(arr);
return st;
}
static void reverse(int[] arr){
int i = 0;
int j = arr.length-1;
while(i<j){
int temp = arr[i];
arr[i] = arr[j];
arr[j] = temp;
i++;
j--;
}
}
static boolean isprime(int n){
if(n==1) return false;
if(n==3 || n==2) return true;
if(n%2==0 || n%3==0) return false;
for(int i = 5;i*i<=n;i+= 6){
if(n%i== 0 || n%(i+2)==0){
return false;
}
}
return true;
}
static class Pair implements Comparable<Pair>{
int val;
int ind;
Pair(int v,int f){
val = v;
ind = f;
}
public int compareTo(Pair p){
return this.ind - p.ind;
}
}
public static void main(String[] args) {
out = new PrintWriter(new BufferedOutputStream(System.out));
int t = sc.nextInt();
// int t= 1;
while(t-- >0){
solve();
}
// Stop writing your solution here. -------------------------------------
out.close();
}
//-----------PrintWriter for faster output---------------------------------
public static PrintWriter out;
//-----------MyScanner class for faster input----------
public static class MyScanner {
BufferedReader br;
StringTokenizer st;
public MyScanner() {
br = new BufferedReader(new InputStreamReader(System.in));
}
String next() {
while (st == null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
public int[] readIntArray(int n){
int arr[] = new int[n];
for(int i = 0;i<n;i++){
arr[i] = Integer.parseInt(next());
}
return arr;
}
int[] reverse(int arr[]){
int n= arr.length;
int i = 0;
int j = n-1;
while(i<j){
int temp = arr[i];
arr[i] = arr[j];
arr[j] = temp;
j--;i++;
}
return arr;
}
long[] readLongArray(int n){
long arr[] = new long[n];
for(int i = 0;i<n;i++){
arr[i] = Long.parseLong(next());
}
return arr;
}
int nextInt() {
return Integer.parseInt(next());
}
long nextLong() {
return Long.parseLong(next());
}
double nextDouble() {
return Double.parseDouble(next());
}
String nextLine() {
String str = "";
try {
str = br.readLine();
} catch (IOException e) {
e.printStackTrace();
}
return str;
}
}
private static void sort(int[] arr) {
List<Integer> list = new ArrayList<>();
for (int i=0; i<arr.length; i++){
list.add(arr[i]);
}
Collections.sort(list); // collections.sort uses nlogn in backend
for (int i = 0; i < arr.length; i++){
arr[i] = list.get(i);
}
}
} | Java | ["4\n\n5\n\n1 1 1 1 1\n\n5\n\n2 1 1 1 2\n\n6\n\n1 1 2 3 3 4\n\n6\n\n1 2 1 4 5 4"] | 1 second | ["2\n1\n2\n0"] | NoteIn the first test case, we can select $$$i=2$$$ and $$$x=2$$$ to form $$$[1, 2, 2, 1, 1]$$$. Then, we can select $$$i=3$$$ and $$$x=3$$$ to form $$$[1, 2, 3, 3, 1]$$$.In the second test case, we can select $$$i=3$$$ and $$$x=100$$$ to form $$$[2, 1, 100, 100, 2]$$$. | Java 11 | standard input | [
"constructive algorithms",
"greedy",
"implementation"
] | a91be662101762bcb2c58b8db9ff61e0 | Each test contains multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains an integer $$$n$$$ ($$$2 \le n \le 2 \cdot 10 ^ 5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10^9$$$) — elements of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10 ^ 5$$$ | 1,100 | For each test case, print the minimum number of operations needed. | standard output | |
PASSED | 1cdd9e573cbbc775abde1cbb8bb96cb7 | train_108.jsonl | 1650722700 | You are given an array $$$a$$$ of length $$$n$$$. We define the equality of the array as the number of indices $$$1 \le i \le n - 1$$$ such that $$$a_i = a_{i + 1}$$$. We are allowed to do the following operation: Select two integers $$$i$$$ and $$$x$$$ such that $$$1 \le i \le n - 1$$$ and $$$1 \le x \le 10^9$$$. Then, set $$$a_i$$$ and $$$a_{i + 1}$$$ to be equal to $$$x$$$. Find the minimum number of operations needed such that the equality of the array is less than or equal to $$$1$$$. | 256 megabytes | import java.util.Scanner;
public class UnequalArray {
public static void main (String[] args) {
Scanner sc = new Scanner(System.in);
int t = sc.nextInt();
while (t-- > 0) {
int n = sc.nextInt();
int[] a = new int[n];
for (int i = 0; i < n; i++) {
a[i] = sc.nextInt();
}
int first = 0;
int second = 0;
boolean flag = false;
for (int i = 0; i < n - 1; i++) {
if (a[i] == a[i + 1]) {
first = i + 1;
flag = true;
break;
}
}
for (int i = n - 1; i >= 1; i--) {
if (a[i] == a[i - 1]) {
second = i - 1;
flag = true;
break;
}
}
if (flag) {
if (second < first) {
System.out.println(0);
} else if (first == second) {
System.out.println(1);
} else {
System.out.println(second - first);
}
} else {
System.out.println(0);
}
}
}
} | Java | ["4\n\n5\n\n1 1 1 1 1\n\n5\n\n2 1 1 1 2\n\n6\n\n1 1 2 3 3 4\n\n6\n\n1 2 1 4 5 4"] | 1 second | ["2\n1\n2\n0"] | NoteIn the first test case, we can select $$$i=2$$$ and $$$x=2$$$ to form $$$[1, 2, 2, 1, 1]$$$. Then, we can select $$$i=3$$$ and $$$x=3$$$ to form $$$[1, 2, 3, 3, 1]$$$.In the second test case, we can select $$$i=3$$$ and $$$x=100$$$ to form $$$[2, 1, 100, 100, 2]$$$. | Java 11 | standard input | [
"constructive algorithms",
"greedy",
"implementation"
] | a91be662101762bcb2c58b8db9ff61e0 | Each test contains multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains an integer $$$n$$$ ($$$2 \le n \le 2 \cdot 10 ^ 5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10^9$$$) — elements of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10 ^ 5$$$ | 1,100 | For each test case, print the minimum number of operations needed. | standard output | |
PASSED | da8580e2ecfda7f40e7c49204a3d42d1 | train_108.jsonl | 1650722700 | You are given an array $$$a$$$ of length $$$n$$$. We define the equality of the array as the number of indices $$$1 \le i \le n - 1$$$ such that $$$a_i = a_{i + 1}$$$. We are allowed to do the following operation: Select two integers $$$i$$$ and $$$x$$$ such that $$$1 \le i \le n - 1$$$ and $$$1 \le x \le 10^9$$$. Then, set $$$a_i$$$ and $$$a_{i + 1}$$$ to be equal to $$$x$$$. Find the minimum number of operations needed such that the equality of the array is less than or equal to $$$1$$$. | 256 megabytes | import java.io.*;
import java.util.*;
import java.util.HashSet;
import java.util.StringTokenizer;
public class cf799 {
static class FastReader{
BufferedReader br;
StringTokenizer st;
public FastReader(){
br=new BufferedReader(new InputStreamReader(System.in));
}
String next(){
while(st==null || !st.hasMoreTokens()){
try {
st=new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt(){
return Integer.parseInt(next());
}
long nextLong(){
return Long.parseLong(next());
}
double nextDouble(){
return Double.parseDouble(next());
}
String nextLine(){
String str="";
try {
str=br.readLine().trim();
} catch (Exception e) {
e.printStackTrace();
}
return str;
}
}
static class FastWriter {
private final BufferedWriter bw;
public FastWriter() {
this.bw = new BufferedWriter(new OutputStreamWriter(System.out));
}
public void print(Object object) throws IOException {
bw.append("" + object);
}
public void println(Object object) throws IOException {
print(object);
bw.append("\n");
}
public void close() throws IOException {
bw.close();
}
}
public static void main(String[] args) {
try {
FastReader sc = new FastReader();
FastWriter out = new FastWriter();
//Scanner sc=new Scanner(System.in);
int t = sc.nextInt();
while (t-- > 0) {
int n = sc.nextInt();
int[] arr = new int[n];
for(int i=0;i<n;i++){
arr[i]=sc.nextInt();
}
int ff=-1;
int ss=-1;
for(int i=0;i<n-1;i++){
if(arr[i]==arr[i+1]) {
if (ff == -1) {
ff=i;
}
ss=i;
}
}
if(ff==ss){
System.out.println(0);
continue;
}
System.out.println(Math.max(1,ss-ff-1));
}
}catch (Exception e) {
return;
}
}
private static void ff(ArrayList<Integer> ar , ArrayList<Integer> ar2) {
for (Integer integer : ar2) {
int count = 0;
int num = integer;
while (num % 2 == 0) {
count++;
num /= 2;
}
ar.add(count);
}
}
}
| Java | ["4\n\n5\n\n1 1 1 1 1\n\n5\n\n2 1 1 1 2\n\n6\n\n1 1 2 3 3 4\n\n6\n\n1 2 1 4 5 4"] | 1 second | ["2\n1\n2\n0"] | NoteIn the first test case, we can select $$$i=2$$$ and $$$x=2$$$ to form $$$[1, 2, 2, 1, 1]$$$. Then, we can select $$$i=3$$$ and $$$x=3$$$ to form $$$[1, 2, 3, 3, 1]$$$.In the second test case, we can select $$$i=3$$$ and $$$x=100$$$ to form $$$[2, 1, 100, 100, 2]$$$. | Java 11 | standard input | [
"constructive algorithms",
"greedy",
"implementation"
] | a91be662101762bcb2c58b8db9ff61e0 | Each test contains multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains an integer $$$n$$$ ($$$2 \le n \le 2 \cdot 10 ^ 5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10^9$$$) — elements of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10 ^ 5$$$ | 1,100 | For each test case, print the minimum number of operations needed. | standard output | |
PASSED | 5e44aaece775ded1e464513fffefe4c7 | train_108.jsonl | 1650722700 | You are given an array $$$a$$$ of length $$$n$$$. We define the equality of the array as the number of indices $$$1 \le i \le n - 1$$$ such that $$$a_i = a_{i + 1}$$$. We are allowed to do the following operation: Select two integers $$$i$$$ and $$$x$$$ such that $$$1 \le i \le n - 1$$$ and $$$1 \le x \le 10^9$$$. Then, set $$$a_i$$$ and $$$a_{i + 1}$$$ to be equal to $$$x$$$. Find the minimum number of operations needed such that the equality of the array is less than or equal to $$$1$$$. | 256 megabytes | import java.util.*;
import java.io.*;
// you can compare with output.txt and expected out
public class RoundGlobal20C {
MyPrintWriter out;
MyScanner in;
// final static long FIXED_RANDOM;
// static {
// FIXED_RANDOM = System.currentTimeMillis();
// }
final static String IMPOSSIBLE = "IMPOSSIBLE";
final static String POSSIBLE = "POSSIBLE";
final static String YES = "YES";
final static String NO = "NO";
private void initIO(boolean isFileIO) {
if (System.getProperty("ONLINE_JUDGE") == null && isFileIO) {
try{
in = new MyScanner(new FileInputStream("input.txt"));
out = new MyPrintWriter(new FileOutputStream("output.txt"));
}
catch(FileNotFoundException e){
e.printStackTrace();
}
}
else{
in = new MyScanner(System.in);
out = new MyPrintWriter(new BufferedOutputStream(System.out));
}
}
public static void main(String[] args){
// Scanner in = new Scanner(new BufferedReader(new InputStreamReader(System.in)));
RoundGlobal20C sol = new RoundGlobal20C();
sol.run();
}
private void run() {
boolean isDebug = false;
boolean isFileIO = true;
boolean hasMultipleTests = true;
initIO(isFileIO);
int t = hasMultipleTests? in.nextInt() : 1;
for (int i = 1; i <= t; ++i) {
int n = in.nextInt();
int[] a = in.nextIntArray(n);
if(isDebug){
out.printf("Test %d\n", i);
}
int ans = solve(a);
out.println(ans);
if(isDebug)
out.flush();
}
in.close();
out.close();
}
private int solve(int[] a) {
int n = a.length;
int first = -1;
for(int i=0; i<n-1; i++) {
if(a[i] == a[i+1]) {
first = i;
break;
}
}
if(first == -1)
return 0;
int last = -1;
for(int i=n-2; i>=0; i--) {
if(a[i] == a[i+1]) {
last = i;
break;
}
}
if(first == last)
return 0;
// a[first] = a[first+1] ........ = a[last] = a[last+1]
// need to on first+1 to last-1
// last-1-first-1+1 = last-first-1
if(last == first+1)
return 1;
else
return last-first-1;
}
public static class MyScanner {
BufferedReader br;
StringTokenizer st;
// 32768?
public MyScanner(InputStream is, int bufferSize) {
br = new BufferedReader(new InputStreamReader(is), bufferSize);
}
public MyScanner(InputStream is) {
br = new BufferedReader(new InputStreamReader(is));
// br = new BufferedReader(new InputStreamReader(System.in));
// br = new BufferedReader(new InputStreamReader(new FileInputStream("input.txt")));
}
public void close() {
try {
br.close();
} catch (IOException e) {
e.printStackTrace();
}
}
String next() {
while (st == null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt() {
return Integer.parseInt(next());
}
long nextLong() {
return Long.parseLong(next());
}
double nextDouble() {
return Double.parseDouble(next());
}
String nextLine(){
String str = "";
try {
str = br.readLine();
} catch (IOException e) {
e.printStackTrace();
}
return str;
}
int[][] nextTreeEdges(int n, int offset){
int[][] e = new int[n-1][2];
for(int i=0; i<n-1; i++){
e[i][0] = nextInt()+offset;
e[i][1] = nextInt()+offset;
}
return e;
}
int[][] nextMatrix(int n, int m) {
return nextMatrix(n, m, 0);
}
int[][] nextMatrix(int n, int m, int offset) {
int[][] mat = new int[n][m];
for(int i=0; i<n; i++) {
for(int j=0; j<m; j++) {
mat[i][j] = nextInt()+offset;
}
}
return mat;
}
int[][] nextPairs(int n){
return nextPairs(n, 0);
}
int[][] nextPairs(int n, int offset) {
int[][] xy = new int[2][n];
for(int i=0; i<n; i++) {
xy[0][i] = nextInt() + offset;
xy[1][i] = nextInt() + offset;
}
return xy;
}
int[][] nextGraphEdges(){
return nextGraphEdges(0);
}
int[][] nextGraphEdges(int offset) {
int m = nextInt();
int[][] e = new int[m][2];
for(int i=0; i<m; i++){
e[i][0] = nextInt()+offset;
e[i][1] = nextInt()+offset;
}
return e;
}
int[] nextIntArray(int len) {
return nextIntArray(len, 0);
}
int[] nextIntArray(int len, int offset){
int[] a = new int[len];
for(int j=0; j<len; j++)
a[j] = nextInt()+offset;
return a;
}
long[] nextLongArray(int len) {
return nextLongArray(len, 0);
}
long[] nextLongArray(int len, int offset){
long[] a = new long[len];
for(int j=0; j<len; j++)
a[j] = nextLong()+offset;
return a;
}
}
public static class MyPrintWriter extends PrintWriter{
public MyPrintWriter(OutputStream os) {
super(os);
}
public void printlnAns(boolean ans) {
if(ans)
println(YES);
else
println(NO);
}
public void printAns(long[] arr){
if(arr != null && arr.length > 0){
print(arr[0]);
for(int i=1; i<arr.length; i++){
print(" ");
print(arr[i]);
}
}
}
public void printlnAns(long[] arr){
printAns(arr);
println();
}
public void printAns(int[] arr){
if(arr != null && arr.length > 0){
print(arr[0]);
for(int i=1; i<arr.length; i++){
print(" ");
print(arr[i]);
}
}
}
public void printlnAns(int[] arr){
printAns(arr);
println();
}
public <T> void printAns(ArrayList<T> arr){
if(arr != null && arr.size() > 0){
print(arr.get(0));
for(int i=1; i<arr.size(); i++){
print(" ");
print(arr.get(i));
}
}
}
public <T> void printlnAns(ArrayList<T> arr){
printAns(arr);
println();
}
public void printAns(int[] arr, int add){
if(arr != null && arr.length > 0){
print(arr[0]+add);
for(int i=1; i<arr.length; i++){
print(" ");
print(arr[i]+add);
}
}
}
public void printlnAns(int[] arr, int add){
printAns(arr, add);
println();
}
public void printAns(ArrayList<Integer> arr, int add) {
if(arr != null && arr.size() > 0){
print(arr.get(0)+add);
for(int i=1; i<arr.size(); i++){
print(" ");
print(arr.get(i)+add);
}
}
}
public void printlnAns(ArrayList<Integer> arr, int add){
printAns(arr, add);
println();
}
public void printlnAnsSplit(long[] arr, int split){
if(arr != null){
for(int i=0; i<arr.length; i+=split){
print(arr[i]);
for(int j=i+1; j<i+split; j++){
print(" ");
print(arr[j]);
}
println();
}
}
}
public void printlnAnsSplit(int[] arr, int split){
if(arr != null){
for(int i=0; i<arr.length; i+=split){
print(arr[i]);
for(int j=i+1; j<i+split; j++){
print(" ");
print(arr[j]);
}
println();
}
}
}
public <T> void printlnAnsSplit(ArrayList<T> arr, int split){
if(arr != null && !arr.isEmpty()){
for(int i=0; i<arr.size(); i+=split){
print(arr.get(i));
for(int j=i+1; j<i+split; j++){
print(" ");
print(arr.get(j));
}
println();
}
}
}
}
static private void permutateAndSort(int[] a) {
int n = a.length;
Random R = new Random(System.currentTimeMillis());
for(int i=0; i<n; i++) {
int t = R.nextInt(n-i);
int temp = a[n-1-i];
a[n-1-i] = a[t];
a[t] = temp;
}
Arrays.sort(a);
}
static private int[][] constructChildren(int n, int[] parent, int parentRoot){
int[][] childrens = new int[n][];
int[] numChildren = new int[n];
for(int i=0; i<parent.length; i++) {
if(parent[i] != parentRoot)
numChildren[parent[i]]++;
}
for(int i=0; i<n; i++) {
childrens[i] = new int[numChildren[i]];
}
int[] idx = new int[n];
for(int i=0; i<parent.length; i++) {
if(parent[i] != parentRoot)
childrens[parent[i]][idx[parent[i]]++] = i;
}
return childrens;
}
static private int[][][] constructDirectedNeighborhood(int n, int[][] e){
int[] inDegree = new int[n];
int[] outDegree = new int[n];
for(int i=0; i<e.length; i++) {
int u = e[i][0];
int v = e[i][1];
outDegree[u]++;
inDegree[v]++;
}
int[][] inNeighbors = new int[n][];
int[][] outNeighbors = new int[n][];
for(int i=0; i<n; i++) {
inNeighbors[i] = new int[inDegree[i]];
outNeighbors[i] = new int[outDegree[i]];
}
int[] inIdx = new int[n];
int[] outIdx = new int[n];
for(int i=0; i<e.length; i++) {
int u = e[i][0];
int v = e[i][1];
outNeighbors[u][outIdx[u]++] = v;
inNeighbors[v][inIdx[v]++] = u;
}
return new int[][][] {inNeighbors, outNeighbors};
}
static private int[][] constructNeighborhood(int n, int[][] e) {
int[] degree = new int[n];
for(int i=0; i<e.length; i++) {
int u = e[i][0];
int v = e[i][1];
degree[u]++;
degree[v]++;
}
int[][] neighbors = new int[n][];
for(int i=0; i<n; i++)
neighbors[i] = new int[degree[i]];
int[] idx = new int[n];
for(int i=0; i<e.length; i++) {
int u = e[i][0];
int v = e[i][1];
neighbors[u][idx[u]++] = v;
neighbors[v][idx[v]++] = u;
}
return neighbors;
}
static private void makeDotUndirected(int[][] e) {
MyPrintWriter out2 = null;
try {
out2 = new MyPrintWriter(new FileOutputStream("graph.dot"));
} catch (FileNotFoundException e1) {
// TODO Auto-generated catch block
e1.printStackTrace();
}
out2.println("strict graph {");
for(int i=0; i<e.length; i++){
out2.println(e[i][0] + "--" + e[i][1] + ";");
}
out2.println("}");
out2.close();
}
static private void makeDotDirected(int[][] e) {
MyPrintWriter out2 = null;
try {
out2 = new MyPrintWriter(new FileOutputStream("graph.dot"));
} catch (FileNotFoundException e1) {
// TODO Auto-generated catch block
e1.printStackTrace();
}
out2.println("strict digraph {");
for(int i=0; i<e.length; i++){
out2.println(e[i][0] + "->" + e[i][1] + ";");
}
out2.println("}");
out2.close();
}
}
| Java | ["4\n\n5\n\n1 1 1 1 1\n\n5\n\n2 1 1 1 2\n\n6\n\n1 1 2 3 3 4\n\n6\n\n1 2 1 4 5 4"] | 1 second | ["2\n1\n2\n0"] | NoteIn the first test case, we can select $$$i=2$$$ and $$$x=2$$$ to form $$$[1, 2, 2, 1, 1]$$$. Then, we can select $$$i=3$$$ and $$$x=3$$$ to form $$$[1, 2, 3, 3, 1]$$$.In the second test case, we can select $$$i=3$$$ and $$$x=100$$$ to form $$$[2, 1, 100, 100, 2]$$$. | Java 11 | standard input | [
"constructive algorithms",
"greedy",
"implementation"
] | a91be662101762bcb2c58b8db9ff61e0 | Each test contains multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains an integer $$$n$$$ ($$$2 \le n \le 2 \cdot 10 ^ 5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10^9$$$) — elements of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10 ^ 5$$$ | 1,100 | For each test case, print the minimum number of operations needed. | standard output | |
PASSED | 786f4c11f3e1fc23ca3f7cfb9a39520d | train_108.jsonl | 1650722700 | You are given an array $$$a$$$ of length $$$n$$$. We define the equality of the array as the number of indices $$$1 \le i \le n - 1$$$ such that $$$a_i = a_{i + 1}$$$. We are allowed to do the following operation: Select two integers $$$i$$$ and $$$x$$$ such that $$$1 \le i \le n - 1$$$ and $$$1 \le x \le 10^9$$$. Then, set $$$a_i$$$ and $$$a_{i + 1}$$$ to be equal to $$$x$$$. Find the minimum number of operations needed such that the equality of the array is less than or equal to $$$1$$$. | 256 megabytes | import java.io.*;
import java.util.*;
public class Main {
public static void main(String[]args) throws Exception{
Scanner sc = new Scanner(System.in) ;
PrintWriter pw = new PrintWriter(System.out) ;
int t=sc.nextInt();
while(t-->0){
int n =sc.nextInt() ;
int [] arr = sc.nextIntArray(n) ;
int l = -1 ;
int r= -1 ;
for(int i=0 ; i<arr.length - 1 ; i++){
if(arr[i]==arr[i+1]){
if(l==-1){
l=i ;
}
r=i ;
}
}
int res ;
if(l==r){
res = 0 ;
}
else{
res = Math.max(1,r-l-1) ;
}
pw.println(res) ;
}
pw.flush();
pw.close();
}
static class Scanner
{
StringTokenizer st;
BufferedReader br;
public Scanner(InputStream s){ br = new BufferedReader(new InputStreamReader(s));}
public String next() throws IOException
{
while (st == null || !st.hasMoreTokens())
st = new StringTokenizer(br.readLine());
return st.nextToken();
}
public int nextInt() throws IOException {return Integer.parseInt(next());}
public long nextLong() throws IOException {return Long.parseLong(next());}
public String nextLine() throws IOException {return br.readLine();}
public double nextDouble() throws IOException
{
String x = next();
StringBuilder sb = new StringBuilder("0");
double res = 0, f = 1;
boolean dec = false, neg = false;
int start = 0;
if(x.charAt(0) == '-')
{
neg = true;
start++;
}
for(int i = start; i < x.length(); i++)
if(x.charAt(i) == '.')
{
res = Long.parseLong(sb.toString());
sb = new StringBuilder("0");
dec = true;
}
else
{
sb.append(x.charAt(i));
if(dec)
f *= 10;
}
res += Long.parseLong(sb.toString()) / f;
return res * (neg?-1:1);
}
public Long[] nextLongArray(int n) throws IOException {
Long[] a = new Long[n];
for (int i = 0; i < n; i++)
a[i] = nextLong();
return a;
}
public int[] nextIntArray(int n) throws IOException {
int[] a = new int[n];
for (int i = 0; i < n; i++)
a[i] = nextInt();
return a;
}
public Integer[] nextIntegerArray(int n) throws IOException {
Integer[] a = new Integer[n];
for (int i = 0; i < n; i++)
a[i] = nextInt();
return a;
}
public boolean ready() throws IOException {return br.ready();}
}
static class Pair {
long x ;
long y ;
public Pair(long x, long y) {
super();
this.x = x;
this.y = y;
}
@Override
public int hashCode() {
final int prime = 31;
int result = 1;
result = prime * result + (int) (x ^ (x >>> 32));
result = prime * result + (int) (y ^ (y >>> 32));
return result;
}
@Override
public boolean equals(Object obj) {
if (this == obj)
return true;
if (obj == null)
return false;
if (getClass() != obj.getClass())
return false;
Pair other = (Pair) obj;
if (x != other.x)
return false;
if (y != other.y)
return false;
return true;
}
public String toString() {
return x + " " + y;
}
}
static class tuble implements Comparable<tuble> {
int x;
int y;
int z;
public tuble(int x, int y, int z) {
this.x = x;
this.y = y;
this.z = z;
}
public String toString() {
return x + " " + y + " " + z;
}
public int compareTo(tuble other) {
if (this.x == other.x) {
if (this.y == other.y) {
return this.z - other.z;
}
return this.y - other.y;
} else {
return this.x - other.x;
}
}
}
}
| Java | ["4\n\n5\n\n1 1 1 1 1\n\n5\n\n2 1 1 1 2\n\n6\n\n1 1 2 3 3 4\n\n6\n\n1 2 1 4 5 4"] | 1 second | ["2\n1\n2\n0"] | NoteIn the first test case, we can select $$$i=2$$$ and $$$x=2$$$ to form $$$[1, 2, 2, 1, 1]$$$. Then, we can select $$$i=3$$$ and $$$x=3$$$ to form $$$[1, 2, 3, 3, 1]$$$.In the second test case, we can select $$$i=3$$$ and $$$x=100$$$ to form $$$[2, 1, 100, 100, 2]$$$. | Java 8 | standard input | [
"constructive algorithms",
"greedy",
"implementation"
] | a91be662101762bcb2c58b8db9ff61e0 | Each test contains multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains an integer $$$n$$$ ($$$2 \le n \le 2 \cdot 10 ^ 5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10^9$$$) — elements of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10 ^ 5$$$ | 1,100 | For each test case, print the minimum number of operations needed. | standard output | |
PASSED | 0742dbb88cd55b21b81b896631ce27f1 | train_108.jsonl | 1650722700 | You are given an array $$$a$$$ of length $$$n$$$. We define the equality of the array as the number of indices $$$1 \le i \le n - 1$$$ such that $$$a_i = a_{i + 1}$$$. We are allowed to do the following operation: Select two integers $$$i$$$ and $$$x$$$ such that $$$1 \le i \le n - 1$$$ and $$$1 \le x \le 10^9$$$. Then, set $$$a_i$$$ and $$$a_{i + 1}$$$ to be equal to $$$x$$$. Find the minimum number of operations needed such that the equality of the array is less than or equal to $$$1$$$. | 256 megabytes | import java.io.*;
import java.util.*;
public class Main {
public static void main(String[]args) throws Exception{
Scanner sc = new Scanner(System.in) ;
PrintWriter pw = new PrintWriter(System.out) ;
int t=sc.nextInt();
while(t-->0){
int n =sc.nextInt() ;
int [] arr = sc.nextIntArray(n) ;
int l = 0 ;
int r= n-1 ;
int newNum = -1 ;
int res = 0;
while(r-l>=2){
if(arr[r]!=arr[r-1]){
r-- ;
}
else{
if(arr[l]!=arr[l+1]){
l++ ;
}
else{
arr[l+1] = newNum ;
arr[l+2] = newNum-- ;
res++ ;
}
}
}
pw.println(res) ;
}
pw.flush();
pw.close();
}
static class Scanner
{
StringTokenizer st;
BufferedReader br;
public Scanner(InputStream s){ br = new BufferedReader(new InputStreamReader(s));}
public String next() throws IOException
{
while (st == null || !st.hasMoreTokens())
st = new StringTokenizer(br.readLine());
return st.nextToken();
}
public int nextInt() throws IOException {return Integer.parseInt(next());}
public long nextLong() throws IOException {return Long.parseLong(next());}
public String nextLine() throws IOException {return br.readLine();}
public double nextDouble() throws IOException
{
String x = next();
StringBuilder sb = new StringBuilder("0");
double res = 0, f = 1;
boolean dec = false, neg = false;
int start = 0;
if(x.charAt(0) == '-')
{
neg = true;
start++;
}
for(int i = start; i < x.length(); i++)
if(x.charAt(i) == '.')
{
res = Long.parseLong(sb.toString());
sb = new StringBuilder("0");
dec = true;
}
else
{
sb.append(x.charAt(i));
if(dec)
f *= 10;
}
res += Long.parseLong(sb.toString()) / f;
return res * (neg?-1:1);
}
public Long[] nextLongArray(int n) throws IOException {
Long[] a = new Long[n];
for (int i = 0; i < n; i++)
a[i] = nextLong();
return a;
}
public int[] nextIntArray(int n) throws IOException {
int[] a = new int[n];
for (int i = 0; i < n; i++)
a[i] = nextInt();
return a;
}
public Integer[] nextIntegerArray(int n) throws IOException {
Integer[] a = new Integer[n];
for (int i = 0; i < n; i++)
a[i] = nextInt();
return a;
}
public boolean ready() throws IOException {return br.ready();}
}
static class Pair {
long x ;
long y ;
public Pair(long x, long y) {
super();
this.x = x;
this.y = y;
}
@Override
public int hashCode() {
final int prime = 31;
int result = 1;
result = prime * result + (int) (x ^ (x >>> 32));
result = prime * result + (int) (y ^ (y >>> 32));
return result;
}
@Override
public boolean equals(Object obj) {
if (this == obj)
return true;
if (obj == null)
return false;
if (getClass() != obj.getClass())
return false;
Pair other = (Pair) obj;
if (x != other.x)
return false;
if (y != other.y)
return false;
return true;
}
public String toString() {
return x + " " + y;
}
}
static class tuble implements Comparable<tuble> {
int x;
int y;
int z;
public tuble(int x, int y, int z) {
this.x = x;
this.y = y;
this.z = z;
}
public String toString() {
return x + " " + y + " " + z;
}
public int compareTo(tuble other) {
if (this.x == other.x) {
if (this.y == other.y) {
return this.z - other.z;
}
return this.y - other.y;
} else {
return this.x - other.x;
}
}
}
}
| Java | ["4\n\n5\n\n1 1 1 1 1\n\n5\n\n2 1 1 1 2\n\n6\n\n1 1 2 3 3 4\n\n6\n\n1 2 1 4 5 4"] | 1 second | ["2\n1\n2\n0"] | NoteIn the first test case, we can select $$$i=2$$$ and $$$x=2$$$ to form $$$[1, 2, 2, 1, 1]$$$. Then, we can select $$$i=3$$$ and $$$x=3$$$ to form $$$[1, 2, 3, 3, 1]$$$.In the second test case, we can select $$$i=3$$$ and $$$x=100$$$ to form $$$[2, 1, 100, 100, 2]$$$. | Java 8 | standard input | [
"constructive algorithms",
"greedy",
"implementation"
] | a91be662101762bcb2c58b8db9ff61e0 | Each test contains multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains an integer $$$n$$$ ($$$2 \le n \le 2 \cdot 10 ^ 5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10^9$$$) — elements of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10 ^ 5$$$ | 1,100 | For each test case, print the minimum number of operations needed. | standard output | |
PASSED | dc162ce08564b6da879efa0049de9c8f | train_108.jsonl | 1650722700 | You are given an array $$$a$$$ of length $$$n$$$. We define the equality of the array as the number of indices $$$1 \le i \le n - 1$$$ such that $$$a_i = a_{i + 1}$$$. We are allowed to do the following operation: Select two integers $$$i$$$ and $$$x$$$ such that $$$1 \le i \le n - 1$$$ and $$$1 \le x \le 10^9$$$. Then, set $$$a_i$$$ and $$$a_{i + 1}$$$ to be equal to $$$x$$$. Find the minimum number of operations needed such that the equality of the array is less than or equal to $$$1$$$. | 256 megabytes | import java.io.*;
import java.util.*;
public class B {
//--------------------------INPUT READER---------------------------------//
static class fs {
public BufferedReader br;
StringTokenizer st = new StringTokenizer("");
public fs() { this(System.in); }
public fs(InputStream is) {
br = new BufferedReader(new InputStreamReader(is));
}
String next() {
while (!st.hasMoreTokens()) {
try { st = new StringTokenizer(br.readLine()); }
catch (IOException e) { e.printStackTrace(); }
}
return st.nextToken();
}
int ni() { return Integer.parseInt(next()); }
long nl() { return Long.parseLong(next()); }
double nd() { return Double.parseDouble(next()); }
String ns() { return next(); }
int[] na(long nn) {
int n = (int) nn;
int[] a = new int[n];
for (int i = 0; i < n; i++) a[i] = ni();
return a;
}
long[] nal(long nn) {
int n = (int) nn;
long[] l = new long[n];
for(int i = 0; i < n; i++) l[i] = nl();
return l;
}
}
//-----------------------------------------------------------------------//
//---------------------------PRINTER-------------------------------------//
static class Printer {
static PrintWriter w;
public Printer() {this(System.out);}
public Printer(OutputStream os) {
w = new PrintWriter(os);
}
public void p(int i) {w.println(i);}
public void p(long l) {w.println(l);}
public void p(double d) {w.println(d);}
public void p(String s) { w.println(s);}
public void pr(int i) {w.print(i);}
public void pr(long l) {w.print(l);}
public void pr(double d) {w.print(d);}
public void pr(String s) { w.print(s);}
public void pl() {w.println();}
public void close() {w.close();}
}
//-----------------------------------------------------------------------//
//--------------------------VARIABLES------------------------------------//
static fs sc = new fs();
static OutputStream outputStream = System.out;
static Printer w = new Printer(outputStream);
static long lma = Long.MAX_VALUE, lmi = Long.MIN_VALUE;
static int ima = Integer.MAX_VALUE, imi = Integer.MIN_VALUE;
static long mod = 1000000007;
//-----------------------------------------------------------------------//
//--------------------------ADMIN_MODE-----------------------------------//
private static void ADMIN_MODE() throws IOException {
if (System.getProperty("ONLINE_JUDGE") == null) {
w = new Printer(new FileOutputStream("output.txt"));
sc = new fs(new FileInputStream("input.txt"));
}
}
//-----------------------------------------------------------------------//
//----------------------------START--------------------------------------//
public static void main(String[] args)
throws IOException {
ADMIN_MODE();
int t = sc.ni();while(t-->0)
solve();
w.close();
}
static void solve() throws IOException {
int n = sc.ni();
int[] arr = sc.na(n);
int sames = 0;
int st = -1;
int end = -1;
for(int i = 0; i < n-1; i++) {
if(arr[i] == arr[i+1]) {
if(st == -1) st = i;
end = i+1;
sames++;
}
}
if(sames <= 1) {
w.p(0);
return;
}
int elems = end-st+1;
if(elems == 3) {
w.p(1);
return;
}
elems-=3;
w.p(elems);
}
} | Java | ["4\n\n5\n\n1 1 1 1 1\n\n5\n\n2 1 1 1 2\n\n6\n\n1 1 2 3 3 4\n\n6\n\n1 2 1 4 5 4"] | 1 second | ["2\n1\n2\n0"] | NoteIn the first test case, we can select $$$i=2$$$ and $$$x=2$$$ to form $$$[1, 2, 2, 1, 1]$$$. Then, we can select $$$i=3$$$ and $$$x=3$$$ to form $$$[1, 2, 3, 3, 1]$$$.In the second test case, we can select $$$i=3$$$ and $$$x=100$$$ to form $$$[2, 1, 100, 100, 2]$$$. | Java 8 | standard input | [
"constructive algorithms",
"greedy",
"implementation"
] | a91be662101762bcb2c58b8db9ff61e0 | Each test contains multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains an integer $$$n$$$ ($$$2 \le n \le 2 \cdot 10 ^ 5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10^9$$$) — elements of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10 ^ 5$$$ | 1,100 | For each test case, print the minimum number of operations needed. | standard output | |
PASSED | 52517d454e67d14de45118d735f630df | train_108.jsonl | 1650722700 | You are given an array $$$a$$$ of length $$$n$$$. We define the equality of the array as the number of indices $$$1 \le i \le n - 1$$$ such that $$$a_i = a_{i + 1}$$$. We are allowed to do the following operation: Select two integers $$$i$$$ and $$$x$$$ such that $$$1 \le i \le n - 1$$$ and $$$1 \le x \le 10^9$$$. Then, set $$$a_i$$$ and $$$a_{i + 1}$$$ to be equal to $$$x$$$. Find the minimum number of operations needed such that the equality of the array is less than or equal to $$$1$$$. | 256 megabytes | import java.util.Scanner;
public class UnequalArray {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int testcases = sc.nextInt();
for (int testcase = 0; testcase < testcases; testcase++) {
int n = sc.nextInt();
int[] arr = new int[n];
for (int i = 0; i < n; i++) {
arr[i] = sc.nextInt();
}
int totalPairs = 0;
for (int i = 0; i < n-1; i++) {
if (arr[i] == arr[i+1]) {
totalPairs++;
}
}
int firstLeft = 999999;
for (int i = 0; i < n-1; i++) {
if (arr[i] == arr[i+1]) {
firstLeft = i+1;
break;
}
}
int firstRight = -999999;
for (int i = n-2; i >= 0; i--) {
if (arr[i] == arr[i+1]) {
firstRight = i;
break;
}
}
if (totalPairs <= 1) {
System.out.println(0);
continue;
}
if (firstRight == firstLeft) {
System.out.println(1);
continue;
}
System.out.println(firstRight - firstLeft);
}
}
}
| Java | ["4\n\n5\n\n1 1 1 1 1\n\n5\n\n2 1 1 1 2\n\n6\n\n1 1 2 3 3 4\n\n6\n\n1 2 1 4 5 4"] | 1 second | ["2\n1\n2\n0"] | NoteIn the first test case, we can select $$$i=2$$$ and $$$x=2$$$ to form $$$[1, 2, 2, 1, 1]$$$. Then, we can select $$$i=3$$$ and $$$x=3$$$ to form $$$[1, 2, 3, 3, 1]$$$.In the second test case, we can select $$$i=3$$$ and $$$x=100$$$ to form $$$[2, 1, 100, 100, 2]$$$. | Java 8 | standard input | [
"constructive algorithms",
"greedy",
"implementation"
] | a91be662101762bcb2c58b8db9ff61e0 | Each test contains multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains an integer $$$n$$$ ($$$2 \le n \le 2 \cdot 10 ^ 5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10^9$$$) — elements of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10 ^ 5$$$ | 1,100 | For each test case, print the minimum number of operations needed. | standard output | |
PASSED | 7bb279dabd169920b7ba84364ce91b4d | train_108.jsonl | 1650722700 | You are given an array $$$a$$$ of length $$$n$$$. We define the equality of the array as the number of indices $$$1 \le i \le n - 1$$$ such that $$$a_i = a_{i + 1}$$$. We are allowed to do the following operation: Select two integers $$$i$$$ and $$$x$$$ such that $$$1 \le i \le n - 1$$$ and $$$1 \le x \le 10^9$$$. Then, set $$$a_i$$$ and $$$a_{i + 1}$$$ to be equal to $$$x$$$. Find the minimum number of operations needed such that the equality of the array is less than or equal to $$$1$$$. | 256 megabytes | import java.util.Scanner;
/**
*
* @author Acer
*/
public class NewClass_C {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int T = sc.nextInt();
while(T-- > 0){
int n = sc.nextInt();
int arr[] = new int[n];
for (int i = 0; i < n; i++) {
arr[i] = sc.nextInt();
}
int index1 = -1;
for (int i = 0; i < n-1; i++) {
if(arr[i] == arr[i+1]){
index1 = i+1;
break;
}
}
int index2 = -1;
for (int i = n-1; i >= 1; i--) {
if(arr[i] == arr[i-1]){
index2 = i-1;
break;
}
}
//System.out.println(index1+" "+index2);
if(index1 > index2 || index1 == -1){
System.out.println(0);
}
else if(index2 == index1){
System.out.println(1);
}
else{
System.out.println(index2-index1);
}
}
}
} | Java | ["4\n\n5\n\n1 1 1 1 1\n\n5\n\n2 1 1 1 2\n\n6\n\n1 1 2 3 3 4\n\n6\n\n1 2 1 4 5 4"] | 1 second | ["2\n1\n2\n0"] | NoteIn the first test case, we can select $$$i=2$$$ and $$$x=2$$$ to form $$$[1, 2, 2, 1, 1]$$$. Then, we can select $$$i=3$$$ and $$$x=3$$$ to form $$$[1, 2, 3, 3, 1]$$$.In the second test case, we can select $$$i=3$$$ and $$$x=100$$$ to form $$$[2, 1, 100, 100, 2]$$$. | Java 8 | standard input | [
"constructive algorithms",
"greedy",
"implementation"
] | a91be662101762bcb2c58b8db9ff61e0 | Each test contains multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains an integer $$$n$$$ ($$$2 \le n \le 2 \cdot 10 ^ 5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10^9$$$) — elements of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10 ^ 5$$$ | 1,100 | For each test case, print the minimum number of operations needed. | standard output | |
PASSED | 3ce70965f680a9d881b6aa3d99e65e51 | train_108.jsonl | 1650722700 | You are given an array $$$a$$$ of length $$$n$$$. We define the equality of the array as the number of indices $$$1 \le i \le n - 1$$$ such that $$$a_i = a_{i + 1}$$$. We are allowed to do the following operation: Select two integers $$$i$$$ and $$$x$$$ such that $$$1 \le i \le n - 1$$$ and $$$1 \le x \le 10^9$$$. Then, set $$$a_i$$$ and $$$a_{i + 1}$$$ to be equal to $$$x$$$. Find the minimum number of operations needed such that the equality of the array is less than or equal to $$$1$$$. | 256 megabytes | import java.util.ArrayList;
import java.util.Collections;
import java.util.HashMap;
import java.util.HashSet;
import java.util.Scanner;
public class Cf22 {
public static boolean helper(int x,int n,int[] ar) {
for(int i =x;i<n-1;i++) {
if(ar[i]==ar[i+1])
return true;
}
return false;
}
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int k = sc.nextInt();
for(int i =0;i<k;i++) {
int res =0;
int l = sc.nextInt();
int[] ar = new int[l];
int firsti =0;
int num=0;
int lasti=0;
for(int j=0;j<l;j++) {
ar[j] = sc.nextInt();
}
for(int j =0;j<l-1;j++) {
if(ar[j]==ar[j+1]) {
if(num==0) {
firsti=j+1;
}
lasti =j;
num++;
}
}
if(num<2) {
System.out.println(0);
}
else {
if(lasti==firsti)
System.out.println(1);
else {
System.out.println(lasti-firsti);
}
}
}
}
}
//
//
//
//public static void 3(String[] args) {
// Scanner sc = new Scanner(System.in);
// int k = sc.nextInt();
// for(int i =0;i<k;i++) {
// int n = sc.nextInt();
// int min = Integer.MAX_VALUE;
// ArrayList<Integer> ar = new ArrayList<>();
// for(int j=0;j<n;j++) {
// int u = sc.nextInt();
// min = Math.min(u, min);
// ar.add(u);
// }
// int res =0;
// for(int j=0;j<n;j++) {
// res+= ar.get(j)-min;
// }
// System.out.println(res);
//
//
// }
//
//}
| Java | ["4\n\n5\n\n1 1 1 1 1\n\n5\n\n2 1 1 1 2\n\n6\n\n1 1 2 3 3 4\n\n6\n\n1 2 1 4 5 4"] | 1 second | ["2\n1\n2\n0"] | NoteIn the first test case, we can select $$$i=2$$$ and $$$x=2$$$ to form $$$[1, 2, 2, 1, 1]$$$. Then, we can select $$$i=3$$$ and $$$x=3$$$ to form $$$[1, 2, 3, 3, 1]$$$.In the second test case, we can select $$$i=3$$$ and $$$x=100$$$ to form $$$[2, 1, 100, 100, 2]$$$. | Java 8 | standard input | [
"constructive algorithms",
"greedy",
"implementation"
] | a91be662101762bcb2c58b8db9ff61e0 | Each test contains multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains an integer $$$n$$$ ($$$2 \le n \le 2 \cdot 10 ^ 5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10^9$$$) — elements of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10 ^ 5$$$ | 1,100 | For each test case, print the minimum number of operations needed. | standard output | |
PASSED | c034ca539cbd50bd862bbcf794d10de7 | train_108.jsonl | 1650722700 | You are given an array $$$a$$$ of length $$$n$$$. We define the equality of the array as the number of indices $$$1 \le i \le n - 1$$$ such that $$$a_i = a_{i + 1}$$$. We are allowed to do the following operation: Select two integers $$$i$$$ and $$$x$$$ such that $$$1 \le i \le n - 1$$$ and $$$1 \le x \le 10^9$$$. Then, set $$$a_i$$$ and $$$a_{i + 1}$$$ to be equal to $$$x$$$. Find the minimum number of operations needed such that the equality of the array is less than or equal to $$$1$$$. | 256 megabytes |
import java.io.*;
import java.math.*;
import java.security.*;
import java.text.*;
import java.util.*;
import java.util.concurrent.*;
import java.util.function.*;
import java.util.regex.*;
import java.util.stream.*;
import java.util.Comparator;
import javax.lang.model.util.ElementScanner6;
import static java.util.stream.Collectors.joining;
import static java.util.stream.Collectors.toList;
public class Solution {
static HashMap<Integer, Integer> createHashMap(int arr[]) {
HashMap<Integer, Integer> hmap = new HashMap<Integer, Integer>();
for (int i = 0; i < arr.length; i++) {
Integer c = hmap.get(arr[i]);
if (hmap.get(arr[i]) == null) {
hmap.put(arr[i], 1);
}
else {
hmap.put(arr[i], ++c);
}
}
return hmap;
}
static int parseInt(String str) {
return Integer.parseInt(str);
}
static String noZeros(char[] num) {
String str = "";
for (int i = 0; i < num.length; i++) {
if (num[i] == '0')
continue;
else
str += num[i];
}
return str;
}
public static void Sort2DArrayBasedOnColumnNumber(int[][] array, final int columnNumber) {
Arrays.sort(array, new Comparator<int[]>() {
@Override
public int compare(int[] first, int[] second) {
if (first[columnNumber - 1] > second[columnNumber - 1])
return 1;
else
return -1;
}
});
}
public static void main(String[] args) throws Exception {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
StringTokenizer tk;
StringTokenizer tk2;
int t = parseInt(br.readLine());
int n;
int count;
int moves;
int[] arr;
ArrayList<Integer> indeces = new ArrayList<Integer>();
String str;
int p;
for (int i = 1; i <= t; i++) {
count = 0;
moves = 0;
n = parseInt(br.readLine());
str = br.readLine();
tk = new StringTokenizer(str);
arr = new int[n + 2];
arr[n] = 0;
arr[0] = 0;
arr[1] = parseInt(tk.nextToken());
for (int j = 2; j <= n; j++) {
arr[j] = parseInt(tk.nextToken());
}
for (int x = 1; x < n + 2; x++) {
if (arr[x] == arr[x - 1]) {
count++;
}
else if (count == 0)
continue;
else {
indeces.add(Math.abs((count + 1) - x));
count = 0;
break;
}
}
for (int y = n; y > 0; y--) {
if (arr[y] == arr[y - 1])
count++;
else if (count == 0)
continue;
else {
indeces.add(y + count);
break;
}
}
if (indeces.size() == 0 || (indeces.size() == 2 && indeces.get(1) == 1))
System.out.println(moves);
else if ((indeces.get(0) + 1 == indeces.get(1) - 1) || Math.abs(indeces.get(1) - indeces.get(0)) == 1) {
moves = Math.abs(indeces.get(0) + 1 - indeces.get(1));
System.out.println(moves);
}
else {
moves = Math.abs(indeces.get(1) - 1 - (indeces.get(0) + 1));
System.out.println(moves);
}
indeces.clear();
}
}
}
/*
*
*
* count=2
* count+1-x
*
*
*
*
* indeces={0,1,6,7}
* 5-1=4
*/
| Java | ["4\n\n5\n\n1 1 1 1 1\n\n5\n\n2 1 1 1 2\n\n6\n\n1 1 2 3 3 4\n\n6\n\n1 2 1 4 5 4"] | 1 second | ["2\n1\n2\n0"] | NoteIn the first test case, we can select $$$i=2$$$ and $$$x=2$$$ to form $$$[1, 2, 2, 1, 1]$$$. Then, we can select $$$i=3$$$ and $$$x=3$$$ to form $$$[1, 2, 3, 3, 1]$$$.In the second test case, we can select $$$i=3$$$ and $$$x=100$$$ to form $$$[2, 1, 100, 100, 2]$$$. | Java 8 | standard input | [
"constructive algorithms",
"greedy",
"implementation"
] | a91be662101762bcb2c58b8db9ff61e0 | Each test contains multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains an integer $$$n$$$ ($$$2 \le n \le 2 \cdot 10 ^ 5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10^9$$$) — elements of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10 ^ 5$$$ | 1,100 | For each test case, print the minimum number of operations needed. | standard output | |
PASSED | de9c4c007485166896ccca0831aa1d60 | train_108.jsonl | 1650722700 | You are given an array $$$a$$$ of length $$$n$$$. We define the equality of the array as the number of indices $$$1 \le i \le n - 1$$$ such that $$$a_i = a_{i + 1}$$$. We are allowed to do the following operation: Select two integers $$$i$$$ and $$$x$$$ such that $$$1 \le i \le n - 1$$$ and $$$1 \le x \le 10^9$$$. Then, set $$$a_i$$$ and $$$a_{i + 1}$$$ to be equal to $$$x$$$. Find the minimum number of operations needed such that the equality of the array is less than or equal to $$$1$$$. | 256 megabytes | import java.io.*;
import java.util.*;
public class Unequal_Array {
public static BufferedReader b = new BufferedReader(new InputStreamReader(System.in));
public static void main(String[] args) throws IOException {
int counts = Integer.parseInt(b.readLine());
for(int i = 0;i < counts;i++){
test();
}
}
static void test() throws IOException{
int number = Integer.parseInt(b.readLine());
int []arr = new int[number];
String []tem = b.readLine().split(" ");
for(int i = 0 ;i<number;i++){
arr[i] = Integer.parseInt(tem[i]);
}
if(isUnequal(arr)){
System.out.println(0);
return;
}
int length = getMaxIndex(arr) -getMinIndex(arr)+1;
if(length == 3){
System.out.println(1);
return;
}
System.out.println(length-3);
}
static boolean isUnequal(int []arr){
int counts = 0;
for(int i = 0;i < arr.length-1;i++){
if(arr[i] == arr[i+1]){
counts++;
}
if(counts > 1){
return false;
}
}
return true;
}
static int getMinIndex(int []arr){
for(int i = 0; i < arr.length-1;i++){
if(arr[i] == arr[i+1]){
return i;
}
}
return -1;
}
static int getMaxIndex(int []arr){
for(int i = arr.length-1;i >= 0 ;i--){
if(arr[i] == arr[i-1]){
return i;
}
}
return -1;
}
}
| Java | ["4\n\n5\n\n1 1 1 1 1\n\n5\n\n2 1 1 1 2\n\n6\n\n1 1 2 3 3 4\n\n6\n\n1 2 1 4 5 4"] | 1 second | ["2\n1\n2\n0"] | NoteIn the first test case, we can select $$$i=2$$$ and $$$x=2$$$ to form $$$[1, 2, 2, 1, 1]$$$. Then, we can select $$$i=3$$$ and $$$x=3$$$ to form $$$[1, 2, 3, 3, 1]$$$.In the second test case, we can select $$$i=3$$$ and $$$x=100$$$ to form $$$[2, 1, 100, 100, 2]$$$. | Java 8 | standard input | [
"constructive algorithms",
"greedy",
"implementation"
] | a91be662101762bcb2c58b8db9ff61e0 | Each test contains multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains an integer $$$n$$$ ($$$2 \le n \le 2 \cdot 10 ^ 5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10^9$$$) — elements of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10 ^ 5$$$ | 1,100 | For each test case, print the minimum number of operations needed. | standard output | |
PASSED | 03a51f0a1c6796566c0e14172820f7c4 | train_108.jsonl | 1650722700 | You are given an array $$$a$$$ of length $$$n$$$. We define the equality of the array as the number of indices $$$1 \le i \le n - 1$$$ such that $$$a_i = a_{i + 1}$$$. We are allowed to do the following operation: Select two integers $$$i$$$ and $$$x$$$ such that $$$1 \le i \le n - 1$$$ and $$$1 \le x \le 10^9$$$. Then, set $$$a_i$$$ and $$$a_{i + 1}$$$ to be equal to $$$x$$$. Find the minimum number of operations needed such that the equality of the array is less than or equal to $$$1$$$. | 256 megabytes | import java.util.*;
import static java.lang.System.out;
public class app {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int t = sc.nextInt();
while (t-- > 0) {
int n = sc.nextInt();
int[] arr = new int[n];
int first = 0, last = 0;
for (int i = 0; i < n; i++) {
arr[i] = sc.nextInt();
if (i > 0 && arr[i] == arr[i - 1]) {
if (first == 0)
first = i;
last = i;
}
}
int ans = last - first - 1;
if (ans <= 0)
System.out.println(ans + 1);
else
System.out.println(ans);
}
}
} | Java | ["4\n\n5\n\n1 1 1 1 1\n\n5\n\n2 1 1 1 2\n\n6\n\n1 1 2 3 3 4\n\n6\n\n1 2 1 4 5 4"] | 1 second | ["2\n1\n2\n0"] | NoteIn the first test case, we can select $$$i=2$$$ and $$$x=2$$$ to form $$$[1, 2, 2, 1, 1]$$$. Then, we can select $$$i=3$$$ and $$$x=3$$$ to form $$$[1, 2, 3, 3, 1]$$$.In the second test case, we can select $$$i=3$$$ and $$$x=100$$$ to form $$$[2, 1, 100, 100, 2]$$$. | Java 8 | standard input | [
"constructive algorithms",
"greedy",
"implementation"
] | a91be662101762bcb2c58b8db9ff61e0 | Each test contains multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains an integer $$$n$$$ ($$$2 \le n \le 2 \cdot 10 ^ 5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10^9$$$) — elements of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10 ^ 5$$$ | 1,100 | For each test case, print the minimum number of operations needed. | standard output | |
PASSED | 27c7e09657a9ae11cd6ce86f4dabea93 | train_108.jsonl | 1650722700 | You are given an array $$$a$$$ of length $$$n$$$. We define the equality of the array as the number of indices $$$1 \le i \le n - 1$$$ such that $$$a_i = a_{i + 1}$$$. We are allowed to do the following operation: Select two integers $$$i$$$ and $$$x$$$ such that $$$1 \le i \le n - 1$$$ and $$$1 \le x \le 10^9$$$. Then, set $$$a_i$$$ and $$$a_{i + 1}$$$ to be equal to $$$x$$$. Find the minimum number of operations needed such that the equality of the array is less than or equal to $$$1$$$. | 256 megabytes | import java.io.*;
import java.util.*;
public class C {
public static PrintWriter out;
public static void main(String[] args)throws IOException{
Scanner sc=new Scanner();
out=new PrintWriter(System.out);
int t=sc.nextInt();
while(t-->0) {
int n=sc.nextInt();
int a[]=new int[n];
for(int i=0;i<n;i++) {
a[i]=sc.nextInt();
}
if(a.length==1||a.length==2) {
out.println(0);
continue;
}
int firstPair=-1;
int lastPair=-1;
for(int i=0;i<n-1;i++) {
if(a[i]==a[i+1]) {
if(firstPair==-1) {
firstPair=i;
}else {
lastPair=i;
}
}
}
if(lastPair==-1) {
out.println(0);
continue;
}
if(lastPair-firstPair==1) {
out.println(1);
}else {
out.println(lastPair-firstPair-1);
}
}
out.close();
}
public static class Scanner {
BufferedReader br;
StringTokenizer st;
public Scanner() {
br = new BufferedReader(new InputStreamReader(System.in));
}
String next() {
while (st == null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt() {
return Integer.parseInt(next());
}
long nextLong() {
return Long.parseLong(next());
}
double nextDouble() {
return Double.parseDouble(next());
}
String nextLine(){
String str = "";
try {
str = br.readLine();
} catch (IOException e) {
e.printStackTrace();
}
return str;
}
}
}
| Java | ["4\n\n5\n\n1 1 1 1 1\n\n5\n\n2 1 1 1 2\n\n6\n\n1 1 2 3 3 4\n\n6\n\n1 2 1 4 5 4"] | 1 second | ["2\n1\n2\n0"] | NoteIn the first test case, we can select $$$i=2$$$ and $$$x=2$$$ to form $$$[1, 2, 2, 1, 1]$$$. Then, we can select $$$i=3$$$ and $$$x=3$$$ to form $$$[1, 2, 3, 3, 1]$$$.In the second test case, we can select $$$i=3$$$ and $$$x=100$$$ to form $$$[2, 1, 100, 100, 2]$$$. | Java 8 | standard input | [
"constructive algorithms",
"greedy",
"implementation"
] | a91be662101762bcb2c58b8db9ff61e0 | Each test contains multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains an integer $$$n$$$ ($$$2 \le n \le 2 \cdot 10 ^ 5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10^9$$$) — elements of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10 ^ 5$$$ | 1,100 | For each test case, print the minimum number of operations needed. | standard output | |
PASSED | 5a152f0d17e05ba3ebe5f4a41a973251 | train_108.jsonl | 1650722700 | You are given an array $$$a$$$ of length $$$n$$$. We define the equality of the array as the number of indices $$$1 \le i \le n - 1$$$ such that $$$a_i = a_{i + 1}$$$. We are allowed to do the following operation: Select two integers $$$i$$$ and $$$x$$$ such that $$$1 \le i \le n - 1$$$ and $$$1 \le x \le 10^9$$$. Then, set $$$a_i$$$ and $$$a_{i + 1}$$$ to be equal to $$$x$$$. Find the minimum number of operations needed such that the equality of the array is less than or equal to $$$1$$$. | 256 megabytes |
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.*;
public class Main
{
public static long check(long a[])
{
long count=0;
for (int i = 0; i <a.length-1; i++) {
if(a[i]==a[i+1])
{
count++;
}
}
return count;
}
public static void main(String args[])
{
FastScanner input = new FastScanner();
int tc = input.nextInt();
work:
while (tc-- > 0) {
int n = input.nextInt();
long a[] = new long[n];
for (int i = 0; i <n; i++) {
a[i] = input.nextLong();
}
long ans = 0;
long count=0;
int start =-1,end =-1;
for (int i = 0; i <a.length-1; i++) {
if(a[i]==a[i+1])
{
count++;
end= i+1;
}
if(start==-1&&a[i]==a[i+1])
{
start = i;
}
}
if(count<=1L)
{
System.out.println("0");
continue work;
}
else
{
System.out.println(Math.max((end-start+1-3), 1));
}
}
}
static class FastScanner
{
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
StringTokenizer st = new StringTokenizer("");
String next()
{
while (!st.hasMoreTokens()) {
try {
st = new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt()
{
return Integer.parseInt(next());
}
long nextLong()
{
return Long.parseLong(next());
}
double nextDouble()
{
return Double.parseDouble(next());
}
String nextLine() throws IOException
{
return br.readLine();
}
}
}
| Java | ["4\n\n5\n\n1 1 1 1 1\n\n5\n\n2 1 1 1 2\n\n6\n\n1 1 2 3 3 4\n\n6\n\n1 2 1 4 5 4"] | 1 second | ["2\n1\n2\n0"] | NoteIn the first test case, we can select $$$i=2$$$ and $$$x=2$$$ to form $$$[1, 2, 2, 1, 1]$$$. Then, we can select $$$i=3$$$ and $$$x=3$$$ to form $$$[1, 2, 3, 3, 1]$$$.In the second test case, we can select $$$i=3$$$ and $$$x=100$$$ to form $$$[2, 1, 100, 100, 2]$$$. | Java 8 | standard input | [
"constructive algorithms",
"greedy",
"implementation"
] | a91be662101762bcb2c58b8db9ff61e0 | Each test contains multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains an integer $$$n$$$ ($$$2 \le n \le 2 \cdot 10 ^ 5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10^9$$$) — elements of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10 ^ 5$$$ | 1,100 | For each test case, print the minimum number of operations needed. | standard output | |
PASSED | ca67edbad13b49be789d13bee86ea704 | train_108.jsonl | 1650722700 | You are given an array $$$a$$$ of length $$$n$$$. We define the equality of the array as the number of indices $$$1 \le i \le n - 1$$$ such that $$$a_i = a_{i + 1}$$$. We are allowed to do the following operation: Select two integers $$$i$$$ and $$$x$$$ such that $$$1 \le i \le n - 1$$$ and $$$1 \le x \le 10^9$$$. Then, set $$$a_i$$$ and $$$a_{i + 1}$$$ to be equal to $$$x$$$. Find the minimum number of operations needed such that the equality of the array is less than or equal to $$$1$$$. | 256 megabytes | //---#ON_MY_WAY---
//---#THE_SILENT_ONE---
import static java.lang.Math.*;
import java.io.*;
import java.math.*;
import java.util.*;
public class apples {
static FastReader x = new FastReader();
static OutputStream outputStream = System.out;
static PrintWriter out = new PrintWriter(outputStream);
/*---------------------------------------CODE STARTS HERE-------------------------*/
public static void main(String[] args) throws NumberFormatException, IOException {
long startTime = System.nanoTime();
int mod = 1000000007;
int t = x.nextInt();
StringBuilder str = new StringBuilder();
while (t > 0) {
int n = x.nextInt();
int a[] = new int[n];
HashSet<Integer> hs = new HashSet<>();
for (int i = 0; i < n; i++) {
a[i] = x.nextInt();
hs.add(a[i]);
}
if(hs.size()==n) str.append(0);
else {
int st = -1, la = -1;
for (int i = 0; i < n-1; i++) {
if(a[i]==a[i+1]) {
st = i;
break;
}
}
for(int i = n-1; i > 0; i--) {
if(a[i]==a[i-1]) {
la = i;
break;
}
}
if(st==-1||la==-1||st==la-1) str.append(0);
else str.append(max(1, (la-st+1-3)));
}
str.append("\n");
t--;
}
out.println(str);
out.flush();
long endTime = System.nanoTime();
//System.out.println((endTime-startTime)/1000000000.0);
}
/*--------------------------------------------FAST I/O-------------------------------*/
static class FastReader {
BufferedReader br;
StringTokenizer st;
public FastReader() {
br = new BufferedReader(new InputStreamReader(System.in));
}
String next() {
while (st == null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt() {
return Integer.parseInt(next());
}
long nextLong() {
return Long.parseLong(next());
}
double nextDouble() {
return Double.parseDouble(next());
}
String nextLine() {
String str = "";
try {
str = br.readLine();
} catch (IOException e) {
e.printStackTrace();
}
return str;
}
char nextchar() {
char ch = ' ';
try {
ch = (char) br.read();
} catch (IOException e) {
e.printStackTrace();
}
return ch;
}
}
/*--------------------------------------------HELPER---------------------------------*/
static class pair implements Comparable<pair> {
int x, y;
public pair(int a, int b) {
x = a;
y = b;
}
@Override
public int hashCode() {
int hash = 3;
hash = 47 * hash + this.x;
hash = 47 * hash + this.y;
return hash;
}
@Override
public boolean equals(Object obj) {
if (obj == null) {
return false;
}
if (getClass() != obj.getClass()) {
return false;
}
final pair other = (pair) obj;
if (this.x != other.x) {
return false;
}
if (this.y != other.y) {
return false;
}
return true;
}
@Override
public int compareTo(apples.pair o) {
if(this.x==o.x) return this.y-o.y;
return this.x-o.x;
}
}
/*--------------------------------------------BOILER PLATE---------------------------*/
static int[] readarr(int n) {
int arr[] = new int[n];
for (int i = 0; i < n; i++) {
arr[i] = x.nextInt();
}
return arr;
}
static int[] sortint(int a[]) {
ArrayList<Integer> al = new ArrayList<>();
for (int i : a) {
al.add(i);
}
Collections.sort(al);
for (int i = 0; i < a.length; i++) {
a[i] = al.get(i);
}
return a;
}
static long[] sortlong(long a[]) {
ArrayList<Long> al = new ArrayList<>();
for (long i : a) {
al.add(i);
}
Collections.sort(al);
for (int i = 0; i < al.size(); i++) {
a[i] = al.get(i);
}
return a;
}
static long pow(long x, long y) {
long result = 1;
while (y > 0) {
if (y % 2 == 0) {
x = x * x;
y = y / 2;
} else {
result = result * x;
y = y - 1;
}
}
return result;
}
static long pow(long x, long y, long mod) {
long result = 1;
x %= mod;
while (y > 0) {
if (y % 2 == 0) {
x = (x % mod * x % mod) % mod;
y /= 2;
} else {
result = (result % mod * x % mod) % mod;
y--;
}
}
return result;
}
static int[] revsort(int a[]) {
ArrayList<Integer> al = new ArrayList<>();
for (int i : a) {
al.add(i);
}
Collections.sort(al, Comparator.reverseOrder());
for (int i = 0; i < a.length; i++) {
a[i] = al.get(i);
}
return a;
}
static int[] gcd(int a, int b, int ar[]) {
if (b == 0) {
ar[0] = a;
ar[1] = 1;
ar[2] = 0;
return ar;
}
ar = gcd(b, a % b, ar);
int t = ar[1];
ar[1] = ar[2];
ar[2] = t - (a / b) * ar[2];
return ar;
}
static boolean[] esieve(int n) {
boolean p[] = new boolean[n + 1];
Arrays.fill(p, true);
for (int i = 2; i * i <= n; i++) {
if (p[i] == true) {
for (int j = i * i; j <= n; j += i) {
p[j] = false;
}
}
}
return p;
}
static ArrayList<Integer> primes(int n) {
boolean p[] = new boolean[n + 1];
ArrayList<Integer> al = new ArrayList<>();
Arrays.fill(p, true);
int i = 0;
for (i = 2; i * i <= n; i++) {
if (p[i] == true) {
al.add(i);
for (int j = i * i; j <= n; j += i) {
p[j] = false;
}
}
}
for (i = i; i <= n; i++) {
if (p[i] == true) {
al.add(i);
}
}
return al;
}
static int gcd(int a, int b) {
if (a == 0) {
return b;
}
return gcd(b % a, a);
}
static long gcd(long a, long b) {
if (a == 0) {
return b;
}
return gcd(b % a, a);
}
} | Java | ["4\n\n5\n\n1 1 1 1 1\n\n5\n\n2 1 1 1 2\n\n6\n\n1 1 2 3 3 4\n\n6\n\n1 2 1 4 5 4"] | 1 second | ["2\n1\n2\n0"] | NoteIn the first test case, we can select $$$i=2$$$ and $$$x=2$$$ to form $$$[1, 2, 2, 1, 1]$$$. Then, we can select $$$i=3$$$ and $$$x=3$$$ to form $$$[1, 2, 3, 3, 1]$$$.In the second test case, we can select $$$i=3$$$ and $$$x=100$$$ to form $$$[2, 1, 100, 100, 2]$$$. | Java 8 | standard input | [
"constructive algorithms",
"greedy",
"implementation"
] | a91be662101762bcb2c58b8db9ff61e0 | Each test contains multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains an integer $$$n$$$ ($$$2 \le n \le 2 \cdot 10 ^ 5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10^9$$$) — elements of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10 ^ 5$$$ | 1,100 | For each test case, print the minimum number of operations needed. | standard output |
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