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 | b9045d37d71fbc863775450e057652fe | train_108.jsonl | 1645367700 | You are given an array $$$a$$$ of size $$$n$$$. Each element in this array is an integer between $$$1$$$ and $$$10^9$$$.You can perform several operations to this array. During an operation, you can replace an element in the array with any integer between $$$1$$$ and $$$10^9$$$. Output the minimum number of operations needed such that the resulting array doesn't contain any local maximums, and the resulting array after the operations.An element $$$a_i$$$ is a local maximum if it is strictly larger than both of its neighbors (that is, $$$a_i > a_{i - 1}$$$ and $$$a_i > a_{i + 1}$$$). Since $$$a_1$$$ and $$$a_n$$$ have only one neighbor each, they will never be a local maximum. | 256 megabytes | import java.util.Scanner;
public class Main{
public static void main(String[] args){
Scanner read = new Scanner(System.in);
int cases = read.nextInt();
int moves =0,cont=0;
for(int c=0;c<cases;c++){
int a = read.nextInt();
int[]array = new int[a];
for(int i=0;i<a;i++){
array[i]= read.nextInt();
}
if(a<3){
System.out.println(moves);
for(int print :array){
System.out.print(print+" ");
}
System.out.println("");
continue;
}
for(int i=1;i<a;i++){
//System.out.println("I. "+i);
if(i==(a-1) && (array[(a-2)]>array[(a-3)] && array[(a-2)]>array[(a-1)])){
array[(a-1)]=array[(a-2)];
moves++;
break;
}else if(i==(a-1)){
break;
}
if(array[i]>array[(i-1)] && array[i]>array[(i+1)]){
replaceNumber(array,i);
moves++;
}
}
//System.out.println("array: "+Arrays.toString(array));
System.out.println(moves);
for(int print :array){
System.out.print(print+" ");
}
System.out.println("");
moves=0;
}
}
/////////////FUNCTIONS////////////FUNCTIONS//////////FUNCTIONS///////////////
public static void replaceNumber(int[]arr,int i){
try{
arr[(i+1)] = arr[(i+2)];
if(arr[(i+2)]<arr[i]){
arr[(i+1)] = arr[i];
}
}catch(Exception e){
arr[(i+1)]=arr[i];
}
}
} | Java | ["5\n3\n2 1 2\n4\n1 2 3 1\n5\n1 2 1 2 1\n9\n1 2 1 3 2 3 1 2 1\n9\n2 1 3 1 3 1 3 1 3"] | 2 seconds | ["0\n2 1 2\n1\n1 3 3 1\n1\n1 2 2 2 1\n2\n1 2 3 3 2 3 3 2 1\n2\n2 1 3 3 3 1 1 1 3"] | NoteIn the first example, the array contains no local maximum, so we don't need to perform operations.In the second example, we can change $$$a_2$$$ to $$$3$$$, then the array don't have local maximums. | Java 8 | standard input | [
"greedy"
] | 255abf8750541f54d8ff0f1f80e1f8e7 | Each test contains multiple test cases. The first line will contain a single integer $$$t$$$ $$$(1 \leq t \leq 10000)$$$ — the number of test cases. Then $$$t$$$ test cases follow. The first line of each test case contains a single integer $$$n$$$ $$$(2 \leq n \leq 2 \cdot 10^5)$$$ — the size of the array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots ,a_n$$$ $$$(1 \leq a_i \leq 10^9)$$$, the elements of array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, first output a line containing a single integer $$$m$$$ — minimum number of operations required. Then ouput a line consist of $$$n$$$ integers — the resulting array after the operations. Note that this array should differ in exactly $$$m$$$ elements from the initial array. If there are multiple answers, print any. | standard output | |
PASSED | d6caab17c99d38e5115e9b592841ae23 | train_108.jsonl | 1645367700 | You are given an array $$$a$$$ of size $$$n$$$. Each element in this array is an integer between $$$1$$$ and $$$10^9$$$.You can perform several operations to this array. During an operation, you can replace an element in the array with any integer between $$$1$$$ and $$$10^9$$$. Output the minimum number of operations needed such that the resulting array doesn't contain any local maximums, and the resulting array after the operations.An element $$$a_i$$$ is a local maximum if it is strictly larger than both of its neighbors (that is, $$$a_i > a_{i - 1}$$$ and $$$a_i > a_{i + 1}$$$). Since $$$a_1$$$ and $$$a_n$$$ have only one neighbor each, they will never be a local maximum. | 256 megabytes | import java.io.*;
import java.util.*;
import javax.lang.model.element.Element;
public class Main {
public static void main(String[] args) {
Scanner scn = new Scanner(System.in);
int testCases = scn.nextInt();
List<List<Integer>> arrays = new ArrayList<List<Integer>>();
for (int i = 0; i < testCases; i++) {
int s = scn.nextInt();
List<Integer> temp = new ArrayList<Integer>();
for (int j = 0; j < s; j++) {
int input = scn.nextInt();
temp.add(input);
}
arrays.add(temp);
}
for (int i = 0; i < testCases; i++) {
List<Integer> list = arrays.get(i);
int m = 0;
for (int j = 1; j < list.size() - 1; j++) {
if (list.get(j) > list.get(j + 1) && list.get(j) > list.get(j - 1)) {
int val = j + 2 > list.size() - 1 ? list.size() - 1 : j + 2;
if (list.get(j) > list.get(val))
list.set(j + 1, list.get(j));
else
list.set(j + 1, list.get(val));
m++;
}
}
System.out.println(m);
printArray(list);
}
scn.close();
}
public static void printArray(List<Integer> list) {
for (int i : list) {
System.out.print(i + " ");
}
System.out.println();
}
}
/*
*
* 5
*
* 3
* 2 1 2
* 4
* 1 2 3 1
* 5
* 1 2 1 2 1
* 9
* 1 2 1 3 2 3 1 2 1
* 9
* 2 1 3 1 3 1 3 1 3
*
*
* 0
* 2 1 2
* 1
* 1 2 3 3
* 1
* 1 2 2 2 1
* 3
* 1 2 2 3 3 3 1 2 2
* 2
* 2 1 3 3 3 1 3 3 3
*
*
* 0
* 2 1 2
* 1
* 1 3 3 1
* 1
* 1 2 2 2 1
* 2
* 1 2 3 3 2 3 3 2 1
* 2
* 2 1 3 3 3 1 1 1 3
*
*/ | Java | ["5\n3\n2 1 2\n4\n1 2 3 1\n5\n1 2 1 2 1\n9\n1 2 1 3 2 3 1 2 1\n9\n2 1 3 1 3 1 3 1 3"] | 2 seconds | ["0\n2 1 2\n1\n1 3 3 1\n1\n1 2 2 2 1\n2\n1 2 3 3 2 3 3 2 1\n2\n2 1 3 3 3 1 1 1 3"] | NoteIn the first example, the array contains no local maximum, so we don't need to perform operations.In the second example, we can change $$$a_2$$$ to $$$3$$$, then the array don't have local maximums. | Java 8 | standard input | [
"greedy"
] | 255abf8750541f54d8ff0f1f80e1f8e7 | Each test contains multiple test cases. The first line will contain a single integer $$$t$$$ $$$(1 \leq t \leq 10000)$$$ — the number of test cases. Then $$$t$$$ test cases follow. The first line of each test case contains a single integer $$$n$$$ $$$(2 \leq n \leq 2 \cdot 10^5)$$$ — the size of the array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots ,a_n$$$ $$$(1 \leq a_i \leq 10^9)$$$, the elements of array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, first output a line containing a single integer $$$m$$$ — minimum number of operations required. Then ouput a line consist of $$$n$$$ integers — the resulting array after the operations. Note that this array should differ in exactly $$$m$$$ elements from the initial array. If there are multiple answers, print any. | standard output | |
PASSED | 56646493ecd10738709751827e4210c7 | train_108.jsonl | 1645367700 | You are given an array $$$a$$$ of size $$$n$$$. Each element in this array is an integer between $$$1$$$ and $$$10^9$$$.You can perform several operations to this array. During an operation, you can replace an element in the array with any integer between $$$1$$$ and $$$10^9$$$. Output the minimum number of operations needed such that the resulting array doesn't contain any local maximums, and the resulting array after the operations.An element $$$a_i$$$ is a local maximum if it is strictly larger than both of its neighbors (that is, $$$a_i > a_{i - 1}$$$ and $$$a_i > a_{i + 1}$$$). Since $$$a_1$$$ and $$$a_n$$$ have only one neighbor each, they will never be a local maximum. | 256 megabytes | import java.io.*;
import java.util.*;
import java.lang.Math;
import static java.lang.Math.max;
import static java.lang.Math.min;
import static java.lang.Math.abs;
public class Main {
static PrintWriter pw;
static Scanner sc;
static StringBuilder ans;
static long mod = 1000000000+7;
static void pn(final Object arg) { pw.print(arg); pw.flush(); }
/*-------------- for input in an value ---------------------*/
static int ni() { return sc.nextInt(); }
static long nl() { return sc.nextLong(); }
static double nd() { return sc.nextDouble(); }
static String ns() { return sc.next(); }
static String nLine() { return sc.nextLine(); }
static void ap(int arg) { ans.append(arg); }
static void ap(long arg) { ans.append(arg); }
static void ap(String arg) { ans.append(arg); }
static void ap(StringBuilder arg) { ans.append(arg); }
static void apn() { ans.append("\n"); }
static void apn(int arg) { ans.append(arg+"\n"); }
static void apn(long arg) { ans.append(arg+"\n"); }
static void apn(String arg) { ans.append(arg+"\n"); }
static void apn(StringBuilder arg) { ans.append(arg+"\n"); }
static void yes() { ap("Yes\n"); }
static void no() { ap("No\n"); }
/* for Dubuggin */
static void printArr(int ar[], int st , int end) { for(int i = st; i<=end; i++ ) ap(ar[i]+ "\n"); ap("\n"); }
static void printArr(long ar[], int st , int end) { for(int i = st; i<=end; i++ ) ap(ar[i]+ "\n"); ap("\n"); }
static void printArr(String ar[], int st , int end) { for(int i = st; i<=end; i++ ) ap(ar[i]+ "\n"); ap("\n"); }
static void printIntegerList(List<Integer> ar, int st , int end) { for(int i = st; i<=end; i++ ) ap(ar.get(i)+ " "); ap("\n"); }
static void printLongList(List<Long> ar, int st , int end) { for(int i = st; i<=end; i++ ) ap(ar.get(i)+ " "); ap("\n"); }
static void printStringList(List<String> ar, int st , int end) { for(int i = st; i<=end; i++ ) ap(ar.get(i)+ " "); ap("\n"); }
/*-------------- for input in an array ---------------------*/
static void readArray(int arr[]){
for(int i=0; i<arr.length; i++)arr[i] = ni();
}
static void readArray(long arr[]){
for(int i=0; i<arr.length; i++)arr[i] = nl();
}
static void readArray(String arr[]){
for(int i=0; i<arr.length; i++)arr[i] = ns();
}
static void readArray(double arr[]){
for(int i=0; i<arr.length; i++)arr[i] = nd();
}
/*-------------- File vs Input ---------------------*/
static void runFile() throws Exception {
sc = new Scanner(new FileReader("input.txt"));
pw = new PrintWriter(new BufferedWriter(new FileWriter("output.txt")));
}
static void runIo() throws Exception {
pw =new PrintWriter(System.out);
sc = new Scanner(System.in);
}
static int log2(int n) { return (int)(Math.log(n) / Math.log(2)); }
static boolean isPowerOfTwo (long x) { return x!=0 && ((x&(x-1)) == 0);}
static long gcd(long a, long b) { if (b == 0) return a; return gcd(b, a % b); }
static long nCr(long n, long r) { // Combinations
if (n < r)
return 0;
if (r > n - r) { // because nCr(n, r) == nCr(n, n - r)
r = n - r;
}
long ans = 1L;
for (long i = 0; i < r; i++) {
ans *= (n - i);
ans /= (i + 1);
}
return ans;
}
static int countDigit(long n){return (int)Math.floor(Math.log10(n) + 1);}
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 (long i = 5; i * i <= n; i = i + 6)
if (n % i == 0 || n % (i + 2) == 0)
return false;
return true;
}
static boolean sv[] = new boolean[10002];
static void seive() {
//true -> not prime
// false->prime
sv[0] = sv[1] = true;
sv[2] = false;
for(int i = 0; i< sv.length; i++) {
if( !sv[i] && (long)i*(long)i < sv.length ) {
for ( int j = i*i; j<sv.length ; j += i ) {
sv[j] = true;
}
}
}
}
static long kadensAlgo(long ar[]) {
int n = ar.length;
long pre = ar[0];
long ans = ar[0];
for(int i = 1; i<n; i++) {
pre = Math.max(pre + ar[i], ar[i]);
ans = Math.max(pre, ans);
}
return ans;
}
static long binpow( long a, long b) {
long res = 1;
while (b > 0) {
if ( (b & 1) > 0){
res = (res * a);
}
a = (a * a);
b >>= 1;
}
return res;
}
static long factorial(long n) {
long res = 1, i;
for (i = 2; i <= n; i++){
res = ((res%mod) * (i%mod))%mod;
}
return res;
}
static int getCountPrime(int n) {
int ans = 0;
while (n%2==0) {
ans ++;
n /= 2;
}
for (int i = 3; i *i<= n; i+= 2) {
while (n%i == 0) {
ans ++;
n /= i;
}
}
if(n > 1 )
ans++;
return ans;
}
static void sort(int[] arr) {
/* because Arrays.sort() uses quicksort which is dumb
Collections.sort() uses merge sort */
ArrayList<Integer> ls = new ArrayList<Integer>();
for(int x: arr)
ls.add(x);
Collections.sort(ls);
for(int i=0; i < arr.length; i++)
arr[i] = ls.get(i);
}
static void sort(long[] arr) {
/* because Arrays.sort() uses quicksort which is dumb
Collections.sort() uses merge sort */
ArrayList<Long> ls = new ArrayList<Long>();
for(Long x: arr)
ls.add(x);
Collections.sort(ls);
for(int i=0; i < arr.length; i++)
arr[i] = ls.get(i);
}
static int lowerBound(ArrayList<Integer> ar, int k, int l, int r) {
while (l <= r) {
int m = (l + r) >> 1 ;
if (ar.get(m) >= k) {
r = m - 1 ;
} else {
l = m + 1 ;
}
}
return l ;
}
static int upperBound(long ar[], long k, int l, int r) {
while (l <= r) {
int m = (l + r) >> 1 ;
if (ar[m] > k) {
r = m - 1 ;
} else {
l = m + 1 ;
}
}
return l ;
}
static int lowerBound(int ar[], int k, int l, int r) {
while (l <= r) {
int m = (l + r) >> 1 ;
if (ar[m] >= k) {
r = m - 1 ;
} else {
l = m + 1 ;
}
}
return l ;
}
static void findDivisor(int n, ArrayList<Integer> al) {
for(int i = 1; i*i <= n; i++){
if( n % i == 0){
if(n/i==i){
al.add(i);
}
else{
al.add(n/i);
al.add(i);
}
}
}
}
static class Pair{
int a;
int b;
Pair( int a, int b ){
this.a = a;
this.b = b;
}
@Override
public boolean equals(Object o) {
if (this == o) return true;
if (!(o instanceof Pair)) return false;
Pair pair = (Pair) o;
return a == pair.a && b == pair.b;
}
@Override
public int hashCode() {
int result = a;
result = 31 * result + b;
return result;
}
}
public static void main(String[] args) throws Exception {
// runFile();
runIo();
int t;
t = 1;
t = sc.nextInt();
ans = new StringBuilder();
while( t-- > 0 ) {
solve();
}
pn(ans+"");
}
public static void solve ( ) {
int n = ni();
int ar[] = new int[n];
readArray(ar);
int ans = 0;
for(int i = 1; i<n-1; i++) {
if( ar[i-1] < ar[i] && ar[i] > ar[i+1]) {
if( i+2 < n ) {
ar[i+1] = max(ar[i], ar[i+2]);
}
else
ar[i+1] = 1000000000;
ans++;
}
}
apn(ans);
for(int i : ar)
ap(i+" ");
apn();
}
} | Java | ["5\n3\n2 1 2\n4\n1 2 3 1\n5\n1 2 1 2 1\n9\n1 2 1 3 2 3 1 2 1\n9\n2 1 3 1 3 1 3 1 3"] | 2 seconds | ["0\n2 1 2\n1\n1 3 3 1\n1\n1 2 2 2 1\n2\n1 2 3 3 2 3 3 2 1\n2\n2 1 3 3 3 1 1 1 3"] | NoteIn the first example, the array contains no local maximum, so we don't need to perform operations.In the second example, we can change $$$a_2$$$ to $$$3$$$, then the array don't have local maximums. | Java 8 | standard input | [
"greedy"
] | 255abf8750541f54d8ff0f1f80e1f8e7 | Each test contains multiple test cases. The first line will contain a single integer $$$t$$$ $$$(1 \leq t \leq 10000)$$$ — the number of test cases. Then $$$t$$$ test cases follow. The first line of each test case contains a single integer $$$n$$$ $$$(2 \leq n \leq 2 \cdot 10^5)$$$ — the size of the array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots ,a_n$$$ $$$(1 \leq a_i \leq 10^9)$$$, the elements of array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, first output a line containing a single integer $$$m$$$ — minimum number of operations required. Then ouput a line consist of $$$n$$$ integers — the resulting array after the operations. Note that this array should differ in exactly $$$m$$$ elements from the initial array. If there are multiple answers, print any. | standard output | |
PASSED | e9203d536bc8ee0923beb8b5b7687481 | train_108.jsonl | 1645367700 | You are given an array $$$a$$$ of size $$$n$$$. Each element in this array is an integer between $$$1$$$ and $$$10^9$$$.You can perform several operations to this array. During an operation, you can replace an element in the array with any integer between $$$1$$$ and $$$10^9$$$. Output the minimum number of operations needed such that the resulting array doesn't contain any local maximums, and the resulting array after the operations.An element $$$a_i$$$ is a local maximum if it is strictly larger than both of its neighbors (that is, $$$a_i > a_{i - 1}$$$ and $$$a_i > a_{i + 1}$$$). Since $$$a_1$$$ and $$$a_n$$$ have only one neighbor each, they will never be a local maximum. | 256 megabytes | //package com.company;
import java.util.*;
import java.lang.*;
import java.io.*;
public class Rough_Work {
public static void main(String[] args) throws IOException {
FastReader sc = new FastReader();
PrintWriter out = new PrintWriter(System.out);
int t = sc.nextInt();
while (t-->0) {
int n = sc.nextInt();
int [] arr = sc.nextIntArray(n);
if (n == 2) {
out.println(0+"\n"+arr[0]+" "+arr[1]);
continue;
}
ArrayList<Integer> x = new ArrayList<>();
for (int i = 1; i < n-1; i++) {
if (arr[i-1] < arr[i] && arr[i] > arr[i+1]) x.add(i);
}
// for (int z: x) out.print(z+" ");
// out.println();
int count = 0;
for (int i = 0; i < x.size(); i++) {
if (x.size()-1 != i && x.get(i+1) - x.get(i) == 2) {
arr[x.get(i)+1] = Math.max(arr[x.get(i)],arr[x.get(i)+2]);
i++;
}
else arr[x.get(i)] = Math.max(arr[x.get(i)+1],arr[x.get(i)-1]);
count++;
}
out.println(count);
for (int z: arr) out.print(z+" ");
out.println();
}
sc.close();
out.close();
}
static class FastReader {
final private int BUFFER_SIZE = 1 << 17;
private DataInputStream din;
private byte[] buffer;
private int bufferPointer, bytesRead;
public FastReader()
{
din = new DataInputStream(System.in);
buffer = new byte[BUFFER_SIZE];
bufferPointer = bytesRead = 0;
}
public FastReader(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[128]; // 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;
}
public int[] nextIntArray(int size) throws IOException {
int[] arr = new int[size];
for (int i = 0; i < size; i++) {
arr[i] = nextInt();
}
return arr;
}
public long[] nextLongArray(int size) throws IOException {
long[] arr = new long[size];
for (int i = 0; i < size; i++) {
arr[i] = nextLong();
}
return arr;
}
public Double[] nextDoubleArray(int size) throws IOException {
Double[] arr = new Double[size];
for (int i = 0; i < size; i++) {
arr[i] = nextDouble();
}
return arr;
}
public int[][] nextIntMatrix(int n, int m) throws IOException {
int[][] arr = new int[n][m];
for (int i = 0; i < n; i++) {
arr[i] = nextIntArray(m);
}
return arr;
}
public long[][] nextLongMatrix(int n, int m) throws IOException {
long[][] arr = new long[n][m];
for (int i = 0; i < n; i++) {
arr[i] = nextLongArray(m);
}
return arr;
}
public ArrayList<Integer> nextIntList(int size) throws IOException {
ArrayList<Integer> arrayList = new ArrayList<>(size);
for (int i = 0; i < size; i++) {
arrayList.add(nextInt());
}
return arrayList;
}
public ArrayList<Long> nextLongList(int size) throws IOException {
ArrayList<Long> arrayList = new ArrayList<>(size);
for (int i = 0; i < size; i++) {
arrayList.add(nextLong());
}
return arrayList;
}
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();
}
}
}
| Java | ["5\n3\n2 1 2\n4\n1 2 3 1\n5\n1 2 1 2 1\n9\n1 2 1 3 2 3 1 2 1\n9\n2 1 3 1 3 1 3 1 3"] | 2 seconds | ["0\n2 1 2\n1\n1 3 3 1\n1\n1 2 2 2 1\n2\n1 2 3 3 2 3 3 2 1\n2\n2 1 3 3 3 1 1 1 3"] | NoteIn the first example, the array contains no local maximum, so we don't need to perform operations.In the second example, we can change $$$a_2$$$ to $$$3$$$, then the array don't have local maximums. | Java 8 | standard input | [
"greedy"
] | 255abf8750541f54d8ff0f1f80e1f8e7 | Each test contains multiple test cases. The first line will contain a single integer $$$t$$$ $$$(1 \leq t \leq 10000)$$$ — the number of test cases. Then $$$t$$$ test cases follow. The first line of each test case contains a single integer $$$n$$$ $$$(2 \leq n \leq 2 \cdot 10^5)$$$ — the size of the array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots ,a_n$$$ $$$(1 \leq a_i \leq 10^9)$$$, the elements of array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, first output a line containing a single integer $$$m$$$ — minimum number of operations required. Then ouput a line consist of $$$n$$$ integers — the resulting array after the operations. Note that this array should differ in exactly $$$m$$$ elements from the initial array. If there are multiple answers, print any. | standard output | |
PASSED | 701c33f8235645352136be9294067ac4 | train_108.jsonl | 1645367700 | You are given an array $$$a$$$ of size $$$n$$$. Each element in this array is an integer between $$$1$$$ and $$$10^9$$$.You can perform several operations to this array. During an operation, you can replace an element in the array with any integer between $$$1$$$ and $$$10^9$$$. Output the minimum number of operations needed such that the resulting array doesn't contain any local maximums, and the resulting array after the operations.An element $$$a_i$$$ is a local maximum if it is strictly larger than both of its neighbors (that is, $$$a_i > a_{i - 1}$$$ and $$$a_i > a_{i + 1}$$$). Since $$$a_1$$$ and $$$a_n$$$ have only one neighbor each, they will never be a local maximum. | 256 megabytes | import java.util.ArrayList;
import java.util.Collections;
import java.util.Scanner;
import static java.lang.Math.max;
public class main {
public static void main(String[] args){
Scanner input = new Scanner(System.in);
int n = input.nextInt();
for (int i = 0; i<n;i++){
int m = input.nextInt();
int res = 0;
ArrayList<Integer> arr = new ArrayList<Integer>();
for (int j=0;j<m;j++){
arr.add(input.nextInt());
}
/* for (int j=1;j<m-1;j++)
{
if (arr.get(j) > arr.get(j - 1)){
if (arr.get(j) > arr.get(j + 1)) {
locmax.set(j, 1);
}
}
}*/
for (int j = 1; j<m;j++){
boolean locmax1 = false;
boolean locmax2 = false;
boolean locmax3 = false;
if (j < m-1 && arr.get(j) > arr.get(j - 1) && arr.get(j) > arr.get(j + 1))
locmax1 = true;
if (j > 2 && arr.get(j-2) > arr.get(j - 3) && arr.get(j-2) > arr.get(j -1))
locmax2 = true;
if (j < m-3 && arr.get(j+2) > arr.get(j +1) && arr.get(j+2) > arr.get(j +3))
locmax3 = true;
//System.out.println(locmax1 + " " + locmax2 + " " + locmax3);
if (locmax1 && locmax2){
res++;
if (arr.get(j - 2) >= arr.get(j)){
arr.set(j - 1, arr.get(j - 2));
}
else {
arr.set(j - 1, arr.get(j));
}
}
else if (locmax1 && !locmax3){
res++;
arr.set(j, max(arr.get(j+1), arr.get(j - 1)));
}
}
System.out.println(res);
for (int j=0;j<m;j++){
System.out.print(arr.get(j) + " ");
}
System.out.println();
}
}
}
| Java | ["5\n3\n2 1 2\n4\n1 2 3 1\n5\n1 2 1 2 1\n9\n1 2 1 3 2 3 1 2 1\n9\n2 1 3 1 3 1 3 1 3"] | 2 seconds | ["0\n2 1 2\n1\n1 3 3 1\n1\n1 2 2 2 1\n2\n1 2 3 3 2 3 3 2 1\n2\n2 1 3 3 3 1 1 1 3"] | NoteIn the first example, the array contains no local maximum, so we don't need to perform operations.In the second example, we can change $$$a_2$$$ to $$$3$$$, then the array don't have local maximums. | Java 8 | standard input | [
"greedy"
] | 255abf8750541f54d8ff0f1f80e1f8e7 | Each test contains multiple test cases. The first line will contain a single integer $$$t$$$ $$$(1 \leq t \leq 10000)$$$ — the number of test cases. Then $$$t$$$ test cases follow. The first line of each test case contains a single integer $$$n$$$ $$$(2 \leq n \leq 2 \cdot 10^5)$$$ — the size of the array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots ,a_n$$$ $$$(1 \leq a_i \leq 10^9)$$$, the elements of array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, first output a line containing a single integer $$$m$$$ — minimum number of operations required. Then ouput a line consist of $$$n$$$ integers — the resulting array after the operations. Note that this array should differ in exactly $$$m$$$ elements from the initial array. If there are multiple answers, print any. | standard output | |
PASSED | b04b7aa3ee5bb2791eaf66b18d2e1bcd | train_108.jsonl | 1645367700 | You are given an array $$$a$$$ of size $$$n$$$. Each element in this array is an integer between $$$1$$$ and $$$10^9$$$.You can perform several operations to this array. During an operation, you can replace an element in the array with any integer between $$$1$$$ and $$$10^9$$$. Output the minimum number of operations needed such that the resulting array doesn't contain any local maximums, and the resulting array after the operations.An element $$$a_i$$$ is a local maximum if it is strictly larger than both of its neighbors (that is, $$$a_i > a_{i - 1}$$$ and $$$a_i > a_{i + 1}$$$). Since $$$a_1$$$ and $$$a_n$$$ have only one neighbor each, they will never be a local maximum. | 256 megabytes | import java.util.*;
import java.io.*;
// cd C:\Users\Lenovo\Desktop\New
//ArrayList<Integer> a=new ArrayList<>();
//List<Integer> lis=new ArrayList<>();
//StringBuilder ans = new StringBuilder();
//HashMap<Integer,Integer> map=new HashMap<>();
public class cf {
static FastReader in=new FastReader();
static final Random random=new Random();
static long mod=1000000007L;
public static void main(String args[]) throws IOException {
FastReader sc=new FastReader();
//Scanner s=new Scanner(System.in);
int tt=sc.nextInt();
//int tt=1;
while(tt-->0){
//StringBuilder ar = new StringBuilder();
int n=sc.nextInt();
int arr[]=new int[n];
int max=0;
long sum=0;
for(int i=0;i<n;i++){
arr[i]=sc.nextInt();
}
int ans=0;
for(int i=1;i<n-1;i++){
if(arr[i]>arr[i-1] && arr[i]>arr[i+1]){
ans++;
if(i!=n-2){
arr[i+1]=Math.max(arr[i],arr[i+2]);
}
else{
arr[i+1]=arr[i];
}
}
}
System.out.println(ans);
for(int i=0;i<n;i++){
System.out.print(arr[i]+" ");
}
System.out.println();
}
}
static int countprimefactors(int n){
int ans=0;
int z=(int)Math.sqrt(n);
for(int i=2;i<=z;i++){
while(n%i==0){
ans++;
n=n/i;
}
}
if(n>1){
ans++;
}
return ans;
}
static int count(int arr[],int idx,long sum,int []dp){
if(idx>=arr.length){
return 0;
}
if(dp[idx]!=-1){
return dp[idx];
}
if(arr[idx]<0){
return dp[idx]=Math.max(1+count(arr,idx+1,sum+arr[idx],dp),count(arr,idx+1,sum,dp));
}
else{
return dp[idx]=1+count(arr,idx+1,sum+arr[idx],dp);
}
}
static String reverse(String s){
String ans="";
for(int i=s.length()-1;i>=0;i--){
ans+=s.charAt(i);
}
return ans;
}
static int find_max(int []check,int y){
int max=0;
for(int i=y;i>=0;i--){
if(check[i]!=0){
max=i;
break;
}
}
return max;
}
static void dfs(int [][]arr,int row,int col,boolean [][]seen,int n){
if(row<0 || col<0 || row==n || col==n){
return;
}
if(arr[row][col]==1){
return;
}
if(seen[row][col]==true){
return;
}
seen[row][col]=true;
dfs(arr,row+1,col,seen,n);
dfs(arr,row,col+1,seen,n);
dfs(arr,row-1,col,seen,n);
dfs(arr,row,col-1,seen,n);
}
static int msb(int x){
int ans=0;
while(x!=0){
x=x/2;
ans++;
}
return ans;
}
static void ruffleSort(int[] a) {
int n=a.length;
for (int i=0; i<n; i++) {
int oi=random.nextInt(n), temp=a[oi];
a[oi]=a[i]; a[i]=temp;
}
Arrays.sort(a);
}
static int gcd(int a,int b)
{
if(b==0)
{
return a;
}
return gcd(b,a%b);
}
/* static boolean checkprime(int n1)
{
if(n1%2==0||n1%3==0)
return false;
else
{
for(int i=5;i*i<=n1;i+=6)
{
if(n1%i==0||n1%(i+2)==0)
return false;
}
return true;
}
}
*/
/* Iterator<Map.Entry<Integer, Integer>> iterator = map.entrySet().iterator();
while(iterator.hasNext()){
Map.Entry<Integer, Integer> entry = iterator.next();
int value = entry.getValue();
if(value==1){
iterator.remove();
}
else{
entry.setValue(value-1);
}
}
*/
static class Pair implements Comparable
{
int a,b;
public String toString()
{
return a+" " + b;
}
public Pair(int x , int y)
{
a=x;b=y;
}
@Override
public int compareTo(Object o) {
Pair p = (Pair)o;
if(p.a!=a){
return a-p.a;//in
}
else{
return b-p.b;//
}
}
}
/* public static boolean checkAP(List<Integer> lis){
for(int i=1;i<lis.size()-1;i++){
if(2*lis.get(i)!=lis.get(i-1)+lis.get(i+1)){
return false;
}
}
return true;
}
public static int minBS(int[]arr,int val){
int l=-1;
int r=arr.length;
while(r>l+1){
int mid=(l+r)/2;
if(arr[mid]>=val){
r=mid;
}
else{
l=mid;
}
}
return r;
}
public static int maxBS(int[]arr,int val){
int l=-1;
int r=arr.length;
while(r>l+1){
int mid=(l+r)/2;
if(arr[mid]<=val){
l=mid;
}
else{
r=mid;
}
}
return l;
}
static int lcm(int a, int b)
{
return (a / gcd(a, b)) * b;
}*/
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 | ["5\n3\n2 1 2\n4\n1 2 3 1\n5\n1 2 1 2 1\n9\n1 2 1 3 2 3 1 2 1\n9\n2 1 3 1 3 1 3 1 3"] | 2 seconds | ["0\n2 1 2\n1\n1 3 3 1\n1\n1 2 2 2 1\n2\n1 2 3 3 2 3 3 2 1\n2\n2 1 3 3 3 1 1 1 3"] | NoteIn the first example, the array contains no local maximum, so we don't need to perform operations.In the second example, we can change $$$a_2$$$ to $$$3$$$, then the array don't have local maximums. | Java 8 | standard input | [
"greedy"
] | 255abf8750541f54d8ff0f1f80e1f8e7 | Each test contains multiple test cases. The first line will contain a single integer $$$t$$$ $$$(1 \leq t \leq 10000)$$$ — the number of test cases. Then $$$t$$$ test cases follow. The first line of each test case contains a single integer $$$n$$$ $$$(2 \leq n \leq 2 \cdot 10^5)$$$ — the size of the array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots ,a_n$$$ $$$(1 \leq a_i \leq 10^9)$$$, the elements of array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, first output a line containing a single integer $$$m$$$ — minimum number of operations required. Then ouput a line consist of $$$n$$$ integers — the resulting array after the operations. Note that this array should differ in exactly $$$m$$$ elements from the initial array. If there are multiple answers, print any. | standard output | |
PASSED | ae68dd47dd44c2979fdcd197f79a887b | train_108.jsonl | 1645367700 | You are given an array $$$a$$$ of size $$$n$$$. Each element in this array is an integer between $$$1$$$ and $$$10^9$$$.You can perform several operations to this array. During an operation, you can replace an element in the array with any integer between $$$1$$$ and $$$10^9$$$. Output the minimum number of operations needed such that the resulting array doesn't contain any local maximums, and the resulting array after the operations.An element $$$a_i$$$ is a local maximum if it is strictly larger than both of its neighbors (that is, $$$a_i > a_{i - 1}$$$ and $$$a_i > a_{i + 1}$$$). Since $$$a_1$$$ and $$$a_n$$$ have only one neighbor each, they will never be a local maximum. | 256 megabytes |
import java.io.*;
import java.util.*;
public class Feb20P2 {
public static void main(String[] args) {
MyScanner sc = new MyScanner();
out = new PrintWriter(new BufferedOutputStream(System.out));
int t = sc.nextInt();
for (int i = 0; i < t; i++) {
int n = sc.nextInt();
long[] o = new long[n];
long[] o1 = new long[n];
for (int j = 0; j < n; j++) {
o[j] = sc.nextLong();
o1[j] = o[j];
}
ArrayList<Integer> om = new ArrayList<Integer>();
for (int j = 1; j < n-1; j++) {
if(o[j] > o[j-1] && o[j] > o[j+1])
om.add(j);
}
long count = 0;
int ind = -1;
boolean swi = true;
for (int j = 1; j < n - 1; j++) {
if (o[j] > o[j - 1] && o[j] > o[j + 1]) {
ind++;
if(om.size() > ind+1)
{
int dist = om.get(ind+1) - om.get(ind) - 1;
if(dist == 1)
{
//out.println("YESA" + dist + " " + j + " " + Math.max(o[om.get(ind+1)], o[om.get(ind)]));
for(int k = j+1; k <= j + dist; k++)
{
o[k] = Math.max(o[om.get(ind+1)], o[om.get(ind)]);
}
count += dist;
ind++;
}
else
{
o[j] = Math.max(o[j-1], o[j+1]);
count++;
}
}
else
{
count++;
o[j] = Math.max(o[j-1], o[j+1]);
}
}
}
out.println(count);
for (int j = 0; j < n; j++) {
out.print(o[j] + " ");
}
out.println();
}
// Start writing your solution here. -------------------------------------
/*
* int n = sc.nextInt(); // read input as integer long k = sc.nextLong(); //
* read input as long double d = sc.nextDouble(); // read input as double String
* str = sc.next(); // read input as String String s = sc.nextLine(); // read
* whole line as String
*
* int result = 3*n; out.println(result); // print via PrintWriter
*/
// 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();
}
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 | ["5\n3\n2 1 2\n4\n1 2 3 1\n5\n1 2 1 2 1\n9\n1 2 1 3 2 3 1 2 1\n9\n2 1 3 1 3 1 3 1 3"] | 2 seconds | ["0\n2 1 2\n1\n1 3 3 1\n1\n1 2 2 2 1\n2\n1 2 3 3 2 3 3 2 1\n2\n2 1 3 3 3 1 1 1 3"] | NoteIn the first example, the array contains no local maximum, so we don't need to perform operations.In the second example, we can change $$$a_2$$$ to $$$3$$$, then the array don't have local maximums. | Java 8 | standard input | [
"greedy"
] | 255abf8750541f54d8ff0f1f80e1f8e7 | Each test contains multiple test cases. The first line will contain a single integer $$$t$$$ $$$(1 \leq t \leq 10000)$$$ — the number of test cases. Then $$$t$$$ test cases follow. The first line of each test case contains a single integer $$$n$$$ $$$(2 \leq n \leq 2 \cdot 10^5)$$$ — the size of the array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots ,a_n$$$ $$$(1 \leq a_i \leq 10^9)$$$, the elements of array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, first output a line containing a single integer $$$m$$$ — minimum number of operations required. Then ouput a line consist of $$$n$$$ integers — the resulting array after the operations. Note that this array should differ in exactly $$$m$$$ elements from the initial array. If there are multiple answers, print any. | standard output | |
PASSED | d15d6a0e01fc45563086c6fffef6cefa | train_108.jsonl | 1645367700 | You are given an array $$$a$$$ of size $$$n$$$. Each element in this array is an integer between $$$1$$$ and $$$10^9$$$.You can perform several operations to this array. During an operation, you can replace an element in the array with any integer between $$$1$$$ and $$$10^9$$$. Output the minimum number of operations needed such that the resulting array doesn't contain any local maximums, and the resulting array after the operations.An element $$$a_i$$$ is a local maximum if it is strictly larger than both of its neighbors (that is, $$$a_i > a_{i - 1}$$$ and $$$a_i > a_{i + 1}$$$). Since $$$a_1$$$ and $$$a_n$$$ have only one neighbor each, they will never be a local maximum. | 256 megabytes | import java.io.*;
import java.util.*;
public class B {
public static void main(String args[]){
FScanner in = new FScanner();
PrintWriter out = new PrintWriter(System.out);
int t = in.nextInt();
while(t-->0) {
int n=in.nextInt();
int arr[]=in.readArray(n);
ArrayList<Integer> list =new ArrayList<>();
int c=0;
for(int i=1;i<n-1;i++)
{
if(arr[i]>arr[i+1]&&arr[i]>arr[i-1])
{
//c++;
list.add(i);
}
}
int v=-1;
int count=0;
for(int i=0;i<list.size();i++)
{
int a=list.get(i);
if(i==list.size()-1)
{
count++;
arr[a]=Math.max(arr[a-1],arr[a+1]);
break;
}
int b=list.get(i+1);
v=b;
if(b-a==2)
{
arr[b-1]=Math.max(arr[b],arr[a]);
count++;
i++;
}
else
{
arr[a]=Math.max(arr[a-1],arr[a+1]);
count++;
}
}
out.println(count);
for(int i=0;i<n;i++)
out.print(arr[i]+" ");
out.println();
}
out.close();
}
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 boolean checkprime(int n1)
{
if(n1%2==0||n1%3==0)
return false;
else
{
for(int i=5;i*i<=n1;i+=6)
{
if(n1%i==0||n1%(i+2)==0)
return false;
}
return true;
}
}
static class FScanner {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
StringTokenizer sb = new StringTokenizer("");
String next(){
while(!sb.hasMoreTokens()){
try{
sb = new StringTokenizer(br.readLine());
} catch(IOException e){ }
}
return sb.nextToken();
}
String nextLine(){
try{ return br.readLine(); }
catch(IOException e) { } return "";
}
int nextInt(){
return Integer.parseInt(next());
}
long nextLong() {
return Long.parseLong(next());
}
int[] readArray(int n) {
int a[] = new int[n];
for(int i=0;i<n;i++) a[i] = nextInt();
return a;
}
float nextFloat(){
return Float.parseFloat(next());
}
double nextDouble(){
return Double.parseDouble(next());
}
}
}
| Java | ["5\n3\n2 1 2\n4\n1 2 3 1\n5\n1 2 1 2 1\n9\n1 2 1 3 2 3 1 2 1\n9\n2 1 3 1 3 1 3 1 3"] | 2 seconds | ["0\n2 1 2\n1\n1 3 3 1\n1\n1 2 2 2 1\n2\n1 2 3 3 2 3 3 2 1\n2\n2 1 3 3 3 1 1 1 3"] | NoteIn the first example, the array contains no local maximum, so we don't need to perform operations.In the second example, we can change $$$a_2$$$ to $$$3$$$, then the array don't have local maximums. | Java 8 | standard input | [
"greedy"
] | 255abf8750541f54d8ff0f1f80e1f8e7 | Each test contains multiple test cases. The first line will contain a single integer $$$t$$$ $$$(1 \leq t \leq 10000)$$$ — the number of test cases. Then $$$t$$$ test cases follow. The first line of each test case contains a single integer $$$n$$$ $$$(2 \leq n \leq 2 \cdot 10^5)$$$ — the size of the array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots ,a_n$$$ $$$(1 \leq a_i \leq 10^9)$$$, the elements of array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, first output a line containing a single integer $$$m$$$ — minimum number of operations required. Then ouput a line consist of $$$n$$$ integers — the resulting array after the operations. Note that this array should differ in exactly $$$m$$$ elements from the initial array. If there are multiple answers, print any. | standard output | |
PASSED | 79450d038c6830d4a0d731e7f97d3053 | train_108.jsonl | 1645367700 | You are given an array $$$a$$$ of size $$$n$$$. Each element in this array is an integer between $$$1$$$ and $$$10^9$$$.You can perform several operations to this array. During an operation, you can replace an element in the array with any integer between $$$1$$$ and $$$10^9$$$. Output the minimum number of operations needed such that the resulting array doesn't contain any local maximums, and the resulting array after the operations.An element $$$a_i$$$ is a local maximum if it is strictly larger than both of its neighbors (that is, $$$a_i > a_{i - 1}$$$ and $$$a_i > a_{i + 1}$$$). Since $$$a_1$$$ and $$$a_n$$$ have only one neighbor each, they will never be a local maximum. | 256 megabytes | import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.*;
public class inversion_graph {
static BufferedReader bf;
public static void main(String[]args)throws IOException{
bf = new BufferedReader(new InputStreamReader(System.in));
int t = nextInt();
while(t-->0){
int n = nextInt();
int[]arr = nextIntArray(n);
int count = 0;
for(int i = 1;i<arr.length-1;i++){
if(arr[i]>arr[i-1] && arr[i]>arr[i+1]){
count ++;
if(i+2<arr.length){
arr[i+1] = Math.max(arr[i],arr[i+2]);
}
else{
arr[i] = Math.max(arr[i-1],arr[i+1]);
}
}
}
System.out.println( count);
for(int i = 0;i<arr.length;i++){
System.out.print(arr[i] );
System.out.print(" ");
}
System.out.println();
}
}
// code for input
public static int nextInt()throws IOException{
return Integer.parseInt(bf.readLine());
}
public static long nextLong()throws IOException{
return Long.parseLong(bf.readLine());
}
public static String nextString()throws IOException{
return bf.readLine();
}
public static int[] nextIntArray(int n)throws IOException{
String[]str = bf.readLine().split(" ");
int[]arr = new int[n];
for(int i =0;i<n;i++){
arr[i] = Integer.parseInt(str[i]);
}
return arr;
}
public static long[] nextLongArray(int n)throws IOException{
String[]str = bf.readLine().split(" ");
long[]arr = new long[n];
for(int i =0;i<n;i++){
arr[i] = Long.parseLong(str[i]);
}
return arr;
}
public static int[][] newIntMatrix(int r,int c)throws IOException{
int[][]arr = new int[r][c];
for(int i =0;i<r;i++){
String[]str = bf.readLine().split(" ");
for(int j =0;j<c;j++){
arr[i][j] = Integer.parseInt(str[j]);
}
}
return arr;
}
public static long[][] newLongMatrix(int r,int c)throws IOException{
long[][]arr = new long[r][c];
for(int i =0;i<r;i++){
String[]str = bf.readLine().split(" ");
for(int j =0;j<c;j++){
arr[i][j] = Long.parseLong(str[j]);
}
}
return arr;
}
public static void print(String s){
System.out.println(s);
}
public static void print(int a){
System.out.println(a);
}
public static void print(long a ){
System.out.println(a);
}
} | Java | ["5\n3\n2 1 2\n4\n1 2 3 1\n5\n1 2 1 2 1\n9\n1 2 1 3 2 3 1 2 1\n9\n2 1 3 1 3 1 3 1 3"] | 2 seconds | ["0\n2 1 2\n1\n1 3 3 1\n1\n1 2 2 2 1\n2\n1 2 3 3 2 3 3 2 1\n2\n2 1 3 3 3 1 1 1 3"] | NoteIn the first example, the array contains no local maximum, so we don't need to perform operations.In the second example, we can change $$$a_2$$$ to $$$3$$$, then the array don't have local maximums. | Java 8 | standard input | [
"greedy"
] | 255abf8750541f54d8ff0f1f80e1f8e7 | Each test contains multiple test cases. The first line will contain a single integer $$$t$$$ $$$(1 \leq t \leq 10000)$$$ — the number of test cases. Then $$$t$$$ test cases follow. The first line of each test case contains a single integer $$$n$$$ $$$(2 \leq n \leq 2 \cdot 10^5)$$$ — the size of the array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots ,a_n$$$ $$$(1 \leq a_i \leq 10^9)$$$, the elements of array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, first output a line containing a single integer $$$m$$$ — minimum number of operations required. Then ouput a line consist of $$$n$$$ integers — the resulting array after the operations. Note that this array should differ in exactly $$$m$$$ elements from the initial array. If there are multiple answers, print any. | standard output | |
PASSED | 89903a16ff5208461bd638246d703d29 | train_108.jsonl | 1645367700 | You are given an array $$$a$$$ of size $$$n$$$. Each element in this array is an integer between $$$1$$$ and $$$10^9$$$.You can perform several operations to this array. During an operation, you can replace an element in the array with any integer between $$$1$$$ and $$$10^9$$$. Output the minimum number of operations needed such that the resulting array doesn't contain any local maximums, and the resulting array after the operations.An element $$$a_i$$$ is a local maximum if it is strictly larger than both of its neighbors (that is, $$$a_i > a_{i - 1}$$$ and $$$a_i > a_{i + 1}$$$). Since $$$a_1$$$ and $$$a_n$$$ have only one neighbor each, they will never be a local maximum. | 256 megabytes | import java.io.*;
import java.util.*;
public class B
{
public static void main(String[] args) throws IOException
{
int N = 200010;
int[] a = new int[N];
boolean[] u = new boolean[N];
in.nextToken();
int t = (int) in.nval;
while (t-- > 0)
{
in.nextToken();
int n = (int) in.nval;
for (int i = 0; i < n; i++)
{
in.nextToken();
a[i] = (int) in.nval;
u[i] = false;
}
for (int i = 1; i < n - 1; i++)
{
if (a[i] > a[i + 1] && a[i] > a[i - 1])
{
u[i] = true;
}
}
int sum = 0;
for (int i = 1; i < n - 1; i++)
{
if (u[i])
{
if (i + 2 < n)
{
a[i + 1] = Math.max(a[i], a[i + 2]);
u[i] = false;
u[i + 2] = false;
} else
{
a[i + 1] = a[i];
u[i] = false;
}
sum++;
}
}
out.println(sum);
for (int i = 0; i < n; i++)
out.print(a[i] + " ");
out.println();
out.flush();
}
out.close();
}
// static Scanner in = new Scanner(System.in);
// static BufferedReader in = new BufferedReader(new
// InputStreamReader(System.in));
static StreamTokenizer in = new StreamTokenizer(new BufferedReader(new InputStreamReader(System.in)));
static PrintWriter out = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));
} | Java | ["5\n3\n2 1 2\n4\n1 2 3 1\n5\n1 2 1 2 1\n9\n1 2 1 3 2 3 1 2 1\n9\n2 1 3 1 3 1 3 1 3"] | 2 seconds | ["0\n2 1 2\n1\n1 3 3 1\n1\n1 2 2 2 1\n2\n1 2 3 3 2 3 3 2 1\n2\n2 1 3 3 3 1 1 1 3"] | NoteIn the first example, the array contains no local maximum, so we don't need to perform operations.In the second example, we can change $$$a_2$$$ to $$$3$$$, then the array don't have local maximums. | Java 8 | standard input | [
"greedy"
] | 255abf8750541f54d8ff0f1f80e1f8e7 | Each test contains multiple test cases. The first line will contain a single integer $$$t$$$ $$$(1 \leq t \leq 10000)$$$ — the number of test cases. Then $$$t$$$ test cases follow. The first line of each test case contains a single integer $$$n$$$ $$$(2 \leq n \leq 2 \cdot 10^5)$$$ — the size of the array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots ,a_n$$$ $$$(1 \leq a_i \leq 10^9)$$$, the elements of array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, first output a line containing a single integer $$$m$$$ — minimum number of operations required. Then ouput a line consist of $$$n$$$ integers — the resulting array after the operations. Note that this array should differ in exactly $$$m$$$ elements from the initial array. If there are multiple answers, print any. | standard output | |
PASSED | 080f80275dc8e70e2059959a92cb192a | train_108.jsonl | 1645367700 | You are given an array $$$a$$$ of size $$$n$$$. Each element in this array is an integer between $$$1$$$ and $$$10^9$$$.You can perform several operations to this array. During an operation, you can replace an element in the array with any integer between $$$1$$$ and $$$10^9$$$. Output the minimum number of operations needed such that the resulting array doesn't contain any local maximums, and the resulting array after the operations.An element $$$a_i$$$ is a local maximum if it is strictly larger than both of its neighbors (that is, $$$a_i > a_{i - 1}$$$ and $$$a_i > a_{i + 1}$$$). Since $$$a_1$$$ and $$$a_n$$$ have only one neighbor each, they will never be a local maximum. | 256 megabytes | import java.util.*;
import java.io.*;
public class test {
public static void main(String[] args) throws IOException {
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 = new int[n];
for (int i = 0; i < arr.length; i++) {
arr[i] = sc.nextInt();
}
int c = 0;
for (int i = 1; i < arr.length-1; i++) {
if(arr[i]>arr[i-1]&&arr[i]>arr[i+1]) {
if(i==arr.length-2) {
c++;
arr[i+1] = arr[i];
}
else {
if(arr[i+2]>=arr[i]) {
c++;
arr[i+1] = arr[i+2];
}
else {
c++;
arr[i+1] = arr[i];
}
}
}
}
pw.println(c);
StringBuilder res = new StringBuilder();
for (int i = 0; i < arr.length; i++) {
res.append(arr[i]);
if(i!=arr.length-1)
res.append(" ");
}
pw.println(res);
}
pw.flush();
}
}
class item{
int val;
int c;
int index;
public item(int val , int index) {
this.val = val;
this.index = index;
}
}
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 boolean ready() throws IOException {
return br.ready();
}
} | Java | ["5\n3\n2 1 2\n4\n1 2 3 1\n5\n1 2 1 2 1\n9\n1 2 1 3 2 3 1 2 1\n9\n2 1 3 1 3 1 3 1 3"] | 2 seconds | ["0\n2 1 2\n1\n1 3 3 1\n1\n1 2 2 2 1\n2\n1 2 3 3 2 3 3 2 1\n2\n2 1 3 3 3 1 1 1 3"] | NoteIn the first example, the array contains no local maximum, so we don't need to perform operations.In the second example, we can change $$$a_2$$$ to $$$3$$$, then the array don't have local maximums. | Java 8 | standard input | [
"greedy"
] | 255abf8750541f54d8ff0f1f80e1f8e7 | Each test contains multiple test cases. The first line will contain a single integer $$$t$$$ $$$(1 \leq t \leq 10000)$$$ — the number of test cases. Then $$$t$$$ test cases follow. The first line of each test case contains a single integer $$$n$$$ $$$(2 \leq n \leq 2 \cdot 10^5)$$$ — the size of the array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots ,a_n$$$ $$$(1 \leq a_i \leq 10^9)$$$, the elements of array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, first output a line containing a single integer $$$m$$$ — minimum number of operations required. Then ouput a line consist of $$$n$$$ integers — the resulting array after the operations. Note that this array should differ in exactly $$$m$$$ elements from the initial array. If there are multiple answers, print any. | standard output | |
PASSED | 034191e8dac6b202ffae195d7b02b208 | train_108.jsonl | 1645367700 | You are given an array $$$a$$$ of size $$$n$$$. Each element in this array is an integer between $$$1$$$ and $$$10^9$$$.You can perform several operations to this array. During an operation, you can replace an element in the array with any integer between $$$1$$$ and $$$10^9$$$. Output the minimum number of operations needed such that the resulting array doesn't contain any local maximums, and the resulting array after the operations.An element $$$a_i$$$ is a local maximum if it is strictly larger than both of its neighbors (that is, $$$a_i > a_{i - 1}$$$ and $$$a_i > a_{i + 1}$$$). Since $$$a_1$$$ and $$$a_n$$$ have only one neighbor each, they will never be a local maximum. | 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
*/
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);
TaskB solver = new TaskB();
int testCount = Integer.parseInt(in.next());
for (int i = 1; i <= testCount; i++)
solver.solve(i, in, out);
out.close();
}
static class TaskB {
public void solve(int testNumber, InputReader in, PrintWriter out) {
int a[] = in.parseInt1D(in.nextInt());
int ct = 0;
boolean changed[] = new boolean[a.length];
for (int i = 1; i < a.length - 1; i++) {
int p = a[i - 1];
int c = a[i];
int n = a[i + 1];
if (c > p && c > n) {
if (changed[i - 1]) {
a[i - 1] = c;
} else {
a[i + 1] = c;
changed[i + 1] = true;
ct++;
}
}
}
out.println(ct);
for (int v : a) {
out.print(v + " ");
}
out.println();
}
}
static class InputReader {
private InputStream stream;
private byte[] buf = new byte[1024];
private int curChar;
private int numChars;
private InputReader.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 readInt() {
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 readString() {
int c = read();
while (isSpaceChar(c)) {
c = read();
}
StringBuilder res = new StringBuilder();
do {
if (Character.isValidCodePoint(c)) {
res.appendCodePoint(c);
}
c = read();
} while (!isSpaceChar(c));
return res.toString();
}
public boolean isSpaceChar(int c) {
if (filter != null) {
return filter.isSpaceChar(c);
}
return isWhitespace(c);
}
public static boolean isWhitespace(int c) {
return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
}
public String next() {
return readString();
}
public int nextInt() {
return readInt();
}
public int[] parseInt1D(int n) {
int r[] = new int[n];
for (int i = 0; i < n; i++) {
r[i] = nextInt();
}
return r;
}
public interface SpaceCharFilter {
public boolean isSpaceChar(int ch);
}
}
}
| Java | ["5\n3\n2 1 2\n4\n1 2 3 1\n5\n1 2 1 2 1\n9\n1 2 1 3 2 3 1 2 1\n9\n2 1 3 1 3 1 3 1 3"] | 2 seconds | ["0\n2 1 2\n1\n1 3 3 1\n1\n1 2 2 2 1\n2\n1 2 3 3 2 3 3 2 1\n2\n2 1 3 3 3 1 1 1 3"] | NoteIn the first example, the array contains no local maximum, so we don't need to perform operations.In the second example, we can change $$$a_2$$$ to $$$3$$$, then the array don't have local maximums. | Java 8 | standard input | [
"greedy"
] | 255abf8750541f54d8ff0f1f80e1f8e7 | Each test contains multiple test cases. The first line will contain a single integer $$$t$$$ $$$(1 \leq t \leq 10000)$$$ — the number of test cases. Then $$$t$$$ test cases follow. The first line of each test case contains a single integer $$$n$$$ $$$(2 \leq n \leq 2 \cdot 10^5)$$$ — the size of the array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots ,a_n$$$ $$$(1 \leq a_i \leq 10^9)$$$, the elements of array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, first output a line containing a single integer $$$m$$$ — minimum number of operations required. Then ouput a line consist of $$$n$$$ integers — the resulting array after the operations. Note that this array should differ in exactly $$$m$$$ elements from the initial array. If there are multiple answers, print any. | standard output | |
PASSED | c4034348928b948fb862997a770fcf17 | train_108.jsonl | 1645367700 | You are given an array $$$a$$$ of size $$$n$$$. Each element in this array is an integer between $$$1$$$ and $$$10^9$$$.You can perform several operations to this array. During an operation, you can replace an element in the array with any integer between $$$1$$$ and $$$10^9$$$. Output the minimum number of operations needed such that the resulting array doesn't contain any local maximums, and the resulting array after the operations.An element $$$a_i$$$ is a local maximum if it is strictly larger than both of its neighbors (that is, $$$a_i > a_{i - 1}$$$ and $$$a_i > a_{i + 1}$$$). Since $$$a_1$$$ and $$$a_n$$$ have only one neighbor each, they will never be a local maximum. | 256 megabytes | // 😀
import java.util.*;
import java.io.*;
public class B {
static InputReader in;
static PrintWriter out;
static int mod = 1000_000_007;
public static void main(String args[]) {
new Thread(null, new Runnable() {
public void run() {
try {
solve();
out.flush();
out.close();
} catch (Exception e) {
e.printStackTrace();
}
}
}, "1", 1 << 26).start();
}
public static void solve() {
in = new InputReader(System.in);
out = new PrintWriter(System.out);
int t = in.nextInt();
while (t-- > 0) {
int n = in.nextInt();
int a[] = in.nextIntArray(n);
if(n == 2){
out.println(0);
out.println(a[0]+" "+a[1]);
continue;
}
int op=0;
for (int i = 1; i < n-2; i++) {
if(a[i] > a[i-1] && a[i] > a[i+1])
{
op++;
a[i+1] = Math.max(a[i],a[i+2]);
}
}
if(a[n-2] > a[n-3] && a[n-2] > a[n-1] )
{
op++;
a[n-1] = a[n-2];
}
out.println(op);
for(int x:a)
out.print(x+" ");
out.println();
}
}
static class InputReader {
private final InputStream stream;
private final byte[] buf = new byte[8192];
private int curChar, 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 String nextLine() {
int c = read();
while (isSpaceChar(c)) {
c = read();
}
StringBuilder res = new StringBuilder();
do {
res.appendCodePoint(c);
c = read();
} while (!isEndOfLine(c));
return res.toString();
}
public String readString() {
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 long nextLong() {
int c = read();
while (isSpaceChar(c)) {
c = read();
}
int sgn = 1;
if (c == '-') {
sgn = -1;
c = read();
}
long 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 int nextInt() {
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 int[] nextIntArray(int n) {
int[] arr = new int[n];
for (int i = 0; i < n; i++) {
arr[i] = nextInt();
}
return arr;
}
public long[] nextLongArray(int n) {
long[] arr = new long[n];
for (int i = 0; i < n; i++) {
arr[i] = nextLong();
}
return arr;
}
public boolean isSpaceChar(int c) {
if (filter != null)
return filter.isSpaceChar(c);
return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
}
private boolean isEndOfLine(int c) {
return c == '\n' || c == '\r' || c == -1;
}
public interface SpaceCharFilter {
public boolean isSpaceChar(int ch);
}
}
public static long inv(long a, long b) {
return 1 < a ? b - inv(b % a, a) * b / a : 1;
}
public static int[] sort(int[] arr) {
List<Integer> temp = new ArrayList();
for (int i : arr) temp.add(i);
Collections.sort(temp);
int start = 0;
for (int i : temp) arr[start++] = i;
return arr;
}
public static long[] sort(long[] arr) {
List<Long> temp = new ArrayList();
for (long i : arr) temp.add(i);
Collections.sort(temp);
int start = 0;
for (long i : temp) arr[start++] = i;
return arr;
}
}
| Java | ["5\n3\n2 1 2\n4\n1 2 3 1\n5\n1 2 1 2 1\n9\n1 2 1 3 2 3 1 2 1\n9\n2 1 3 1 3 1 3 1 3"] | 2 seconds | ["0\n2 1 2\n1\n1 3 3 1\n1\n1 2 2 2 1\n2\n1 2 3 3 2 3 3 2 1\n2\n2 1 3 3 3 1 1 1 3"] | NoteIn the first example, the array contains no local maximum, so we don't need to perform operations.In the second example, we can change $$$a_2$$$ to $$$3$$$, then the array don't have local maximums. | Java 8 | standard input | [
"greedy"
] | 255abf8750541f54d8ff0f1f80e1f8e7 | Each test contains multiple test cases. The first line will contain a single integer $$$t$$$ $$$(1 \leq t \leq 10000)$$$ — the number of test cases. Then $$$t$$$ test cases follow. The first line of each test case contains a single integer $$$n$$$ $$$(2 \leq n \leq 2 \cdot 10^5)$$$ — the size of the array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots ,a_n$$$ $$$(1 \leq a_i \leq 10^9)$$$, the elements of array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, first output a line containing a single integer $$$m$$$ — minimum number of operations required. Then ouput a line consist of $$$n$$$ integers — the resulting array after the operations. Note that this array should differ in exactly $$$m$$$ elements from the initial array. If there are multiple answers, print any. | standard output | |
PASSED | b75febccde851598df4bfcc0e3bd00f4 | train_108.jsonl | 1645367700 | You are given an array $$$a$$$ of size $$$n$$$. Each element in this array is an integer between $$$1$$$ and $$$10^9$$$.You can perform several operations to this array. During an operation, you can replace an element in the array with any integer between $$$1$$$ and $$$10^9$$$. Output the minimum number of operations needed such that the resulting array doesn't contain any local maximums, and the resulting array after the operations.An element $$$a_i$$$ is a local maximum if it is strictly larger than both of its neighbors (that is, $$$a_i > a_{i - 1}$$$ and $$$a_i > a_{i + 1}$$$). Since $$$a_1$$$ and $$$a_n$$$ have only one neighbor each, they will never be a local maximum. | 256 megabytes | import java.util.*;
import java.io.*;
public class q2 {
static FastScanner fs = new FastScanner();
static PrintWriter pw = new PrintWriter(System.out);
public static void main(String[] args) {
int T = fs.nextInt();
for (int t=0;t<T;t++){
int n = fs.nextInt(), cnt = 0;
int[] pp = fs.readArray(n);
for (int i=1;i<n-2;i++){
if (pp[i - 1] < pp[i] && pp[i] > pp[i + 1]){
pp[i + 1] = Math.max(pp[i], pp[i + 2]);
cnt++;
}
}
if (n >= 3 && pp[n - 3] < pp[n - 2] && pp[n - 2] > pp[n - 1]){
pp[n - 1] = pp[n - 2];
cnt++;
}
pw.println(cnt);
for (int i : pp) pw.print(i + " ");
pw.println();
}
pw.close();
}
// ----------input function----------
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 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());
}
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 nextLong() {
return Long.parseLong(next());
}
}
} | Java | ["5\n3\n2 1 2\n4\n1 2 3 1\n5\n1 2 1 2 1\n9\n1 2 1 3 2 3 1 2 1\n9\n2 1 3 1 3 1 3 1 3"] | 2 seconds | ["0\n2 1 2\n1\n1 3 3 1\n1\n1 2 2 2 1\n2\n1 2 3 3 2 3 3 2 1\n2\n2 1 3 3 3 1 1 1 3"] | NoteIn the first example, the array contains no local maximum, so we don't need to perform operations.In the second example, we can change $$$a_2$$$ to $$$3$$$, then the array don't have local maximums. | Java 8 | standard input | [
"greedy"
] | 255abf8750541f54d8ff0f1f80e1f8e7 | Each test contains multiple test cases. The first line will contain a single integer $$$t$$$ $$$(1 \leq t \leq 10000)$$$ — the number of test cases. Then $$$t$$$ test cases follow. The first line of each test case contains a single integer $$$n$$$ $$$(2 \leq n \leq 2 \cdot 10^5)$$$ — the size of the array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots ,a_n$$$ $$$(1 \leq a_i \leq 10^9)$$$, the elements of array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, first output a line containing a single integer $$$m$$$ — minimum number of operations required. Then ouput a line consist of $$$n$$$ integers — the resulting array after the operations. Note that this array should differ in exactly $$$m$$$ elements from the initial array. If there are multiple answers, print any. | standard output | |
PASSED | 16b10061f7a48ce70268d6a65b6b113f | train_108.jsonl | 1645367700 | You are given an array $$$a$$$ of size $$$n$$$. Each element in this array is an integer between $$$1$$$ and $$$10^9$$$.You can perform several operations to this array. During an operation, you can replace an element in the array with any integer between $$$1$$$ and $$$10^9$$$. Output the minimum number of operations needed such that the resulting array doesn't contain any local maximums, and the resulting array after the operations.An element $$$a_i$$$ is a local maximum if it is strictly larger than both of its neighbors (that is, $$$a_i > a_{i - 1}$$$ and $$$a_i > a_{i + 1}$$$). Since $$$a_1$$$ and $$$a_n$$$ have only one neighbor each, they will never be a local maximum. | 256 megabytes | import java.util.*;
import java.io.*;
public class B1635 {
public static void main(String[] args) throws IOException {
Scanner sc = new Scanner(System.in);
PrintWriter pw = new PrintWriter(System.out);
int t = sc.nextInt();
while (t-- > 0) {
int n = sc.nextInt();
int[] a = sc.nextIntArr(n);
int ans = 0;
for (int i = 1; i < a.length - 1; i++) {
if (a[i] == a[i - 1] || (i != a.length - 1 && a[i] == a[i + 1]))
continue;
while (i < a.length && a[i] < a[i - 1]) {
i++;
}
if (i < a.length && a[i] == a[i - 1]) {
continue;
}
while (i < a.length && a[i] > a[i - 1]) {
i++;
}
if (i < a.length && a[i] == a[i - 1]) {
continue;
}
if (i < a.length) {
if (i == a.length - 1)
a[i] = a[i - 1];
else
a[i] = Math.max(a[i - 1], a[i + 1]);
ans++;
}
}
pw.println(ans);
for (int x : a) {
pw.print(x + " ");
}
pw.println();
}
pw.close();
}
static class Scanner {
BufferedReader br;
StringTokenizer st;
public Scanner(InputStream s) {
br = new BufferedReader(new InputStreamReader(s));
}
public Scanner(FileReader f) {
br = new BufferedReader(f);
}
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 double nextDouble() throws IOException {
return Double.parseDouble(next());
}
public int[] nextIntArr(int n) throws IOException {
int[] arr = new int[n];
for (int i = 0; i < n; i++) {
arr[i] = Integer.parseInt(next());
}
return arr;
}
}
}
| Java | ["5\n3\n2 1 2\n4\n1 2 3 1\n5\n1 2 1 2 1\n9\n1 2 1 3 2 3 1 2 1\n9\n2 1 3 1 3 1 3 1 3"] | 2 seconds | ["0\n2 1 2\n1\n1 3 3 1\n1\n1 2 2 2 1\n2\n1 2 3 3 2 3 3 2 1\n2\n2 1 3 3 3 1 1 1 3"] | NoteIn the first example, the array contains no local maximum, so we don't need to perform operations.In the second example, we can change $$$a_2$$$ to $$$3$$$, then the array don't have local maximums. | Java 8 | standard input | [
"greedy"
] | 255abf8750541f54d8ff0f1f80e1f8e7 | Each test contains multiple test cases. The first line will contain a single integer $$$t$$$ $$$(1 \leq t \leq 10000)$$$ — the number of test cases. Then $$$t$$$ test cases follow. The first line of each test case contains a single integer $$$n$$$ $$$(2 \leq n \leq 2 \cdot 10^5)$$$ — the size of the array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots ,a_n$$$ $$$(1 \leq a_i \leq 10^9)$$$, the elements of array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, first output a line containing a single integer $$$m$$$ — minimum number of operations required. Then ouput a line consist of $$$n$$$ integers — the resulting array after the operations. Note that this array should differ in exactly $$$m$$$ elements from the initial array. If there are multiple answers, print any. | standard output | |
PASSED | 9f92bdbf100446ab127f840f6ab8ef11 | train_108.jsonl | 1645367700 | You are given an array $$$a$$$ of size $$$n$$$. Each element in this array is an integer between $$$1$$$ and $$$10^9$$$.You can perform several operations to this array. During an operation, you can replace an element in the array with any integer between $$$1$$$ and $$$10^9$$$. Output the minimum number of operations needed such that the resulting array doesn't contain any local maximums, and the resulting array after the operations.An element $$$a_i$$$ is a local maximum if it is strictly larger than both of its neighbors (that is, $$$a_i > a_{i - 1}$$$ and $$$a_i > a_{i + 1}$$$). Since $$$a_1$$$ and $$$a_n$$$ have only one neighbor each, they will never be a local maximum. | 256 megabytes | /*
Rating: 1367
Date: 20-02-2022
Time: 20-10-40
Author: Kartik Papney
Linkedin: https://www.linkedin.com/in/kartik-papney-4951161a6/
Leetcode: https://leetcode.com/kartikpapney/
Codechef: https://www.codechef.com/users/kartikpapney
*/
import java.util.*;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
public class B_Avoid_Local_Maximums {
public static boolean debug = false;
static void debug(String st) {
if(debug) p.writeln(st);
}
public static void s() {
int n = sc.nextInt();
int[] arr = sc.readArray(n);
int cnt = 0;
for(int i=1; i<n-1; i++ ) {
if(arr[i] > arr[i-1] && arr[i] > arr[i+1]) {
if(i+2 < n-1 && arr[i+2] > arr[i+1] && arr[i+2] > arr[i+3]) {
arr[i+1] = Math.max(arr[i], arr[i+2]);
} else {
arr[i+1] = arr[i];
}
cnt++;
}
}
p.writeln(cnt);
p.writes(arr);
p.writeln();
}
public static void main(String[] args) {
int t = 1;
t = sc.nextInt();
while (t-- != 0) {
s();
}
p.print();
}
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 | ["5\n3\n2 1 2\n4\n1 2 3 1\n5\n1 2 1 2 1\n9\n1 2 1 3 2 3 1 2 1\n9\n2 1 3 1 3 1 3 1 3"] | 2 seconds | ["0\n2 1 2\n1\n1 3 3 1\n1\n1 2 2 2 1\n2\n1 2 3 3 2 3 3 2 1\n2\n2 1 3 3 3 1 1 1 3"] | NoteIn the first example, the array contains no local maximum, so we don't need to perform operations.In the second example, we can change $$$a_2$$$ to $$$3$$$, then the array don't have local maximums. | Java 8 | standard input | [
"greedy"
] | 255abf8750541f54d8ff0f1f80e1f8e7 | Each test contains multiple test cases. The first line will contain a single integer $$$t$$$ $$$(1 \leq t \leq 10000)$$$ — the number of test cases. Then $$$t$$$ test cases follow. The first line of each test case contains a single integer $$$n$$$ $$$(2 \leq n \leq 2 \cdot 10^5)$$$ — the size of the array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots ,a_n$$$ $$$(1 \leq a_i \leq 10^9)$$$, the elements of array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, first output a line containing a single integer $$$m$$$ — minimum number of operations required. Then ouput a line consist of $$$n$$$ integers — the resulting array after the operations. Note that this array should differ in exactly $$$m$$$ elements from the initial array. If there are multiple answers, print any. | standard output | |
PASSED | 3f51c707fda945c0f9b1e1ab4d0e14c8 | train_108.jsonl | 1645367700 | You are given an array $$$a$$$ of size $$$n$$$. Each element in this array is an integer between $$$1$$$ and $$$10^9$$$.You can perform several operations to this array. During an operation, you can replace an element in the array with any integer between $$$1$$$ and $$$10^9$$$. Output the minimum number of operations needed such that the resulting array doesn't contain any local maximums, and the resulting array after the operations.An element $$$a_i$$$ is a local maximum if it is strictly larger than both of its neighbors (that is, $$$a_i > a_{i - 1}$$$ and $$$a_i > a_{i + 1}$$$). Since $$$a_1$$$ and $$$a_n$$$ have only one neighbor each, they will never be a local maximum. | 256 megabytes | import java.util.Scanner;
public class Main {
public static void main(String[] args) {
Scanner cin = new Scanner(System.in);
int t = cin.nextInt();
for (int i = 0; i < t; i++) {
int n = cin.nextInt();
int[] a = new int[n];
for (int j = 0; j < n; j++) {
a[j] = cin.nextInt();
}
StringBuilder sb = new StringBuilder();
int[] ans = new Solution().func(a);
for (int e : ans) {
sb.append(e);
sb.append(' ');
}
System.out.println(sb);
}
}
}
class Solution {
public int[] func(int[] a) {
int cnt = 0;
for (int i = 1; i < a.length - 1; i++) {
if (a[i] > a[i - 1] && a[i] > a[i + 1]) {
int t = 0;
if (i < a.length - 2) {
t = a[i + 2];
}
a[i + 1] = Math.max(a[i], t);
cnt++;
}
}
System.out.println(cnt);
return a;
}
} | Java | ["5\n3\n2 1 2\n4\n1 2 3 1\n5\n1 2 1 2 1\n9\n1 2 1 3 2 3 1 2 1\n9\n2 1 3 1 3 1 3 1 3"] | 2 seconds | ["0\n2 1 2\n1\n1 3 3 1\n1\n1 2 2 2 1\n2\n1 2 3 3 2 3 3 2 1\n2\n2 1 3 3 3 1 1 1 3"] | NoteIn the first example, the array contains no local maximum, so we don't need to perform operations.In the second example, we can change $$$a_2$$$ to $$$3$$$, then the array don't have local maximums. | Java 8 | standard input | [
"greedy"
] | 255abf8750541f54d8ff0f1f80e1f8e7 | Each test contains multiple test cases. The first line will contain a single integer $$$t$$$ $$$(1 \leq t \leq 10000)$$$ — the number of test cases. Then $$$t$$$ test cases follow. The first line of each test case contains a single integer $$$n$$$ $$$(2 \leq n \leq 2 \cdot 10^5)$$$ — the size of the array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots ,a_n$$$ $$$(1 \leq a_i \leq 10^9)$$$, the elements of array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, first output a line containing a single integer $$$m$$$ — minimum number of operations required. Then ouput a line consist of $$$n$$$ integers — the resulting array after the operations. Note that this array should differ in exactly $$$m$$$ elements from the initial array. If there are multiple answers, print any. | standard output | |
PASSED | 5bf09fa2803f0adb0a6c0a60b0c3d4da | train_108.jsonl | 1645367700 | You are given an array $$$a$$$ of size $$$n$$$. Each element in this array is an integer between $$$1$$$ and $$$10^9$$$.You can perform several operations to this array. During an operation, you can replace an element in the array with any integer between $$$1$$$ and $$$10^9$$$. Output the minimum number of operations needed such that the resulting array doesn't contain any local maximums, and the resulting array after the operations.An element $$$a_i$$$ is a local maximum if it is strictly larger than both of its neighbors (that is, $$$a_i > a_{i - 1}$$$ and $$$a_i > a_{i + 1}$$$). Since $$$a_1$$$ and $$$a_n$$$ have only one neighbor each, they will never be a local maximum. | 256 megabytes | import java.util.*;
import java.io.*;
public class Hiking {
static Scanner sc;
static PrintWriter pw;
public static void main(String[] args) throws IOException, InterruptedException {
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);
int c=0;
int [] ori=arr.clone();
LinkedList<Integer> ff=new LinkedList<>();
for(int j=1;j<n-1;j++) {
if(arr[j]>arr[j-1] && arr[j]>arr[j+1]) {
ff.add(j);
}
}
while(ff.size()>1) {
int x=ff.pollFirst();
if(ff.getFirst()-x==2) {
arr[x+1]=Math.max(arr[x], arr[ff.getFirst()]);
ff.pollFirst();
}
else arr[x]=Math.max(arr[x-1], arr[x+1]);
}
if(ff.size()==1) {
int x=ff.poll();
arr[x]=Math.max(arr[x-1], arr[x+1]);
}
for(int i=0;i<n;i++) {
if(arr[i]!=ori[i])
c++;
}
pw.println(c);
for(int i:arr)
pw.print(i+" ");
pw.println();
}
pw.flush();
}
static class FenwickTree { // one-based DS
int n;
long[] ft;
FenwickTree(int size) {
n = size;
ft = new long[n + 1];
}
long rsq(int b) // O(log n)
{
b++;
long sum = 0;
while (b > 0) {
sum += ft[b];
b -= b & -b;
} // min?
return sum;
}
long rsq(int a, int b) {
b = Math.min(b, n - 1);
return rsq(b) - rsq(a - 1);
}
void point_update(int k, long val) // O(log n), update = increment
{
k++;
while (k <= n) {
ft[k] += val;
k += k & -k;
}
}
}
static class pair {
// long x,y;
int x, y;
public pair(int x, int y) {
this.x = x;
this.y = y;
}
public String toString() {
return x + " " + y;
}
public boolean equals(Object o) {
if (o instanceof pair) {
pair p = (pair) o;
return p.x == x && p.y == y;
}
return false;
}
public int hashCode() {
return new Long(x).hashCode() * 31 + new Long(y).hashCode();
}
}
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 | ["5\n3\n2 1 2\n4\n1 2 3 1\n5\n1 2 1 2 1\n9\n1 2 1 3 2 3 1 2 1\n9\n2 1 3 1 3 1 3 1 3"] | 2 seconds | ["0\n2 1 2\n1\n1 3 3 1\n1\n1 2 2 2 1\n2\n1 2 3 3 2 3 3 2 1\n2\n2 1 3 3 3 1 1 1 3"] | NoteIn the first example, the array contains no local maximum, so we don't need to perform operations.In the second example, we can change $$$a_2$$$ to $$$3$$$, then the array don't have local maximums. | Java 8 | standard input | [
"greedy"
] | 255abf8750541f54d8ff0f1f80e1f8e7 | Each test contains multiple test cases. The first line will contain a single integer $$$t$$$ $$$(1 \leq t \leq 10000)$$$ — the number of test cases. Then $$$t$$$ test cases follow. The first line of each test case contains a single integer $$$n$$$ $$$(2 \leq n \leq 2 \cdot 10^5)$$$ — the size of the array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots ,a_n$$$ $$$(1 \leq a_i \leq 10^9)$$$, the elements of array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, first output a line containing a single integer $$$m$$$ — minimum number of operations required. Then ouput a line consist of $$$n$$$ integers — the resulting array after the operations. Note that this array should differ in exactly $$$m$$$ elements from the initial array. If there are multiple answers, print any. | standard output | |
PASSED | 4b81ec1e3adf99bab56b6e44e33c7350 | train_108.jsonl | 1645367700 | You are given an array $$$a$$$ of size $$$n$$$. Each element in this array is an integer between $$$1$$$ and $$$10^9$$$.You can perform several operations to this array. During an operation, you can replace an element in the array with any integer between $$$1$$$ and $$$10^9$$$. Output the minimum number of operations needed such that the resulting array doesn't contain any local maximums, and the resulting array after the operations.An element $$$a_i$$$ is a local maximum if it is strictly larger than both of its neighbors (that is, $$$a_i > a_{i - 1}$$$ and $$$a_i > a_{i + 1}$$$). Since $$$a_1$$$ and $$$a_n$$$ have only one neighbor each, they will never be a local maximum. | 256 megabytes | import java.io.BufferedWriter;
import java.io.IOException;
import java.io.InputStream;
import java.io.OutputStreamWriter;
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.LinkedList;
import java.util.PriorityQueue;
import java.util.Map.Entry;
import java.util.Random;
import java.util.TreeSet;
public final class CF_772_D2_B
{
static boolean verb=true;
static void log(Object X){if (verb) System.err.println(X);}
static void log(Object[] X){if (verb) {for (Object U:X) System.err.print(U+" ");System.err.println("");}}
static void log(int[] X){if (verb) {for (int U:X) System.err.print(U+" ");System.err.println("");}}
static void logWln(int[] X){if (verb) {for (int U:X) System.err.print(U+" ");}}
static void log(int[] X,int L){if (verb) {for (int i=0;i<L;i++) System.err.print(X[i]+" ");System.err.println("");}}
static void log(long[] X){if (verb) {for (long U:X) System.err.print(U+" ");System.err.println("");}}
static void logWln(Object X){if (verb) System.err.print(X);}
static void info(Object o){ System.out.println(o);}
static void output(Object o){outputWln(""+o+"\n"); }
static void outputFlush(Object o){try {out.write(""+ o+"\n");out.flush();} catch (Exception e) {}}
static void outputWln(Object o){try {out.write(""+ o);} catch (Exception e) {}}
// Global vars
static BufferedWriter out;
static InputReader reader;
static void test() {
log("testing");
Random r=new Random();
int NTESTS=1000000;
int NMAX=22;
int VMAX=10;
for (int t=0;t<NTESTS;t++) {
}
log("testing done");
}
static class Composite implements Comparable<Composite>{
int a;
int b;
public int compareTo(Composite X) {
int diff=b-a;
int Xdiff=X.b-X.a;
if (diff!=Xdiff)
return diff-Xdiff;
return a-X.a;
}
public Composite(int a, int b) {
this.a = a;
this.b = b;
}
}
static void process() throws Exception {
out = new BufferedWriter(new OutputStreamWriter(System.out));
reader = new InputReader(System.in);
log("hello");
int T=reader.readInt();
for (int t=0;t<T;t++) {
int n=reader.readInt();
int[] a=new int[n];
int[] b=new int[n];
for (int i=0;i<n;i++) {
a[i]=reader.readInt();
b[i]=a[i];
}
int cnt=0;
int[] lst=new int[n];
int st=0;
boolean[] changed=new boolean[n];
for (int i=1;i+1<n;i++) {
if (a[i]>a[i-1] && a[i]>a[i+1]) {
if (changed[i-1]) {
a[i-1]=a[i];
} else {
changed[i+1]=true;
a[i+1]=a[i];
cnt++;
}
}
}
output(cnt);
StringBuffer sb=new StringBuffer();
for (int x:a)
sb.append(x+" ");
output(sb);
/*
int zn=0;
for (int i=0;i<n;i++) {
if (a[i]!=b[i]) {
zn++;
}
}
for (int i=1;i+1<n;i++) {
if (a[i]>a[i-1] && a[i]>a[i+1]) {
log("issue");
}
}
if (zn!=cnt) {
log("Error");
}
*/
}
try {
out.close();
} catch (Exception Ex) {
}
}
public static void main(String[] args) throws Exception {
process();
}
static final class InputReader {
private final InputStream stream;
private final byte[] buf = new byte[1024];
private int curChar;
private int numChars;
public InputReader(InputStream stream) {
this.stream = stream;
}
private int read() throws IOException {
if (curChar >= numChars) {
curChar = 0;
numChars = stream.read(buf);
if (numChars <= 0) {
return -1;
}
}
return buf[curChar++];
}
public final String readString() throws IOException {
int c = read();
while (isSpaceChar(c)) {
c = read();
}
StringBuilder res = new StringBuilder();
do {
res.append((char) c);
c = read();
} while (!isSpaceChar(c));
return res.toString();
}
public final String readString(int L) throws IOException {
int c = read();
while (isSpaceChar(c)) {
c = read();
}
StringBuilder res = new StringBuilder(L);
do {
res.append((char) c);
c = read();
} while (!isSpaceChar(c));
return res.toString();
}
public final int readInt() throws IOException {
int c = read();
boolean neg = false;
while (isSpaceChar(c)) {
c = read();
}
char d = (char) c;
// log("d:"+d);
if (d == '-') {
neg = true;
c = read();
}
int res = 0;
do {
res *= 10;
res += c - '0';
c = read();
} while (!isSpaceChar(c));
// log("res:"+res);
if (neg)
return -res;
return res;
}
public final long readLong() throws IOException {
int c = read();
boolean neg = false;
while (isSpaceChar(c)) {
c = read();
}
char d = (char) c;
// log("d:"+d);
if (d == '-') {
neg = true;
c = read();
}
long res = 0;
do {
res *= 10;
res += c - '0';
c = read();
} while (!isSpaceChar(c));
// log("res:"+res);
if (neg)
return -res;
return res;
}
private boolean isSpaceChar(int c) {
return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
}
}
}
/*
1
10
1 0 0 0 0 1 0 0 1 1
1
12
1 0 0 1 0 0 1 0 0 0 0 1
1
12
1 0 0 1 0 0 1 0 0 0 0 1
3
12
1 0 0 1 0 0 1 0 0 0 0 1
10
1 0 0 0 0 1 0 0 1 1
12
1 0 0 1 0 0 1 0 0 0 0 1
1
11
1 0 0 0 0 0 0 0 1 1 1
19
1 1 0 1 1 1 1 1 1 0 1 0 0 0 0 1 0 1 1
19
1 0 0 1 1 0 0 1 1 0 0 0 1 1 0 0 0 1 0
19
1 0 0 0 1 1 0 0 1 1 0 0 0 1 1 0 0 1 0
1
20
1 1 0 0 0 1 0 1 1 1 0 1 1 1 1 1 0 0 1 0
*/ | Java | ["5\n3\n2 1 2\n4\n1 2 3 1\n5\n1 2 1 2 1\n9\n1 2 1 3 2 3 1 2 1\n9\n2 1 3 1 3 1 3 1 3"] | 2 seconds | ["0\n2 1 2\n1\n1 3 3 1\n1\n1 2 2 2 1\n2\n1 2 3 3 2 3 3 2 1\n2\n2 1 3 3 3 1 1 1 3"] | NoteIn the first example, the array contains no local maximum, so we don't need to perform operations.In the second example, we can change $$$a_2$$$ to $$$3$$$, then the array don't have local maximums. | Java 8 | standard input | [
"greedy"
] | 255abf8750541f54d8ff0f1f80e1f8e7 | Each test contains multiple test cases. The first line will contain a single integer $$$t$$$ $$$(1 \leq t \leq 10000)$$$ — the number of test cases. Then $$$t$$$ test cases follow. The first line of each test case contains a single integer $$$n$$$ $$$(2 \leq n \leq 2 \cdot 10^5)$$$ — the size of the array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots ,a_n$$$ $$$(1 \leq a_i \leq 10^9)$$$, the elements of array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, first output a line containing a single integer $$$m$$$ — minimum number of operations required. Then ouput a line consist of $$$n$$$ integers — the resulting array after the operations. Note that this array should differ in exactly $$$m$$$ elements from the initial array. If there are multiple answers, print any. | standard output | |
PASSED | e99533ce6e95479655d88ec169e758c7 | train_108.jsonl | 1645367700 | You are given an array $$$a$$$ of size $$$n$$$. Each element in this array is an integer between $$$1$$$ and $$$10^9$$$.You can perform several operations to this array. During an operation, you can replace an element in the array with any integer between $$$1$$$ and $$$10^9$$$. Output the minimum number of operations needed such that the resulting array doesn't contain any local maximums, and the resulting array after the operations.An element $$$a_i$$$ is a local maximum if it is strictly larger than both of its neighbors (that is, $$$a_i > a_{i - 1}$$$ and $$$a_i > a_{i + 1}$$$). Since $$$a_1$$$ and $$$a_n$$$ have only one neighbor each, they will never be a local maximum. | 256 megabytes |
import java.util.Scanner;
public class cfContest1635 {
public static void main(String[] args) {
Scanner scan = new Scanner(System.in);
int t = scan.nextInt();
StringBuilder sb = new StringBuilder();
while (t-- > 0) {
int n = scan.nextInt();
int[] a = new int[n];
for (int i = 0; i < n; i++) {
a[i] = scan.nextInt();
}
int c = 0;
for (int i = 0; i < n - 2; i++) {
if (a[i + 1] > a[i] && a[i + 1] > a[i + 2]) {
if (i + 3 < n && a[i + 1] <= a[i + 3]) {
a[i + 2] = a[i + 3];
} else {
a[i + 2] = a[i + 1];
}
++c;
}
}
sb.append(c + "\n");
for (int i = 0; i < n; i++) {
sb.append(a[i] + " ");
}
sb.append("\n");
}
System.out.println(sb);
}
}
| Java | ["5\n3\n2 1 2\n4\n1 2 3 1\n5\n1 2 1 2 1\n9\n1 2 1 3 2 3 1 2 1\n9\n2 1 3 1 3 1 3 1 3"] | 2 seconds | ["0\n2 1 2\n1\n1 3 3 1\n1\n1 2 2 2 1\n2\n1 2 3 3 2 3 3 2 1\n2\n2 1 3 3 3 1 1 1 3"] | NoteIn the first example, the array contains no local maximum, so we don't need to perform operations.In the second example, we can change $$$a_2$$$ to $$$3$$$, then the array don't have local maximums. | Java 8 | standard input | [
"greedy"
] | 255abf8750541f54d8ff0f1f80e1f8e7 | Each test contains multiple test cases. The first line will contain a single integer $$$t$$$ $$$(1 \leq t \leq 10000)$$$ — the number of test cases. Then $$$t$$$ test cases follow. The first line of each test case contains a single integer $$$n$$$ $$$(2 \leq n \leq 2 \cdot 10^5)$$$ — the size of the array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots ,a_n$$$ $$$(1 \leq a_i \leq 10^9)$$$, the elements of array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, first output a line containing a single integer $$$m$$$ — minimum number of operations required. Then ouput a line consist of $$$n$$$ integers — the resulting array after the operations. Note that this array should differ in exactly $$$m$$$ elements from the initial array. If there are multiple answers, print any. | standard output | |
PASSED | 2f9a0708dad6617ffe8d302bcf464f65 | train_108.jsonl | 1645367700 | You are given an array $$$a$$$ of size $$$n$$$. Each element in this array is an integer between $$$1$$$ and $$$10^9$$$.You can perform several operations to this array. During an operation, you can replace an element in the array with any integer between $$$1$$$ and $$$10^9$$$. Output the minimum number of operations needed such that the resulting array doesn't contain any local maximums, and the resulting array after the operations.An element $$$a_i$$$ is a local maximum if it is strictly larger than both of its neighbors (that is, $$$a_i > a_{i - 1}$$$ and $$$a_i > a_{i + 1}$$$). Since $$$a_1$$$ and $$$a_n$$$ have only one neighbor each, they will never be a local maximum. | 256 megabytes | //import java.lang.reflect.Array;
import java.util.*;
import java.lang.reflect.Array;
import javax.print.DocFlavor.STRING;
import javax.swing.Popup;
import javax.swing.plaf.synth.SynthStyleFactory;
public class Simple{
public static class Pair implements Comparable<Pair>{
int val;
int freq = 0;
Pair prev ;
Pair next;
boolean bool = false;
public Pair(int val,int freq){
this.val = val;
this.freq= freq;
}
public int compareTo(Pair p){
// if(p.freq == this.freq){
// return this.val - p.freq;
// };
// if(this.freq > p.freq)return -1;
// return 1;
return (p.freq-p.val) - (this.freq - this.val);
}
}
public static long factorial(long n)
{
// single line to find factorial
return (n == 1 || n == 0) ? 1 : n * factorial(n - 1);
}
static long m = 998244353;
// Function to return the GCD of given numbers
static long gcd(int a, int b)
{
if (a == 0)
return b;
return gcd(b % a, a);
}
// Recursive function to return (x ^ n) % m
static long modexp(long x, long n)
{
if (n == 0) {
return 1;
}
else if (n % 2 == 0) {
return modexp((x * x) % m, n / 2);
}
else {
return (x * modexp((x * x) % m, (n - 1) / 2) % m);
}
}
// Function to return the fraction modulo mod
// static long getFractionModulo(long a, long b)
// {
// long c = gcd(a, b);
// a = a / c;
// b = b / c;
// // (b ^ m-2) % m
// long d = modexp(b, m - 2);
// // Final answer
// long ans = ((a % m) * (d % m)) % m;
// return ans;
// }
// public static long bs(long lo,long hi,long fact,long num){
// long help = num/fact;
// long ans = hi;
// while(lo<=hi){
// long mid = (lo+hi)/2;
// if(mid/)
// }
// }
// public static boolean isPrime(int n)
// {
// // Check if number is less than
// // equal to 1
// if (n <= 1)
// return false;
// // Check if number is 2
// else if (n == 2)
// return true;
// // Check if n is a multiple of 2
// else if (n % 2 == 0)
// return false;
// // If not, then just check the odds
// for (int i = 3; i <= sqrt(n); i += 2)
// {
// if (n % i == 0)
// return false;
// }
// return true;
// }
// public static int countDigit(long n)
// {
// int count = 0;
// while (n != 0) {
// n = n / 10;
// ++count;
// }
// return count;
// }
// static ArrayList<Long> al ;
// static boolean checkperfectsquare(long n)
// {
// // If ceil and floor are equal
// // the number is a perfect
// // square
// if (processesh.ceil((double)processesh.sqrt(n)) ==
// processesh.floor((double)processesh.sqrt(n)))
// {
// return true;
// }
// else
// {
// return false;
// }
// }
public static void decToBinary(int n,int arr[],int j)
{
// Size of an integer is assumed to be 32 bits
for (int i = 31; i >= 0; i--) {
int k = n >> i;
if ((k & 1) > 0){
arr[j]++;
}
j++;
}
}
public static class Node{
//int u;
long v;
long w;
public Node(long v,long w){
//this.u = u;
this.v=v;
this.w=w;
}
}
public static boolean[] sieve(int n){
boolean isPrime[] = new boolean[n+1];
Arrays.fill(isPrime, true);
isPrime[0] = false;
isPrime[1] = false;
for(int i=2;i*i<=n;i++){
if(!isPrime[i])continue;
for(int j= i*2;j<=n;j=j+i){
isPrime[j] = false;
}
}
return isPrime;
}
static long power(long x, long y, long p)
{
long res = 1; // Initialize result
x = x % p; // Update x if it is more
// than or equal to p
while (y > 0)
{
// If y is odd, multiply
// x with result
if ((y & 1) > 0)
res = (res * x) % p;
// y must be even now
y = y >> 1; // y = y/2
x = (x * x) % p;
}
return res;
}
// static int modInverse(int a, int p)
// {
// return power(a, p - 2, p);
// }
public static void bfs(ArrayList<ArrayList<Integer>> adj,boolean vis[],int dis[]){
vis[0] = true;
Queue<Integer> q = new LinkedList<>();
q.add(0);
int count = 0;
while(!q.isEmpty()){
int size = q.size();
for(int j = 0;j<size;j++){
int poll = q.poll();
dis[poll] = count;
vis[poll] = true;
for(Integer x : adj.get(poll)){
if(!vis[x]){
q.add(x);
}
}
}
count++;
}
}
static void swap(int[] arr, int i, int j)
{
int temp = arr[i];
arr[i] = arr[j];
arr[j] = temp;
}
public static boolean isSorted(int arr[],int[] copy,int n){
for(int i=0;i<n;i++){
if(arr[i]!=copy[i])return false;
}
return true;
}
public static class DSU{
int par[];
int rank[];//rank denotes the height of graph
public DSU(int n){
par = new int[n];
for(int i=0;i<n;i++){
par[i] = i;
}
rank = new int[n];
}
public int findPar(int node){
if(par[node]==node)return node;
par[node] = findPar(par[node]);
return par[node];
}
public void union(int u,int v){
u = findPar(u);
v = findPar(v);
if(rank[v]>rank[u]){
par[u] = v;
}
else if(rank[v]<rank[u]){
par[v] = u;
}
else{
rank[u]++;
par[v] = u;
}
}
}
public static long modFact(int n,int k)
{
long M = 998244353;
long f = 1;
for (int i = 1; i <= n; i++){
if(i==k+1)continue;
f = (f%M*i%m) % M; // Now f never can
} // exceed 10^9+7
return f;
}
public static class Item{
int val;
int index;
public Item(int val,int index){
this.val = val;
this.index = index;
}
}
public static void mergeSortCount(Item[] itemArr,int count[],int n,int i,int j){
if(i>=j)return;
int mid = (i+j)/2;
mergeSortCount(itemArr, count, n, i, mid);
mergeSortCount(itemArr, count, n, mid+1, j);
mergeCount(itemArr,count,i,mid+1,mid,j);
}
public static void mergeCount(Item[] itemArr,int count[],int i1,int i2,int j1,int j2){
int size = j2 - i1+1;
Item[] sortedItems = new Item[size];
int index = 0;
int i11 = i1;
//int j11 = j2;
int i22 = i2;
//int j22 = j2;
while(i11<=j1 && i22 <=j2 ){
if(itemArr[i11].val < itemArr[i22].val){
count[itemArr[i11].index]+= (j2-i22+1);
sortedItems[index++] = itemArr[i11++];
}
else{
sortedItems[index++] = itemArr[i22++];
}
}
while(i11<=j1){
sortedItems[index++] = itemArr[i11++];
}
while(i22<=j2){
sortedItems[index++] = itemArr[i22++];
}
index=0;
for(int i=i1;i<=j2;i++){
itemArr[i] = sortedItems[index++];
}
}
public static void main(String args[]){
//System.out.println("Hello Java");
Scanner s = new Scanner(System.in);
int t = s.nextInt();
for(int t1 = 1;t1<=t;t1++){
int n =s.nextInt();
int arr[]= new int[n];
for(int i=0;i<n;i++){
arr[i] = s.nextInt();
}
int ans = 0;
int e9 = 1000000000;
for(int i=1;i<n-1;i++){
if(arr[i]>arr[i-1] && arr[i]>arr[i+1]){
if(i+2==n){
arr[i+1] = e9;
}
else{
arr[i+1]=Math.max(arr[i],arr[i+2]);
}
ans++;
}
}
System.out.println(ans);
for(int i=0;i<n;i++){
System.out.print(arr[i]+" ");
}
System.out.println();
}
}
}
/*
1 2 3 4 5 6
*/
| Java | ["5\n3\n2 1 2\n4\n1 2 3 1\n5\n1 2 1 2 1\n9\n1 2 1 3 2 3 1 2 1\n9\n2 1 3 1 3 1 3 1 3"] | 2 seconds | ["0\n2 1 2\n1\n1 3 3 1\n1\n1 2 2 2 1\n2\n1 2 3 3 2 3 3 2 1\n2\n2 1 3 3 3 1 1 1 3"] | NoteIn the first example, the array contains no local maximum, so we don't need to perform operations.In the second example, we can change $$$a_2$$$ to $$$3$$$, then the array don't have local maximums. | Java 8 | standard input | [
"greedy"
] | 255abf8750541f54d8ff0f1f80e1f8e7 | Each test contains multiple test cases. The first line will contain a single integer $$$t$$$ $$$(1 \leq t \leq 10000)$$$ — the number of test cases. Then $$$t$$$ test cases follow. The first line of each test case contains a single integer $$$n$$$ $$$(2 \leq n \leq 2 \cdot 10^5)$$$ — the size of the array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots ,a_n$$$ $$$(1 \leq a_i \leq 10^9)$$$, the elements of array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, first output a line containing a single integer $$$m$$$ — minimum number of operations required. Then ouput a line consist of $$$n$$$ integers — the resulting array after the operations. Note that this array should differ in exactly $$$m$$$ elements from the initial array. If there are multiple answers, print any. | standard output | |
PASSED | f8c6bbee360169c70f83dde03d741994 | train_108.jsonl | 1645367700 | You are given an array $$$a$$$ of size $$$n$$$. Each element in this array is an integer between $$$1$$$ and $$$10^9$$$.You can perform several operations to this array. During an operation, you can replace an element in the array with any integer between $$$1$$$ and $$$10^9$$$. Output the minimum number of operations needed such that the resulting array doesn't contain any local maximums, and the resulting array after the operations.An element $$$a_i$$$ is a local maximum if it is strictly larger than both of its neighbors (that is, $$$a_i > a_{i - 1}$$$ and $$$a_i > a_{i + 1}$$$). Since $$$a_1$$$ and $$$a_n$$$ have only one neighbor each, they will never be a local maximum. | 256 megabytes | import java.util.*;
import java.io.*;
public class Main {
static long startTime = System.currentTimeMillis();
// for global initializations and methods starts here
// global initialisations and methods end here
static void run() {
boolean tc = true;
AdityaFastIO r = new AdityaFastIO();
//FastReader r = new FastReader();
try (OutputStream out = new BufferedOutputStream(System.out)) {
//long startTime = System.currentTimeMillis();
int testcases = tc ? r.ni() : 1;
int tcCounter = 1;
// Hold Here Sparky------------------->>>
// Solution Starts Here
start:
while (testcases-- > 0) {
int n = r.ni();
List<Long> al = readLongList(n, r);
long count = 0;
if (n > 2) {
int i = 1;
while (i < n - 1) {
if (al.get(i) > al.get(i - 1) && al.get(i + 1) < al.get(i)) {
long get = Long.MIN_VALUE;
if (i + 2 < n) get = al.get(i + 2);
al.set(i + 1, Math.max(al.get(i), get));
count++;
}
i++;
}
}
out.write((count + " ").getBytes());
out.write(("\n").getBytes());
for (long ele : al) out.write((ele + " ").getBytes());
out.write(("\n").getBytes());
}
// Solution Ends Here
} catch (IOException e) {
e.printStackTrace();
}
}
static class AdityaFastIO {
final private int BUFFER_SIZE = 1 << 16;
private final DataInputStream din;
private final byte[] buffer;
private int bufferPointer, bytesRead;
public BufferedReader br;
public StringTokenizer st;
public AdityaFastIO() {
br = new BufferedReader(new InputStreamReader(System.in));
din = new DataInputStream(System.in);
buffer = new byte[BUFFER_SIZE];
bufferPointer = bytesRead = 0;
}
public AdityaFastIO(String file_name) throws IOException {
din = new DataInputStream(new FileInputStream(file_name));
buffer = new byte[BUFFER_SIZE];
bufferPointer = bytesRead = 0;
}
public String word() {
while (st == null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
public String line() {
String str = "";
try {
str = br.readLine();
} catch (IOException e) {
e.printStackTrace();
}
return str;
}
public String readLine() throws IOException {
byte[] buf = new byte[100000001]; // line length
int cnt = 0, c;
while ((c = read()) != -1) {
if (c == '\n') break;
buf[cnt++] = (byte) c;
}
return new String(buf, 0, cnt);
}
public int ni() 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 nl() 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 nd() 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();
}
}
public static void main(String[] args) throws Exception {
run();
}
static int[] readIntArr(int n, AdityaFastIO r) throws IOException {
int[] arr = new int[n];
for (int i = 0; i < n; i++) arr[i] = r.ni();
return arr;
}
static long[] readLongArr(int n, AdityaFastIO r) throws IOException {
long[] arr = new long[n];
for (int i = 0; i < n; i++) arr[i] = r.nl();
return arr;
}
static List<Integer> readIntList(int n, AdityaFastIO r) throws IOException {
List<Integer> al = new ArrayList<>();
for (int i = 0; i < n; i++) al.add(r.ni());
return al;
}
static List<Long> readLongList(int n, AdityaFastIO r) throws IOException {
List<Long> al = new ArrayList<>();
for (int i = 0; i < n; i++) al.add(r.nl());
return al;
}
static long mod = 998244353;
static long modInv(long base, long e) {
long result = 1;
base %= mod;
while (e > 0) {
if ((e & 1) > 0) result = result * base % mod;
base = base * base % mod;
e >>= 1;
}
return result;
}
static class FastReader {
BufferedReader br;
StringTokenizer st;
public FastReader() {
br = new BufferedReader(new InputStreamReader(System.in));
}
String word() {
while (st == null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
String line() {
String str = "";
try {
str = br.readLine();
} catch (IOException e) {
e.printStackTrace();
}
return str;
}
int ni() {
return Integer.parseInt(word());
}
long nl() {
return Long.parseLong(word());
}
double nd() {
return Double.parseDouble(word());
}
}
static int MOD = (int) (1e9 + 7);
static long powerLL(long x, long n) {
long result = 1;
while (n > 0) {
if (n % 2 == 1) result = result * x % MOD;
n = n / 2;
x = x * x % MOD;
}
return result;
}
static long powerStrings(int i1, int i2) {
String sa = String.valueOf(i1);
String sb = String.valueOf(i2);
long a = 0, b = 0;
for (int i = 0; i < sa.length(); i++) a = (a * 10 + (sa.charAt(i) - '0')) % MOD;
for (int i = 0; i < sb.length(); i++) b = (b * 10 + (sb.charAt(i) - '0')) % (MOD - 1);
return powerLL(a, b);
}
static long gcd(long a, long b) {
if (a == 0) return b;
else return gcd(b % a, a);
}
static long lcm(long a, long b) {
return (a * b) / gcd(a, b);
}
static long lower_bound(int[] arr, int x) {
int l = -1, r = arr.length;
while (l + 1 < r) {
int m = (l + r) >>> 1;
if (arr[m] >= x) r = m;
else l = m;
}
return r;
}
static int upper_bound(int[] arr, int x) {
int l = -1, r = arr.length;
while (l + 1 < r) {
int m = (l + r) >>> 1;
if (arr[m] <= x) l = m;
else r = m;
}
return l + 1;
}
static void addEdge(ArrayList<ArrayList<Integer>> graph, int edge1, int edge2) {
graph.get(edge1).add(edge2);
graph.get(edge2).add(edge1);
}
public static class Pair implements Comparable<Pair> {
int first;
int second;
public Pair(int first, int second) {
this.first = first;
this.second = second;
}
public String toString() {
return "(" + first + "," + second + ")";
}
public int compareTo(Pair o) {
// TODO Auto-generated method stub
if (this.first != o.first)
return (int) (this.first - o.first);
else return (int) (this.second - o.second);
}
}
public static class PairC<X, Y> implements Comparable<PairC> {
X first;
Y second;
public PairC(X first, Y second) {
this.first = first;
this.second = second;
}
public String toString() {
return "(" + first + "," + second + ")";
}
public int compareTo(PairC o) {
// TODO Auto-generated method stub
return o.compareTo((PairC) first);
}
}
static boolean isCollectionsSorted(List<Long> list) {
if (list.size() == 0 || list.size() == 1) return true;
for (int i = 1; i < list.size(); i++) if (list.get(i) <= list.get(i - 1)) return false;
return true;
}
static boolean isCollectionsSortedReverseOrder(List<Long> list) {
if (list.size() == 0 || list.size() == 1) return true;
for (int i = 1; i < list.size(); i++) if (list.get(i) >= list.get(i - 1)) return false;
return true;
}
} | Java | ["5\n3\n2 1 2\n4\n1 2 3 1\n5\n1 2 1 2 1\n9\n1 2 1 3 2 3 1 2 1\n9\n2 1 3 1 3 1 3 1 3"] | 2 seconds | ["0\n2 1 2\n1\n1 3 3 1\n1\n1 2 2 2 1\n2\n1 2 3 3 2 3 3 2 1\n2\n2 1 3 3 3 1 1 1 3"] | NoteIn the first example, the array contains no local maximum, so we don't need to perform operations.In the second example, we can change $$$a_2$$$ to $$$3$$$, then the array don't have local maximums. | Java 8 | standard input | [
"greedy"
] | 255abf8750541f54d8ff0f1f80e1f8e7 | Each test contains multiple test cases. The first line will contain a single integer $$$t$$$ $$$(1 \leq t \leq 10000)$$$ — the number of test cases. Then $$$t$$$ test cases follow. The first line of each test case contains a single integer $$$n$$$ $$$(2 \leq n \leq 2 \cdot 10^5)$$$ — the size of the array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots ,a_n$$$ $$$(1 \leq a_i \leq 10^9)$$$, the elements of array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, first output a line containing a single integer $$$m$$$ — minimum number of operations required. Then ouput a line consist of $$$n$$$ integers — the resulting array after the operations. Note that this array should differ in exactly $$$m$$$ elements from the initial array. If there are multiple answers, print any. | standard output | |
PASSED | aa9c271ce6cd46d91c3a5649118122ae | train_108.jsonl | 1645367700 | You are given an array $$$a$$$ of size $$$n$$$. Each element in this array is an integer between $$$1$$$ and $$$10^9$$$.You can perform several operations to this array. During an operation, you can replace an element in the array with any integer between $$$1$$$ and $$$10^9$$$. Output the minimum number of operations needed such that the resulting array doesn't contain any local maximums, and the resulting array after the operations.An element $$$a_i$$$ is a local maximum if it is strictly larger than both of its neighbors (that is, $$$a_i > a_{i - 1}$$$ and $$$a_i > a_{i + 1}$$$). Since $$$a_1$$$ and $$$a_n$$$ have only one neighbor each, they will never be a local maximum. | 256 megabytes | import java.util.*;
import java.io.*;
public class tr0 {
static PrintWriter out;
static StringBuilder sb;
static long mod = (long) 998244353;
static long inf = (long) 1e16;
static int n, l, k;
static TreeSet<Integer>[] ad, ad1, ad2;
static ArrayList<int[]>[] quer;
static int[][] remove, add;
static int[][] memo;
static boolean vis[];
static long[] inv, f, ncr[];
static HashMap<Integer, Integer> hm;
static int[] pre, suf, Smax[], Smin[];
static int idmax, idmin;
static ArrayList<Integer> av;
static HashMap<Integer, Integer> mm;
static boolean[] msks;
static int[] lazy[], lazyCount;
static int[] a, b;
static long ans, tem, c;
static ArrayList<Integer> gl;
static int[] ar;
static String s;
public static void main(String[] args) throws Exception {
Scanner sc = new Scanner(System.in);
out = new PrintWriter(System.out);
int t = sc.nextInt();
while (t-- > 0) {
int n = sc.nextInt();
int moves = 0;
int[] a = sc.nextArrInt(n);
for (int i = 1; i < n - 1; i++) {
if (a[i] > a[i - 1] && a[i] > a[i + 1]) {
moves++;
int max = a[i];
if (i + 2 < n)
max = Math.max(max, a[i + 2]);
a[i + 1] = max;
}
}
out.println(moves);
for (int k : a)
out.print(k + " ");
out.println();
}
out.flush();
}
static class Scanner {
StringTokenizer st;
BufferedReader br;
public Scanner(InputStream system) {
br = new BufferedReader(new InputStreamReader(system));
}
public Scanner(String file) throws Exception {
br = new BufferedReader(new FileReader(file));
}
public String next() throws IOException {
while (st == null || !st.hasMoreTokens())
st = new StringTokenizer(br.readLine());
return st.nextToken();
}
public String nextLine() throws IOException {
return br.readLine();
}
public int nextInt() throws IOException {
return Integer.parseInt(next());
}
public double nextDouble() throws IOException {
return Double.parseDouble(next());
}
public char nextChar() throws IOException {
return next().charAt(0);
}
public Long nextLong() throws IOException {
return Long.parseLong(next());
}
public int[] nextArrInt(int n) throws IOException {
int[] a = new int[n];
for (int i = 0; i < n; i++)
a[i] = nextInt();
return a;
}
public long[] nextArrLong(int n) throws IOException {
long[] a = new long[n];
for (int i = 0; i < n; i++)
a[i] = nextLong();
return a;
}
public int[] nextArrIntSorted(int n) throws IOException {
int[] a = new int[n];
Integer[] a1 = new Integer[n];
for (int i = 0; i < n; i++)
a1[i] = nextInt();
Arrays.sort(a1);
for (int i = 0; i < n; i++)
a[i] = a1[i].intValue();
return a;
}
public long[] nextArrLongSorted(int n) throws IOException {
long[] a = new long[n];
Long[] a1 = new Long[n];
for (int i = 0; i < n; i++)
a1[i] = nextLong();
Arrays.sort(a1);
for (int i = 0; i < n; i++)
a[i] = a1[i].longValue();
return a;
}
public boolean ready() throws IOException {
return br.ready();
}
public void waitForInput() throws InterruptedException {
Thread.sleep(3000);
}
}
} | Java | ["5\n3\n2 1 2\n4\n1 2 3 1\n5\n1 2 1 2 1\n9\n1 2 1 3 2 3 1 2 1\n9\n2 1 3 1 3 1 3 1 3"] | 2 seconds | ["0\n2 1 2\n1\n1 3 3 1\n1\n1 2 2 2 1\n2\n1 2 3 3 2 3 3 2 1\n2\n2 1 3 3 3 1 1 1 3"] | NoteIn the first example, the array contains no local maximum, so we don't need to perform operations.In the second example, we can change $$$a_2$$$ to $$$3$$$, then the array don't have local maximums. | Java 8 | standard input | [
"greedy"
] | 255abf8750541f54d8ff0f1f80e1f8e7 | Each test contains multiple test cases. The first line will contain a single integer $$$t$$$ $$$(1 \leq t \leq 10000)$$$ — the number of test cases. Then $$$t$$$ test cases follow. The first line of each test case contains a single integer $$$n$$$ $$$(2 \leq n \leq 2 \cdot 10^5)$$$ — the size of the array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots ,a_n$$$ $$$(1 \leq a_i \leq 10^9)$$$, the elements of array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, first output a line containing a single integer $$$m$$$ — minimum number of operations required. Then ouput a line consist of $$$n$$$ integers — the resulting array after the operations. Note that this array should differ in exactly $$$m$$$ elements from the initial array. If there are multiple answers, print any. | standard output | |
PASSED | c068a9c806a8eeb671fc967bd143ff07 | train_108.jsonl | 1645367700 | You are given an array $$$a$$$ of size $$$n$$$. Each element in this array is an integer between $$$1$$$ and $$$10^9$$$.You can perform several operations to this array. During an operation, you can replace an element in the array with any integer between $$$1$$$ and $$$10^9$$$. Output the minimum number of operations needed such that the resulting array doesn't contain any local maximums, and the resulting array after the operations.An element $$$a_i$$$ is a local maximum if it is strictly larger than both of its neighbors (that is, $$$a_i > a_{i - 1}$$$ and $$$a_i > a_{i + 1}$$$). Since $$$a_1$$$ and $$$a_n$$$ have only one neighbor each, they will never be a local maximum. | 256 megabytes | /*
Setting up my ambitions
Check 'em one at a time, yeah, I made it
Now it's time for ignition
I'm the start, heat it up
They follow, burning up a chain reaction, oh-oh-oh
Go fire it up, oh, oh, no
Assemble the crowd now, now
Line 'em up on the ground
No leaving anyone behind
*/
import static java.lang.Math.*;
import java.util.*;
import java.io.*;
import java.math.*;
public class x1635B
{
public static void main(String hi[]) throws Exception
{
BufferedReader infile = new BufferedReader(new InputStreamReader(System.in));
StringTokenizer st = new StringTokenizer(infile.readLine());
int T = Integer.parseInt(st.nextToken());
StringBuilder sb = new StringBuilder();
while(T-->0)
{
st = new StringTokenizer(infile.readLine());
int N = Integer.parseInt(st.nextToken());
int[] arr = readArr(N, infile, st);
ArrayList<Integer> ls = new ArrayList<Integer>();
for(int i=1; i < N-1; i++)
if(arr[i-1] < arr[i] && arr[i] > arr[i+1])
ls.add(i);
int res = 0;
int curr = 0;
while(curr < ls.size())
{
int dex = ls.get(curr++);
if(curr == ls.size())
{
arr[dex] = max(arr[dex-1], arr[dex+1]);
res++;
}
else
{
res++;
if(dex+2 == ls.get(curr))
{
arr[dex+1] = max(arr[dex], arr[dex+2]);
curr++;
}
else
arr[dex] = max(arr[dex-1], arr[dex+1]);
}
}
sb.append(res+"\n");
for(int x: arr)
sb.append(x+" ");
sb.append("\n");
}
System.out.print(sb);
}
public static int[] readArr(int N, BufferedReader infile, StringTokenizer st) throws Exception
{
int[] arr = new int[N];
st = new StringTokenizer(infile.readLine());
for(int i=0; i < N; i++)
arr[i] = Integer.parseInt(st.nextToken());
return arr;
}
} | Java | ["5\n3\n2 1 2\n4\n1 2 3 1\n5\n1 2 1 2 1\n9\n1 2 1 3 2 3 1 2 1\n9\n2 1 3 1 3 1 3 1 3"] | 2 seconds | ["0\n2 1 2\n1\n1 3 3 1\n1\n1 2 2 2 1\n2\n1 2 3 3 2 3 3 2 1\n2\n2 1 3 3 3 1 1 1 3"] | NoteIn the first example, the array contains no local maximum, so we don't need to perform operations.In the second example, we can change $$$a_2$$$ to $$$3$$$, then the array don't have local maximums. | Java 8 | standard input | [
"greedy"
] | 255abf8750541f54d8ff0f1f80e1f8e7 | Each test contains multiple test cases. The first line will contain a single integer $$$t$$$ $$$(1 \leq t \leq 10000)$$$ — the number of test cases. Then $$$t$$$ test cases follow. The first line of each test case contains a single integer $$$n$$$ $$$(2 \leq n \leq 2 \cdot 10^5)$$$ — the size of the array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots ,a_n$$$ $$$(1 \leq a_i \leq 10^9)$$$, the elements of array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, first output a line containing a single integer $$$m$$$ — minimum number of operations required. Then ouput a line consist of $$$n$$$ integers — the resulting array after the operations. Note that this array should differ in exactly $$$m$$$ elements from the initial array. If there are multiple answers, print any. | standard output | |
PASSED | aa4a6c594639c1281156dc505c6016bf | train_108.jsonl | 1645367700 | You are given an array $$$a$$$ of size $$$n$$$. Each element in this array is an integer between $$$1$$$ and $$$10^9$$$.You can perform several operations to this array. During an operation, you can replace an element in the array with any integer between $$$1$$$ and $$$10^9$$$. Output the minimum number of operations needed such that the resulting array doesn't contain any local maximums, and the resulting array after the operations.An element $$$a_i$$$ is a local maximum if it is strictly larger than both of its neighbors (that is, $$$a_i > a_{i - 1}$$$ and $$$a_i > a_{i + 1}$$$). Since $$$a_1$$$ and $$$a_n$$$ have only one neighbor each, they will never be a local maximum. | 256 megabytes | import java.util.*;
import java.io.*;
import java.util.stream.*;
public class App {
static BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
static BufferedWriter bw = new BufferedWriter(new OutputStreamWriter(System.out));
public static void main(String[] args) throws Exception {
if (System.getProperty("ONLINE_JUDGE") == null) {
File inputFile = new File("/Users/vipinjain/self/cp/input.txt");
File outputFile = new File("/Users/vipinjain/self/cp/output.txt");
br = new BufferedReader(new FileReader(inputFile));
bw = new BufferedWriter(new FileWriter(outputFile));
}
int tests;
tests = Integer.parseInt(br.readLine());
//tests = 1;
while (tests-- > 0) {
solve();
}
bw.flush();
bw.close();
br.close();
}
static void solve() throws Exception {
// String[] tmp = br.readLine().split(" ");
// int a = Integer.parseInt(tmp[0]);
// int b = Integer.parseInt(tmp[1]);
// int c = Integer.parseInt(tmp[2]);
int n = Integer.parseInt(br.readLine());
int[] arr = Stream.of(br.readLine().split(" ")).mapToInt(Integer::parseInt).toArray();
int count = 0;
for (int i = 1; i < n - 1; ++i) {
if (arr[i] > arr[i - 1] && arr[i] > arr[i + 1]) {
if (i + 2 < n) {
arr[i + 1] = Math.max(arr[i + 2], arr[i]);
} else {
arr[i + 1] = arr[i];
}
count++;
}
}
bw.write(count + "\n");
printArray(arr);
bw.write("\n");
}
static void printArray(int[] arr) throws IOException{
StringBuilder sb = new StringBuilder();
for (int num : arr) {
sb.append(num + " ");
}
bw.write(sb.toString().trim());
}
static void shuffleArray(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;
}
}
static int gcd(int a, int b) {
if (a == 0) {
return b;
}
return gcd(b % a, a);
}
static boolean isPrime(long n) {
for (long i = 2; i * i <= n; i++) {
if (n % i == 0) {
return false;
}
}
return true;
}
static void debug(Object... obj) {
System.err.println(Arrays.deepToString(obj));
}
} | Java | ["5\n3\n2 1 2\n4\n1 2 3 1\n5\n1 2 1 2 1\n9\n1 2 1 3 2 3 1 2 1\n9\n2 1 3 1 3 1 3 1 3"] | 2 seconds | ["0\n2 1 2\n1\n1 3 3 1\n1\n1 2 2 2 1\n2\n1 2 3 3 2 3 3 2 1\n2\n2 1 3 3 3 1 1 1 3"] | NoteIn the first example, the array contains no local maximum, so we don't need to perform operations.In the second example, we can change $$$a_2$$$ to $$$3$$$, then the array don't have local maximums. | Java 8 | standard input | [
"greedy"
] | 255abf8750541f54d8ff0f1f80e1f8e7 | Each test contains multiple test cases. The first line will contain a single integer $$$t$$$ $$$(1 \leq t \leq 10000)$$$ — the number of test cases. Then $$$t$$$ test cases follow. The first line of each test case contains a single integer $$$n$$$ $$$(2 \leq n \leq 2 \cdot 10^5)$$$ — the size of the array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots ,a_n$$$ $$$(1 \leq a_i \leq 10^9)$$$, the elements of array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, first output a line containing a single integer $$$m$$$ — minimum number of operations required. Then ouput a line consist of $$$n$$$ integers — the resulting array after the operations. Note that this array should differ in exactly $$$m$$$ elements from the initial array. If there are multiple answers, print any. | standard output | |
PASSED | 94db74b92ae3cb2e3da26820ecba63a4 | train_108.jsonl | 1645367700 | You are given an array $$$a$$$ of size $$$n$$$. Each element in this array is an integer between $$$1$$$ and $$$10^9$$$.You can perform several operations to this array. During an operation, you can replace an element in the array with any integer between $$$1$$$ and $$$10^9$$$. Output the minimum number of operations needed such that the resulting array doesn't contain any local maximums, and the resulting array after the operations.An element $$$a_i$$$ is a local maximum if it is strictly larger than both of its neighbors (that is, $$$a_i > a_{i - 1}$$$ and $$$a_i > a_{i + 1}$$$). Since $$$a_1$$$ and $$$a_n$$$ have only one neighbor each, they will never be a local maximum. | 256 megabytes | import java.io.DataInputStream;
import java.io.FileInputStream;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.*;
import java.util.StringTokenizer;
public class Avoid_Local_Maximums
{
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[64]; // 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();
}
}
public static void main(String Args[]) throws java.lang.Exception
{
try
{
Reader obj = new Reader();
int t = obj.nextInt();
while (t > 0)
{
t--;
int n = obj.nextInt();
int arr[] = new int[n];
int i, c = 0;
for (i=0;i<n;i++)
{
arr[i] = obj.nextInt();
}
for (i=1;i<n-1;i++)
{
if (arr[i] > arr[i-1] && arr[i] > arr[i+1])
{
c++;
if ((i+2) < n)
{
arr[i+1] = Math.max (arr[i], arr[i+2]);
}
else
{
arr[i+1] = arr[i];
}
}
}
System.out.println (c);
for (i=0;i<n;i++)
{
System.out.print (arr[i] + " ");
}
System.out.println ();
}
}catch(Exception e){
return;
}
}
} | Java | ["5\n3\n2 1 2\n4\n1 2 3 1\n5\n1 2 1 2 1\n9\n1 2 1 3 2 3 1 2 1\n9\n2 1 3 1 3 1 3 1 3"] | 2 seconds | ["0\n2 1 2\n1\n1 3 3 1\n1\n1 2 2 2 1\n2\n1 2 3 3 2 3 3 2 1\n2\n2 1 3 3 3 1 1 1 3"] | NoteIn the first example, the array contains no local maximum, so we don't need to perform operations.In the second example, we can change $$$a_2$$$ to $$$3$$$, then the array don't have local maximums. | Java 8 | standard input | [
"greedy"
] | 255abf8750541f54d8ff0f1f80e1f8e7 | Each test contains multiple test cases. The first line will contain a single integer $$$t$$$ $$$(1 \leq t \leq 10000)$$$ — the number of test cases. Then $$$t$$$ test cases follow. The first line of each test case contains a single integer $$$n$$$ $$$(2 \leq n \leq 2 \cdot 10^5)$$$ — the size of the array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots ,a_n$$$ $$$(1 \leq a_i \leq 10^9)$$$, the elements of array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, first output a line containing a single integer $$$m$$$ — minimum number of operations required. Then ouput a line consist of $$$n$$$ integers — the resulting array after the operations. Note that this array should differ in exactly $$$m$$$ elements from the initial array. If there are multiple answers, print any. | standard output | |
PASSED | 697d69da8345a0136e6fc9437ffe5c68 | train_108.jsonl | 1645367700 | You are given an array $$$a$$$ of size $$$n$$$. Each element in this array is an integer between $$$1$$$ and $$$10^9$$$.You can perform several operations to this array. During an operation, you can replace an element in the array with any integer between $$$1$$$ and $$$10^9$$$. Output the minimum number of operations needed such that the resulting array doesn't contain any local maximums, and the resulting array after the operations.An element $$$a_i$$$ is a local maximum if it is strictly larger than both of its neighbors (that is, $$$a_i > a_{i - 1}$$$ and $$$a_i > a_{i + 1}$$$). Since $$$a_1$$$ and $$$a_n$$$ have only one neighbor each, they will never be a local maximum. | 256 megabytes | import java.util.Scanner;
public class B1635 {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
int T = in.nextInt();
StringBuilder output = new StringBuilder();
for (int t=0; t<T; t++) {
int N = in.nextInt();
int[] A = new int[N];
for (int n=0; n<N; n++) {
A[n] = in.nextInt();
}
int answer = 0;
for (int n=1; n<N-1; n++) {
if (A[n] > A[n-1] && A[n] > A[n+1]) {
answer++;
A[n+1] = (n+2 < N) ? Math.max(A[n], A[n+2]) : A[n];
}
}
output.append(answer).append('\n');
for (int a : A) {
output.append(a).append(' ');
}
output.append('\n');
}
System.out.print(output);
}
}
| Java | ["5\n3\n2 1 2\n4\n1 2 3 1\n5\n1 2 1 2 1\n9\n1 2 1 3 2 3 1 2 1\n9\n2 1 3 1 3 1 3 1 3"] | 2 seconds | ["0\n2 1 2\n1\n1 3 3 1\n1\n1 2 2 2 1\n2\n1 2 3 3 2 3 3 2 1\n2\n2 1 3 3 3 1 1 1 3"] | NoteIn the first example, the array contains no local maximum, so we don't need to perform operations.In the second example, we can change $$$a_2$$$ to $$$3$$$, then the array don't have local maximums. | Java 8 | standard input | [
"greedy"
] | 255abf8750541f54d8ff0f1f80e1f8e7 | Each test contains multiple test cases. The first line will contain a single integer $$$t$$$ $$$(1 \leq t \leq 10000)$$$ — the number of test cases. Then $$$t$$$ test cases follow. The first line of each test case contains a single integer $$$n$$$ $$$(2 \leq n \leq 2 \cdot 10^5)$$$ — the size of the array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots ,a_n$$$ $$$(1 \leq a_i \leq 10^9)$$$, the elements of array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, first output a line containing a single integer $$$m$$$ — minimum number of operations required. Then ouput a line consist of $$$n$$$ integers — the resulting array after the operations. Note that this array should differ in exactly $$$m$$$ elements from the initial array. If there are multiple answers, print any. | standard output | |
PASSED | d4dc42fd4258e00ac1a9e0e465d5eea7 | train_108.jsonl | 1645367700 | You are given an array $$$a$$$ of size $$$n$$$. Each element in this array is an integer between $$$1$$$ and $$$10^9$$$.You can perform several operations to this array. During an operation, you can replace an element in the array with any integer between $$$1$$$ and $$$10^9$$$. Output the minimum number of operations needed such that the resulting array doesn't contain any local maximums, and the resulting array after the operations.An element $$$a_i$$$ is a local maximum if it is strictly larger than both of its neighbors (that is, $$$a_i > a_{i - 1}$$$ and $$$a_i > a_{i + 1}$$$). Since $$$a_1$$$ and $$$a_n$$$ have only one neighbor each, they will never be a local maximum. | 256 megabytes |
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.*;
import javafx.scene.layout.Priority;
public class Main
{
public static void main(String args[])
{
FastScanner input = new FastScanner();
int tc = input.nextInt();
work:
while (tc-- > 0) {
int n = input.nextInt();
int a[] = new int[n];
for (int i = 0; i <n; i++) {
a[i] = input.nextInt();
}
int count =0;
if(n!=2)
{
for (int i = 1; i <n-1; i++) {
if(a[i-1]<a[i]&&a[i]>a[i+1])
{
if(i+3<n&&a[i+1]<a[i+2]&&a[i+2]>a[i+3])
{
a[i+1]= Math.max(a[i], a[i+2]);
i+=3;
}
else
{
a[i] = Math.max(a[i-1], a[i+1]);
}
count++;
}
}
}
System.out.println(count);
for (int i : a) {
System.out.print(i+" ");
}
System.out.println("");
}
}
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 | ["5\n3\n2 1 2\n4\n1 2 3 1\n5\n1 2 1 2 1\n9\n1 2 1 3 2 3 1 2 1\n9\n2 1 3 1 3 1 3 1 3"] | 2 seconds | ["0\n2 1 2\n1\n1 3 3 1\n1\n1 2 2 2 1\n2\n1 2 3 3 2 3 3 2 1\n2\n2 1 3 3 3 1 1 1 3"] | NoteIn the first example, the array contains no local maximum, so we don't need to perform operations.In the second example, we can change $$$a_2$$$ to $$$3$$$, then the array don't have local maximums. | Java 8 | standard input | [
"greedy"
] | 255abf8750541f54d8ff0f1f80e1f8e7 | Each test contains multiple test cases. The first line will contain a single integer $$$t$$$ $$$(1 \leq t \leq 10000)$$$ — the number of test cases. Then $$$t$$$ test cases follow. The first line of each test case contains a single integer $$$n$$$ $$$(2 \leq n \leq 2 \cdot 10^5)$$$ — the size of the array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots ,a_n$$$ $$$(1 \leq a_i \leq 10^9)$$$, the elements of array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, first output a line containing a single integer $$$m$$$ — minimum number of operations required. Then ouput a line consist of $$$n$$$ integers — the resulting array after the operations. Note that this array should differ in exactly $$$m$$$ elements from the initial array. If there are multiple answers, print any. | standard output | |
PASSED | 1fc2b25d410625269b4026e0c4d3a5a9 | train_108.jsonl | 1645367700 | You are given an array $$$a$$$ of size $$$n$$$. Each element in this array is an integer between $$$1$$$ and $$$10^9$$$.You can perform several operations to this array. During an operation, you can replace an element in the array with any integer between $$$1$$$ and $$$10^9$$$. Output the minimum number of operations needed such that the resulting array doesn't contain any local maximums, and the resulting array after the operations.An element $$$a_i$$$ is a local maximum if it is strictly larger than both of its neighbors (that is, $$$a_i > a_{i - 1}$$$ and $$$a_i > a_{i + 1}$$$). Since $$$a_1$$$ and $$$a_n$$$ have only one neighbor each, they will never be a local maximum. | 256 megabytes | import java.awt.*;
import java.util.*;
import java.io.*;
public class Codeforces {
static FScanner sc = new FScanner();
static PrintWriter out = new PrintWriter(System.out);
static final Random random = new Random();
static long mod = 1000000007L;
static HashMap<String, Integer> map = new HashMap<>();
static boolean[] sieve = new boolean[1000000];
static double[] fib = new double[1000000];
public static void main(String args[]) throws IOException {
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 cnt=0;
for(int i=1;i<n-1;i++)
{
if(a[i]>a[i-1]&&a[i]>a[i+1])
{
if(i+1==n-1)
a[i+1]=a[i];
else
a[i+1]=max(a[i],a[i+2]);
cnt++;
}
}
out.println(cnt);
for(int i=0;i<n;i++)
out.print(a[i]+" ");
out.println();
}
out.close();
}
// TemplateCode
static void fib() {
fib[0] = 0;
fib[1] = 1;
for (int i = 2; i < fib.length; i++)
fib[i] = fib[i - 1] + fib[i - 2];
}
static void primeSieve() {
for (int i = 0; i < sieve.length; i++)
sieve[i] = true;
for (int i = 2; i * i <= sieve.length; i++) {
if (sieve[i]) {
for (int j = i * i; j < sieve.length; j += i) {
sieve[j] = false;
}
}
}
}
static int max(int a, int b) {
if (a < b)
return b;
return a;
}
static int min(int a, int b) {
if (a < b)
return a;
return b;
}
static void ruffleSort(int[] a) {
int n = a.length;
for (int i = 0; i < n; i++) {
int oi = random.nextInt(n), temp = a[oi];
a[oi] = a[i];
a[i] = temp;
}
Arrays.sort(a);
}
static <E> void print(E res) {
System.out.println(res);
}
static int gcd(int a, int b) {
if (b == 0) {
return a;
}
return gcd(b, a % b);
}
static int lcm(int a, int b) {
return (a / gcd(a, b)) * b;
}
static int abs(int a) {
if (a < 0)
return -1 * a;
return a;
}
static class FScanner {
BufferedReader br;
StringTokenizer st;
public FScanner() {
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;
}
int[] readintarray(int n) {
int res[] = new int[n];
for (int i = 0; i < n; i++) res[i] = nextInt();
return res;
}
long[] readlongarray(int n) {
long res[] = new long[n];
for (int i = 0; i < n; i++) res[i] = nextLong();
return res;
}
}
}
| Java | ["5\n3\n2 1 2\n4\n1 2 3 1\n5\n1 2 1 2 1\n9\n1 2 1 3 2 3 1 2 1\n9\n2 1 3 1 3 1 3 1 3"] | 2 seconds | ["0\n2 1 2\n1\n1 3 3 1\n1\n1 2 2 2 1\n2\n1 2 3 3 2 3 3 2 1\n2\n2 1 3 3 3 1 1 1 3"] | NoteIn the first example, the array contains no local maximum, so we don't need to perform operations.In the second example, we can change $$$a_2$$$ to $$$3$$$, then the array don't have local maximums. | Java 8 | standard input | [
"greedy"
] | 255abf8750541f54d8ff0f1f80e1f8e7 | Each test contains multiple test cases. The first line will contain a single integer $$$t$$$ $$$(1 \leq t \leq 10000)$$$ — the number of test cases. Then $$$t$$$ test cases follow. The first line of each test case contains a single integer $$$n$$$ $$$(2 \leq n \leq 2 \cdot 10^5)$$$ — the size of the array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots ,a_n$$$ $$$(1 \leq a_i \leq 10^9)$$$, the elements of array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, first output a line containing a single integer $$$m$$$ — minimum number of operations required. Then ouput a line consist of $$$n$$$ integers — the resulting array after the operations. Note that this array should differ in exactly $$$m$$$ elements from the initial array. If there are multiple answers, print any. | standard output | |
PASSED | a50214e2780da6e7534ba342cecbcb37 | train_108.jsonl | 1645367700 | You are given an array $$$a$$$ of size $$$n$$$. Each element in this array is an integer between $$$1$$$ and $$$10^9$$$.You can perform several operations to this array. During an operation, you can replace an element in the array with any integer between $$$1$$$ and $$$10^9$$$. Output the minimum number of operations needed such that the resulting array doesn't contain any local maximums, and the resulting array after the operations.An element $$$a_i$$$ is a local maximum if it is strictly larger than both of its neighbors (that is, $$$a_i > a_{i - 1}$$$ and $$$a_i > a_{i + 1}$$$). Since $$$a_1$$$ and $$$a_n$$$ have only one neighbor each, they will never be a local maximum. | 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 Code
{
public static void main (String[] args) throws java.lang.Exception
{
Scanner sc=new Scanner(System.in);
int t=sc.nextInt();
while(t>0){
t--;
int n=sc.nextInt();
int ans=0;
int a[]=new int[n];
for(int i=0;i<n;i++){
a[i]=sc.nextInt();
}
int count=0;
for(int i=1;i<n-1;i++){
if (a[i] > a[i - 1] && a[i] > a[i + 1]) {
if (i + 2 < n) {
a[i + 1] = Math.max(a[i], a[i + 2]);
} else {
a[i + 1] = a[i];
}
count++;}}
System.out.println(count);
for(int i=0;i<n;i++){
System.out.print(a[i]+" ");}
System.out.println();
}
}
}
| Java | ["5\n3\n2 1 2\n4\n1 2 3 1\n5\n1 2 1 2 1\n9\n1 2 1 3 2 3 1 2 1\n9\n2 1 3 1 3 1 3 1 3"] | 2 seconds | ["0\n2 1 2\n1\n1 3 3 1\n1\n1 2 2 2 1\n2\n1 2 3 3 2 3 3 2 1\n2\n2 1 3 3 3 1 1 1 3"] | NoteIn the first example, the array contains no local maximum, so we don't need to perform operations.In the second example, we can change $$$a_2$$$ to $$$3$$$, then the array don't have local maximums. | Java 8 | standard input | [
"greedy"
] | 255abf8750541f54d8ff0f1f80e1f8e7 | Each test contains multiple test cases. The first line will contain a single integer $$$t$$$ $$$(1 \leq t \leq 10000)$$$ — the number of test cases. Then $$$t$$$ test cases follow. The first line of each test case contains a single integer $$$n$$$ $$$(2 \leq n \leq 2 \cdot 10^5)$$$ — the size of the array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots ,a_n$$$ $$$(1 \leq a_i \leq 10^9)$$$, the elements of array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, first output a line containing a single integer $$$m$$$ — minimum number of operations required. Then ouput a line consist of $$$n$$$ integers — the resulting array after the operations. Note that this array should differ in exactly $$$m$$$ elements from the initial array. If there are multiple answers, print any. | standard output | |
PASSED | 677f586e48d58b13bc8b3418b70639ca | train_108.jsonl | 1645367700 | You are given an array $$$a$$$ of size $$$n$$$. Each element in this array is an integer between $$$1$$$ and $$$10^9$$$.You can perform several operations to this array. During an operation, you can replace an element in the array with any integer between $$$1$$$ and $$$10^9$$$. Output the minimum number of operations needed such that the resulting array doesn't contain any local maximums, and the resulting array after the operations.An element $$$a_i$$$ is a local maximum if it is strictly larger than both of its neighbors (that is, $$$a_i > a_{i - 1}$$$ and $$$a_i > a_{i + 1}$$$). Since $$$a_1$$$ and $$$a_n$$$ have only one neighbor each, they will never be a local maximum. | 256 megabytes |
import java.io.*;
import java.util.*;
public final class Main {
//int 2e9 - long 9e18
static PrintWriter out = new PrintWriter(System.out);
static FastReader in = new FastReader();
static Pair[] moves = new Pair[]{new Pair(-1, 0), new Pair(0, 1), new Pair(1, 0), new Pair(0, -1)};
static int mod = (int) (1e9 + 7);
static int mod2 = 998244353;
public static void main(String[] args) {
int tt = i();
while (tt-- > 0) {
solve();
}
out.flush();
}
public static void solve() {
int n = i();
int[] a = input(n);
int ans = 0;
for (int i = 1; i < n - 1; i++) {
if (a[i] > a[i - 1] && a[i] > a[i + 1]) {
a[i + 1] = Math.max(a[i], ((i + 2 < n) ? a[i + 2] : 0));
ans++;
}
}
out.println(ans);
print(a);
}
// (10,5) = 2 ,(11,5) = 3
static long upperDiv(long a, long b) {
return (a / b) + ((a % b == 0) ? 0 : 1);
}
static long sum(int[] a) {
long sum = 0;
for (int x : a) {
sum += x;
}
return sum;
}
static int[] preint(int[] a) {
int[] pre = new int[a.length + 1];
pre[0] = 0;
for (int i = 0; i < a.length; i++) {
pre[i + 1] = pre[i] + a[i];
}
return pre;
}
static long[] pre(int[] a) {
long[] pre = new long[a.length + 1];
pre[0] = 0;
for (int i = 0; i < a.length; i++) {
pre[i + 1] = pre[i] + a[i];
}
return pre;
}
static long[] post(int[] a) {
long[] post = new long[a.length + 1];
post[0] = 0;
for (int i = 0; i < a.length; i++) {
post[i + 1] = post[i] + a[a.length - 1 - i];
}
return post;
}
static long[] pre(long[] a) {
long[] pre = new long[a.length + 1];
pre[0] = 0;
for (int i = 0; i < a.length; i++) {
pre[i + 1] = pre[i] + a[i];
}
return pre;
}
static void print(char A[]) {
for (char c : A) {
out.print(c);
}
out.println();
}
static void print(boolean A[]) {
for (boolean c : A) {
out.print(c + " ");
}
out.println();
}
static void print(int A[]) {
for (int c : A) {
out.print(c + " ");
}
out.println();
}
static void print(long A[]) {
for (long i : A) {
out.print(i + " ");
}
out.println();
}
static void print(List<Integer> A) {
for (int a : A) {
out.print(a + " ");
}
}
static int i() {
return in.nextInt();
}
static long l() {
return in.nextLong();
}
static double d() {
return in.nextDouble();
}
static String s() {
return in.nextLine();
}
static String c() {
return in.next();
}
static int[][] inputWithIdx(int N) {
int A[][] = new int[N][2];
for (int i = 0; i < N; i++) {
A[i] = new int[]{i, in.nextInt()};
}
return A;
}
static int[] input(int N) {
int A[] = new int[N];
for (int i = 0; i < N; i++) {
A[i] = in.nextInt();
}
return A;
}
static long[] inputLong(int N) {
long A[] = new long[N];
for (int i = 0; i < A.length; i++) {
A[i] = in.nextLong();
}
return A;
}
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 long LCM(int a, int b) {
return (long) a / GCD(a, b) * b;
}
static long LCM(long a, long b) {
return a / GCD(a, b) * b;
}
// find highest i which satisfy a[i]<=x
static int lowerbound(int[] a, int x) {
int l = 0;
int r = a.length - 1;
while (l < r) {
int m = (l + r + 1) / 2;
if (a[m] <= x) {
l = m;
} else {
r = m - 1;
}
}
return l;
}
static void shuffle(int[] arr) {
for (int i = 0; i < arr.length; i++) {
int rand = (int) (Math.random() * arr.length);
int temp = arr[rand];
arr[rand] = arr[i];
arr[i] = temp;
}
}
static void shuffleAndSort(int[] arr) {
for (int i = 0; i < arr.length; i++) {
int rand = (int) (Math.random() * arr.length);
int temp = arr[rand];
arr[rand] = arr[i];
arr[i] = temp;
}
Arrays.sort(arr);
}
static void shuffleAndSort(int[][] arr, Comparator<? super int[]> comparator) {
for (int i = 0; i < arr.length; i++) {
int rand = (int) (Math.random() * arr.length);
int[] temp = arr[rand];
arr[rand] = arr[i];
arr[i] = temp;
}
Arrays.sort(arr, comparator);
}
static void shuffleAndSort(long[] arr) {
for (int i = 0; i < arr.length; i++) {
int rand = (int) (Math.random() * arr.length);
long temp = arr[rand];
arr[rand] = arr[i];
arr[i] = temp;
}
Arrays.sort(arr);
}
static boolean isPerfectSquare(double number) {
double sqrt = Math.sqrt(number);
return ((sqrt - Math.floor(sqrt)) == 0);
}
static void swap(int A[], int a, int b) {
int t = A[a];
A[a] = A[b];
A[b] = t;
}
static void swap(char A[], int a, int b) {
char t = A[a];
A[a] = A[b];
A[b] = t;
}
static long pow(long a, long b, int mod) {
long pow = 1;
long x = a;
while (b != 0) {
if ((b & 1) != 0) {
pow = (pow * x) % mod;
}
x = (x * x) % mod;
b /= 2;
}
return pow;
}
static long pow(long a, long b) {
long pow = 1;
long x = a;
while (b != 0) {
if ((b & 1) != 0) {
pow *= x;
}
x = x * x;
b /= 2;
}
return pow;
}
static long modInverse(long x, int mod) {
return pow(x, mod - 2, mod);
}
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;
}
public static String reverse(String str) {
if (str == null) {
return null;
}
return new StringBuilder(str).reverse().toString();
}
public static void reverse(int[] arr) {
for (int i = 0; i < arr.length / 2; i++) {
int tmp = arr[i];
arr[arr.length - 1 - i] = tmp;
arr[i] = arr[arr.length - 1 - i];
}
}
public static String repeat(char ch, int repeat) {
if (repeat <= 0) {
return "";
}
final char[] buf = new char[repeat];
for (int i = repeat - 1; i >= 0; i--) {
buf[i] = ch;
}
return new String(buf);
}
public static int[] manacher(String s) {
char[] chars = s.toCharArray();
int n = s.length();
int[] d1 = new int[n];
for (int i = 0, l = 0, r = -1; i < n; i++) {
int k = (i > r) ? 1 : Math.min(d1[l + r - i], r - i + 1);
while (0 <= i - k && i + k < n && chars[i - k] == chars[i + k]) {
k++;
}
d1[i] = k--;
if (i + k > r) {
l = i - k;
r = i + k;
}
}
return d1;
}
public static int[] kmp(String s) {
int n = s.length();
int[] res = new int[n];
for (int i = 1; i < n; ++i) {
int j = res[i - 1];
while (j > 0 && s.charAt(i) != s.charAt(j)) {
j = res[j - 1];
}
if (s.charAt(i) == s.charAt(j)) {
++j;
}
res[i] = j;
}
return res;
}
}
class Pair {
int i;
int j;
Pair(int i, int j) {
this.i = i;
this.j = j;
}
@Override
public boolean equals(Object o) {
if (this == o) {
return true;
}
if (o == null || getClass() != o.getClass()) {
return false;
}
Pair pair = (Pair) o;
return i == pair.i && j == pair.j;
}
@Override
public int hashCode() {
return Objects.hash(i, j);
}
}
class ThreePair {
int i;
int j;
int k;
ThreePair(int i, int j, int k) {
this.i = i;
this.j = j;
this.k = k;
}
@Override
public boolean equals(Object o) {
if (this == o) {
return true;
}
if (o == null || getClass() != o.getClass()) {
return false;
}
ThreePair pair = (ThreePair) o;
return i == pair.i && j == pair.j && k == pair.k;
}
@Override
public int hashCode() {
return Objects.hash(i, j);
}
}
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;
}
}
class Node {
int val;
public Node(int val) {
this.val = val;
}
}
class ST {
int n;
Node[] st;
ST(int n) {
this.n = n;
st = new Node[4 * Integer.highestOneBit(n)];
}
void build(Node[] nodes) {
build(0, 0, n - 1, nodes);
}
private void build(int id, int l, int r, Node[] nodes) {
if (l == r) {
st[id] = nodes[l];
return;
}
int mid = (l + r) >> 1;
build((id << 1) + 1, l, mid, nodes);
build((id << 1) + 2, mid + 1, r, nodes);
st[id] = comb(st[(id << 1) + 1], st[(id << 1) + 2]);
}
void update(int i, Node node) {
update(0, 0, n - 1, i, node);
}
private void update(int id, int l, int r, int i, Node node) {
if (i < l || r < i) {
return;
}
if (l == r) {
st[id] = node;
return;
}
int mid = (l + r) >> 1;
update((id << 1) + 1, l, mid, i, node);
update((id << 1) + 2, mid + 1, r, i, node);
st[id] = comb(st[(id << 1) + 1], st[(id << 1) + 2]);
}
Node get(int x, int y) {
return get(0, 0, n - 1, x, y);
}
private Node get(int id, int l, int r, int x, int y) {
if (x > r || y < l) {
return new Node(0);
}
if (x <= l && r <= y) {
return st[id];
}
int mid = (l + r) >> 1;
return comb(get((id << 1) + 1, l, mid, x, y), get((id << 1) + 2, mid + 1, r, x, y));
}
Node comb(Node a, Node b) {
if (a == null) {
return b;
}
if (b == null) {
return a;
}
return new Node(GCD(a.val, b.val));
}
static int GCD(int a, int b) {
if (b == 0) {
return a;
} else {
return GCD(b, a % b);
}
}
} | Java | ["5\n3\n2 1 2\n4\n1 2 3 1\n5\n1 2 1 2 1\n9\n1 2 1 3 2 3 1 2 1\n9\n2 1 3 1 3 1 3 1 3"] | 2 seconds | ["0\n2 1 2\n1\n1 3 3 1\n1\n1 2 2 2 1\n2\n1 2 3 3 2 3 3 2 1\n2\n2 1 3 3 3 1 1 1 3"] | NoteIn the first example, the array contains no local maximum, so we don't need to perform operations.In the second example, we can change $$$a_2$$$ to $$$3$$$, then the array don't have local maximums. | Java 8 | standard input | [
"greedy"
] | 255abf8750541f54d8ff0f1f80e1f8e7 | Each test contains multiple test cases. The first line will contain a single integer $$$t$$$ $$$(1 \leq t \leq 10000)$$$ — the number of test cases. Then $$$t$$$ test cases follow. The first line of each test case contains a single integer $$$n$$$ $$$(2 \leq n \leq 2 \cdot 10^5)$$$ — the size of the array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots ,a_n$$$ $$$(1 \leq a_i \leq 10^9)$$$, the elements of array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, first output a line containing a single integer $$$m$$$ — minimum number of operations required. Then ouput a line consist of $$$n$$$ integers — the resulting array after the operations. Note that this array should differ in exactly $$$m$$$ elements from the initial array. If there are multiple answers, print any. | standard output | |
PASSED | 133678bef8041d7b7f71604a52f8b73c | train_108.jsonl | 1645367700 | You are given an array $$$a$$$ of size $$$n$$$. Each element in this array is an integer between $$$1$$$ and $$$10^9$$$.You can perform several operations to this array. During an operation, you can replace an element in the array with any integer between $$$1$$$ and $$$10^9$$$. Output the minimum number of operations needed such that the resulting array doesn't contain any local maximums, and the resulting array after the operations.An element $$$a_i$$$ is a local maximum if it is strictly larger than both of its neighbors (that is, $$$a_i > a_{i - 1}$$$ and $$$a_i > a_{i + 1}$$$). Since $$$a_1$$$ and $$$a_n$$$ have only one neighbor each, they will never be a local maximum. | 256 megabytes |
import java.util.Scanner;
public class AvoidLocalMaximums {
public static void main(String[] args) {
Scanner scanner = new Scanner(System.in);
int t = scanner.nextInt();
for (int i = 0; i < t; i++) {
int n = scanner.nextInt();
int arr[] = new int[n];
for (int j = 0; j < arr.length; j++) {
arr[j] = scanner.nextInt();
}
int start = 0, end = 2, count = 0;
while (end < n) {
count += updateData(start, end, arr);
++start;
++end;
}
System.out.println(count);
for (int j = 0; j < n; j++) {
System.out.print(arr[j] + " ");
}
System.out.println();
}
}
public static int updateData(int start, int end, int arr[]) {
if (arr[start + 1] > arr[start] && arr[start + 1] > arr[end]) {
if (end + 1 < arr.length) {
arr[end] = Math.max(arr[end + 1], arr[start + 1]);
return 1;
}
arr[end] = arr[start + 1];
return 1;
}
return 0;
}
}
| Java | ["5\n3\n2 1 2\n4\n1 2 3 1\n5\n1 2 1 2 1\n9\n1 2 1 3 2 3 1 2 1\n9\n2 1 3 1 3 1 3 1 3"] | 2 seconds | ["0\n2 1 2\n1\n1 3 3 1\n1\n1 2 2 2 1\n2\n1 2 3 3 2 3 3 2 1\n2\n2 1 3 3 3 1 1 1 3"] | NoteIn the first example, the array contains no local maximum, so we don't need to perform operations.In the second example, we can change $$$a_2$$$ to $$$3$$$, then the array don't have local maximums. | Java 8 | standard input | [
"greedy"
] | 255abf8750541f54d8ff0f1f80e1f8e7 | Each test contains multiple test cases. The first line will contain a single integer $$$t$$$ $$$(1 \leq t \leq 10000)$$$ — the number of test cases. Then $$$t$$$ test cases follow. The first line of each test case contains a single integer $$$n$$$ $$$(2 \leq n \leq 2 \cdot 10^5)$$$ — the size of the array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots ,a_n$$$ $$$(1 \leq a_i \leq 10^9)$$$, the elements of array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, first output a line containing a single integer $$$m$$$ — minimum number of operations required. Then ouput a line consist of $$$n$$$ integers — the resulting array after the operations. Note that this array should differ in exactly $$$m$$$ elements from the initial array. If there are multiple answers, print any. | standard output | |
PASSED | 1cb262de95e3ee9e26f7575d321ad122 | train_108.jsonl | 1645367700 | You are given an array $$$a$$$ of size $$$n$$$. Each element in this array is an integer between $$$1$$$ and $$$10^9$$$.You can perform several operations to this array. During an operation, you can replace an element in the array with any integer between $$$1$$$ and $$$10^9$$$. Output the minimum number of operations needed such that the resulting array doesn't contain any local maximums, and the resulting array after the operations.An element $$$a_i$$$ is a local maximum if it is strictly larger than both of its neighbors (that is, $$$a_i > a_{i - 1}$$$ and $$$a_i > a_{i + 1}$$$). Since $$$a_1$$$ and $$$a_n$$$ have only one neighbor each, they will never be a local maximum. | 256 megabytes | import java.util.*;
import java.io.*;
public class Main {
static FastReader scan = new FastReader();
// static Scanner scan= new Scanner(System.in);
static PrintWriter out = new PrintWriter(System.out);
public static void main(String[] args) {
int n = scan.nextInt();while (n-->0)
A();
out.close();
}
static void A(){
int n = scan.nextInt();
int arr[] = new int[n];
int op = 0;
for (int i = 0; i < n; i++) {
arr[i] = scan.nextInt();
}
for (int i = 1; i < n-1; i++) {
if(arr[i] > arr[i-1] && arr[i] > arr[i+1]){
if(i+2 < n)
arr[i+1] = Math.max(arr[i] , arr[i+2]);
else arr[i+1] = arr[i];
op++;
}
}
out.println(op);
for (int m: arr) out.print(m+" ");
out.println();
}
}
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 | ["5\n3\n2 1 2\n4\n1 2 3 1\n5\n1 2 1 2 1\n9\n1 2 1 3 2 3 1 2 1\n9\n2 1 3 1 3 1 3 1 3"] | 2 seconds | ["0\n2 1 2\n1\n1 3 3 1\n1\n1 2 2 2 1\n2\n1 2 3 3 2 3 3 2 1\n2\n2 1 3 3 3 1 1 1 3"] | NoteIn the first example, the array contains no local maximum, so we don't need to perform operations.In the second example, we can change $$$a_2$$$ to $$$3$$$, then the array don't have local maximums. | Java 8 | standard input | [
"greedy"
] | 255abf8750541f54d8ff0f1f80e1f8e7 | Each test contains multiple test cases. The first line will contain a single integer $$$t$$$ $$$(1 \leq t \leq 10000)$$$ — the number of test cases. Then $$$t$$$ test cases follow. The first line of each test case contains a single integer $$$n$$$ $$$(2 \leq n \leq 2 \cdot 10^5)$$$ — the size of the array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots ,a_n$$$ $$$(1 \leq a_i \leq 10^9)$$$, the elements of array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, first output a line containing a single integer $$$m$$$ — minimum number of operations required. Then ouput a line consist of $$$n$$$ integers — the resulting array after the operations. Note that this array should differ in exactly $$$m$$$ elements from the initial array. If there are multiple answers, print any. | standard output | |
PASSED | d8a47957fafb764bf86ed046c924e8d9 | train_108.jsonl | 1645367700 | You are given an array $$$a$$$ of size $$$n$$$. Each element in this array is an integer between $$$1$$$ and $$$10^9$$$.You can perform several operations to this array. During an operation, you can replace an element in the array with any integer between $$$1$$$ and $$$10^9$$$. Output the minimum number of operations needed such that the resulting array doesn't contain any local maximums, and the resulting array after the operations.An element $$$a_i$$$ is a local maximum if it is strictly larger than both of its neighbors (that is, $$$a_i > a_{i - 1}$$$ and $$$a_i > a_{i + 1}$$$). Since $$$a_1$$$ and $$$a_n$$$ have only one neighbor each, they will never be a local maximum. | 256 megabytes | import java.util.*;
public class code {
static class pair{
int x,y;
public pair(int x, int y) {
this.x = x;
this.y = y;
}
}
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
long t = sc.nextLong();
while (t-- > 0) {
int n=sc.nextInt();
long[] a=new long[n];
for (int i = 0; i < n; i++) {
a[i]=sc.nextLong();
}
long ans=0;
for (int i = 1; i+1 <n ; i++) {
if(a[i]>a[i-1] && a[i]>a[i+1]){
if (i + 2 < n) {
a[i + 1] = Math.max(a[i], a[i + 2]);
} else {
a[i + 1] = a[i];
}
ans++;
}
}
System.out.println(ans);
for (int i = 0; i < n; i++) {
System.out.print(a[i]+" ");
}
System.out.println();
}
}
}
| Java | ["5\n3\n2 1 2\n4\n1 2 3 1\n5\n1 2 1 2 1\n9\n1 2 1 3 2 3 1 2 1\n9\n2 1 3 1 3 1 3 1 3"] | 2 seconds | ["0\n2 1 2\n1\n1 3 3 1\n1\n1 2 2 2 1\n2\n1 2 3 3 2 3 3 2 1\n2\n2 1 3 3 3 1 1 1 3"] | NoteIn the first example, the array contains no local maximum, so we don't need to perform operations.In the second example, we can change $$$a_2$$$ to $$$3$$$, then the array don't have local maximums. | Java 8 | standard input | [
"greedy"
] | 255abf8750541f54d8ff0f1f80e1f8e7 | Each test contains multiple test cases. The first line will contain a single integer $$$t$$$ $$$(1 \leq t \leq 10000)$$$ — the number of test cases. Then $$$t$$$ test cases follow. The first line of each test case contains a single integer $$$n$$$ $$$(2 \leq n \leq 2 \cdot 10^5)$$$ — the size of the array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots ,a_n$$$ $$$(1 \leq a_i \leq 10^9)$$$, the elements of array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, first output a line containing a single integer $$$m$$$ — minimum number of operations required. Then ouput a line consist of $$$n$$$ integers — the resulting array after the operations. Note that this array should differ in exactly $$$m$$$ elements from the initial array. If there are multiple answers, print any. | standard output | |
PASSED | 52bb10222239b15a0fdb55f12c613a65 | train_108.jsonl | 1645367700 | You are given an array $$$a$$$ of size $$$n$$$. Each element in this array is an integer between $$$1$$$ and $$$10^9$$$.You can perform several operations to this array. During an operation, you can replace an element in the array with any integer between $$$1$$$ and $$$10^9$$$. Output the minimum number of operations needed such that the resulting array doesn't contain any local maximums, and the resulting array after the operations.An element $$$a_i$$$ is a local maximum if it is strictly larger than both of its neighbors (that is, $$$a_i > a_{i - 1}$$$ and $$$a_i > a_{i + 1}$$$). Since $$$a_1$$$ and $$$a_n$$$ have only one neighbor each, they will never be a local maximum. | 256 megabytes | import java.io.*;
public class Main {
static BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
static BufferedWriter out = new BufferedWriter(new OutputStreamWriter(System.out));
private static int T, n;
private static int[] a;
public static void main(String[] args) throws IOException {
T = Integer.parseInt(in.readLine());
while(T-- > 0) {
n = Integer.parseInt(in.readLine());
a = new int[n + 1];
String[] strs = in.readLine().split(" ");
for(int i = 1; i <= n; i++) {
a[i] = Integer.parseInt(strs[i - 1]);
}
solve();
}
in.close();
out.flush();
out.close();
}
public static void solve() throws IOException {
int[] mark = new int[n + 1];
for(int i = 2; i < n; i++) {
if(a[i] > a[i-1] && a[i] > a[i+1]) {
mark[i] = 1;
}
}
int res = 0;
for(int i = 2; i < n; i++) {
if(mark[i-1] == 1 && mark[i] == 0 && mark[i + 1] == 1) {
a[i] = Math.max(a[i - 1], a[i + 1]);
mark[i + 1] = 0;
res++;
}
}
for(int i = 2; i < n; i++) {
if(a[i] > a[i+1] && a[i] > a[i-1]) {
a[i] = Math.max(a[i - 1], a[i + 1]);
res++;
}
}
out.write(res + "\n");
for(int i = 1; i <= n; i++) {
out.write(a[i] + " ");
}
out.write("\n");
}
} | Java | ["5\n3\n2 1 2\n4\n1 2 3 1\n5\n1 2 1 2 1\n9\n1 2 1 3 2 3 1 2 1\n9\n2 1 3 1 3 1 3 1 3"] | 2 seconds | ["0\n2 1 2\n1\n1 3 3 1\n1\n1 2 2 2 1\n2\n1 2 3 3 2 3 3 2 1\n2\n2 1 3 3 3 1 1 1 3"] | NoteIn the first example, the array contains no local maximum, so we don't need to perform operations.In the second example, we can change $$$a_2$$$ to $$$3$$$, then the array don't have local maximums. | Java 8 | standard input | [
"greedy"
] | 255abf8750541f54d8ff0f1f80e1f8e7 | Each test contains multiple test cases. The first line will contain a single integer $$$t$$$ $$$(1 \leq t \leq 10000)$$$ — the number of test cases. Then $$$t$$$ test cases follow. The first line of each test case contains a single integer $$$n$$$ $$$(2 \leq n \leq 2 \cdot 10^5)$$$ — the size of the array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots ,a_n$$$ $$$(1 \leq a_i \leq 10^9)$$$, the elements of array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, first output a line containing a single integer $$$m$$$ — minimum number of operations required. Then ouput a line consist of $$$n$$$ integers — the resulting array after the operations. Note that this array should differ in exactly $$$m$$$ elements from the initial array. If there are multiple answers, print any. | standard output | |
PASSED | f0fb85130ebc40b987e00e6d26e8aa13 | train_108.jsonl | 1645367700 | You are given an array $$$a$$$ of size $$$n$$$. Each element in this array is an integer between $$$1$$$ and $$$10^9$$$.You can perform several operations to this array. During an operation, you can replace an element in the array with any integer between $$$1$$$ and $$$10^9$$$. Output the minimum number of operations needed such that the resulting array doesn't contain any local maximums, and the resulting array after the operations.An element $$$a_i$$$ is a local maximum if it is strictly larger than both of its neighbors (that is, $$$a_i > a_{i - 1}$$$ and $$$a_i > a_{i + 1}$$$). Since $$$a_1$$$ and $$$a_n$$$ have only one neighbor each, they will never be a local maximum. | 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;
}
}
public static void main (String[] args) throws java.lang.Exception
{
FastReader sc=new FastReader();
int t=sc.nextInt();
while(t-->0)
{
int n=sc.nextInt(),a[]=new int[n],c=0;
for(int i=0;i<n;i++)
a[i]=sc.nextInt();
for(int i=1;i<n-1;i++)
{
if(a[i]>a[i-1] && a[i]>a[i+1])
{
if(i+2<n)
a[i+1]=Math.max(a[i],a[i+2]);
else
a[i+1]=a[i];
c++;
}
}
System.out.println(c);
for(int i=0;i<n;i++)
System.out.print(a[i]+" ");
System.out.println();
}
}
}
| Java | ["5\n3\n2 1 2\n4\n1 2 3 1\n5\n1 2 1 2 1\n9\n1 2 1 3 2 3 1 2 1\n9\n2 1 3 1 3 1 3 1 3"] | 2 seconds | ["0\n2 1 2\n1\n1 3 3 1\n1\n1 2 2 2 1\n2\n1 2 3 3 2 3 3 2 1\n2\n2 1 3 3 3 1 1 1 3"] | NoteIn the first example, the array contains no local maximum, so we don't need to perform operations.In the second example, we can change $$$a_2$$$ to $$$3$$$, then the array don't have local maximums. | Java 8 | standard input | [
"greedy"
] | 255abf8750541f54d8ff0f1f80e1f8e7 | Each test contains multiple test cases. The first line will contain a single integer $$$t$$$ $$$(1 \leq t \leq 10000)$$$ — the number of test cases. Then $$$t$$$ test cases follow. The first line of each test case contains a single integer $$$n$$$ $$$(2 \leq n \leq 2 \cdot 10^5)$$$ — the size of the array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots ,a_n$$$ $$$(1 \leq a_i \leq 10^9)$$$, the elements of array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, first output a line containing a single integer $$$m$$$ — minimum number of operations required. Then ouput a line consist of $$$n$$$ integers — the resulting array after the operations. Note that this array should differ in exactly $$$m$$$ elements from the initial array. If there are multiple answers, print any. | standard output | |
PASSED | a52464f2ada085e8c1c5a2ba674dbedf | train_108.jsonl | 1645367700 | You are given an array $$$a$$$ of size $$$n$$$. Each element in this array is an integer between $$$1$$$ and $$$10^9$$$.You can perform several operations to this array. During an operation, you can replace an element in the array with any integer between $$$1$$$ and $$$10^9$$$. Output the minimum number of operations needed such that the resulting array doesn't contain any local maximums, and the resulting array after the operations.An element $$$a_i$$$ is a local maximum if it is strictly larger than both of its neighbors (that is, $$$a_i > a_{i - 1}$$$ and $$$a_i > a_{i + 1}$$$). Since $$$a_1$$$ and $$$a_n$$$ have only one neighbor each, they will never be a local maximum. | 256 megabytes | import java.util.*;
public class Avoid_local_maximums {
public static void main(String[] args) {
// TODO Auto-generated method stub
Scanner sc=new Scanner(System.in);
int t=sc.nextInt();
while(t-->0)
{
int n=sc.nextInt();
long []arr=new long[n];
for(int i=0;i<n;i++)
arr[i]=sc.nextLong();
long c=0;
for(int i=1;i<n-1;i++)
{
if(arr[i]>arr[i-1] && arr[i]>arr[i+1])
{
c++;
long max=arr[i];
if(i+2<n)
max=Math.max(arr[i],arr[i+2]);
arr[i+1]=max;
}
}
System.out.println(c);
for(long x:arr)
System.out.print(x+" ");
System.out.println();
}
}
}
| Java | ["5\n3\n2 1 2\n4\n1 2 3 1\n5\n1 2 1 2 1\n9\n1 2 1 3 2 3 1 2 1\n9\n2 1 3 1 3 1 3 1 3"] | 2 seconds | ["0\n2 1 2\n1\n1 3 3 1\n1\n1 2 2 2 1\n2\n1 2 3 3 2 3 3 2 1\n2\n2 1 3 3 3 1 1 1 3"] | NoteIn the first example, the array contains no local maximum, so we don't need to perform operations.In the second example, we can change $$$a_2$$$ to $$$3$$$, then the array don't have local maximums. | Java 8 | standard input | [
"greedy"
] | 255abf8750541f54d8ff0f1f80e1f8e7 | Each test contains multiple test cases. The first line will contain a single integer $$$t$$$ $$$(1 \leq t \leq 10000)$$$ — the number of test cases. Then $$$t$$$ test cases follow. The first line of each test case contains a single integer $$$n$$$ $$$(2 \leq n \leq 2 \cdot 10^5)$$$ — the size of the array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots ,a_n$$$ $$$(1 \leq a_i \leq 10^9)$$$, the elements of array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, first output a line containing a single integer $$$m$$$ — minimum number of operations required. Then ouput a line consist of $$$n$$$ integers — the resulting array after the operations. Note that this array should differ in exactly $$$m$$$ elements from the initial array. If there are multiple answers, print any. | standard output | |
PASSED | 79d1aa4c55f958e4f132cf5de86e2bfe | train_108.jsonl | 1645367700 | You are given an array $$$a$$$ of size $$$n$$$. Each element in this array is an integer between $$$1$$$ and $$$10^9$$$.You can perform several operations to this array. During an operation, you can replace an element in the array with any integer between $$$1$$$ and $$$10^9$$$. Output the minimum number of operations needed such that the resulting array doesn't contain any local maximums, and the resulting array after the operations.An element $$$a_i$$$ is a local maximum if it is strictly larger than both of its neighbors (that is, $$$a_i > a_{i - 1}$$$ and $$$a_i > a_{i + 1}$$$). Since $$$a_1$$$ and $$$a_n$$$ have only one neighbor each, they will never be a local maximum. | 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.HashMap;
import java.util.HashSet;
import java.util.Random;
import java.util.Stack;
import java.util.StringTokenizer;
// B -> CodeForcesProblemSet
public final class B {
static FastReader fr = new FastReader();
static PrintWriter out = new PrintWriter(System.out);
static final int gigamod = 1000000007;
static int t = 1;
static double epsilon = 0.00000001;
static boolean[] isPrime;
static int[] smallestFactorOf;
@SuppressWarnings("unused")
public static void main(String[] args) {
t = fr.nextInt();
OUTER:
for (int tc = 0; tc < t; tc++) {
int n = fr.nextInt();
long[] arr = fr.nextLongArray(n);
boolean[] is = new boolean[n];
for (int i = 1; i < n - 1; i++)
if (arr[i] > arr[i - 1] && arr[i] > arr[i + 1])
is[i] = true;
long oper = 0;
for (int i = 1; i < n - 1; i++)
if (is[i - 1] && is[i + 1]) {
oper++;
arr[i] = Math.max(arr[i - 1], arr[i + 1]);
is[i - 1] = false;
is[i + 1] = false;
}
is = new boolean[n];
for (int i = 1; i < n - 1; i++)
if (arr[i] > arr[i - 1] && arr[i] > arr[i + 1])
is[i] = true;
for (int i = 1; i < n - 1; i++)
if (is[i]) {
oper++;
arr[i] = Math.max(arr[i - 1], arr[i + 1]);
}
out.println(oper);
out.println(toString(arr));
}
out.close();
}
static class UnionFind {
// Uses weighted quick-union with path compression.
private int[] parent; // parent[i] = parent of i
private int[] size; // size[i] = number of sites in tree rooted at i
// Note: not necessarily correct if i is not a root node
private int count; // number of components
public UnionFind(int n) {
count = n;
parent = new int[n];
size = new int[n];
for (int i = 0; i < n; i++) {
parent[i] = i;
size[i] = 1;
}
}
// Number of connected components.
public int count() {
return count;
}
// Find the root of p.
public int find(int p) {
while (p != parent[p])
p = parent[p];
return p;
}
public boolean connected(int p, int q) {
return find(p) == find(q);
}
public int numConnectedTo(int node) {
return size[find(node)];
}
// Weighted union.
public void union(int p, int q) {
int rootP = find(p);
int rootQ = find(q);
if (rootP == rootQ) return;
// make smaller root point to larger one
if (size[rootP] < size[rootQ]) {
parent[rootP] = rootQ;
size[rootQ] += size[rootP];
}
else {
parent[rootQ] = rootP;
size[rootP] += size[rootQ];
}
count--;
}
public static int[] connectedComponents(UnionFind uf) {
// We can do this in nlogn.
int n = uf.size.length;
int[] compoColors = new int[n];
for (int i = 0; i < n; i++)
compoColors[i] = uf.find(i);
HashMap<Integer, Integer> oldToNew = new HashMap<>();
int newCtr = 0;
for (int i = 0; i < n; i++) {
int thisOldColor = compoColors[i];
Integer thisNewColor = oldToNew.get(thisOldColor);
if (thisNewColor == null)
thisNewColor = newCtr++;
oldToNew.put(thisOldColor, thisNewColor);
compoColors[i] = thisNewColor;
}
return compoColors;
}
}
static class UGraph {
// Adjacency list.
private HashSet<Integer>[] adj;
private static final String NEWLINE = "\n";
private int E;
@SuppressWarnings("unchecked")
public UGraph(int V) {
adj = (HashSet<Integer>[]) new HashSet[V];
E = 0;
for (int i = 0; i < V; i++)
adj[i] = new HashSet<Integer>();
}
public void addEdge(int from, int to) {
if (adj[from].contains(to)) return;
E++;
adj[from].add(to);
adj[to].add(from);
}
public HashSet<Integer> adj(int from) {
return adj[from];
}
public int degree(int v) {
return adj[v].size();
}
public int V() {
return adj.length;
}
public int E() {
return E;
}
public String toString() {
StringBuilder s = new StringBuilder();
s.append(V() + " vertices, " + E() + " edges " + NEWLINE);
for (int v = 0; v < V(); v++) {
s.append(v + ": ");
for (int w : adj[v]) {
s.append(w + " ");
}
s.append(NEWLINE);
}
return s.toString();
}
public static void dfsMark(int current, boolean[] marked, UGraph g) {
if (marked[current]) return;
marked[current] = true;
Iterable<Integer> adj = g.adj(current);
for (int adjc : adj)
dfsMark(adjc, marked, g);
}
public static void dfsMark(int current, int from, long[] distTo, boolean[] marked, UGraph g, ArrayList<Integer> endPoints) {
if (marked[current]) return;
marked[current] = true;
if (from != -1)
distTo[current] = distTo[from] + 1;
HashSet<Integer> adj = g.adj(current);
int alreadyMarkedCtr = 0;
for (int adjc : adj) {
if (marked[adjc]) alreadyMarkedCtr++;
dfsMark(adjc, current, distTo, marked, g, endPoints);
}
if (alreadyMarkedCtr == adj.size())
endPoints.add(current);
}
public static void bfsOrder(int current, UGraph g) {
}
public static void dfsMark(int current, int[] colorIds, int color, UGraph g) {
if (colorIds[current] != -1) return;
colorIds[current] = color;
Iterable<Integer> adj = g.adj(current);
for (int adjc : adj)
dfsMark(adjc, colorIds, color, g);
}
public static int[] connectedComponents(UGraph g) {
int n = g.V();
int[] componentId = new int[n];
Arrays.fill(componentId, -1);
int colorCtr = 0;
for (int i = 0; i < n; i++) {
if (componentId[i] != -1) continue;
dfsMark(i, componentId, colorCtr, g);
colorCtr++;
}
return componentId;
}
public static boolean hasCycle(UGraph ug) {
int n = ug.V();
boolean[] marked = new boolean[n];
boolean[] hasCycleFirst = new boolean[1];
for (int i = 0; i < n; i++) {
if (marked[i]) continue;
hcDfsMark(i, ug, marked, hasCycleFirst, -1);
}
return hasCycleFirst[0];
}
// Helper for hasCycle.
private static void hcDfsMark(int current, UGraph ug, boolean[] marked, boolean[] hasCycleFirst, int parent) {
if (marked[current]) return;
if (hasCycleFirst[0]) return;
marked[current] = true;
HashSet<Integer> adjc = ug.adj(current);
for (int adj : adjc) {
if (marked[adj] && adj != parent && parent != -1) {
hasCycleFirst[0] = true;
return;
}
hcDfsMark(adj, ug, marked, hasCycleFirst, current);
}
}
}
static class Digraph {
// Adjacency list.
private HashSet<Integer>[] adj;
private static final String NEWLINE = "\n";
private int E;
@SuppressWarnings("unchecked")
public Digraph(int V) {
adj = (HashSet<Integer>[]) new HashSet[V];
E = 0;
for (int i = 0; i < V; i++)
adj[i] = new HashSet<Integer>();
}
public void addEdge(int from, int to) {
if (adj[from].contains(to)) return;
E++;
adj[from].add(to);
}
public HashSet<Integer> adj(int from) {
return adj[from];
}
public int V() {
return adj.length;
}
public int E() {
return E;
}
public Digraph reversed() {
Digraph dg = new Digraph(V());
for (int i = 0; i < V(); i++)
for (int adjVert : adj(i)) dg.addEdge(adjVert, i);
return dg;
}
public String toString() {
StringBuilder s = new StringBuilder();
s.append(V() + " vertices, " + E() + " edges " + NEWLINE);
for (int v = 0; v < V(); v++) {
s.append(v + ": ");
for (int w : adj[v]) {
s.append(w + " ");
}
s.append(NEWLINE);
}
return s.toString();
}
public static int[] KosarajuSharirSCC(Digraph dg) {
int[] id = new int[dg.V()];
Digraph reversed = dg.reversed();
// Gotta perform topological sort on this one to get the stack.
Stack<Integer> revStack = Digraph.topologicalSort(reversed);
// Initializing id and idCtr.
id = new int[dg.V()];
int idCtr = -1;
// Creating a 'marked' array.
boolean[] marked = new boolean[dg.V()];
while (!revStack.isEmpty()) {
int vertex = revStack.pop();
if (!marked[vertex])
sccDFS(dg, vertex, marked, ++idCtr, id);
}
return id;
}
private static void sccDFS(Digraph dg, int source, boolean[] marked, int idCtr, int[] id) {
marked[source] = true;
id[source] = idCtr;
for (Integer adjVertex : dg.adj(source))
if (!marked[adjVertex]) sccDFS(dg, adjVertex, marked, idCtr, id);
}
public static Stack<Integer> topologicalSort(Digraph dg) {
// dg has to be a directed acyclic graph.
// We'll have to run dfs on the digraph and push the deepest nodes on stack first.
// We'll need a Stack<Integer> and a int[] marked.
Stack<Integer> topologicalStack = new Stack<Integer>();
boolean[] marked = new boolean[dg.V()];
// Calling dfs
for (int i = 0; i < dg.V(); i++)
if (!marked[i])
runDfs(dg, topologicalStack, marked, i);
return topologicalStack;
}
static void runDfs(Digraph dg, Stack<Integer> topologicalStack, boolean[] marked, int source) {
marked[source] = true;
for (Integer adjVertex : dg.adj(source))
if (!marked[adjVertex])
runDfs(dg, topologicalStack, marked, adjVertex);
topologicalStack.add(source);
}
}
static class FastReader {
private BufferedReader bfr;
private StringTokenizer st;
public FastReader() {
bfr = new BufferedReader(new InputStreamReader(System.in));
}
String next() {
if (st == null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(bfr.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());
}
char nextChar() {
return next().toCharArray()[0];
}
String nextString() {
return next();
}
int[] nextIntArray(int n) {
int[] arr = new int[n];
for (int i = 0; i < n; i++)
arr[i] = nextInt();
return arr;
}
double[] nextDoubleArray(int n) {
double[] arr = new double[n];
for (int i = 0; i < arr.length; i++)
arr[i] = nextDouble();
return arr;
}
long[] nextLongArray(int n) {
long[] arr = new long[n];
for (int i = 0; i < n; i++)
arr[i] = nextLong();
return arr;
}
int[][] nextIntGrid(int n, int m) {
int[][] grid = new int[n][m];
for (int i = 0; i < n; i++) {
char[] line = fr.next().toCharArray();
for (int j = 0; j < m; j++)
grid[i][j] = line[j] - 48;
}
return grid;
}
}
static class LCA {
int[] height, first, segtree;
ArrayList<Integer> euler;
boolean[] visited;
int n;
LCA(ArrayList<Point>[] outFrom, int root) {
n = outFrom.length;
height = new int[n];
first = new int[n];
euler = new ArrayList<>();
visited = new boolean[n];
dfs(outFrom, root, 0);
int m = euler.size();
segtree = new int[m * 4];
build(1, 0, m - 1);
}
void dfs(ArrayList<Point>[] outFrom, int node, int h) {
visited[node] = true;
height[node] = h;
first[node] = euler.size();
euler.add(node);
for (Point to : outFrom[node]) {
if (!visited[(int) to.x]) {
dfs(outFrom, (int) to.x, h + 1);
euler.add(node);
}
}
}
void build(int node, int b, int e) {
if (b == e) {
segtree[node] = euler.get(b);
} else {
int mid = (b + e) / 2;
build(node << 1, b, mid);
build(node << 1 | 1, mid + 1, e);
int l = segtree[node << 1], r = segtree[node << 1 | 1];
segtree[node] = (height[l] < height[r]) ? l : r;
}
}
int query(int node, int b, int e, int L, int R) {
if (b > R || e < L)
return -1;
if (b >= L && e <= R)
return segtree[node];
int mid = (b + e) >> 1;
int left = query(node << 1, b, mid, L, R);
int right = query(node << 1 | 1, mid + 1, e, L, R);
if (left == -1) return right;
if (right == -1) return left;
return height[left] < height[right] ? left : right;
}
int lca(int u, int v) {
int left = first[u], right = first[v];
if (left > right) {
int temp = left;
left = right;
right = temp;
}
return query(1, 0, euler.size() - 1, left, right);
}
}
static class FenwickTree {
long[] array; // 1-indexed array, In this array We save cumulative information to perform efficient range queries and updates
public FenwickTree(int size) {
array = new long[size + 1];
}
public long rsq(int ind) {
assert ind > 0;
long sum = 0;
while (ind > 0) {
sum += array[ind];
//Extracting the portion up to the first significant one of the binary representation of 'ind' and decrementing ind by that number
ind -= ind & (-ind);
}
return sum;
}
public long rsq(int a, int b) {
assert b >= a && a > 0 && b > 0;
return rsq(b) - rsq(a - 1);
}
public void update(int ind, long value) {
assert ind > 0;
while (ind < array.length) {
array[ind] += value;
//Extracting the portion up to the first significant one of the binary representation of 'ind' and incrementing ind by that number
ind += ind & (-ind);
}
}
public int size() {
return array.length - 1;
}
}
static final class RedBlackCountSet {
// Manual implementation of RB-BST for C++ PBDS functionality.
// Modification required according to requirements.
// Colors for Nodes:
private static final boolean RED = true;
private static final boolean BLACK = false;
// Pointer to the root:
private Node root;
// Public constructor:
public RedBlackCountSet() {
root = null;
}
// Node class for storing key value pairs:
private class Node {
long key;
long value;
Node left, right;
boolean color;
long size; // Size of the subtree rooted at that node.
public Node(long key, long value, boolean color) {
this.key = key;
this.value = value;
this.left = this.right = null;
this.color = color;
this.size = value;
}
}
/***** Invariant Maintenance functions: *****/
// When a temporary 4 node is formed:
private Node flipColors(Node node) {
node.left.color = node.right.color = BLACK;
node.color = RED;
return node;
}
// When there's a right leaning red link and it is not a 3 node:
private Node rotateLeft(Node node) {
Node rn = node.right;
node.right = rn.left;
rn.left = node;
rn.color = node.color;
node.color = RED;
return rn;
}
// When there are 2 red links in a row:
private Node rotateRight(Node node) {
Node ln = node.left;
node.left = ln.right;
ln.right = node;
ln.color = node.color;
node.color = RED;
return ln;
}
/***** Invariant Maintenance functions end *****/
// Public functions:
public void put(long key) {
root = put(root, key);
root.color = BLACK; // One of the invariants.
}
private Node put(Node node, long key) {
// Standard insertion procedure:
if (node == null)
return new Node(key, 1, RED);
// Recursing:
int cmp = Long.compare(key, node.key);
if (cmp < 0)
node.left = put(node.left, key);
else if (cmp > 0)
node.right = put(node.right, key);
else
node.value++;
// Invariant maintenance:
if (node.left != null && node.right != null) {
if (node.right.color = RED && node.left.color == BLACK)
node = rotateLeft(node);
if (node.left.color == RED && node.left.left.color == RED)
node = rotateRight(node);
if (node.left.color == RED && node.right.color == RED)
node = flipColors(node);
}
// Property maintenance:
node.size = node.value + size(node.left) + size(node.right);
return node;
}
private long size(Node node) {
if (node == null)
return 0;
else
return node.size;
}
public long rank(long key) {
return rank(root, key);
}
// Modify according to requirements.
private long rank(Node node, long key) {
if (node == null) return 0;
int cmp = Long.compare(key, node.key);
if (cmp < 0)
return rank(node.left, key);
else if (cmp > 0)
return node.value + size(node.left) + rank(node.right, key);
else
return node.value + size(node.left);
}
}
static class SegmentTree {
private Node[] heap;
private int[] array;
private int size;
public SegmentTree(int[] array) {
this.array = Arrays.copyOf(array, array.length);
//The max size of this array is about 2 * 2 ^ log2(n) + 1
size = (int) (2 * Math.pow(2.0, Math.floor((Math.log((double) array.length) / Math.log(2.0)) + 1)));
heap = new Node[size];
build(1, 0, array.length);
}
public int size() {
return array.length;
}
//Initialize the Nodes of the Segment tree
private void build(int v, int from, int size) {
heap[v] = new Node();
heap[v].from = from;
heap[v].to = from + size - 1;
if (size == 1) {
heap[v].sum = array[from];
heap[v].min = array[from];
} else {
//Build childs
build(2 * v, from, size / 2);
build(2 * v + 1, from + size / 2, size - size / 2);
heap[v].sum = heap[2 * v].sum + heap[2 * v + 1].sum;
//min = min of the children
heap[v].min = Math.min(heap[2 * v].min, heap[2 * v + 1].min);
}
}
public int rsq(int from, int to) {
return rsq(1, from, to);
}
private int rsq(int v, int from, int to) {
Node n = heap[v];
//If you did a range update that contained this node, you can infer the Sum without going down the tree
if (n.pendingVal != null && contains(n.from, n.to, from, to)) {
return (to - from + 1) * n.pendingVal;
}
if (contains(from, to, n.from, n.to)) {
return heap[v].sum;
}
if (intersects(from, to, n.from, n.to)) {
propagate(v);
int leftSum = rsq(2 * v, from, to);
int rightSum = rsq(2 * v + 1, from, to);
return leftSum + rightSum;
}
return 0;
}
public int rMinQ(int from, int to) {
return rMinQ(1, from, to);
}
private int rMinQ(int v, int from, int to) {
Node n = heap[v];
//If you did a range update that contained this node, you can infer the Min value without going down the tree
if (n.pendingVal != null && contains(n.from, n.to, from, to)) {
return n.pendingVal;
}
if (contains(from, to, n.from, n.to)) {
return heap[v].min;
}
if (intersects(from, to, n.from, n.to)) {
propagate(v);
int leftMin = rMinQ(2 * v, from, to);
int rightMin = rMinQ(2 * v + 1, from, to);
return Math.min(leftMin, rightMin);
}
return Integer.MAX_VALUE;
}
public void update(int from, int to, int value) {
update(1, from, to, value);
}
private void update(int v, int from, int to, int value) {
//The Node of the heap tree represents a range of the array with bounds: [n.from, n.to]
Node n = heap[v];
if (contains(from, to, n.from, n.to)) {
change(n, value);
}
if (n.size() == 1) return;
if (intersects(from, to, n.from, n.to)) {
propagate(v);
update(2 * v, from, to, value);
update(2 * v + 1, from, to, value);
n.sum = heap[2 * v].sum + heap[2 * v + 1].sum;
n.min = Math.min(heap[2 * v].min, heap[2 * v + 1].min);
}
}
//Propagate temporal values to children
private void propagate(int v) {
Node n = heap[v];
if (n.pendingVal != null) {
change(heap[2 * v], n.pendingVal);
change(heap[2 * v + 1], n.pendingVal);
n.pendingVal = null; //unset the pending propagation value
}
}
//Save the temporal values that will be propagated lazily
private void change(Node n, int value) {
n.pendingVal = value;
n.sum = n.size() * value;
n.min = value;
array[n.from] = value;
}
//Test if the range1 contains range2
private boolean contains(int from1, int to1, int from2, int to2) {
return from2 >= from1 && to2 <= to1;
}
//check inclusive intersection, test if range1[from1, to1] intersects range2[from2, to2]
private boolean intersects(int from1, int to1, int from2, int to2) {
return from1 <= from2 && to1 >= from2 // (.[..)..] or (.[...]..)
|| from1 >= from2 && from1 <= to2; // [.(..]..) or [..(..)..
}
//The Node class represents a partition range of the array.
static class Node {
int sum;
int min;
//Here We store the value that will be propagated lazily
Integer pendingVal = null;
int from;
int to;
int size() {
return to - from + 1;
}
}
}
@SuppressWarnings("serial")
static class CountMap<T> extends HashMap<T, Integer>{
CountMap() {
}
CountMap(T[] arr) {
this.putCM(arr);
}
public Integer putCM(T key) {
if (super.containsKey(key)) {
return super.put(key, super.get(key) + 1);
} else {
return super.put(key, 1);
}
}
public Integer removeCM(T key) {
Integer count = super.get(key);
if (count == null) return -1;
if (count == 1)
return super.remove(key);
else
return super.put(key, super.get(key) - 1);
}
public Integer getCM(T key) {
Integer count = super.get(key);
if (count == null)
return 0;
return count;
}
public void putCM(T[] arr) {
for (T l : arr)
this.putCM(l);
}
}
static class Point implements Comparable<Point> {
long x;
long y;
long id;
Point() {
x = y = id = 0;
}
Point(Point p) {
this.x = p.x;
this.y = p.y;
this.id = p.id;
}
Point(long a, long b, long id) {
this.x = a;
this.y = b;
this.id = id;
}
Point(long a, long b) {
this.x = a;
this.y = b;
}
@Override
public int compareTo(Point o) {
if (this.x < o.x)
return -1;
if (this.x > o.x)
return 1;
if (this.y < o.y)
return -1;
if (this.y > o.y)
return 1;
return 0;
}
public boolean equals(Point that) {
return this.compareTo(that) == 0;
}
}
static int longestPrefixPalindromeLen(char[] s) {
int n = s.length;
StringBuilder sb = new StringBuilder();
for (int i = 0; i < n; i++)
sb.append(s[i]);
StringBuilder sb2 = new StringBuilder(sb);
sb.append('^');
sb.append(sb2.reverse());
// out.println(sb.toString());
char[] newS = sb.toString().toCharArray();
int m = newS.length;
int[] pi = piCalcKMP(newS);
return pi[m - 1];
}
static int KMPNumOcc(char[] text, char[] pat) {
int n = text.length;
int m = pat.length;
char[] patPlusText = new char[n + m + 1];
for (int i = 0; i < m; i++)
patPlusText[i] = pat[i];
patPlusText[m] = '^'; // Seperator
for (int i = 0; i < n; i++)
patPlusText[m + i] = text[i];
int[] fullPi = piCalcKMP(patPlusText);
int answer = 0;
for (int i = 0; i < n + m + 1; i++)
if (fullPi[i] == m)
answer++;
return answer;
}
static int[] piCalcKMP(char[] s) {
int n = s.length;
int[] pi = new int[n];
for (int i = 1; i < n; i++) {
int j = pi[i - 1];
while (j > 0 && s[i] != s[j])
j = pi[j - 1];
if (s[i] == s[j])
j++;
pi[i] = j;
}
return pi;
}
static String toBinaryString(long num, int bits) {
StringBuilder sb = new StringBuilder(Long.toBinaryString(num));
sb.reverse();
for (int i = sb.length(); i < bits; i++)
sb.append('0');
return sb.reverse().toString();
}
static long modDiv(long a, long b) {
return mod(a * power(b, gigamod - 2, gigamod));
}
static void coprimeGenerator(int m, int n, ArrayList<Point> coprimes, int limit, int numCoprimes) {
if (m > limit) return;
if (m <= limit && n <= limit)
coprimes.add(new Point(m, n));
if (coprimes.size() > numCoprimes) return;
coprimeGenerator(2 * m - n, m, coprimes, limit, numCoprimes);
coprimeGenerator(2 * m + n, m, coprimes, limit, numCoprimes);
coprimeGenerator(m + 2 * n, n, coprimes, limit, numCoprimes);
}
static long hash(long i) {
return (i * 2654435761L % gigamod);
}
static long hash2(long key) {
long h = Long.hashCode(key);
h ^= (h >>> 20) ^ (h >>> 12) ^ (h >>> 7) ^ (h >>> 4);
return h & (gigamod-1);
}
static int mapTo1D(int row, int col, int n, int m) {
// Maps elements in a 2D matrix serially to elements in
// a 1D array.
return row * m + col;
}
static int[] mapTo2D(int idx, int n, int m) {
// Inverse of what the one above does.
int[] rnc = new int[2];
rnc[0] = idx / m;
rnc[1] = idx % m;
return rnc;
}
static double distance(Point p1, Point p2) {
return Math.sqrt(Math.pow(p2.y-p1.y, 2) + Math.pow(p2.x-p1.x, 2));
}
static HashMap<Integer, Integer> smolNumPrimeFactorization(int num) {
if (smallestFactorOf == null)
primeGenerator(200001);
CountMap<Integer> fnps = new CountMap<>();
while (num != 1) {
fnps.putCM(smallestFactorOf[num]);
num /= smallestFactorOf[num];
}
return fnps;
}
static HashMap<Long, Integer> primeFactorization(long num) {
// Returns map of factor and its power in the number.
HashMap<Long, Integer> map = new HashMap<>();
while (num % 2 == 0) {
num /= 2;
Integer pwrCnt = map.get(2L);
map.put(2L, pwrCnt != null ? pwrCnt + 1 : 1);
}
for (long i = 3; i * i <= num; i += 2) {
while (num % i == 0) {
num /= i;
Integer pwrCnt = map.get(i);
map.put(i, pwrCnt != null ? pwrCnt + 1 : 1);
}
}
// If the number is prime, we have to add it to the
// map.
if (num != 1)
map.put(num, 1);
return map;
}
static HashSet<Long> divisors(long num) {
HashSet<Long> divisors = new HashSet<Long>();
divisors.add(1L);
divisors.add(num);
for (long i = 2; i * i <= num; i++) {
if (num % i == 0) {
divisors.add(num/i);
divisors.add(i);
}
}
return divisors;
}
static int bsearch(int[] arr, int val, int lo, int hi) {
// Returns the index of the first element
// larger than or equal to val.
int idx = -1;
while (lo <= hi) {
int mid = lo + (hi - lo)/2;
if (arr[mid] >= val) {
idx = mid;
hi = mid - 1;
} else
lo = mid + 1;
}
return idx;
}
static int bsearch(long[] arr, long val, int lo, int hi) {
int idx = -1;
while (lo <= hi) {
int mid = lo + (hi - lo)/2;
if (arr[mid] >= val) {
idx = mid;
hi = mid - 1;
} else
lo = mid + 1;
}
return idx;
}
static int bsearch(long[] arr, long val, int lo, int hi, boolean sMode) {
// Returns the index of the last element
// smaller than or equal to val.
int idx = -1;
while (lo <= hi) {
int mid = lo + (hi - lo)/2;
if (arr[mid] >= val) {
hi = mid - 1;
} else {
idx = mid;
lo = mid + 1;
}
}
return idx;
}
static int bsearch(int[] arr, long val, int lo, int hi, boolean sMode) {
int idx = -1;
while (lo <= hi) {
int mid = lo + (hi - lo)/2;
if (arr[mid] > val) {
hi = mid - 1;
} else {
idx = mid;
lo = mid + 1;
}
}
return idx;
}
static long factorial(long n) {
if (n <= 1)
return 1;
long factorial = 1;
for (int i = 1; i <= n; i++)
factorial = mod(factorial * i);
return factorial;
}
static long factorialInDivision(long a, long b) {
if (a == b)
return 1;
if (b < a) {
long temp = a;
a = b;
b = temp;
}
long factorial = 1;
for (long i = a + 1; i <= b; i++)
factorial = mod(factorial * i);
return factorial;
}
static long nCr(long n, long r, long[] fac) {
long p = 998244353;
// Base case
if (r == 0)
return 1;
// Fill factorial array so that we
// can find all factorial of r, n
// and n-r
/*long fac[] = new long[(int)n + 1];
fac[0] = 1;
for (int i = 1; i <= n; i++)
fac[i] = fac[i - 1] * i % p;*/
return (fac[(int)n] * modInverse(fac[(int)r], p) % p
* modInverse(fac[(int)n - (int)r], p) % p) % p;
}
static long modInverse(long n, long p) {
return power(n, p - 2, p);
}
static long power(long x, long y, long p) {
long res = 1; // Initialize result
x = x % p; // Update x if it is more than or
// equal to p
while (y > 0) {
// If y is odd, multiply x with result
if ((y & 1)==1)
res = (res * x) % p;
// y must be even now
y = y >> 1; // y = y/2
x = (x * x) % p;
}
return res;
}
static long nPr(long n, long r) {
return factorialInDivision(n, n - r);
}
static int logk(long n, long k) {
return (int)(Math.log(n) / Math.log(k));
}
static boolean[] primeGenerator(int upto) {
// Sieve of Eratosthenes:
isPrime = new boolean[upto + 1];
smallestFactorOf = new int[upto + 1];
Arrays.fill(smallestFactorOf, 1);
Arrays.fill(isPrime, true);
isPrime[1] = isPrime[0] = false;
for (long i = 2; i < upto + 1; i++)
if (isPrime[(int) i]) {
smallestFactorOf[(int) i] = (int) i;
// Mark all the multiples greater than or equal
// to the square of i to be false.
for (long j = i; j * i < upto + 1; j++) {
if (isPrime[(int) j * (int) i]) {
isPrime[(int) j * (int) i] = false;
smallestFactorOf[(int) j * (int) i] = (int) i;
}
}
}
return isPrime;
}
static long gcd(long a, long b) {
if (b == 0) {
return a;
} else {
return gcd(b, a % b);
}
}
static int gcd(int a, int b) {
if (b == 0) {
return a;
} else {
return gcd(b, a % b);
}
}
static long gcd(long[] arr) {
int n = arr.length;
long gcd = arr[0];
for (int i = 1; i < n; i++) {
gcd = gcd(gcd, arr[i]);
}
return gcd;
}
static int gcd(int[] arr) {
int n = arr.length;
int gcd = arr[0];
for (int i = 1; i < n; i++) {
gcd = gcd(gcd, arr[i]);
}
return gcd;
}
static long lcm(long[] arr) {
long lcm = arr[0];
int n = arr.length;
for (int i = 1; i < n; i++) {
lcm = (lcm * arr[i]) / gcd(lcm, arr[i]);
}
return lcm;
}
static long lcm(long a, long b) {
return (a * b)/gcd(a, b);
}
static boolean less(int a, int b) {
return a < b ? true : false;
}
static boolean isSorted(int[] a) {
for (int i = 1; i < a.length; i++) {
if (less(a[i], a[i - 1]))
return false;
}
return true;
}
static boolean isSorted(long[] a) {
for (int i = 1; i < a.length; i++) {
if (a[i] < a[i - 1])
return false;
}
return true;
}
static void swap(int[] a, int i, int j) {
int temp = a[i];
a[i] = a[j];
a[j] = temp;
}
static void swap(long[] a, int i, int j) {
long temp = a[i];
a[i] = a[j];
a[j] = temp;
}
static void swap(double[] a, int i, int j) {
double 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 sort(int[] arr) {
shuffleArray(arr, 0, arr.length - 1);
Arrays.sort(arr);
}
static void sort(char[] arr) {
shuffleArray(arr, 0, arr.length - 1);
Arrays.sort(arr);
}
static void sort(long[] arr) {
shuffleArray(arr, 0, arr.length - 1);
Arrays.sort(arr);
}
static void sort(double[] arr) {
shuffleArray(arr, 0, arr.length - 1);
Arrays.sort(arr);
}
static void reverseSort(int[] arr) {
shuffleArray(arr, 0, arr.length - 1);
Arrays.sort(arr);
int n = arr.length;
for (int i = 0; i < n/2; i++)
swap(arr, i, n - 1 - i);
}
static void reverseSort(char[] arr) {
shuffleArray(arr, 0, arr.length - 1);
Arrays.sort(arr);
int n = arr.length;
for (int i = 0; i < n/2; i++)
swap(arr, i, n - 1 - i);
}
static void reverse(char[] arr) {
int n = arr.length;
for (int i = 0; i < n / 2; i++)
swap(arr, i, n - 1 - i);
}
static void reverseSort(long[] arr) {
shuffleArray(arr, 0, arr.length - 1);
Arrays.sort(arr);
int n = arr.length;
for (int i = 0; i < n/2; i++)
swap(arr, i, n - 1 - i);
}
static void reverseSort(double[] arr) {
shuffleArray(arr, 0, arr.length - 1);
Arrays.sort(arr);
int n = arr.length;
for (int i = 0; i < n/2; i++)
swap(arr, i, n - 1 - i);
}
static void shuffleArray(long[] arr, int startPos, int endPos) {
Random rnd = new Random();
for (int i = startPos; i < endPos; ++i) {
long tmp = arr[i];
int randomPos = i + rnd.nextInt(endPos - i);
arr[i] = arr[randomPos];
arr[randomPos] = tmp;
}
}
static void shuffleArray(int[] arr, int startPos, int endPos) {
Random rnd = new Random();
for (int i = startPos; i < endPos; ++i) {
int tmp = arr[i];
int randomPos = i + rnd.nextInt(endPos - i);
arr[i] = arr[randomPos];
arr[randomPos] = tmp;
}
}
static void shuffleArray(double[] arr, int startPos, int endPos) {
Random rnd = new Random();
for (int i = startPos; i < endPos; ++i) {
double tmp = arr[i];
int randomPos = i + rnd.nextInt(endPos - i);
arr[i] = arr[randomPos];
arr[randomPos] = tmp;
}
}
static void shuffleArray(char[] arr, int startPos, int endPos) {
Random rnd = new Random();
for (int i = startPos; i < endPos; ++i) {
char tmp = arr[i];
int randomPos = i + rnd.nextInt(endPos - i);
arr[i] = arr[randomPos];
arr[randomPos] = tmp;
}
}
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 (long i = 5; i * i <= n; i = i + 6)
if (n % i == 0 || n % (i + 2) == 0)
return false;
return true;
}
static String toString(int[] dp) {
StringBuilder sb = new StringBuilder();
for (int i = 0; i < dp.length; i++)
sb.append(dp[i] + " ");
return sb.toString();
}
static String toString(boolean[] dp) {
StringBuilder sb = new StringBuilder();
for (int i = 0; i < dp.length; i++)
sb.append(dp[i] + " ");
return sb.toString();
}
static String toString(long[] dp) {
StringBuilder sb = new StringBuilder();
for (int i = 0; i < dp.length; i++)
sb.append(dp[i] + " ");
return sb.toString();
}
static String toString(char[] dp) {
StringBuilder sb = new StringBuilder();
for (int i = 0; i < dp.length; i++)
sb.append(dp[i] + "");
return sb.toString();
}
static String toString(int[][] dp) {
StringBuilder sb = new StringBuilder();
for (int i = 0; i < dp.length; i++) {
for (int j = 0; j < dp[i].length; j++) {
sb.append(dp[i][j] + " ");
}
sb.append('\n');
}
return sb.toString();
}
static String toString(long[][] dp) {
StringBuilder sb = new StringBuilder();
for (int i = 0; i < dp.length; i++) {
for (int j = 0; j < dp[i].length; j++) {
sb.append(dp[i][j] + " ");
}
sb.append('\n');
}
return sb.toString();
}
static String toString(double[][] dp) {
StringBuilder sb = new StringBuilder();
for (int i = 0; i < dp.length; i++) {
for (int j = 0; j < dp[i].length; j++) {
sb.append(dp[i][j] + " ");
}
sb.append('\n');
}
return sb.toString();
}
static String toString(char[][] dp) {
StringBuilder sb = new StringBuilder();
for (int i = 0; i < dp.length; i++) {
for (int j = 0; j < dp[i].length; j++) {
sb.append(dp[i][j] + " ");
}
sb.append('\n');
}
return sb.toString();
}
static long mod(long a, long m) {
return (a%m + m) % m;
}
static long mod(long num) {
return (num % gigamod + gigamod) % gigamod;
}
}
// NOTES:
// ASCII VALUE OF 'A': 65
// ASCII VALUE OF 'a': 97
// Range of long: 9 * 10^18
// ASCII VALUE OF '0': 48
// Primes upto 'n' can be given by (n / (logn)). | Java | ["5\n3\n2 1 2\n4\n1 2 3 1\n5\n1 2 1 2 1\n9\n1 2 1 3 2 3 1 2 1\n9\n2 1 3 1 3 1 3 1 3"] | 2 seconds | ["0\n2 1 2\n1\n1 3 3 1\n1\n1 2 2 2 1\n2\n1 2 3 3 2 3 3 2 1\n2\n2 1 3 3 3 1 1 1 3"] | NoteIn the first example, the array contains no local maximum, so we don't need to perform operations.In the second example, we can change $$$a_2$$$ to $$$3$$$, then the array don't have local maximums. | Java 8 | standard input | [
"greedy"
] | 255abf8750541f54d8ff0f1f80e1f8e7 | Each test contains multiple test cases. The first line will contain a single integer $$$t$$$ $$$(1 \leq t \leq 10000)$$$ — the number of test cases. Then $$$t$$$ test cases follow. The first line of each test case contains a single integer $$$n$$$ $$$(2 \leq n \leq 2 \cdot 10^5)$$$ — the size of the array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots ,a_n$$$ $$$(1 \leq a_i \leq 10^9)$$$, the elements of array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, first output a line containing a single integer $$$m$$$ — minimum number of operations required. Then ouput a line consist of $$$n$$$ integers — the resulting array after the operations. Note that this array should differ in exactly $$$m$$$ elements from the initial array. If there are multiple answers, print any. | standard output | |
PASSED | 3b76d83caf4fd2b058bacb7370493b39 | train_108.jsonl | 1645367700 | You are given an array $$$a$$$ of size $$$n$$$. Each element in this array is an integer between $$$1$$$ and $$$10^9$$$.You can perform several operations to this array. During an operation, you can replace an element in the array with any integer between $$$1$$$ and $$$10^9$$$. Output the minimum number of operations needed such that the resulting array doesn't contain any local maximums, and the resulting array after the operations.An element $$$a_i$$$ is a local maximum if it is strictly larger than both of its neighbors (that is, $$$a_i > a_{i - 1}$$$ and $$$a_i > a_{i + 1}$$$). Since $$$a_1$$$ and $$$a_n$$$ have only one neighbor each, they will never be a local maximum. | 256 megabytes | import java.math.BigInteger;
import java.util.*;
import java.io.*;
//code by tishrah_
public class _practise {
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[] ia(int n)
{
int a[]=new int[n];
for(int i=0;i<n;i++)a[i]=nextInt();
return a;
}
int[][] ia(int n , int m)
{
int a[][]=new int[n][m];
for(int i=0;i<n;i++) for(int j=0;j<m ;j++) a[i][j]=nextInt();
return a;
}
long[][] la(int n , int m)
{
long a[][]=new long[n][m];
for(int i=0;i<n;i++) for(int j=0;j<m ;j++) a[i][j]=nextLong();
return a;
}
char[][] ca(int n , int m)
{
char a[][]=new char[n][m];
for(int i=0;i<n;i++)
{
String x =next();
for(int j=0;j<m ;j++) a[i][j]=x.charAt(j);
}
return a;
}
long[] la(int n)
{
long a[]=new long[n];
for(int i=0;i<n;i++)a[i]=nextLong();
return a;
}
String nextLine()
{
String str = "";
try
{
str = br.readLine();
}
catch (IOException e)
{
e.printStackTrace();
}
return str;
}
}
static void sort(long[] a)
{int n=a.length;Random r=new Random();for (int i=0; i<a.length; i++) {long oi=r.nextInt(n), temp=a[i];a[i]=a[(int)oi];a[(int)oi]=temp;}Arrays.sort(a);}
static void sort(int[] a)
{int n=a.length;Random r=new Random();for (int i=0; i<a.length; i++) {int oi=r.nextInt(n), temp=a[i];a[i]=a[oi];a[oi]=temp;}Arrays.sort(a);}
public static long sum(long a[])
{long sum=0; for(long i : a) sum+=i; return(sum);}
public static long count(long a[] , long x)
{long c=0; for(long i : a) if(i==x) c++; return(c);}
public static int sum(int a[])
{ int sum=0; for(int i : a) sum+=i; return(sum);}
public static int count(int a[] ,int x)
{int c=0; for(int i : a) if(i==x) c++; return(c);}
public static int count(String s ,char ch)
{int c=0; char x[] = s.toCharArray(); for(char i : x) if(ch==i) c++; return(c);}
public static boolean prime(int n)
{for(int i=2 ; i<=Math.sqrt(n) ; i++) if(n%i==0) return false; return true;}
public static boolean prime(long n)
{for(long i=2 ; i<=Math.sqrt(n) ; i++) if(n%i==0) return false; return true;}
public static int gcd(int n1, int n2)
{ if (n2 != 0)return gcd(n2, n1 % n2); else return n1;}
public static long gcd(long n1, long n2)
{ if (n2 != 0)return gcd(n2, n1 % n2); else return n1;}
public static int[] freq(int a[], int n)
{ int f[]=new int[n+1]; for(int i:a) f[i]++; return f;}
public static int[] pos(int a[], int n)
{ int f[]=new int[n+1]; for(int i=0; i<n ;i++) f[a[i]]=i; return f;}
public static int[] rev(int a[])
{
for(int i=0 ; i<(a.length+1)/2;i++)
{
int temp=a[i];
a[i]=a[a.length-1-i];
a[a.length-1-i]=temp;
}
return a;
}
public static boolean palin(String s)
{
StringBuilder sb = new StringBuilder();
sb.append(s);
String str=String.valueOf(sb.reverse());
if(s.equals(str))
return true;
else return false;
}
public static String rev(String s)
{
StringBuilder sb = new StringBuilder();
sb.append(s);
return String.valueOf(sb.reverse());
}
public static void main(String args[])
{
FastReader in=new FastReader();
PrintWriter so = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));
_practise ob = new _practise();
int T = in.nextInt();
// int T = 1;
tc : while(T-->0)
{
int n = in.nextInt() ;
int a[] = in.ia(n);
int c=0;
for(int i=1 ; i<n-1 ; i++)
{
if(a[i]>a[i-1] && a[i]>a[i+1])
{
c++;
int mx=a[i];
if(i+2<n) mx=Math.max(a[i], a[i+2]);
a[i+1]=mx;
}
}
so.println(c);
for(int i : a) so.print(i+" ");
so.println();
}
so.flush();
/*String s = in.next();
* Arrays.stream(f).min().getAsInt()
* BigInteger f = new BigInteger("1"); 1 ke jagah koi bhi value ho skta jo aap
* initial value banan chahte
int a[] = new int[n];
Stack<Integer> stack = new Stack<Integer>();
Deque<Integer> q = new LinkedList<>(); or Deque<Integer> q = new ArrayDeque<Integer>();
PriorityQueue<Long> pq = new PriorityQueue<Long>();
ArrayList<Integer> al = new ArrayList<Integer>();
StringBuilder sb = new StringBuilder();
HashSet<Integer> st = new LinkedHashSet<Integer>();
Set<Integer> s = new HashSet<Integer>();
Map<Long,Integer> hm = new HashMap<Long, Integer>(); //<key,value>
for(Map.Entry<Integer, Integer> i :hm.entrySet())
for(int i : hm.values())
for(int i : hm.keySet())
HashMap<Integer, Integer> hmap = new HashMap<Integer, Integer>();
so.println("HELLO");
Arrays.sort(a,Comparator.comparingDouble(o->o[0]))
Arrays.sort(a, (aa, bb) -> Integer.compare(aa[1], bb[1]));
Set<String> ts = new TreeSet<>();*/
}
} | Java | ["5\n3\n2 1 2\n4\n1 2 3 1\n5\n1 2 1 2 1\n9\n1 2 1 3 2 3 1 2 1\n9\n2 1 3 1 3 1 3 1 3"] | 2 seconds | ["0\n2 1 2\n1\n1 3 3 1\n1\n1 2 2 2 1\n2\n1 2 3 3 2 3 3 2 1\n2\n2 1 3 3 3 1 1 1 3"] | NoteIn the first example, the array contains no local maximum, so we don't need to perform operations.In the second example, we can change $$$a_2$$$ to $$$3$$$, then the array don't have local maximums. | Java 8 | standard input | [
"greedy"
] | 255abf8750541f54d8ff0f1f80e1f8e7 | Each test contains multiple test cases. The first line will contain a single integer $$$t$$$ $$$(1 \leq t \leq 10000)$$$ — the number of test cases. Then $$$t$$$ test cases follow. The first line of each test case contains a single integer $$$n$$$ $$$(2 \leq n \leq 2 \cdot 10^5)$$$ — the size of the array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots ,a_n$$$ $$$(1 \leq a_i \leq 10^9)$$$, the elements of array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, first output a line containing a single integer $$$m$$$ — minimum number of operations required. Then ouput a line consist of $$$n$$$ integers — the resulting array after the operations. Note that this array should differ in exactly $$$m$$$ elements from the initial array. If there are multiple answers, print any. | standard output | |
PASSED | f18d5babe54f049bf044684ef6ebd35d | train_108.jsonl | 1645367700 | You are given an array $$$a$$$ of size $$$n$$$. Each element in this array is an integer between $$$1$$$ and $$$10^9$$$.You can perform several operations to this array. During an operation, you can replace an element in the array with any integer between $$$1$$$ and $$$10^9$$$. Output the minimum number of operations needed such that the resulting array doesn't contain any local maximums, and the resulting array after the operations.An element $$$a_i$$$ is a local maximum if it is strictly larger than both of its neighbors (that is, $$$a_i > a_{i - 1}$$$ and $$$a_i > a_{i + 1}$$$). Since $$$a_1$$$ and $$$a_n$$$ have only one neighbor each, they will never be a local maximum. | 256 megabytes | //<———My cp————
import java.util.*;
import java.io.*;
public class B_Avoid_Local_Maximums{
public static void main(String[] args) throws Exception{
FastReader fr = new FastReader(System.in);
int t = fr.nextInt();
while(t-->0){
int n = fr.nextInt();
int[] vals = new int[n];
for(int i=0;i<n;i++){
vals[i] = fr.nextInt();
}
int c = 0;
for(int i = 1;i<n-1;i++){
if(vals[i]>vals[i-1] && vals[i]>vals[i+1]){
c++;
if(i+2<n){
vals[i+1] = Math.max(vals[i],vals[i+2]);
}else{
vals[i+1] =vals[i];
}
}
}
println(c);
for(int i = 0;i<n;i++){
print(vals[i]+" ");
}
println("");
}
}
public static void print(Object val){
System.out.print(val);
}
public static void println(Object val){
System.out.println(val);
}
public static int[] sort(int[] vals){
ArrayList<Integer> values = new ArrayList<>();
for(int i = 0;i<vals.length;i++){
values.add(vals[i]);
}
Collections.sort(values);
for(int i =0;i<values.size();i++){
vals[i] = values.get(i);
}
return vals;
}
public static long[] sort(long[] vals){
ArrayList<Long> values = new ArrayList<>();
for(int i = 0;i<vals.length;i++){
values.add(vals[i]);
}
Collections.sort(values);
for(int i =0;i<values.size();i++){
vals[i] = values.get(i);
}
return vals;
}
public static void reverseArray(long[] vals){
int startIndex = 0;
int endIndex = vals.length-1;
while(startIndex<=endIndex){
long temp = vals[startIndex];
vals[startIndex] = vals[endIndex];
vals[endIndex] = temp;
startIndex++;
endIndex--;
}
}
public static void reverseArray(int[] vals){
int startIndex = 0;
int endIndex = vals.length-1;
while(startIndex<=endIndex){
int temp = vals[startIndex];
vals[startIndex] = vals[endIndex];
vals[endIndex] = temp;
startIndex++;
endIndex--;
}
}
static class FastReader{
byte[] buf = new byte[2048];
int index, total;
InputStream in;
FastReader(InputStream is) {
in = is;
}
int scan() throws IOException {
if (index >= total) {
index = 0;
total = in.read(buf);
if (total <= 0) {
return -1;
}
}
return buf[index++];
}
String next() throws IOException {
int c;
for (c = scan(); c <= 32; c = scan());
StringBuilder sb = new StringBuilder();
for (; c > 32; c = scan()) {
sb.append((char) c);
}
return sb.toString();
}
int nextInt() throws IOException {
int c, val = 0;
for (c = scan(); c <= 32; c = scan());
boolean neg = c == '-';
if (c == '-' || c == '+') {
c = scan();
}
for (; c >= '0' && c <= '9'; c = scan()) {
val = (val << 3) + (val << 1) + (c & 15);
}
return neg ? -val : val;
}
long nextLong() throws IOException {
int c;
long val = 0;
for (c = scan(); c <= 32; c = scan());
boolean neg = c == '-';
if (c == '-' || c == '+') {
c = scan();
}
for (; c >= '0' && c <= '9'; c = scan()) {
val = (val << 3) + (val << 1) + (c & 15);
}
return neg ? -val : val;
}
}
public static int GCD(int numA, int numB){
if(numA==0){
return numB;
}else if(numB==0){
return numA;
}else{
if(numA>numB){
return GCD(numA%numB,numB);
}else{
return GCD(numA,numB%numA);
}
}
}
}
| Java | ["5\n3\n2 1 2\n4\n1 2 3 1\n5\n1 2 1 2 1\n9\n1 2 1 3 2 3 1 2 1\n9\n2 1 3 1 3 1 3 1 3"] | 2 seconds | ["0\n2 1 2\n1\n1 3 3 1\n1\n1 2 2 2 1\n2\n1 2 3 3 2 3 3 2 1\n2\n2 1 3 3 3 1 1 1 3"] | NoteIn the first example, the array contains no local maximum, so we don't need to perform operations.In the second example, we can change $$$a_2$$$ to $$$3$$$, then the array don't have local maximums. | Java 8 | standard input | [
"greedy"
] | 255abf8750541f54d8ff0f1f80e1f8e7 | Each test contains multiple test cases. The first line will contain a single integer $$$t$$$ $$$(1 \leq t \leq 10000)$$$ — the number of test cases. Then $$$t$$$ test cases follow. The first line of each test case contains a single integer $$$n$$$ $$$(2 \leq n \leq 2 \cdot 10^5)$$$ — the size of the array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots ,a_n$$$ $$$(1 \leq a_i \leq 10^9)$$$, the elements of array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, first output a line containing a single integer $$$m$$$ — minimum number of operations required. Then ouput a line consist of $$$n$$$ integers — the resulting array after the operations. Note that this array should differ in exactly $$$m$$$ elements from the initial array. If there are multiple answers, print any. | standard output | |
PASSED | 8de4cd718eaca391b6c2b4c3dfb40c03 | train_108.jsonl | 1645367700 | You are given an array $$$a$$$ of size $$$n$$$. Each element in this array is an integer between $$$1$$$ and $$$10^9$$$.You can perform several operations to this array. During an operation, you can replace an element in the array with any integer between $$$1$$$ and $$$10^9$$$. Output the minimum number of operations needed such that the resulting array doesn't contain any local maximums, and the resulting array after the operations.An element $$$a_i$$$ is a local maximum if it is strictly larger than both of its neighbors (that is, $$$a_i > a_{i - 1}$$$ and $$$a_i > a_{i + 1}$$$). Since $$$a_1$$$ and $$$a_n$$$ have only one neighbor each, they will never be a local maximum. | 256 megabytes | import java.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.StringTokenizer;
import java.util.*;
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);
solve(in, out);
out.close();
}
static String reverse(String s) {
return (new StringBuilder(s)).reverse().toString();
}
static void sieveOfEratosthenes(int n, int factors[], ArrayList<Integer> ar)
{
factors[1]=1;
int p;
for(p = 2; p*p <=n; p++)
{
if(factors[p] == 0)
{
ar.add(p);
factors[p]=p;
for(int i = p*p; i <= n; i += p)
if(factors[i]==0)
factors[i] = p;
}
}
for(;p<=n;p++){
if(factors[p] == 0)
{
factors[p] = p;
ar.add(p);
}
}
}
static void sort(int ar[]) {
int n = ar.length;
ArrayList<Integer> a = new ArrayList<>();
for (int i = 0; i < n; i++)
a.add(ar[i]);
Collections.sort(a);
for (int i = 0; i < n; i++)
ar[i] = a.get(i);
}
static void sort1(long ar[]) {
int n = ar.length;
ArrayList<Long> a = new ArrayList<>();
for (int i = 0; i < n; i++)
a.add(ar[i]);
Collections.sort(a);
for (int i = 0; i < n; i++)
ar[i] = a.get(i);
}
static long ncr(long n, long r, long mod) {
if (r == 0)
return 1;
long val = ncr(n - 1, r - 1, mod);
val = (n * val) % mod;
val = (val * modInverse(r, mod)) % mod;
return val;
}
static int findMax(int a[], int n, int vis[], int i, int d){
if(i>=n)
return 0;
if(vis[i]==1)
return findMax(a, n, vis, i+1, d);
int max = 0;
for(int j=i+1;j<n;j++){
if(Math.abs(a[i]-a[j])>d||vis[j]==1)
continue;
vis[j] = 1;
max = Math.max(max, 1 + findMax(a, n, vis, i+1, d));
vis[j] = 0;
}
return max;
}
static void findSub(ArrayList<ArrayList<Integer>> ar, int n, ArrayList<Integer> a, int i){
if(i==n){
ArrayList<Integer> b = new ArrayList<Integer>();
for(int y:a){
b.add(y);
}
ar.add(b);
return;
}
for(int j=0;j<n;j++){
if(j==i)
continue;
a.set(i,j);
findSub(ar, n, a, i+1);
}
}
// *-------------------code starts here--------------------------------------------------*
public static void solve(InputReader sc, PrintWriter pw){
long mod=(long)1e9+7;
int t=sc.nextInt();
// int t=1;
L : while(--t>=0){
int n=sc.nextInt();
int a[]=sc.readArray(n);
int b[]=a.clone();
int cnt=0;
for(int i=1;i<n-1;i++){
if(a[i]>a[i+1]&&a[i]>a[i-1]){
if(i==n-2){
a[i]=a[i-1];
cnt++;
}
else{
a[i+1]=Math.max(a[i],a[i+2]);
cnt++;
}
}
}
pw.println(cnt);
for(int i=0;i<n;i++){
pw.print(a[i]+" ");
}
pw.println();
}
}
// *-------------------code ends here-----------------------------------------------------*
static void assignAnc(ArrayList<Integer> ar[],int sz[], int pa[] ,int curr, int par){
sz[curr] = 1;
pa[curr] = par;
for(int v:ar[curr]){
if(par==v)
continue;
assignAnc(ar, sz, pa, v, curr);
sz[curr] += sz[v];
}
}
static int findLCA(int a, int b, int par[][], int depth[]){
if(depth[a]>depth[b]){
a = a^b;
b = a^b;
a = a^b;
}
int diff = depth[b] - depth[a];
for(int i=19;i>=0;i--){
if((diff&(1<<i))>0){
b = par[b][i];
}
}
if(a==b)
return a;
for(int i=19;i>=0;i--){
if(par[b][i]!=par[a][i]){
b = par[b][i];
a = par[a][i];
}
}
return par[a][0];
}
static void formArrayForBinaryLifting(int n, int par[][]){
for(int j=1;j<20;j++){
for(int i=0;i<n;i++){
if(par[i][j-1]==-1)
continue;
par[i][j] = par[par[i][j-1]][j-1];
}
}
}
static long lcm(int a, int b){
return a*b/gcd(a,b);
}
static class Pair1 {
long a;
long b;
Pair1(long a, long b) {
this.a = a;
this.b = b;
}
}
static class Pair implements Comparable<Pair> {
int a;
int b;
// int c;
Pair(int a, int b) {
this.a = a;
this.b = b;
// this.c = c;
}
public int compareTo(Pair p) {
return Integer.compare(a,p.a);
}
}
static boolean isPrime(int 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 long gcd(long a, long b) {
if (b == 0)
return a;
return gcd(b, a % b);
}
static long fast_pow(long base, long n, long M) {
if (n == 0)
return 1;
if (n == 1)
return base % M;
long halfn = fast_pow(base, n / 2, M);
if (n % 2 == 0)
return (halfn * halfn) % M;
else
return (((halfn * halfn) % M) * base) % M;
}
static long modInverse(long n, long M) {
return fast_pow(n, M - 2, M);
}
static class InputReader {
public BufferedReader reader;
public StringTokenizer tokenizer;
public InputReader(InputStream stream) {
reader = new BufferedReader(new InputStreamReader(stream), 9992768);
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[] readArray(int n) {
int[] a=new int[n];
for (int i=0; i<n; i++) a[i]=nextInt();
return a;
}
public long[] readarray(int n) {
long[] a=new long[n];
for (int i=0; i<n; i++) a[i]=nextLong();
return a;
}
public int nextInt() {
return Integer.parseInt(next());
}
public long nextLong() {
return Long.parseLong(next());
}
}
} | Java | ["5\n3\n2 1 2\n4\n1 2 3 1\n5\n1 2 1 2 1\n9\n1 2 1 3 2 3 1 2 1\n9\n2 1 3 1 3 1 3 1 3"] | 2 seconds | ["0\n2 1 2\n1\n1 3 3 1\n1\n1 2 2 2 1\n2\n1 2 3 3 2 3 3 2 1\n2\n2 1 3 3 3 1 1 1 3"] | NoteIn the first example, the array contains no local maximum, so we don't need to perform operations.In the second example, we can change $$$a_2$$$ to $$$3$$$, then the array don't have local maximums. | Java 8 | standard input | [
"greedy"
] | 255abf8750541f54d8ff0f1f80e1f8e7 | Each test contains multiple test cases. The first line will contain a single integer $$$t$$$ $$$(1 \leq t \leq 10000)$$$ — the number of test cases. Then $$$t$$$ test cases follow. The first line of each test case contains a single integer $$$n$$$ $$$(2 \leq n \leq 2 \cdot 10^5)$$$ — the size of the array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots ,a_n$$$ $$$(1 \leq a_i \leq 10^9)$$$, the elements of array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, first output a line containing a single integer $$$m$$$ — minimum number of operations required. Then ouput a line consist of $$$n$$$ integers — the resulting array after the operations. Note that this array should differ in exactly $$$m$$$ elements from the initial array. If there are multiple answers, print any. | standard output | |
PASSED | aa3f828b83b4a61b306aceb99fe96ded | train_108.jsonl | 1645367700 | You are given an array $$$a$$$ of size $$$n$$$. Each element in this array is an integer between $$$1$$$ and $$$10^9$$$.You can perform several operations to this array. During an operation, you can replace an element in the array with any integer between $$$1$$$ and $$$10^9$$$. Output the minimum number of operations needed such that the resulting array doesn't contain any local maximums, and the resulting array after the operations.An element $$$a_i$$$ is a local maximum if it is strictly larger than both of its neighbors (that is, $$$a_i > a_{i - 1}$$$ and $$$a_i > a_{i + 1}$$$). Since $$$a_1$$$ and $$$a_n$$$ have only one neighbor each, they will never be a local maximum. | 256 megabytes | import java.util.*;
public class avoidlocmax
{
public static void main(String[] args)
{
Scanner sc=new Scanner(System.in);
int t=sc.nextInt();
while(t-->0)
{
int n=sc.nextInt();
int ar[]=new int[n];int c=0;
for(int i=0;i<n;i++)
{
ar[i]=sc.nextInt();
}
for(int i=1;i<n-1;i++)
{
if(ar[i-1]<ar[i]&&ar[i+1]<ar[i])
{
if((i+2)<n&&ar[i]<ar[i+2])
{
ar[i+1]=ar[i+2];
}else
{
ar[i+1]=ar[i];
}c++;
}
}System.out.println(c);
for(int i=0;i<n;i++)
{
System.out.print(ar[i]+" ");
}System.out.println();
}
}
} | Java | ["5\n3\n2 1 2\n4\n1 2 3 1\n5\n1 2 1 2 1\n9\n1 2 1 3 2 3 1 2 1\n9\n2 1 3 1 3 1 3 1 3"] | 2 seconds | ["0\n2 1 2\n1\n1 3 3 1\n1\n1 2 2 2 1\n2\n1 2 3 3 2 3 3 2 1\n2\n2 1 3 3 3 1 1 1 3"] | NoteIn the first example, the array contains no local maximum, so we don't need to perform operations.In the second example, we can change $$$a_2$$$ to $$$3$$$, then the array don't have local maximums. | Java 8 | standard input | [
"greedy"
] | 255abf8750541f54d8ff0f1f80e1f8e7 | Each test contains multiple test cases. The first line will contain a single integer $$$t$$$ $$$(1 \leq t \leq 10000)$$$ — the number of test cases. Then $$$t$$$ test cases follow. The first line of each test case contains a single integer $$$n$$$ $$$(2 \leq n \leq 2 \cdot 10^5)$$$ — the size of the array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots ,a_n$$$ $$$(1 \leq a_i \leq 10^9)$$$, the elements of array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, first output a line containing a single integer $$$m$$$ — minimum number of operations required. Then ouput a line consist of $$$n$$$ integers — the resulting array after the operations. Note that this array should differ in exactly $$$m$$$ elements from the initial array. If there are multiple answers, print any. | standard output | |
PASSED | 50391e6ae317f8192ed98b5ab3e2f661 | train_108.jsonl | 1645367700 | You are given an array $$$a$$$ of size $$$n$$$. Each element in this array is an integer between $$$1$$$ and $$$10^9$$$.You can perform several operations to this array. During an operation, you can replace an element in the array with any integer between $$$1$$$ and $$$10^9$$$. Output the minimum number of operations needed such that the resulting array doesn't contain any local maximums, and the resulting array after the operations.An element $$$a_i$$$ is a local maximum if it is strictly larger than both of its neighbors (that is, $$$a_i > a_{i - 1}$$$ and $$$a_i > a_{i + 1}$$$). Since $$$a_1$$$ and $$$a_n$$$ have only one neighbor each, they will never be a local maximum. | 256 megabytes | import java.io.*;
public class LocalMaxims {
public static void main(String[] args) throws IOException
{
BufferedReader bf = new BufferedReader(new InputStreamReader(System.in));
int test = Integer.parseInt(bf.readLine());
while(test-- > 0)
{
int n = Integer.parseInt(bf.readLine());
int[] a = new int[n];
String line = bf.readLine();
String[] elements = line.trim().split("\\s+");
for(int i=0;i<n;i++)
{
a[i] = Integer.parseInt(elements[i]);
}
int result = 0;
for(int i=1;i<n-1;i++)
{
if(a[i]>a[i-1]&&a[i]>a[i+1])
{
result++;
a[i+1] = (i+2<n)?Math.max(a[i],a[i+2]):a[i];
}
}
System.out.println(result);
for(int i=0;i<n;i++)
{
System.out.print(a[i]+" ");
}
System.out.println();
}
}
}
| Java | ["5\n3\n2 1 2\n4\n1 2 3 1\n5\n1 2 1 2 1\n9\n1 2 1 3 2 3 1 2 1\n9\n2 1 3 1 3 1 3 1 3"] | 2 seconds | ["0\n2 1 2\n1\n1 3 3 1\n1\n1 2 2 2 1\n2\n1 2 3 3 2 3 3 2 1\n2\n2 1 3 3 3 1 1 1 3"] | NoteIn the first example, the array contains no local maximum, so we don't need to perform operations.In the second example, we can change $$$a_2$$$ to $$$3$$$, then the array don't have local maximums. | Java 8 | standard input | [
"greedy"
] | 255abf8750541f54d8ff0f1f80e1f8e7 | Each test contains multiple test cases. The first line will contain a single integer $$$t$$$ $$$(1 \leq t \leq 10000)$$$ — the number of test cases. Then $$$t$$$ test cases follow. The first line of each test case contains a single integer $$$n$$$ $$$(2 \leq n \leq 2 \cdot 10^5)$$$ — the size of the array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots ,a_n$$$ $$$(1 \leq a_i \leq 10^9)$$$, the elements of array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, first output a line containing a single integer $$$m$$$ — minimum number of operations required. Then ouput a line consist of $$$n$$$ integers — the resulting array after the operations. Note that this array should differ in exactly $$$m$$$ elements from the initial array. If there are multiple answers, print any. | standard output | |
PASSED | 01749ef50555c38ef990749f6f2cd4ff | train_108.jsonl | 1645367700 | You are given an array $$$a$$$ of size $$$n$$$. Each element in this array is an integer between $$$1$$$ and $$$10^9$$$.You can perform several operations to this array. During an operation, you can replace an element in the array with any integer between $$$1$$$ and $$$10^9$$$. Output the minimum number of operations needed such that the resulting array doesn't contain any local maximums, and the resulting array after the operations.An element $$$a_i$$$ is a local maximum if it is strictly larger than both of its neighbors (that is, $$$a_i > a_{i - 1}$$$ and $$$a_i > a_{i + 1}$$$). Since $$$a_1$$$ and $$$a_n$$$ have only one neighbor each, they will never be a local maximum. | 256 megabytes | /*
Challenge 1: Newbie to CM in 1year (Dec 2021 - Nov 2022) 🔥 5* Codechef
Challenge 2: CM to IM in 1 year (Dec 2022 - Nov 2023) 🔥🔥 6* Codechef
Challenge 3: IM to GM in 1 year (Dec 2023 - Nov 2024) 🔥🔥🔥 7* Codechef
Goal: Become better in CP!
Key : Consistency and Discipline
Desire: SDE @ Google USA
*/
import java.util.*;
import java.io.*;
import java.math.*;
public class Coder {
static StringBuffer str=new StringBuffer();
static int n;
static long a[];
static void solve(){
List<Integer> l=new ArrayList<>();
for(int i=1;i<n-1;i++){
if(a[i]>a[i-1] && a[i]>a[i+1]) l.add(i);
}
// case 1: x<z => y=x
// case 2: z<x => y=z
// case 3: x=z => y=x || y=z
long ans=0;
for(int i=0;i<l.size();i++){
long x=a[l.get(i)-1];
long y=a[l.get(i)];
long z=a[l.get(i)+1];
if(i+1<l.size() && l.get(i+1)-l.get(i)==2){
a[l.get(i)+1]=Math.max(a[l.get(i)], a[l.get(i+1)]);
i++;
}else if(x<z){
a[l.get(i)]=z;
}else if(z<x){
a[l.get(i)]=x;
}else{
a[l.get(i)]=x;
}
ans++;
}
str.append(ans).append("\n");
for(int i=0;i<n;i++) str.append(a[i]).append(" ");
str.append("\n");
}
public static void main(String[] args) throws java.lang.Exception {
BufferedReader bf;
PrintWriter pw;
boolean lenv=false;
if(lenv){
bf = new BufferedReader(
new FileReader("input.txt"));
pw=new PrintWriter(new
BufferedWriter(new FileWriter("output.txt")));
}else{
bf = new BufferedReader(new InputStreamReader(System.in));
pw = new PrintWriter(new OutputStreamWriter(System.out));
}
int q = Integer.parseInt(bf.readLine().trim());
while (q-- > 0) {
n=Integer.parseInt(bf.readLine().trim());
String s[]=bf.readLine().trim().split("\\s+");
a=new long[n];
for(int i=0;i<n;i++) a[i]=Long.parseLong(s[i]);
solve();
}
pw.println(str);
pw.flush();
// System.outin.print(str);
}
} | Java | ["5\n3\n2 1 2\n4\n1 2 3 1\n5\n1 2 1 2 1\n9\n1 2 1 3 2 3 1 2 1\n9\n2 1 3 1 3 1 3 1 3"] | 2 seconds | ["0\n2 1 2\n1\n1 3 3 1\n1\n1 2 2 2 1\n2\n1 2 3 3 2 3 3 2 1\n2\n2 1 3 3 3 1 1 1 3"] | NoteIn the first example, the array contains no local maximum, so we don't need to perform operations.In the second example, we can change $$$a_2$$$ to $$$3$$$, then the array don't have local maximums. | Java 8 | standard input | [
"greedy"
] | 255abf8750541f54d8ff0f1f80e1f8e7 | Each test contains multiple test cases. The first line will contain a single integer $$$t$$$ $$$(1 \leq t \leq 10000)$$$ — the number of test cases. Then $$$t$$$ test cases follow. The first line of each test case contains a single integer $$$n$$$ $$$(2 \leq n \leq 2 \cdot 10^5)$$$ — the size of the array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots ,a_n$$$ $$$(1 \leq a_i \leq 10^9)$$$, the elements of array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, first output a line containing a single integer $$$m$$$ — minimum number of operations required. Then ouput a line consist of $$$n$$$ integers — the resulting array after the operations. Note that this array should differ in exactly $$$m$$$ elements from the initial array. If there are multiple answers, print any. | standard output | |
PASSED | 7352528604d44eac176623c7292b5d2a | train_108.jsonl | 1645367700 | You are given an array $$$a$$$ of size $$$n$$$. Each element in this array is an integer between $$$1$$$ and $$$10^9$$$.You can perform several operations to this array. During an operation, you can replace an element in the array with any integer between $$$1$$$ and $$$10^9$$$. Output the minimum number of operations needed such that the resulting array doesn't contain any local maximums, and the resulting array after the operations.An element $$$a_i$$$ is a local maximum if it is strictly larger than both of its neighbors (that is, $$$a_i > a_{i - 1}$$$ and $$$a_i > a_{i + 1}$$$). Since $$$a_1$$$ and $$$a_n$$$ have only one neighbor each, they will never be a local maximum. | 256 megabytes | import java.util.*;
import java.io.*;
public class MyCpClass{
public static void main(String []args) throws IOException{
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
StringBuilder sb = new StringBuilder();
int T = Integer.parseInt(br.readLine().trim());
while(T-- > 0){
int n = Integer.parseInt(br.readLine().trim());
int []a = new int[n];
String []str = br.readLine().trim().split(" ");
for(int i=0; i<n; i++)
a[i] = Integer.parseInt(str[i]);
int cnt = 0, i = 1;
while(i < n-1){
if(a[i]>a[i-1] && a[i]>a[i+1]){
if(i+2 >= n) a[i+1]=a[i];
else
a[i+1] = Math.max(a[i], a[i+2]);
cnt++;
}
i++;
}
sb.append(cnt + "\n");
for(int x : a) sb.append(x+" ");
sb.append("\n");
}
System.out.println(sb);
}
} | Java | ["5\n3\n2 1 2\n4\n1 2 3 1\n5\n1 2 1 2 1\n9\n1 2 1 3 2 3 1 2 1\n9\n2 1 3 1 3 1 3 1 3"] | 2 seconds | ["0\n2 1 2\n1\n1 3 3 1\n1\n1 2 2 2 1\n2\n1 2 3 3 2 3 3 2 1\n2\n2 1 3 3 3 1 1 1 3"] | NoteIn the first example, the array contains no local maximum, so we don't need to perform operations.In the second example, we can change $$$a_2$$$ to $$$3$$$, then the array don't have local maximums. | Java 8 | standard input | [
"greedy"
] | 255abf8750541f54d8ff0f1f80e1f8e7 | Each test contains multiple test cases. The first line will contain a single integer $$$t$$$ $$$(1 \leq t \leq 10000)$$$ — the number of test cases. Then $$$t$$$ test cases follow. The first line of each test case contains a single integer $$$n$$$ $$$(2 \leq n \leq 2 \cdot 10^5)$$$ — the size of the array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots ,a_n$$$ $$$(1 \leq a_i \leq 10^9)$$$, the elements of array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, first output a line containing a single integer $$$m$$$ — minimum number of operations required. Then ouput a line consist of $$$n$$$ integers — the resulting array after the operations. Note that this array should differ in exactly $$$m$$$ elements from the initial array. If there are multiple answers, print any. | standard output | |
PASSED | 3561c2dbf056002ed013d75318d043b9 | train_108.jsonl | 1645367700 | You are given an array $$$a$$$ of size $$$n$$$. Each element in this array is an integer between $$$1$$$ and $$$10^9$$$.You can perform several operations to this array. During an operation, you can replace an element in the array with any integer between $$$1$$$ and $$$10^9$$$. Output the minimum number of operations needed such that the resulting array doesn't contain any local maximums, and the resulting array after the operations.An element $$$a_i$$$ is a local maximum if it is strictly larger than both of its neighbors (that is, $$$a_i > a_{i - 1}$$$ and $$$a_i > a_{i + 1}$$$). Since $$$a_1$$$ and $$$a_n$$$ have only one neighbor each, they will never be a local maximum. | 256 megabytes | //CP- MASS_2701
import java.util.*;
import java.io.*;
public class B_Avoid_Local_Maximums {
static FastReader in=new FastReader();
static final Random random=new Random();
static long mod=1000000007L;
static HashMap<String,Integer>map=new HashMap<>();
public static void main(String args[]) throws IOException {
int t=in.nextInt();
int cse=1;
loop:
while(t-->0)
{
StringBuilder res=new StringBuilder();
//res.append("Hello"+"\n");
int n=in.nextInt();
int[] a=new int[n];
for(int i=0;i<n;i++)
{
a[i]=in.nextInt();
}
int ans=0;
for(int i=1;i<n-1;i++)
{
if(a[i]>a[i+1] && a[i]>a[i-1])
{
if(i<n-2)
{
a[i+1]=Math.max(a[i],a[i+2]);
}
else{
a[i+1]=a[i];
}
ans++;
}
}
println(ans);
for(int i=0;i<n;i++)
{
res.append(a[i]+" ");
}
println(res);
}
}
static int max(int a, int b)
{
if(a<b)
return b;
return a;
}
static void ruffleSort(int[] a) {
int n=a.length;
for (int i=0; i<n; i++) {
int oi=random.nextInt(n), temp=a[oi];
a[oi]=a[i]; a[i]=temp;
}
Arrays.sort(a);
}
static < E > void print(E res)
{
System.out.print(res);
}
static < E > void println(E res)
{
System.out.println(res);
}
static int gcd(int a,int b)
{
if(b==0)
{
return a;
}
return gcd(b,a%b);
}
static int lcm(int a, int b)
{
return (a / gcd(a, b)) * b;
}
static int abs(int a)
{
if(a<0)
return -1*a;
return a;
}
static boolean isPrime(int n)
{
// Corner cases
if (n <= 1)
return false;
if (n <= 3)
return true;
// This is checked so that we can skip
// middle five numbers in below loop
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 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;
}
int [] readintarray(int n) {
int res [] = new int [n];
for(int i = 0; i<n; i++)res[i] = nextInt();
return res;
}
long [] readlongarray(int n) {
long res [] = new long [n];
for(int i = 0; i<n; i++)res[i] = nextLong();
return res;
}
}
}
| Java | ["5\n3\n2 1 2\n4\n1 2 3 1\n5\n1 2 1 2 1\n9\n1 2 1 3 2 3 1 2 1\n9\n2 1 3 1 3 1 3 1 3"] | 2 seconds | ["0\n2 1 2\n1\n1 3 3 1\n1\n1 2 2 2 1\n2\n1 2 3 3 2 3 3 2 1\n2\n2 1 3 3 3 1 1 1 3"] | NoteIn the first example, the array contains no local maximum, so we don't need to perform operations.In the second example, we can change $$$a_2$$$ to $$$3$$$, then the array don't have local maximums. | Java 8 | standard input | [
"greedy"
] | 255abf8750541f54d8ff0f1f80e1f8e7 | Each test contains multiple test cases. The first line will contain a single integer $$$t$$$ $$$(1 \leq t \leq 10000)$$$ — the number of test cases. Then $$$t$$$ test cases follow. The first line of each test case contains a single integer $$$n$$$ $$$(2 \leq n \leq 2 \cdot 10^5)$$$ — the size of the array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots ,a_n$$$ $$$(1 \leq a_i \leq 10^9)$$$, the elements of array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, first output a line containing a single integer $$$m$$$ — minimum number of operations required. Then ouput a line consist of $$$n$$$ integers — the resulting array after the operations. Note that this array should differ in exactly $$$m$$$ elements from the initial array. If there are multiple answers, print any. | standard output | |
PASSED | 227b1281f6ad9dfca30a3d6d92cbcf46 | train_108.jsonl | 1645367700 | You are given an array $$$a$$$ of size $$$n$$$. Each element in this array is an integer between $$$1$$$ and $$$10^9$$$.You can perform several operations to this array. During an operation, you can replace an element in the array with any integer between $$$1$$$ and $$$10^9$$$. Output the minimum number of operations needed such that the resulting array doesn't contain any local maximums, and the resulting array after the operations.An element $$$a_i$$$ is a local maximum if it is strictly larger than both of its neighbors (that is, $$$a_i > a_{i - 1}$$$ and $$$a_i > a_{i + 1}$$$). Since $$$a_1$$$ and $$$a_n$$$ have only one neighbor each, they will never be a local maximum. | 256 megabytes | import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.StringTokenizer;
public class Main {
static int N = 2 * 100010;
static AReader scan = new AReader();
static void slove() {
int n = scan.nextInt();
int[] arr = new int[n];
for(int i = 0;i<n;i++)
arr[i] = scan.nextInt();
int cnt = 0;
for(int i = 1;i<n-1;i++){
if(arr[i] > arr[i-1] && arr[i] > arr[i+1]){
cnt++;
arr[i+1] = arr[i];
if(i+2< n && arr[i+2] > arr[i])
arr[i+1] = arr[i+2];
}
}
System.out.println(cnt);
for(int i = 0;i<n;i++){
System.out.print(arr[i]+" ");
}
System.out.println();
}
public static void main(String[] args) {
int T = scan.nextInt();
while (T-- > 0) {
slove();
}
}
}
class AReader {
private BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
private StringTokenizer tokenizer = new StringTokenizer("");
private String innerNextLine() {
try {
return reader.readLine();
} catch (IOException ex) {
return null;
}
}
public boolean hasNext() {
while (!tokenizer.hasMoreTokens()) {
String nextLine = innerNextLine();
if (nextLine == null) {
return false;
}
tokenizer = new StringTokenizer(nextLine);
}
return true;
}
public String nextLine() {
tokenizer = new StringTokenizer("");
return innerNextLine();
}
public String next() {
hasNext();
return tokenizer.nextToken();
}
public int nextInt() {
return Integer.parseInt(next());
}
public long nextLong() {
return Long.parseLong(next());
}
public double nextDouble() {
return Double.parseDouble(next());
}
}
class Pair {
int x;
int y;
public Pair(int x, int y) {
this.x = x;
this.y = y;
}
}
| Java | ["5\n3\n2 1 2\n4\n1 2 3 1\n5\n1 2 1 2 1\n9\n1 2 1 3 2 3 1 2 1\n9\n2 1 3 1 3 1 3 1 3"] | 2 seconds | ["0\n2 1 2\n1\n1 3 3 1\n1\n1 2 2 2 1\n2\n1 2 3 3 2 3 3 2 1\n2\n2 1 3 3 3 1 1 1 3"] | NoteIn the first example, the array contains no local maximum, so we don't need to perform operations.In the second example, we can change $$$a_2$$$ to $$$3$$$, then the array don't have local maximums. | Java 8 | standard input | [
"greedy"
] | 255abf8750541f54d8ff0f1f80e1f8e7 | Each test contains multiple test cases. The first line will contain a single integer $$$t$$$ $$$(1 \leq t \leq 10000)$$$ — the number of test cases. Then $$$t$$$ test cases follow. The first line of each test case contains a single integer $$$n$$$ $$$(2 \leq n \leq 2 \cdot 10^5)$$$ — the size of the array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots ,a_n$$$ $$$(1 \leq a_i \leq 10^9)$$$, the elements of array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, first output a line containing a single integer $$$m$$$ — minimum number of operations required. Then ouput a line consist of $$$n$$$ integers — the resulting array after the operations. Note that this array should differ in exactly $$$m$$$ elements from the initial array. If there are multiple answers, print any. | standard output | |
PASSED | 4681ef3cc202568cd3bc48875203d45a | train_108.jsonl | 1645367700 | You are given an array $$$a$$$ of size $$$n$$$. Each element in this array is an integer between $$$1$$$ and $$$10^9$$$.You can perform several operations to this array. During an operation, you can replace an element in the array with any integer between $$$1$$$ and $$$10^9$$$. Output the minimum number of operations needed such that the resulting array doesn't contain any local maximums, and the resulting array after the operations.An element $$$a_i$$$ is a local maximum if it is strictly larger than both of its neighbors (that is, $$$a_i > a_{i - 1}$$$ and $$$a_i > a_{i + 1}$$$). Since $$$a_1$$$ and $$$a_n$$$ have only one neighbor each, they will never be a local maximum. | 256 megabytes | import java.io.*;
import java.util.*;
public class cfrcs {
//
// static boolean prime[] = new boolean[n+1];
// static void sieve(int n)
// {
//
// for(int i=0;i<n;i++)
// prime[i] = true;
//
// for(int p = 2; p*p <=n; p++)
// {
//
// if(prime[p] == true)
// {
//
// for(int i = p*p; i <= n; i += p)
// prime[i] = false;
// }
// }
// }
static long gcd(long a, long b) {
if (b == 0)
return a;
return gcd(b, a % b);
}
static long fun(char c[],char cc,int a,int b) {
long ans=0l;
int count=0;
int count2=0;
for(int i=0;i<c.length;i++) {
if(c[i]==cc) {
count++;
}
else {
if(count!=0) {ans+=1;}
count=0;
}
}
if(count>0)ans+=1;
return ans;
}
public static void main(String[] args) throws IOException {
// BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
// int t = Integer.parseInt(br.readLine());
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 count=0;
HashSet<Integer> hs=new HashSet<>();
for (int i=1;i<n-1;i++) {
if(arr[i]>arr[i-1]&&arr[i]>arr[i+1]) {
if(hs.contains(i-1)) {
arr[i-1]=arr[i];
}
else {
arr[i+1]=arr[i];
hs.add(i+1);
}
}
}
System.out.println(hs.size());
for(int i=0;i<n;i++) {
System.out.print(arr[i]+" ");
}
System.out.println();
}
}
} | Java | ["5\n3\n2 1 2\n4\n1 2 3 1\n5\n1 2 1 2 1\n9\n1 2 1 3 2 3 1 2 1\n9\n2 1 3 1 3 1 3 1 3"] | 2 seconds | ["0\n2 1 2\n1\n1 3 3 1\n1\n1 2 2 2 1\n2\n1 2 3 3 2 3 3 2 1\n2\n2 1 3 3 3 1 1 1 3"] | NoteIn the first example, the array contains no local maximum, so we don't need to perform operations.In the second example, we can change $$$a_2$$$ to $$$3$$$, then the array don't have local maximums. | Java 8 | standard input | [
"greedy"
] | 255abf8750541f54d8ff0f1f80e1f8e7 | Each test contains multiple test cases. The first line will contain a single integer $$$t$$$ $$$(1 \leq t \leq 10000)$$$ — the number of test cases. Then $$$t$$$ test cases follow. The first line of each test case contains a single integer $$$n$$$ $$$(2 \leq n \leq 2 \cdot 10^5)$$$ — the size of the array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots ,a_n$$$ $$$(1 \leq a_i \leq 10^9)$$$, the elements of array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, first output a line containing a single integer $$$m$$$ — minimum number of operations required. Then ouput a line consist of $$$n$$$ integers — the resulting array after the operations. Note that this array should differ in exactly $$$m$$$ elements from the initial array. If there are multiple answers, print any. | standard output | |
PASSED | 979836cee815f71218eb48e057c16f57 | train_108.jsonl | 1645367700 | You are given an array $$$a$$$ of size $$$n$$$. Each element in this array is an integer between $$$1$$$ and $$$10^9$$$.You can perform several operations to this array. During an operation, you can replace an element in the array with any integer between $$$1$$$ and $$$10^9$$$. Output the minimum number of operations needed such that the resulting array doesn't contain any local maximums, and the resulting array after the operations.An element $$$a_i$$$ is a local maximum if it is strictly larger than both of its neighbors (that is, $$$a_i > a_{i - 1}$$$ and $$$a_i > a_{i + 1}$$$). Since $$$a_1$$$ and $$$a_n$$$ have only one neighbor each, they will never be a local maximum. | 256 megabytes | import java.io.*;
public class avoid_local_maximums
{
public static void main(String[] args) throws IOException
{
BufferedReader y = new BufferedReader (new InputStreamReader (System.in));
int t = Integer.parseInt(y.readLine());
int loop, j;
while(t-- > 0)
{
int n = Integer.parseInt(y.readLine());
String str = y.readLine();
String s[] = str.split(" ");
int arr[] = new int[n];
for(j = 0; j < n; j++)
{
arr[j] = Integer.parseInt(s[j]);
}
int c = 0;
for(j = 1; j < n - 1; j++)
{
if(arr[j] > arr[j - 1] && arr[j] > arr[j + 1])
{
if(j + 2 < n)
{
arr[j + 1] = Math.max(arr[j], arr[j + 2]);
j = j + 2;
}
else
{
arr[j + 1] = arr[j];
j = j + 1;
}
c++;
}
}
System.out.println(c);
for(j = 0; j < n; j++)
{
System.out.print(arr[j] + " ");
}
System.out.println();
}
}
} | Java | ["5\n3\n2 1 2\n4\n1 2 3 1\n5\n1 2 1 2 1\n9\n1 2 1 3 2 3 1 2 1\n9\n2 1 3 1 3 1 3 1 3"] | 2 seconds | ["0\n2 1 2\n1\n1 3 3 1\n1\n1 2 2 2 1\n2\n1 2 3 3 2 3 3 2 1\n2\n2 1 3 3 3 1 1 1 3"] | NoteIn the first example, the array contains no local maximum, so we don't need to perform operations.In the second example, we can change $$$a_2$$$ to $$$3$$$, then the array don't have local maximums. | Java 11 | standard input | [
"greedy"
] | 255abf8750541f54d8ff0f1f80e1f8e7 | Each test contains multiple test cases. The first line will contain a single integer $$$t$$$ $$$(1 \leq t \leq 10000)$$$ — the number of test cases. Then $$$t$$$ test cases follow. The first line of each test case contains a single integer $$$n$$$ $$$(2 \leq n \leq 2 \cdot 10^5)$$$ — the size of the array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots ,a_n$$$ $$$(1 \leq a_i \leq 10^9)$$$, the elements of array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, first output a line containing a single integer $$$m$$$ — minimum number of operations required. Then ouput a line consist of $$$n$$$ integers — the resulting array after the operations. Note that this array should differ in exactly $$$m$$$ elements from the initial array. If there are multiple answers, print any. | standard output | |
PASSED | 1c2734f82ef3f8324b4dba8b36eb4b64 | train_108.jsonl | 1645367700 | You are given an array $$$a$$$ of size $$$n$$$. Each element in this array is an integer between $$$1$$$ and $$$10^9$$$.You can perform several operations to this array. During an operation, you can replace an element in the array with any integer between $$$1$$$ and $$$10^9$$$. Output the minimum number of operations needed such that the resulting array doesn't contain any local maximums, and the resulting array after the operations.An element $$$a_i$$$ is a local maximum if it is strictly larger than both of its neighbors (that is, $$$a_i > a_{i - 1}$$$ and $$$a_i > a_{i + 1}$$$). Since $$$a_1$$$ and $$$a_n$$$ have only one neighbor each, they will never be a local maximum. | 256 megabytes | import java.io.*;
public class avoid_local_maximums
{
public static void main(String[] args) throws IOException
{
BufferedReader y = new BufferedReader (new InputStreamReader (System.in));
int t = Integer.parseInt(y.readLine());
int loop, j;
while(t-- > 0)
{
int n = Integer.parseInt(y.readLine());
String str = y.readLine();
String s[] = str.split(" ");
int arr[] = new int[n];
for(j = 0; j < n; j++)
{
arr[j] = Integer.parseInt(s[j]);
}
String nstr = arr[0] + " ";
int c = 0;
for(j = 1; j < n - 1; j++)
{
if(arr[j] > arr[j - 1] && arr[j] > arr[j + 1])
{
if(j + 2 < n)
{
arr[j + 1] = Math.max(arr[j], arr[j + 2]);
j = j + 2;
}
else
{
arr[j + 1] = arr[j];
j = j + 1;
}
c++;
}
}
System.out.println(c);
for(j = 0; j < n; j++)
{
System.out.print(arr[j] + " ");
}
System.out.println();
}
}
} | Java | ["5\n3\n2 1 2\n4\n1 2 3 1\n5\n1 2 1 2 1\n9\n1 2 1 3 2 3 1 2 1\n9\n2 1 3 1 3 1 3 1 3"] | 2 seconds | ["0\n2 1 2\n1\n1 3 3 1\n1\n1 2 2 2 1\n2\n1 2 3 3 2 3 3 2 1\n2\n2 1 3 3 3 1 1 1 3"] | NoteIn the first example, the array contains no local maximum, so we don't need to perform operations.In the second example, we can change $$$a_2$$$ to $$$3$$$, then the array don't have local maximums. | Java 11 | standard input | [
"greedy"
] | 255abf8750541f54d8ff0f1f80e1f8e7 | Each test contains multiple test cases. The first line will contain a single integer $$$t$$$ $$$(1 \leq t \leq 10000)$$$ — the number of test cases. Then $$$t$$$ test cases follow. The first line of each test case contains a single integer $$$n$$$ $$$(2 \leq n \leq 2 \cdot 10^5)$$$ — the size of the array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots ,a_n$$$ $$$(1 \leq a_i \leq 10^9)$$$, the elements of array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, first output a line containing a single integer $$$m$$$ — minimum number of operations required. Then ouput a line consist of $$$n$$$ integers — the resulting array after the operations. Note that this array should differ in exactly $$$m$$$ elements from the initial array. If there are multiple answers, print any. | standard output | |
PASSED | 5cd42ae373e57247f4c78e26a1eeea66 | train_108.jsonl | 1645367700 | You are given an array $$$a$$$ of size $$$n$$$. Each element in this array is an integer between $$$1$$$ and $$$10^9$$$.You can perform several operations to this array. During an operation, you can replace an element in the array with any integer between $$$1$$$ and $$$10^9$$$. Output the minimum number of operations needed such that the resulting array doesn't contain any local maximums, and the resulting array after the operations.An element $$$a_i$$$ is a local maximum if it is strictly larger than both of its neighbors (that is, $$$a_i > a_{i - 1}$$$ and $$$a_i > a_{i + 1}$$$). Since $$$a_1$$$ and $$$a_n$$$ have only one neighbor each, they will never be a local maximum. | 256 megabytes | import java.io.*;
public class avoid_local_maximums
{
public static void main(String[] args) throws IOException
{
BufferedReader y = new BufferedReader (new InputStreamReader (System.in));
int t = Integer.parseInt(y.readLine());
int loop, i;
while(t-- > 0)
{
int n = Integer.parseInt(y.readLine());
String str = y.readLine();
String s[] = str.split(" ");
int arr[] = new int[n];
for(i = 0; i < n; i++)
{
arr[i] = Integer.parseInt(s[i]);
}
String nstr = arr[0] + " ";
int c = 0;
for(i = 1; i < n - 1; i++)
{
if(arr[i] > arr[i - 1] && arr[i] > arr[i + 1])
{
if(i + 2 < n)
{
arr[i + 1] = Math.max(arr[i], arr[i + 2]);
i = i + 2;
}
else
{
arr[i + 1] = arr[i];
i = i + 1;
}
c++;
}
}
System.out.println(c);
for(i = 0; i < n; i++)
{
System.out.print(arr[i] + " ");
}
System.out.println();
}
}
}
| Java | ["5\n3\n2 1 2\n4\n1 2 3 1\n5\n1 2 1 2 1\n9\n1 2 1 3 2 3 1 2 1\n9\n2 1 3 1 3 1 3 1 3"] | 2 seconds | ["0\n2 1 2\n1\n1 3 3 1\n1\n1 2 2 2 1\n2\n1 2 3 3 2 3 3 2 1\n2\n2 1 3 3 3 1 1 1 3"] | NoteIn the first example, the array contains no local maximum, so we don't need to perform operations.In the second example, we can change $$$a_2$$$ to $$$3$$$, then the array don't have local maximums. | Java 11 | standard input | [
"greedy"
] | 255abf8750541f54d8ff0f1f80e1f8e7 | Each test contains multiple test cases. The first line will contain a single integer $$$t$$$ $$$(1 \leq t \leq 10000)$$$ — the number of test cases. Then $$$t$$$ test cases follow. The first line of each test case contains a single integer $$$n$$$ $$$(2 \leq n \leq 2 \cdot 10^5)$$$ — the size of the array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots ,a_n$$$ $$$(1 \leq a_i \leq 10^9)$$$, the elements of array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, first output a line containing a single integer $$$m$$$ — minimum number of operations required. Then ouput a line consist of $$$n$$$ integers — the resulting array after the operations. Note that this array should differ in exactly $$$m$$$ elements from the initial array. If there are multiple answers, print any. | standard output | |
PASSED | 5bdeda24edb2b217e97501150842f67e | train_108.jsonl | 1645367700 | You are given an array $$$a$$$ of size $$$n$$$. Each element in this array is an integer between $$$1$$$ and $$$10^9$$$.You can perform several operations to this array. During an operation, you can replace an element in the array with any integer between $$$1$$$ and $$$10^9$$$. Output the minimum number of operations needed such that the resulting array doesn't contain any local maximums, and the resulting array after the operations.An element $$$a_i$$$ is a local maximum if it is strictly larger than both of its neighbors (that is, $$$a_i > a_{i - 1}$$$ and $$$a_i > a_{i + 1}$$$). Since $$$a_1$$$ and $$$a_n$$$ have only one neighbor each, they will never be a local maximum. | 256 megabytes | import java.util.Scanner;
public class AvoidLocalMaximums {
public static void main(String[] args) {
Scanner sc=new Scanner(System.in);
int t=sc.nextInt();
while(t-->0) {
int n=sc.nextInt(),c=0;
int[] arr=new int[n+1];
for(int i=0;i<n;i++) {
arr[i]=sc.nextInt();
}
for(int i=1;i+1<n;i++) {
if(arr[i]>arr[i-1] && arr[i]>arr[i+1]) {
if(i+2<n) {
arr[i+1]=Math.max(arr[i],arr[i+2]);
}else {
arr[i+1]=arr[i];
}
c++;
}
}
System.out.println(c);
for(int i = 0; i < n; i++)
{
System.out.print(arr[i] + " ");
}
System.out.println();
}
}
} | Java | ["5\n3\n2 1 2\n4\n1 2 3 1\n5\n1 2 1 2 1\n9\n1 2 1 3 2 3 1 2 1\n9\n2 1 3 1 3 1 3 1 3"] | 2 seconds | ["0\n2 1 2\n1\n1 3 3 1\n1\n1 2 2 2 1\n2\n1 2 3 3 2 3 3 2 1\n2\n2 1 3 3 3 1 1 1 3"] | NoteIn the first example, the array contains no local maximum, so we don't need to perform operations.In the second example, we can change $$$a_2$$$ to $$$3$$$, then the array don't have local maximums. | Java 11 | standard input | [
"greedy"
] | 255abf8750541f54d8ff0f1f80e1f8e7 | Each test contains multiple test cases. The first line will contain a single integer $$$t$$$ $$$(1 \leq t \leq 10000)$$$ — the number of test cases. Then $$$t$$$ test cases follow. The first line of each test case contains a single integer $$$n$$$ $$$(2 \leq n \leq 2 \cdot 10^5)$$$ — the size of the array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots ,a_n$$$ $$$(1 \leq a_i \leq 10^9)$$$, the elements of array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, first output a line containing a single integer $$$m$$$ — minimum number of operations required. Then ouput a line consist of $$$n$$$ integers — the resulting array after the operations. Note that this array should differ in exactly $$$m$$$ elements from the initial array. If there are multiple answers, print any. | standard output | |
PASSED | f1c5ecf3f46d42c34635e9ce446da3e4 | train_108.jsonl | 1645367700 | You are given an array $$$a$$$ of size $$$n$$$. Each element in this array is an integer between $$$1$$$ and $$$10^9$$$.You can perform several operations to this array. During an operation, you can replace an element in the array with any integer between $$$1$$$ and $$$10^9$$$. Output the minimum number of operations needed such that the resulting array doesn't contain any local maximums, and the resulting array after the operations.An element $$$a_i$$$ is a local maximum if it is strictly larger than both of its neighbors (that is, $$$a_i > a_{i - 1}$$$ and $$$a_i > a_{i + 1}$$$). Since $$$a_1$$$ and $$$a_n$$$ have only one neighbor each, they will never be a local maximum. | 256 megabytes | import java.util.Scanner;
public class AvoidLocalMaximums {
public static void main(String[] args) {
Scanner sc=new Scanner(System.in);
int t=sc.nextInt();
while(t-->0) {
int n=sc.nextInt(),c=0;
int[] arr=new int[n+1];
for(int i=0;i<n;i++) {
arr[i]=sc.nextInt();
}
for(int i=1;i+1<n;i++) {
if(arr[i]>arr[i-1] && arr[i]>arr[i+1]) {
if(i+2<n) {
arr[i+1]=Math.max(arr[i],arr[i+2]);
}else {
arr[i+1]=arr[i];
}
c++;
}
}System.out.println(c);
for(int k=0;k<n;k++) {
System.out.print(arr[k]+" ");
}
System.out.println();
}
}
} | Java | ["5\n3\n2 1 2\n4\n1 2 3 1\n5\n1 2 1 2 1\n9\n1 2 1 3 2 3 1 2 1\n9\n2 1 3 1 3 1 3 1 3"] | 2 seconds | ["0\n2 1 2\n1\n1 3 3 1\n1\n1 2 2 2 1\n2\n1 2 3 3 2 3 3 2 1\n2\n2 1 3 3 3 1 1 1 3"] | NoteIn the first example, the array contains no local maximum, so we don't need to perform operations.In the second example, we can change $$$a_2$$$ to $$$3$$$, then the array don't have local maximums. | Java 11 | standard input | [
"greedy"
] | 255abf8750541f54d8ff0f1f80e1f8e7 | Each test contains multiple test cases. The first line will contain a single integer $$$t$$$ $$$(1 \leq t \leq 10000)$$$ — the number of test cases. Then $$$t$$$ test cases follow. The first line of each test case contains a single integer $$$n$$$ $$$(2 \leq n \leq 2 \cdot 10^5)$$$ — the size of the array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots ,a_n$$$ $$$(1 \leq a_i \leq 10^9)$$$, the elements of array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, first output a line containing a single integer $$$m$$$ — minimum number of operations required. Then ouput a line consist of $$$n$$$ integers — the resulting array after the operations. Note that this array should differ in exactly $$$m$$$ elements from the initial array. If there are multiple answers, print any. | standard output | |
PASSED | ea8ea48671d1920a2a7ec41b5df809b7 | train_108.jsonl | 1645367700 | You are given an array $$$a$$$ of size $$$n$$$. Each element in this array is an integer between $$$1$$$ and $$$10^9$$$.You can perform several operations to this array. During an operation, you can replace an element in the array with any integer between $$$1$$$ and $$$10^9$$$. Output the minimum number of operations needed such that the resulting array doesn't contain any local maximums, and the resulting array after the operations.An element $$$a_i$$$ is a local maximum if it is strictly larger than both of its neighbors (that is, $$$a_i > a_{i - 1}$$$ and $$$a_i > a_{i + 1}$$$). Since $$$a_1$$$ and $$$a_n$$$ have only one neighbor each, they will never be a local maximum. | 256 megabytes | import java.util.*;
import java.io.*;
public class solution{
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());
}
}
static int count;
public static void main(String[] args){
FastReader in = new FastReader();
int t = in.nextInt();
while(t-->0){
int n = in.nextInt();
int[] arr = new int[n];
Arrays.setAll(arr, i->in.nextInt());
count = 0;
AvoidLocalMax(arr,n);
System.out.println(count);
for(int i=0;i<n;i++) System.out.print(arr[i]+" ");
System.out.println();
}
}
public static void AvoidLocalMax(int[] arr, int n){
for(int i=1;i<n-1;i++){
if(arr[i-1]<arr[i] && arr[i+1]<arr[i]){
if(i+2<n) arr[i+1] = Math.max(arr[i],arr[i+2]);
else arr[i+1] =arr[i];
count++;
}
}
}
} | Java | ["5\n3\n2 1 2\n4\n1 2 3 1\n5\n1 2 1 2 1\n9\n1 2 1 3 2 3 1 2 1\n9\n2 1 3 1 3 1 3 1 3"] | 2 seconds | ["0\n2 1 2\n1\n1 3 3 1\n1\n1 2 2 2 1\n2\n1 2 3 3 2 3 3 2 1\n2\n2 1 3 3 3 1 1 1 3"] | NoteIn the first example, the array contains no local maximum, so we don't need to perform operations.In the second example, we can change $$$a_2$$$ to $$$3$$$, then the array don't have local maximums. | Java 11 | standard input | [
"greedy"
] | 255abf8750541f54d8ff0f1f80e1f8e7 | Each test contains multiple test cases. The first line will contain a single integer $$$t$$$ $$$(1 \leq t \leq 10000)$$$ — the number of test cases. Then $$$t$$$ test cases follow. The first line of each test case contains a single integer $$$n$$$ $$$(2 \leq n \leq 2 \cdot 10^5)$$$ — the size of the array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots ,a_n$$$ $$$(1 \leq a_i \leq 10^9)$$$, the elements of array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, first output a line containing a single integer $$$m$$$ — minimum number of operations required. Then ouput a line consist of $$$n$$$ integers — the resulting array after the operations. Note that this array should differ in exactly $$$m$$$ elements from the initial array. If there are multiple answers, print any. | standard output | |
PASSED | b7e83dab9c92ec421a7dd44a2c71472a | train_108.jsonl | 1645367700 | You are given an array $$$a$$$ of size $$$n$$$. Each element in this array is an integer between $$$1$$$ and $$$10^9$$$.You can perform several operations to this array. During an operation, you can replace an element in the array with any integer between $$$1$$$ and $$$10^9$$$. Output the minimum number of operations needed such that the resulting array doesn't contain any local maximums, and the resulting array after the operations.An element $$$a_i$$$ is a local maximum if it is strictly larger than both of its neighbors (that is, $$$a_i > a_{i - 1}$$$ and $$$a_i > a_{i + 1}$$$). Since $$$a_1$$$ and $$$a_n$$$ have only one neighbor each, they will never be a local maximum. | 256 megabytes | import java.util.*;
import java.io.*;
public class solution{
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());
}
}
static int count;
public static void main(String[] args){
FastReader in = new FastReader();
int t = in.nextInt();
while(t-->0){
int n = in.nextInt();
int[] arr = new int[n];
Arrays.setAll(arr, i->in.nextInt());
count = 0;
AvoidLocalMax(arr,n);
System.out.println(count);
for(int i=0;i<n;i++) System.out.print(arr[i]+" ");
System.out.println();
}
}
public static void AvoidLocalMax(int[] arr, int n){
for(int i=1;i<n-1;i++){
if(arr[i-1]<arr[i] && arr[i+1]<arr[i]){
if(i+1==n-1) arr[i] = Math.max(arr[i+1],arr[i-1]);
else{
if(i+2<n) arr[i+1] = arr[i+2];
arr[i+1] = Math.max(arr[i],arr[i+1]);
}
count++;
}
}
}
} | Java | ["5\n3\n2 1 2\n4\n1 2 3 1\n5\n1 2 1 2 1\n9\n1 2 1 3 2 3 1 2 1\n9\n2 1 3 1 3 1 3 1 3"] | 2 seconds | ["0\n2 1 2\n1\n1 3 3 1\n1\n1 2 2 2 1\n2\n1 2 3 3 2 3 3 2 1\n2\n2 1 3 3 3 1 1 1 3"] | NoteIn the first example, the array contains no local maximum, so we don't need to perform operations.In the second example, we can change $$$a_2$$$ to $$$3$$$, then the array don't have local maximums. | Java 11 | standard input | [
"greedy"
] | 255abf8750541f54d8ff0f1f80e1f8e7 | Each test contains multiple test cases. The first line will contain a single integer $$$t$$$ $$$(1 \leq t \leq 10000)$$$ — the number of test cases. Then $$$t$$$ test cases follow. The first line of each test case contains a single integer $$$n$$$ $$$(2 \leq n \leq 2 \cdot 10^5)$$$ — the size of the array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots ,a_n$$$ $$$(1 \leq a_i \leq 10^9)$$$, the elements of array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, first output a line containing a single integer $$$m$$$ — minimum number of operations required. Then ouput a line consist of $$$n$$$ integers — the resulting array after the operations. Note that this array should differ in exactly $$$m$$$ elements from the initial array. If there are multiple answers, print any. | standard output | |
PASSED | 71400fd11ad5d909594fee3b048b8100 | train_108.jsonl | 1645367700 | You are given an array $$$a$$$ of size $$$n$$$. Each element in this array is an integer between $$$1$$$ and $$$10^9$$$.You can perform several operations to this array. During an operation, you can replace an element in the array with any integer between $$$1$$$ and $$$10^9$$$. Output the minimum number of operations needed such that the resulting array doesn't contain any local maximums, and the resulting array after the operations.An element $$$a_i$$$ is a local maximum if it is strictly larger than both of its neighbors (that is, $$$a_i > a_{i - 1}$$$ and $$$a_i > a_{i + 1}$$$). Since $$$a_1$$$ and $$$a_n$$$ have only one neighbor each, they will never be a local maximum. | 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()
{
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;
}
}
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 t = ni();
while(t-- > 0)
{
int n = ni();
int[] arr = new int[n+1];
for(int i=0; i<n; i++) arr[i]=ni();
arr[n] = 0;
int count = 0;
for(int i=1; i<n-1; i++)
{
if(arr[i] > arr[i-1] && arr[i] > arr[i+1])
{
count++;
arr[i+1] = Math.max(arr[i], arr[i+2]);
}
}
out.println(count);
for(int i=0; i<n; i++)
{
out.print(arr[i]+" ");
}
out.println();
}
}
} | Java | ["5\n3\n2 1 2\n4\n1 2 3 1\n5\n1 2 1 2 1\n9\n1 2 1 3 2 3 1 2 1\n9\n2 1 3 1 3 1 3 1 3"] | 2 seconds | ["0\n2 1 2\n1\n1 3 3 1\n1\n1 2 2 2 1\n2\n1 2 3 3 2 3 3 2 1\n2\n2 1 3 3 3 1 1 1 3"] | NoteIn the first example, the array contains no local maximum, so we don't need to perform operations.In the second example, we can change $$$a_2$$$ to $$$3$$$, then the array don't have local maximums. | Java 11 | standard input | [
"greedy"
] | 255abf8750541f54d8ff0f1f80e1f8e7 | Each test contains multiple test cases. The first line will contain a single integer $$$t$$$ $$$(1 \leq t \leq 10000)$$$ — the number of test cases. Then $$$t$$$ test cases follow. The first line of each test case contains a single integer $$$n$$$ $$$(2 \leq n \leq 2 \cdot 10^5)$$$ — the size of the array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots ,a_n$$$ $$$(1 \leq a_i \leq 10^9)$$$, the elements of array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, first output a line containing a single integer $$$m$$$ — minimum number of operations required. Then ouput a line consist of $$$n$$$ integers — the resulting array after the operations. Note that this array should differ in exactly $$$m$$$ elements from the initial array. If there are multiple answers, print any. | standard output | |
PASSED | ddf07f7dd1af286c7454a1e1c39dcda3 | train_108.jsonl | 1645367700 | You are given an array $$$a$$$ of size $$$n$$$. Each element in this array is an integer between $$$1$$$ and $$$10^9$$$.You can perform several operations to this array. During an operation, you can replace an element in the array with any integer between $$$1$$$ and $$$10^9$$$. Output the minimum number of operations needed such that the resulting array doesn't contain any local maximums, and the resulting array after the operations.An element $$$a_i$$$ is a local maximum if it is strictly larger than both of its neighbors (that is, $$$a_i > a_{i - 1}$$$ and $$$a_i > a_{i + 1}$$$). Since $$$a_1$$$ and $$$a_n$$$ have only one neighbor each, they will never be a local maximum. | 256 megabytes |
// Working program with FastReader
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.*;
public class Avoid {
public static void main(String[] args) {
FastReader fs = new FastReader();
StringBuilder sb = new StringBuilder();
int t = fs.nextInt();
for (int i = 0; i < t; i++) {
int n = fs.nextInt();
int[] arr = fs.readArray(n);
int[] dp = new int[n];
for (int j = 1; j < n - 1; j++) {
if (arr[j - 1] < arr[j] && arr[j] > arr[j + 1]) {
dp[j] = 1;
}
}
int count = 0;
for (int j = 1; j < n - 1; j++) {
if (dp[j - 1] == 1 && dp[j] == 0 && dp[j + 1] == 1) {
count++;
dp[j + 1] = 0;
dp[j - 1] = 0;
arr[j] = Math.max(arr[j - 1], arr[j + 1]);
}
}
for (int j = 1; j < n - 1; j++) {
if (dp[j - 1] == 0 && dp[j] == 1 && dp[j + 1] == 0) {
count++;
arr[j] = Math.max(arr[j - 1], arr[j + 1]);
dp[j] = 0;
}
}
sb.append(count).append("\n");
for (int v : arr) {
sb.append(v).append(" ");
}
sb.append("\n");
}
System.out.println(sb);
}
private static long getGcd(long a, long b) {
if (b == 0) return a;
return getGcd(b, a % b);
}
private static long binaryExponentiation(long a, long b) {
long res = 1;
while (b > 0) {
if ((b & 1) == 1) res = res * a;
a = a * a;
b >>= 1;
}
return res;
}
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;
}
int[] readArray(int n) {
int[] arr = new int[n];
for (int i = 0; i < n; i++) arr[i] = Integer.parseInt(next());
return arr;
}
}
}
| Java | ["5\n3\n2 1 2\n4\n1 2 3 1\n5\n1 2 1 2 1\n9\n1 2 1 3 2 3 1 2 1\n9\n2 1 3 1 3 1 3 1 3"] | 2 seconds | ["0\n2 1 2\n1\n1 3 3 1\n1\n1 2 2 2 1\n2\n1 2 3 3 2 3 3 2 1\n2\n2 1 3 3 3 1 1 1 3"] | NoteIn the first example, the array contains no local maximum, so we don't need to perform operations.In the second example, we can change $$$a_2$$$ to $$$3$$$, then the array don't have local maximums. | Java 11 | standard input | [
"greedy"
] | 255abf8750541f54d8ff0f1f80e1f8e7 | Each test contains multiple test cases. The first line will contain a single integer $$$t$$$ $$$(1 \leq t \leq 10000)$$$ — the number of test cases. Then $$$t$$$ test cases follow. The first line of each test case contains a single integer $$$n$$$ $$$(2 \leq n \leq 2 \cdot 10^5)$$$ — the size of the array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots ,a_n$$$ $$$(1 \leq a_i \leq 10^9)$$$, the elements of array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, first output a line containing a single integer $$$m$$$ — minimum number of operations required. Then ouput a line consist of $$$n$$$ integers — the resulting array after the operations. Note that this array should differ in exactly $$$m$$$ elements from the initial array. If there are multiple answers, print any. | standard output | |
PASSED | df1fbf57d013e9804b3f271590819f7b | train_108.jsonl | 1645367700 | You are given an array $$$a$$$ of size $$$n$$$. Each element in this array is an integer between $$$1$$$ and $$$10^9$$$.You can perform several operations to this array. During an operation, you can replace an element in the array with any integer between $$$1$$$ and $$$10^9$$$. Output the minimum number of operations needed such that the resulting array doesn't contain any local maximums, and the resulting array after the operations.An element $$$a_i$$$ is a local maximum if it is strictly larger than both of its neighbors (that is, $$$a_i > a_{i - 1}$$$ and $$$a_i > a_{i + 1}$$$). Since $$$a_1$$$ and $$$a_n$$$ have only one neighbor each, they will never be a local maximum. | 256 megabytes | import java.util.*;
import java.lang.*;
import java.io.*;
public class Main
{
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();
while(t-->0){
solve(sc);
}
}
public static void solve(FastReader sc) throws IOException{
int n = sc.nextInt();
int [] arr = new int[n];
for(int i = 0;i<n;++i){
arr[i] = sc.nextInt();
}
if(n<3){
out.println(0);
out.println(arr[0] + " " + arr[1]);out.flush();return;
}
boolean[] maxi = new boolean[n];
for(int i = 1;i<n-1;++i){
if(arr[i-1]<arr[i]&&arr[i+1]<arr[i]){
maxi[i] = true;
}
}
long opn = 0;
for(int i = 1;i<n-1;++i){
if(maxi[i]){
opn++;
if(i+3<n&&maxi[i+2]){
arr[i+1] = Math.max(arr[i], arr[i+2]);
i+=2;
}else{
arr[i] = Math.max(arr[i-1], arr[i+1]);
}
}
}
out.println(opn);
for(int i = 0;i<n;++i){
out.print(arr[i] + " ");
}
out.println(); out.flush();
}
/*
int [] arr = new int[n];
for(int i = 0;i<n;++i){
arr[i] = sc.nextInt();
}
*/
static void mysort(int[] arr) {
for(int i=0;i<arr.length;i++) {
int rand = (int) (Math.random() * arr.length);
int loc = arr[rand];
arr[rand] = arr[i];
arr[i] = loc;
}
Arrays.sort(arr);
}
static class Pair implements Comparable<Pair>{
int ind;int val;
Pair(int ind, int val){
this.ind=ind;
this.val=val;
}
public int compareTo(Pair o){
return this.val-o.val;
}
}
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 | ["5\n3\n2 1 2\n4\n1 2 3 1\n5\n1 2 1 2 1\n9\n1 2 1 3 2 3 1 2 1\n9\n2 1 3 1 3 1 3 1 3"] | 2 seconds | ["0\n2 1 2\n1\n1 3 3 1\n1\n1 2 2 2 1\n2\n1 2 3 3 2 3 3 2 1\n2\n2 1 3 3 3 1 1 1 3"] | NoteIn the first example, the array contains no local maximum, so we don't need to perform operations.In the second example, we can change $$$a_2$$$ to $$$3$$$, then the array don't have local maximums. | Java 11 | standard input | [
"greedy"
] | 255abf8750541f54d8ff0f1f80e1f8e7 | Each test contains multiple test cases. The first line will contain a single integer $$$t$$$ $$$(1 \leq t \leq 10000)$$$ — the number of test cases. Then $$$t$$$ test cases follow. The first line of each test case contains a single integer $$$n$$$ $$$(2 \leq n \leq 2 \cdot 10^5)$$$ — the size of the array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots ,a_n$$$ $$$(1 \leq a_i \leq 10^9)$$$, the elements of array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, first output a line containing a single integer $$$m$$$ — minimum number of operations required. Then ouput a line consist of $$$n$$$ integers — the resulting array after the operations. Note that this array should differ in exactly $$$m$$$ elements from the initial array. If there are multiple answers, print any. | standard output | |
PASSED | 5522980e7a92a15820c6e3ed04d5febc | train_108.jsonl | 1645367700 | You are given an array $$$a$$$ of size $$$n$$$. Each element in this array is an integer between $$$1$$$ and $$$10^9$$$.You can perform several operations to this array. During an operation, you can replace an element in the array with any integer between $$$1$$$ and $$$10^9$$$. Output the minimum number of operations needed such that the resulting array doesn't contain any local maximums, and the resulting array after the operations.An element $$$a_i$$$ is a local maximum if it is strictly larger than both of its neighbors (that is, $$$a_i > a_{i - 1}$$$ and $$$a_i > a_{i + 1}$$$). Since $$$a_1$$$ and $$$a_n$$$ have only one neighbor each, they will never be a local maximum. | 256 megabytes |
import java.io.*;
import java.util.*;
import java.lang.*;
import static java.lang.Math.*;
public class Codeforces {
static int MAX = Integer.MAX_VALUE,MIN = Integer.MIN_VALUE,MOD = (int)1e9+7;
static long MAXL = Long.MAX_VALUE,MINL = Long.MIN_VALUE;
static void solve() {
int n = fs.nInt();
int[]ar = new int[n];
for(int i=0;i<n;i++)ar[i] = fs.nInt();
int[]cnt = new int[n];
for(int i=1;i<n-1;i++){
if(ar[i] > ar[i-1] && ar[i] > ar[i+1]){
cnt[i]++;
}
}
int ans = 0;
for(int i=1;i<n-2;i++){
if( cnt[i] > 0 && cnt[i+2] > 0 ){
ans++;
ar[i+1] = max(ar[i],ar[i+2]);
cnt[i]=0;
cnt[i+2]=0;
}
}
for(int i=1;i<n-1;i++){
if( ar[i] > ar[i-1] && ar[i] > ar[i+1] ){
ans++;
ar[i]= max(ar[i-1],ar[i+1]);
cnt[i]=0;
}
}
out.println(ans);
for(int i:ar)
out.print(i+" ");
out.println();
}
static class PairI implements Comparable<PairI>{
int f,s,t;
PairI(int f,int s,int t){
this.f = f;
this.s = s;
this.t = t;
}
@Override
public int compareTo(PairI o) {
if(this.s == o.s ){
if(this.f == o.f)
return this.t-o.t;
return this.f - o.f;
}
return this.s - o.s;
}
}
static class Pair implements Comparable<Pair>{
int f,s;
Pair(int f,int s) {
this.f = f;
this.s = s;
}
@Override
public int compareTo(Pair o) {
if(o.f == this.f )
return Integer.compare(this.s,o.s);
return Integer.compare(this.f,o.f);
}
}
static boolean multipleTestCase = true ;static FastScanner fs;static PrintWriter out;
public static void main(String[]args){
try{
out = new PrintWriter(System.out);
fs = new FastScanner();
int tc = multipleTestCase?fs.nInt():1;
while (tc-->0)solve();
out.flush();
out.close();
}catch (Exception e){
e.printStackTrace();
}
}
static class FastScanner extends PrintWriter {
private InputStream stream;
private byte[] buf = new byte[1<<16];
private int curChar, numChars;
public FastScanner() { this(System.in,System.out); }
public FastScanner(InputStream i, OutputStream o) {
super(o);
stream = i;
}
// file input
public FastScanner(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 nInt() { // 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 nLong(){return Long.parseLong(next());}
public double nextDouble() { return Double.parseDouble(next()); }
}
public static void sort(int[] arr){
ArrayList<Integer> ls = new ArrayList<>();
for(int x: arr)
ls.add(x);
Collections.sort(ls);
for(int i=0; i < arr.length; i++)
arr[i] = ls.get(i);
}
public static void sortR(int[] arr){
ArrayList<Integer> ls = new ArrayList<>();
for(int x: arr)
ls.add(x);
Collections.sort(ls,Collections.reverseOrder());
for(int i=0; i < arr.length; i++)
arr[i] = ls.get(i);
}
public static void sort(long[] arr){
ArrayList<Long> ls = new ArrayList<>();
for(long x: arr)
ls.add(x);
Collections.sort(ls);
for(int i=0; i < arr.length; i++)
arr[i] = ls.get(i);
}
public static void sortR(long[] arr){
ArrayList<Long> ls = new ArrayList<>();
for(long x: arr)
ls.add(x);
Collections.sort(ls,Collections.reverseOrder());
for(int i=0; i < arr.length; i++)
arr[i] = ls.get(i);
}
} | Java | ["5\n3\n2 1 2\n4\n1 2 3 1\n5\n1 2 1 2 1\n9\n1 2 1 3 2 3 1 2 1\n9\n2 1 3 1 3 1 3 1 3"] | 2 seconds | ["0\n2 1 2\n1\n1 3 3 1\n1\n1 2 2 2 1\n2\n1 2 3 3 2 3 3 2 1\n2\n2 1 3 3 3 1 1 1 3"] | NoteIn the first example, the array contains no local maximum, so we don't need to perform operations.In the second example, we can change $$$a_2$$$ to $$$3$$$, then the array don't have local maximums. | Java 11 | standard input | [
"greedy"
] | 255abf8750541f54d8ff0f1f80e1f8e7 | Each test contains multiple test cases. The first line will contain a single integer $$$t$$$ $$$(1 \leq t \leq 10000)$$$ — the number of test cases. Then $$$t$$$ test cases follow. The first line of each test case contains a single integer $$$n$$$ $$$(2 \leq n \leq 2 \cdot 10^5)$$$ — the size of the array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots ,a_n$$$ $$$(1 \leq a_i \leq 10^9)$$$, the elements of array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, first output a line containing a single integer $$$m$$$ — minimum number of operations required. Then ouput a line consist of $$$n$$$ integers — the resulting array after the operations. Note that this array should differ in exactly $$$m$$$ elements from the initial array. If there are multiple answers, print any. | standard output | |
PASSED | b7152d305a611583c44a0403f3fc69c7 | train_108.jsonl | 1645367700 | You are given an array $$$a$$$ of size $$$n$$$. Each element in this array is an integer between $$$1$$$ and $$$10^9$$$.You can perform several operations to this array. During an operation, you can replace an element in the array with any integer between $$$1$$$ and $$$10^9$$$. Output the minimum number of operations needed such that the resulting array doesn't contain any local maximums, and the resulting array after the operations.An element $$$a_i$$$ is a local maximum if it is strictly larger than both of its neighbors (that is, $$$a_i > a_{i - 1}$$$ and $$$a_i > a_{i + 1}$$$). Since $$$a_1$$$ and $$$a_n$$$ have only one neighbor each, they will never be a local maximum. | 256 megabytes | import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.StringTokenizer;
import java.io.PrintWriter;
public class B {
static final FastReader sc = new FastReader();
static final PrintWriter out = new PrintWriter(System.out);
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 = 1; i < n - 1; i++) {
if (a[i] > a[i + 1] && a[i] > a[i - 1]) {
if (i < n - 2) {
a[i + 1] = Math.max(a[i], a[i + 2]);
} else {
a[i + 1] = a[i];
}
c++;
}
}
out.println(c);
for (int i : a) {
out.print(i + " ");
}
out.println();
}
out.close();
}
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 | ["5\n3\n2 1 2\n4\n1 2 3 1\n5\n1 2 1 2 1\n9\n1 2 1 3 2 3 1 2 1\n9\n2 1 3 1 3 1 3 1 3"] | 2 seconds | ["0\n2 1 2\n1\n1 3 3 1\n1\n1 2 2 2 1\n2\n1 2 3 3 2 3 3 2 1\n2\n2 1 3 3 3 1 1 1 3"] | NoteIn the first example, the array contains no local maximum, so we don't need to perform operations.In the second example, we can change $$$a_2$$$ to $$$3$$$, then the array don't have local maximums. | Java 11 | standard input | [
"greedy"
] | 255abf8750541f54d8ff0f1f80e1f8e7 | Each test contains multiple test cases. The first line will contain a single integer $$$t$$$ $$$(1 \leq t \leq 10000)$$$ — the number of test cases. Then $$$t$$$ test cases follow. The first line of each test case contains a single integer $$$n$$$ $$$(2 \leq n \leq 2 \cdot 10^5)$$$ — the size of the array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots ,a_n$$$ $$$(1 \leq a_i \leq 10^9)$$$, the elements of array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, first output a line containing a single integer $$$m$$$ — minimum number of operations required. Then ouput a line consist of $$$n$$$ integers — the resulting array after the operations. Note that this array should differ in exactly $$$m$$$ elements from the initial array. If there are multiple answers, print any. | standard output | |
PASSED | 4ee76ba3f176eb1d045b9e023994f367 | train_108.jsonl | 1645367700 | You are given an array $$$a$$$ of size $$$n$$$. Each element in this array is an integer between $$$1$$$ and $$$10^9$$$.You can perform several operations to this array. During an operation, you can replace an element in the array with any integer between $$$1$$$ and $$$10^9$$$. Output the minimum number of operations needed such that the resulting array doesn't contain any local maximums, and the resulting array after the operations.An element $$$a_i$$$ is a local maximum if it is strictly larger than both of its neighbors (that is, $$$a_i > a_{i - 1}$$$ and $$$a_i > a_{i + 1}$$$). Since $$$a_1$$$ and $$$a_n$$$ have only one neighbor each, they will never be a local maximum. | 256 megabytes | import java.util.Scanner;
public class LineUp {
public static void main(String[] args) {
Scanner scn = new Scanner(System.in);
int t = scn.nextInt();
while(t-->0)
{
int n = scn.nextInt();
int[] array = new int[n];
for(int i = 0 ; i< n ; i++) {
array[i] = scn.nextInt();
}
int changes = 0;
for(int i = 1 ; i<n-1 ; i++)
{
if(array[i]>array[i-1] && array[i]>array[i+1]) {
if(i+2<n)
array[i+1] = Math.max(array[i],array[i+2]);
else
array[i+1]= array[i];
changes++;
}
}
System.out.println(changes);
for(int i = 0 ; i<n ; i++)
System.out.print(array[i]+" ");
System.out.println();
}
}
}
| Java | ["5\n3\n2 1 2\n4\n1 2 3 1\n5\n1 2 1 2 1\n9\n1 2 1 3 2 3 1 2 1\n9\n2 1 3 1 3 1 3 1 3"] | 2 seconds | ["0\n2 1 2\n1\n1 3 3 1\n1\n1 2 2 2 1\n2\n1 2 3 3 2 3 3 2 1\n2\n2 1 3 3 3 1 1 1 3"] | NoteIn the first example, the array contains no local maximum, so we don't need to perform operations.In the second example, we can change $$$a_2$$$ to $$$3$$$, then the array don't have local maximums. | Java 11 | standard input | [
"greedy"
] | 255abf8750541f54d8ff0f1f80e1f8e7 | Each test contains multiple test cases. The first line will contain a single integer $$$t$$$ $$$(1 \leq t \leq 10000)$$$ — the number of test cases. Then $$$t$$$ test cases follow. The first line of each test case contains a single integer $$$n$$$ $$$(2 \leq n \leq 2 \cdot 10^5)$$$ — the size of the array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots ,a_n$$$ $$$(1 \leq a_i \leq 10^9)$$$, the elements of array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, first output a line containing a single integer $$$m$$$ — minimum number of operations required. Then ouput a line consist of $$$n$$$ integers — the resulting array after the operations. Note that this array should differ in exactly $$$m$$$ elements from the initial array. If there are multiple answers, print any. | standard output | |
PASSED | cc57d9a8d3b91faa6ade91eb3890d202 | train_108.jsonl | 1645367700 | You are given an array $$$a$$$ of size $$$n$$$. Each element in this array is an integer between $$$1$$$ and $$$10^9$$$.You can perform several operations to this array. During an operation, you can replace an element in the array with any integer between $$$1$$$ and $$$10^9$$$. Output the minimum number of operations needed such that the resulting array doesn't contain any local maximums, and the resulting array after the operations.An element $$$a_i$$$ is a local maximum if it is strictly larger than both of its neighbors (that is, $$$a_i > a_{i - 1}$$$ and $$$a_i > a_{i + 1}$$$). Since $$$a_1$$$ and $$$a_n$$$ have only one neighbor each, they will never be a local maximum. | 256 megabytes | import java.io.*;
import java.util.*;
public class Program{
static BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
static PrintWriter pw = new PrintWriter(System.out);
static StringTokenizer st = new StringTokenizer("");
static int T;
public static void solve() throws IOException {
int n = Integer.parseInt(next());
int[] a = new int[n];
for (int i = 0; i < n; i++) {
int val = Integer.parseInt(next());
a[i] = val;
}
int changed = 0;
ArrayList<Integer> peaks = new ArrayList<Integer>();
for (int i = 1; i < n - 1; i++) {
if (a[i] > a[i - 1] && a[i] > a[i + 1]) {
peaks.add(i);
}
}
for (int i = 0; i < peaks.size() - 1; i++) {
if (Math.abs(peaks.get(i) - peaks.get(i + 1)) == 2) {
int changedVal = Math.max(a[peaks.get(i)], a[peaks.get(i + 1)]);
a[peaks.get(i) + 1] = changedVal;
peaks.remove(i);
peaks.remove(i);
i -= 1;
changed++;
}
}
for (int i = 0; i < peaks.size(); i++) {
int changedVal = Math.max(a[peaks.get(i) - 1], a[peaks.get(i) + 1]);
a[peaks.get(i)] = changedVal;
peaks.remove(i);
changed++;
i--;
}
pw.println(changed);
for (int i = 0; i < n; i++) {
pw.print(a[i] + " ");
}
}
public static void main(String args[]) throws IOException{
T = Integer.parseInt(next());
while(T-- > 0){
solve();
pw.println();
}
br.close(); pw.close();
}
static String next() throws IOException{
while(!st.hasMoreTokens()) st = new StringTokenizer(br.readLine());
return st.nextToken();
}
}
| Java | ["5\n3\n2 1 2\n4\n1 2 3 1\n5\n1 2 1 2 1\n9\n1 2 1 3 2 3 1 2 1\n9\n2 1 3 1 3 1 3 1 3"] | 2 seconds | ["0\n2 1 2\n1\n1 3 3 1\n1\n1 2 2 2 1\n2\n1 2 3 3 2 3 3 2 1\n2\n2 1 3 3 3 1 1 1 3"] | NoteIn the first example, the array contains no local maximum, so we don't need to perform operations.In the second example, we can change $$$a_2$$$ to $$$3$$$, then the array don't have local maximums. | Java 11 | standard input | [
"greedy"
] | 255abf8750541f54d8ff0f1f80e1f8e7 | Each test contains multiple test cases. The first line will contain a single integer $$$t$$$ $$$(1 \leq t \leq 10000)$$$ — the number of test cases. Then $$$t$$$ test cases follow. The first line of each test case contains a single integer $$$n$$$ $$$(2 \leq n \leq 2 \cdot 10^5)$$$ — the size of the array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots ,a_n$$$ $$$(1 \leq a_i \leq 10^9)$$$, the elements of array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, first output a line containing a single integer $$$m$$$ — minimum number of operations required. Then ouput a line consist of $$$n$$$ integers — the resulting array after the operations. Note that this array should differ in exactly $$$m$$$ elements from the initial array. If there are multiple answers, print any. | standard output | |
PASSED | 0a4a63210f9ed7842af5a916241443fa | train_108.jsonl | 1645367700 | You are given an array $$$a$$$ of size $$$n$$$. Each element in this array is an integer between $$$1$$$ and $$$10^9$$$.You can perform several operations to this array. During an operation, you can replace an element in the array with any integer between $$$1$$$ and $$$10^9$$$. Output the minimum number of operations needed such that the resulting array doesn't contain any local maximums, and the resulting array after the operations.An element $$$a_i$$$ is a local maximum if it is strictly larger than both of its neighbors (that is, $$$a_i > a_{i - 1}$$$ and $$$a_i > a_{i + 1}$$$). Since $$$a_1$$$ and $$$a_n$$$ have only one neighbor each, they will never be a local maximum. | 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);
if(n<3)
{
System.out.println("0");
System.out.println(a[0]+" "+a[1]);
continue;
}
int count=0;
for(int i=1;i<n-2;i++)
{
if(a[i-1]<a[i] && a[i]>a[i+1])
{
count++;
a[i+1]=Math.max(a[i],a[i+2]);
}
}
if(a[n-2]>a[n-3] && a[n-2]>a[n-1])
{
count++;
a[n-2]=Math.max(a[n-1],a[n-3]);
}
System.out.println(count);
for(int i=0;i<n;i++)
{
System.out.print(a[i]+" ");
}
System.out.println();
}
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 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 | ["5\n3\n2 1 2\n4\n1 2 3 1\n5\n1 2 1 2 1\n9\n1 2 1 3 2 3 1 2 1\n9\n2 1 3 1 3 1 3 1 3"] | 2 seconds | ["0\n2 1 2\n1\n1 3 3 1\n1\n1 2 2 2 1\n2\n1 2 3 3 2 3 3 2 1\n2\n2 1 3 3 3 1 1 1 3"] | NoteIn the first example, the array contains no local maximum, so we don't need to perform operations.In the second example, we can change $$$a_2$$$ to $$$3$$$, then the array don't have local maximums. | Java 11 | standard input | [
"greedy"
] | 255abf8750541f54d8ff0f1f80e1f8e7 | Each test contains multiple test cases. The first line will contain a single integer $$$t$$$ $$$(1 \leq t \leq 10000)$$$ — the number of test cases. Then $$$t$$$ test cases follow. The first line of each test case contains a single integer $$$n$$$ $$$(2 \leq n \leq 2 \cdot 10^5)$$$ — the size of the array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots ,a_n$$$ $$$(1 \leq a_i \leq 10^9)$$$, the elements of array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, first output a line containing a single integer $$$m$$$ — minimum number of operations required. Then ouput a line consist of $$$n$$$ integers — the resulting array after the operations. Note that this array should differ in exactly $$$m$$$ elements from the initial array. If there are multiple answers, print any. | standard output | |
PASSED | ac45890b981f04a168201a5f837c8bcc | train_108.jsonl | 1645367700 | You are given an array $$$a$$$ of size $$$n$$$. Each element in this array is an integer between $$$1$$$ and $$$10^9$$$.You can perform several operations to this array. During an operation, you can replace an element in the array with any integer between $$$1$$$ and $$$10^9$$$. Output the minimum number of operations needed such that the resulting array doesn't contain any local maximums, and the resulting array after the operations.An element $$$a_i$$$ is a local maximum if it is strictly larger than both of its neighbors (that is, $$$a_i > a_{i - 1}$$$ and $$$a_i > a_{i + 1}$$$). Since $$$a_1$$$ and $$$a_n$$$ have only one neighbor each, they will never be a local maximum. | 256 megabytes | import java.util.*;
public class AvoidLocal{
public static void main(String[] args){
Scanner input=new Scanner(System.in);
int times=input.nextInt();
for(int j=0;j<times;j++){
int n=input.nextInt();
int m=0;
int nums[]=new int[n];
for(int i=0;i<n;i++){
nums[i]=input.nextInt();
}
for(int i=1;i<=n-2;i++){
if(nums[i-1]<nums[i]&&nums[i]>nums[i+1]){
m++;
if(i+2<=n-1&&nums[i+2]>nums[i+1]){
if(nums[i+2]>=nums[i]){
nums[i+1]=nums[i+2];
}else{
nums[i+1]=nums[i];
}
}else{
nums[i+1]=nums[i];
}
}
}
System.out.println(m);
for(int i=0;i<n;i++){
System.out.println(nums[i]);
}
}
input.close();
}
}
| Java | ["5\n3\n2 1 2\n4\n1 2 3 1\n5\n1 2 1 2 1\n9\n1 2 1 3 2 3 1 2 1\n9\n2 1 3 1 3 1 3 1 3"] | 2 seconds | ["0\n2 1 2\n1\n1 3 3 1\n1\n1 2 2 2 1\n2\n1 2 3 3 2 3 3 2 1\n2\n2 1 3 3 3 1 1 1 3"] | NoteIn the first example, the array contains no local maximum, so we don't need to perform operations.In the second example, we can change $$$a_2$$$ to $$$3$$$, then the array don't have local maximums. | Java 11 | standard input | [
"greedy"
] | 255abf8750541f54d8ff0f1f80e1f8e7 | Each test contains multiple test cases. The first line will contain a single integer $$$t$$$ $$$(1 \leq t \leq 10000)$$$ — the number of test cases. Then $$$t$$$ test cases follow. The first line of each test case contains a single integer $$$n$$$ $$$(2 \leq n \leq 2 \cdot 10^5)$$$ — the size of the array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots ,a_n$$$ $$$(1 \leq a_i \leq 10^9)$$$, the elements of array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, first output a line containing a single integer $$$m$$$ — minimum number of operations required. Then ouput a line consist of $$$n$$$ integers — the resulting array after the operations. Note that this array should differ in exactly $$$m$$$ elements from the initial array. If there are multiple answers, print any. | standard output | |
PASSED | 8003e11ab63ae9b35202a0e4c4c94d90 | train_108.jsonl | 1645367700 | You are given an array $$$a$$$ of size $$$n$$$. Each element in this array is an integer between $$$1$$$ and $$$10^9$$$.You can perform several operations to this array. During an operation, you can replace an element in the array with any integer between $$$1$$$ and $$$10^9$$$. Output the minimum number of operations needed such that the resulting array doesn't contain any local maximums, and the resulting array after the operations.An element $$$a_i$$$ is a local maximum if it is strictly larger than both of its neighbors (that is, $$$a_i > a_{i - 1}$$$ and $$$a_i > a_{i + 1}$$$). Since $$$a_1$$$ and $$$a_n$$$ have only one neighbor each, they will never be a local maximum. | 256 megabytes | import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.util.*;
public class Main{
public static void main(String[] args) throws Exception{
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
int testCases = Integer.parseInt(br.readLine());
StringBuilder result = new StringBuilder();
StringTokenizer str;
for(int i = 1; i <= testCases; i++){
int n = Integer.parseInt(br.readLine());
str = new StringTokenizer (br.readLine());
long num[] = new long[n];
int min = 0;
for(int j = 0; j < n; j++){
num[j] = Long.parseLong(str.nextToken());
}
long smallest[] = new long[n];
smallest[0] = 0;
for(int j = 1; j < n; j++){
smallest[j] = num[j] - num[j - 1];
}
smallest[n - 1] = 0;
for(int j = n - 2; j > 0; j--){
smallest[j] = Math.min(smallest[j], num[j] - num[j + 1]);
}
int count = 0;
for(int j = 1; j < n - 1; ){
if(smallest[j] > 0){
if((j + 2) < n && smallest[j + 2] > 0){
num[j + 1] = Math.max(num[j],num[j + 2]);
j +=2;
}else{
num[j] = Math.max(num[j - 1], num[j + 1]);
}
count++;
}
j++;
}
result.append(count + "\n");
for(long number : num){
result.append(number + " ");
}
result.append("\n");
}
System.out.println(result);
}
} | Java | ["5\n3\n2 1 2\n4\n1 2 3 1\n5\n1 2 1 2 1\n9\n1 2 1 3 2 3 1 2 1\n9\n2 1 3 1 3 1 3 1 3"] | 2 seconds | ["0\n2 1 2\n1\n1 3 3 1\n1\n1 2 2 2 1\n2\n1 2 3 3 2 3 3 2 1\n2\n2 1 3 3 3 1 1 1 3"] | NoteIn the first example, the array contains no local maximum, so we don't need to perform operations.In the second example, we can change $$$a_2$$$ to $$$3$$$, then the array don't have local maximums. | Java 11 | standard input | [
"greedy"
] | 255abf8750541f54d8ff0f1f80e1f8e7 | Each test contains multiple test cases. The first line will contain a single integer $$$t$$$ $$$(1 \leq t \leq 10000)$$$ — the number of test cases. Then $$$t$$$ test cases follow. The first line of each test case contains a single integer $$$n$$$ $$$(2 \leq n \leq 2 \cdot 10^5)$$$ — the size of the array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots ,a_n$$$ $$$(1 \leq a_i \leq 10^9)$$$, the elements of array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, first output a line containing a single integer $$$m$$$ — minimum number of operations required. Then ouput a line consist of $$$n$$$ integers — the resulting array after the operations. Note that this array should differ in exactly $$$m$$$ elements from the initial array. If there are multiple answers, print any. | standard output | |
PASSED | 360824686fbc151a50e5e17e0ce1a8ac | train_108.jsonl | 1645367700 | You are given an array $$$a$$$ of size $$$n$$$. Each element in this array is an integer between $$$1$$$ and $$$10^9$$$.You can perform several operations to this array. During an operation, you can replace an element in the array with any integer between $$$1$$$ and $$$10^9$$$. Output the minimum number of operations needed such that the resulting array doesn't contain any local maximums, and the resulting array after the operations.An element $$$a_i$$$ is a local maximum if it is strictly larger than both of its neighbors (that is, $$$a_i > a_{i - 1}$$$ and $$$a_i > a_{i + 1}$$$). Since $$$a_1$$$ and $$$a_n$$$ have only one neighbor each, they will never be a local maximum. | 256 megabytes | import java.util.Scanner;
public class Problems2 {
public static int max1(int[] arr,int len){
int max=arr[0];
for(int i=0;i<len;i++){
if(max<arr[i]){
max=arr[i];
}
}
return max;
}
public static int minimum(int[] arr,int len){
int max=arr[0];
for(int i=0;i>len;i++){
if(max>arr[i]){
max=arr[i];
}
}
return max;
}
public static int index(int[] arr,int len){
int max=arr[0];
int inid=0;
for(int i=1;i<len;i++){
if(max<arr[i]){
max=arr[i];
inid=i;
}
}
return inid;
}
public static void main(String[] args) {
Scanner s=new Scanner(System.in);
int t=s.nextInt();
for(int i=0;i<t;i++){
int len=s.nextInt();
int[] arr=new int[len];
for(int j=0;j<len;j++){
arr[j]=s.nextInt();
}
int count=0;
int maxi1=max1(arr,len);//maximum
for(int k=1;k<(len-1);k++){
if(arr[k]>arr[k-1] && arr[k]>arr[k+1]){
count++;
if(k!=(len-2)){
if(arr[k+2]>arr[k]){
arr[k+1]=arr[k+2];
}
else{
arr[k+1]=arr[k];
}
}
else{
arr[k+1]=arr[k];
}
}
}
System.out.println(count);
for(int j=0;j<len;j++){
System.out.print(arr[j]+" ");
}
System.out.println();
}
}
}
| Java | ["5\n3\n2 1 2\n4\n1 2 3 1\n5\n1 2 1 2 1\n9\n1 2 1 3 2 3 1 2 1\n9\n2 1 3 1 3 1 3 1 3"] | 2 seconds | ["0\n2 1 2\n1\n1 3 3 1\n1\n1 2 2 2 1\n2\n1 2 3 3 2 3 3 2 1\n2\n2 1 3 3 3 1 1 1 3"] | NoteIn the first example, the array contains no local maximum, so we don't need to perform operations.In the second example, we can change $$$a_2$$$ to $$$3$$$, then the array don't have local maximums. | Java 11 | standard input | [
"greedy"
] | 255abf8750541f54d8ff0f1f80e1f8e7 | Each test contains multiple test cases. The first line will contain a single integer $$$t$$$ $$$(1 \leq t \leq 10000)$$$ — the number of test cases. Then $$$t$$$ test cases follow. The first line of each test case contains a single integer $$$n$$$ $$$(2 \leq n \leq 2 \cdot 10^5)$$$ — the size of the array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots ,a_n$$$ $$$(1 \leq a_i \leq 10^9)$$$, the elements of array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, first output a line containing a single integer $$$m$$$ — minimum number of operations required. Then ouput a line consist of $$$n$$$ integers — the resulting array after the operations. Note that this array should differ in exactly $$$m$$$ elements from the initial array. If there are multiple answers, print any. | standard output | |
PASSED | da3c3a328928ec089a05c619bb89d6ab | train_108.jsonl | 1645367700 | You are given an array $$$a$$$ of size $$$n$$$. Each element in this array is an integer between $$$1$$$ and $$$10^9$$$.You can perform several operations to this array. During an operation, you can replace an element in the array with any integer between $$$1$$$ and $$$10^9$$$. Output the minimum number of operations needed such that the resulting array doesn't contain any local maximums, and the resulting array after the operations.An element $$$a_i$$$ is a local maximum if it is strictly larger than both of its neighbors (that is, $$$a_i > a_{i - 1}$$$ and $$$a_i > a_{i + 1}$$$). Since $$$a_1$$$ and $$$a_n$$$ have only one neighbor each, they will never be a local maximum. | 256 megabytes |
import java.util.*;
import java.math.*;
import java.io.*;
public class Solution {
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 sort(int a[]){ // int -> long
ArrayList<Integer> arr=new ArrayList<>(); // Integer -> Long
for(int i=0;i<a.length;i++)
arr.add(a[i]);
Collections.sort(arr);
for(int i=0;i<a.length;i++)
a[i]=arr.get(i);
}
private static long gcd(long a, long b){
if(b==0)return a;
return gcd(b,a%b);
}
private static long pow(long x,long y){
if(y==0)return 1;
long temp = pow(x, y/2);
if(y%2==1){
return x*temp*temp;
}
else{
return temp*temp;
}
}
static long powM(long a, long b) {
if (b == 0) return 1;
long res = pow(a, b / 2);
res = (res%mod * res%mod) % 1_000_000_007;
if (b % 2 == 1) {
res = (res%mod * a%mod) % 1_000_000_007;
}
return res%mod;
}
static int log(long n){
int res = 0;
while(n>0){
res++;
n/=2;
}
return res;
}
static int mod = (int)1e9+7;
static PrintWriter out;
static FastReader sc ;
public static void main(String[] args) throws IOException {
sc = new FastReader();
out = new PrintWriter(System.out);
// primes();
// ________________________________
// int test = 1;
StringBuilder output = new StringBuilder();
int test =sc.nextInt();
while (test-- > 0) {
//System.out.println(mdh+" "+nhc);
int n = sc.nextInt();
int a [] = new int[n];
for(int i =0;i<n;i++)
{
a[i] = sc.nextInt();
;
}
solver(n,a);
}
// solver(s);
// int n = sc.nextInt();
// out.println(solver());
// ________________________________
out.flush();
}
static class Edge
{
int u, v;
Edge(int u, int v)
{
this.u = u;
this.v = v;
}
}
static class Pair implements Comparable <Pair>{
int l, r;
Pair(int l, int r)
{
this.l = l;
this.r = r;
}
public int compareTo(Pair o)
{
return this.l-o.l;
}
}
static ArrayList<Integer>adj [] ;
static boolean [] vst ;
public static void solver( int n, int a []) {
int temp [] = new int[n];
int cnt =0;
int max =0;
if(n<=2)
{
System.out.println(cnt);
for(int i : a)
System.out.print(i+" ");
return;
}
for(int i =1;i<n-1;i++)
{
if(a[i]>a[i-1]&&a[i]>a[i+1])
{
int p1 = a[i];
int p2 = Integer.MIN_VALUE;
if(i+2<n)
{
p2 = a[i+2];
}
a[i+1] =Math.max(p1, p2);
cnt++;
}
}
System.out.println(cnt);
for(int i : a)
System.out.print(i+" ");
System.out.println();
}
static void dfs(int cur) {
if(vst[cur]) return ;
vst[cur]=true;
for(int c:adj[cur]) dfs(c);
}
public static int Mex(int []a, int i, int j)
{
boolean [] vis = new boolean[a.length+2];
int mex =0;
for(int k=i;k<j;k++)
{
if(a[k]<vis.length)
{
vis[a[k]] = true;
}
}
while(vis[mex]==true)
{
mex++;
}
return mex;
}
public static long log2(long N)
{
// calculate log2 N indirectly
// using log() method
long result = (long)(Math.log(N) / Math.log(2));
return result;
}
static long highestPowerof2(long 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);
}
public
static void rotate(int [] arr, int s, int n)
{
int x = arr[n], i;
for (i = n; i > s; i--)
arr[i] = arr[i-1];
arr[s] = x;
// for(int j=s;j<=n;j++)
// System.out.print(arr[j]+" ");
// System.out.println();
}
static int lower_bound(int[] a , long x)
{
int i = 0;
int j = a.length-1;
//if(arr[i] > key)return -1;
if(a[j] < x)return a.length;
while(i<j)
{
int mid = (i+j)/2;
if(a[mid] == x)
{
j = mid;
}
else if(a[mid] < x)
{
i = mid+1;
}
else
j = mid-1;
}
return i;
}
int upper_bound(int [] arr , int key)
{
int i = 0;
int j = arr.length-1;
if(arr[j] <= key)return j+1;
while(i<j)
{
int mid = (i+j)/2;
if(arr[mid] <= key)
{
i = mid+1;
}
else
j = mid;
}
return i;
}
static void reverseArray(int arr[], int start,
int end)
{
while (start < end)
{
int temp = arr[start];
arr[start] = arr[end];
arr[end] = temp;
start++;
end--;
}
}
} | Java | ["5\n3\n2 1 2\n4\n1 2 3 1\n5\n1 2 1 2 1\n9\n1 2 1 3 2 3 1 2 1\n9\n2 1 3 1 3 1 3 1 3"] | 2 seconds | ["0\n2 1 2\n1\n1 3 3 1\n1\n1 2 2 2 1\n2\n1 2 3 3 2 3 3 2 1\n2\n2 1 3 3 3 1 1 1 3"] | NoteIn the first example, the array contains no local maximum, so we don't need to perform operations.In the second example, we can change $$$a_2$$$ to $$$3$$$, then the array don't have local maximums. | Java 11 | standard input | [
"greedy"
] | 255abf8750541f54d8ff0f1f80e1f8e7 | Each test contains multiple test cases. The first line will contain a single integer $$$t$$$ $$$(1 \leq t \leq 10000)$$$ — the number of test cases. Then $$$t$$$ test cases follow. The first line of each test case contains a single integer $$$n$$$ $$$(2 \leq n \leq 2 \cdot 10^5)$$$ — the size of the array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots ,a_n$$$ $$$(1 \leq a_i \leq 10^9)$$$, the elements of array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, first output a line containing a single integer $$$m$$$ — minimum number of operations required. Then ouput a line consist of $$$n$$$ integers — the resulting array after the operations. Note that this array should differ in exactly $$$m$$$ elements from the initial array. If there are multiple answers, print any. | standard output | |
PASSED | aee91ce937fc4aff53b40d7fe5e87963 | train_108.jsonl | 1645367700 | You are given an array $$$a$$$ of size $$$n$$$. Each element in this array is an integer between $$$1$$$ and $$$10^9$$$.You can perform several operations to this array. During an operation, you can replace an element in the array with any integer between $$$1$$$ and $$$10^9$$$. Output the minimum number of operations needed such that the resulting array doesn't contain any local maximums, and the resulting array after the operations.An element $$$a_i$$$ is a local maximum if it is strictly larger than both of its neighbors (that is, $$$a_i > a_{i - 1}$$$ and $$$a_i > a_{i + 1}$$$). Since $$$a_1$$$ and $$$a_n$$$ have only one neighbor each, they will never be a local maximum. | 256 megabytes | import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.lang.reflect.Array;
import java.util.*;
public class Main {
static int test;
//TODO: create char array input
static final int MOD = 1000000007;
static class Input {
static BufferedReader bufferedReader = new BufferedReader(new InputStreamReader(System.in));
static StringTokenizer stringTokenizer = new StringTokenizer("");
Input() throws IOException {
}
static int nextInt() throws IOException {
if(stringTokenizer.hasMoreTokens())
return Integer.parseInt(stringTokenizer.nextToken());
stringTokenizer = new StringTokenizer(bufferedReader.readLine());
return Integer.parseInt(stringTokenizer.nextToken());
}
public static long nextLong() throws IOException {
if(stringTokenizer.hasMoreTokens())
return Long.parseLong(stringTokenizer.nextToken());
stringTokenizer = new StringTokenizer(bufferedReader.readLine());
return Long.parseLong(stringTokenizer.nextToken());
}
static int[] getArray(int n) throws IOException {
StringTokenizer stringTokenizer = new StringTokenizer(bufferedReader.readLine());
int[] ar = new int[n];
for (int i = 0; i < n; i++) {
ar[i] = Integer.parseInt(stringTokenizer.nextToken());
}
return ar;
}
static List<Integer> getList() throws IOException {
stringTokenizer = new StringTokenizer(bufferedReader.readLine());
List<Integer> list = new ArrayList<>();
while (stringTokenizer.hasMoreTokens()){
list.add(Integer.parseInt(stringTokenizer.nextToken()));
}
return list;
}
static long[] getLongArray(int n) throws IOException {
StringTokenizer stringTokenizer = new StringTokenizer(bufferedReader.readLine());
long[] ar = new long[n];
for (int i = 0; i < n; i++) {
ar[i] = Long.parseLong(stringTokenizer.nextToken());
}
return ar;
}
static String nextString() throws IOException {
if(stringTokenizer.hasMoreTokens())
return stringTokenizer.nextToken();
stringTokenizer = new StringTokenizer(bufferedReader.readLine());
return stringTokenizer.nextToken();
}
static int[] getArrayDontKnowSize() throws IOException {
StringTokenizer stringTokenizer = new StringTokenizer(bufferedReader.readLine());
List<Integer> array = new ArrayList<>();
while (stringTokenizer.hasMoreTokens()) {
array.add(Integer.parseInt(stringTokenizer.nextToken()));
}
return array.stream().mapToInt(value -> value).toArray();
}
}
static long modPower(long x, int y) {
long ans = 1;
x = x % MOD;
while (y > 0) {
if (y % 2 == 1)
ans = (ans * x) % MOD;
y = y >> 1;
x = (x * x) % MOD;
}
return ans;
}
static int getSetBits(Integer n){
int count = 0 ;
while(n > 0){
count += n&1;
n>>= 1;
}
return count;
}
static int upperBound(int[] ar, int x) {
int mid, n = ar.length;
int start = 0;
int end = n;
if (x < ar[0])
return 0;
if (x > ar[n - 1])
return n;
while (start < end) {
mid = start + (end - start) / 2;
if (x >= ar[mid]) {
start = mid + 1;
} else {
end = mid;
}
}
return start;
}
/*
All good things must come to an end.
Sample case:
ans:
corner case:
ans:
Time complexity:
*/
static void googleSout(Object ans, int t){
System.out.println("Case #"+ ( test-t) + ": " + ans);
}
public static void main(String[] args) throws IOException {
int t = Input.nextInt();
while (t-- > 0){
int n = Input.nextInt();
int[] ar = Input.getArray(n);
int count = 0;
for(int i = 1; i< n-1; i++){
if(ar[i] > ar[i-1] && ar[i] > ar[i+1]){
ar[i+1] = i+2 < n ? Math.max(ar[i],ar[i+2]) : ar[i];
count++;
}
}
System.out.println(count);
for(int i = 0; i < n ; i++){
System.out.print(ar[i] + " ");
}
System.out.println();
}
}
static class Pair{
int u;
int v;
public Pair(int u, int v) {
this.u = u;
this.v = v;
}
}
}
| Java | ["5\n3\n2 1 2\n4\n1 2 3 1\n5\n1 2 1 2 1\n9\n1 2 1 3 2 3 1 2 1\n9\n2 1 3 1 3 1 3 1 3"] | 2 seconds | ["0\n2 1 2\n1\n1 3 3 1\n1\n1 2 2 2 1\n2\n1 2 3 3 2 3 3 2 1\n2\n2 1 3 3 3 1 1 1 3"] | NoteIn the first example, the array contains no local maximum, so we don't need to perform operations.In the second example, we can change $$$a_2$$$ to $$$3$$$, then the array don't have local maximums. | Java 11 | standard input | [
"greedy"
] | 255abf8750541f54d8ff0f1f80e1f8e7 | Each test contains multiple test cases. The first line will contain a single integer $$$t$$$ $$$(1 \leq t \leq 10000)$$$ — the number of test cases. Then $$$t$$$ test cases follow. The first line of each test case contains a single integer $$$n$$$ $$$(2 \leq n \leq 2 \cdot 10^5)$$$ — the size of the array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots ,a_n$$$ $$$(1 \leq a_i \leq 10^9)$$$, the elements of array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, first output a line containing a single integer $$$m$$$ — minimum number of operations required. Then ouput a line consist of $$$n$$$ integers — the resulting array after the operations. Note that this array should differ in exactly $$$m$$$ elements from the initial array. If there are multiple answers, print any. | standard output | |
PASSED | adfe5f62a34caad9d16e15683085b26e | train_108.jsonl | 1645367700 | You are given an array $$$a$$$ of size $$$n$$$. Each element in this array is an integer between $$$1$$$ and $$$10^9$$$.You can perform several operations to this array. During an operation, you can replace an element in the array with any integer between $$$1$$$ and $$$10^9$$$. Output the minimum number of operations needed such that the resulting array doesn't contain any local maximums, and the resulting array after the operations.An element $$$a_i$$$ is a local maximum if it is strictly larger than both of its neighbors (that is, $$$a_i > a_{i - 1}$$$ and $$$a_i > a_{i + 1}$$$). Since $$$a_1$$$ and $$$a_n$$$ have only one neighbor each, they will never be a local maximum. | 256 megabytes | import java.util.*;
import java.io.*;
public class Main {
static MyScanner sc;
static PrintWriter out;
static {
sc = new MyScanner();
out = new PrintWriter(System.out);
}
public static void solve() {
int n = sc.nextInt();
int[] a = new int[n];
int[] ans = new int[n];
for(int i = 0; i < n; i++)
ans[i] = a[i] = sc.nextInt();
boolean[] status = new boolean[n];
for(int i = 1; i < n - 1; i++) {
if(a[i] > a[i - 1] && a[i] > a[i + 1])
status[i] = true;
}
int c = 0;
for(int i = 0; i < n; i++) {
if(status[i]) {
if(i + 2 < n && status[i + 2]) {
ans[i + 1] = Math.max(a[i + 2], a[i]);
status[i + 2] = false;
}
else
ans[i] = Math.max(ans[i + 1], ans[i - 1]);
c++;
}
}
out.println(c);
for(int i = 0; i < n; i++)
out.print(ans[i] + " ");
out.println();
}
public static void main(String[] args) {
int t = sc.nextInt();
while(t-- > 0)
solve();
out.flush();
}
}
class MyScanner {
BufferedReader br;
StringTokenizer tok;
MyScanner() {
try { br = new BufferedReader(new InputStreamReader(System.in)); }
catch(Exception e) { System.out.println(e); }
tok = new StringTokenizer("");
}
public String next() {
try {
while(!tok.hasMoreTokens()) tok = new StringTokenizer(br.readLine());
}
catch(Exception e) { System.out.println(e); }
return tok.nextToken();
}
public int nextInt() {
return Integer.parseInt(next());
}
public long nextLong() {
return Long.parseLong(next());
}
public double nextDouble() {
return Double.parseDouble(next());
}
}
| Java | ["5\n3\n2 1 2\n4\n1 2 3 1\n5\n1 2 1 2 1\n9\n1 2 1 3 2 3 1 2 1\n9\n2 1 3 1 3 1 3 1 3"] | 2 seconds | ["0\n2 1 2\n1\n1 3 3 1\n1\n1 2 2 2 1\n2\n1 2 3 3 2 3 3 2 1\n2\n2 1 3 3 3 1 1 1 3"] | NoteIn the first example, the array contains no local maximum, so we don't need to perform operations.In the second example, we can change $$$a_2$$$ to $$$3$$$, then the array don't have local maximums. | Java 11 | standard input | [
"greedy"
] | 255abf8750541f54d8ff0f1f80e1f8e7 | Each test contains multiple test cases. The first line will contain a single integer $$$t$$$ $$$(1 \leq t \leq 10000)$$$ — the number of test cases. Then $$$t$$$ test cases follow. The first line of each test case contains a single integer $$$n$$$ $$$(2 \leq n \leq 2 \cdot 10^5)$$$ — the size of the array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots ,a_n$$$ $$$(1 \leq a_i \leq 10^9)$$$, the elements of array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, first output a line containing a single integer $$$m$$$ — minimum number of operations required. Then ouput a line consist of $$$n$$$ integers — the resulting array after the operations. Note that this array should differ in exactly $$$m$$$ elements from the initial array. If there are multiple answers, print any. | standard output | |
PASSED | 4a7e4810f8f5c7b59016a1abf0f882b8 | train_108.jsonl | 1645367700 | You are given an array $$$a$$$ of size $$$n$$$. Each element in this array is an integer between $$$1$$$ and $$$10^9$$$.You can perform several operations to this array. During an operation, you can replace an element in the array with any integer between $$$1$$$ and $$$10^9$$$. Output the minimum number of operations needed such that the resulting array doesn't contain any local maximums, and the resulting array after the operations.An element $$$a_i$$$ is a local maximum if it is strictly larger than both of its neighbors (that is, $$$a_i > a_{i - 1}$$$ and $$$a_i > a_{i + 1}$$$). Since $$$a_1$$$ and $$$a_n$$$ have only one neighbor each, they will never be a local maximum. | 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 Main
{
public static void main (String[] args) throws java.lang.Exception
{
Scanner cin = new Scanner(System.in);
int t = cin.nextInt();
while(t!=0){
int n = cin.nextInt();
int arr[] = new int[n];
for(int i=0;i<n;i++){
arr[i] = cin.nextInt();
}
int ans = 0;
for(int i=1;i<n-1;i++){
if(arr[i]>arr[i-1] && arr[i]>arr[i+1]){
if(i+2<n){
arr[i+1] = Math.max(arr[i],arr[i+2]);
}else{
arr[i+1] = arr[i];
}
ans++;
}
}
System.out.println(ans);
for(int i=0;i<n;i++){
System.out.print(arr[i]+" ");
}
System.out.println();
t--;
}
}
}
| Java | ["5\n3\n2 1 2\n4\n1 2 3 1\n5\n1 2 1 2 1\n9\n1 2 1 3 2 3 1 2 1\n9\n2 1 3 1 3 1 3 1 3"] | 2 seconds | ["0\n2 1 2\n1\n1 3 3 1\n1\n1 2 2 2 1\n2\n1 2 3 3 2 3 3 2 1\n2\n2 1 3 3 3 1 1 1 3"] | NoteIn the first example, the array contains no local maximum, so we don't need to perform operations.In the second example, we can change $$$a_2$$$ to $$$3$$$, then the array don't have local maximums. | Java 11 | standard input | [
"greedy"
] | 255abf8750541f54d8ff0f1f80e1f8e7 | Each test contains multiple test cases. The first line will contain a single integer $$$t$$$ $$$(1 \leq t \leq 10000)$$$ — the number of test cases. Then $$$t$$$ test cases follow. The first line of each test case contains a single integer $$$n$$$ $$$(2 \leq n \leq 2 \cdot 10^5)$$$ — the size of the array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots ,a_n$$$ $$$(1 \leq a_i \leq 10^9)$$$, the elements of array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, first output a line containing a single integer $$$m$$$ — minimum number of operations required. Then ouput a line consist of $$$n$$$ integers — the resulting array after the operations. Note that this array should differ in exactly $$$m$$$ elements from the initial array. If there are multiple answers, print any. | standard output | |
PASSED | 718dfa468eb310e3209ab09b5fa52b09 | train_108.jsonl | 1645367700 | You are given an array $$$a$$$ of size $$$n$$$. Each element in this array is an integer between $$$1$$$ and $$$10^9$$$.You can perform several operations to this array. During an operation, you can replace an element in the array with any integer between $$$1$$$ and $$$10^9$$$. Output the minimum number of operations needed such that the resulting array doesn't contain any local maximums, and the resulting array after the operations.An element $$$a_i$$$ is a local maximum if it is strictly larger than both of its neighbors (that is, $$$a_i > a_{i - 1}$$$ and $$$a_i > a_{i + 1}$$$). Since $$$a_1$$$ and $$$a_n$$$ have only one neighbor each, they will never be a local maximum. | 256 megabytes | import java.util.*;
public class f{
static class pair{
int x,y;
pair(int x,int y){
this.x=x;
this.y=y;
}
}
public static void main(String ...asada){
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 c=0;
for(int i=1;i<n-1;i++){
if(a[i]>a[i-1] && a[i]>a[i+1] && i+3<n && a[i+2]>a[i+1] && a[i+2]>a[i+3]){
a[i+1]=Math.max(a[i],a[i+2]);
c++;
}
else if(a[i]>a[i-1] && a[i]>a[i+1]){
a[i+1]=a[i];c++;}
}
System.out.println(c);
for(int i:a)
System.out.print(i+" ");
System.out.println();
}
}
} | Java | ["5\n3\n2 1 2\n4\n1 2 3 1\n5\n1 2 1 2 1\n9\n1 2 1 3 2 3 1 2 1\n9\n2 1 3 1 3 1 3 1 3"] | 2 seconds | ["0\n2 1 2\n1\n1 3 3 1\n1\n1 2 2 2 1\n2\n1 2 3 3 2 3 3 2 1\n2\n2 1 3 3 3 1 1 1 3"] | NoteIn the first example, the array contains no local maximum, so we don't need to perform operations.In the second example, we can change $$$a_2$$$ to $$$3$$$, then the array don't have local maximums. | Java 11 | standard input | [
"greedy"
] | 255abf8750541f54d8ff0f1f80e1f8e7 | Each test contains multiple test cases. The first line will contain a single integer $$$t$$$ $$$(1 \leq t \leq 10000)$$$ — the number of test cases. Then $$$t$$$ test cases follow. The first line of each test case contains a single integer $$$n$$$ $$$(2 \leq n \leq 2 \cdot 10^5)$$$ — the size of the array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots ,a_n$$$ $$$(1 \leq a_i \leq 10^9)$$$, the elements of array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, first output a line containing a single integer $$$m$$$ — minimum number of operations required. Then ouput a line consist of $$$n$$$ integers — the resulting array after the operations. Note that this array should differ in exactly $$$m$$$ elements from the initial array. If there are multiple answers, print any. | standard output | |
PASSED | 17be8915a33344709f3adc93bdd6d103 | train_108.jsonl | 1645367700 | You are given an array $$$a$$$ of size $$$n$$$. Each element in this array is an integer between $$$1$$$ and $$$10^9$$$.You can perform several operations to this array. During an operation, you can replace an element in the array with any integer between $$$1$$$ and $$$10^9$$$. Output the minimum number of operations needed such that the resulting array doesn't contain any local maximums, and the resulting array after the operations.An element $$$a_i$$$ is a local maximum if it is strictly larger than both of its neighbors (that is, $$$a_i > a_{i - 1}$$$ and $$$a_i > a_{i + 1}$$$). Since $$$a_1$$$ and $$$a_n$$$ have only one neighbor each, they will never be a local maximum. | 256 megabytes | // 🔥🔥🔥🔥 Author: Aman Bhatt (Codeforces Handle: bhattaman0001) 🔥🔥🔥🔥 //
import java.io.*;
import java.util.*;
public class Practice extends PrintWriter {
Practice() {
super(System.out);
}
static Scanner sc = new Scanner(System.in);
class Pair {
int first, second;
public Pair(int first, int second) {
this.first = first;
this.second = second;
}
public Pair() {}
}
int binary(int n) {
if (n == 1) {
return 1;
}
int temp = binary(n / 2);
String str = temp + "";
if (n % 2 == 0) str += '0'; else str += '1';
return Integer.parseInt(str);
}
int isSorted(long[] arr, long n) {
if (n == 1 || n == 0) {
return 1;
}
if (arr[(int) (n - 1)] < arr[(int) (n - 2)]) {
return 0;
}
return isSorted(arr, n - 1);
}
int findIndex(int arr[], int t) {
if (arr == null) {
return -1;
}
int len = arr.length;
int i = 0;
while (i < len) {
if (arr[i] == t) {
return i;
} else {
i = i + 1;
}
}
return -1;
}
boolean isSorted(ArrayList<Integer> list) {
int last = Integer.MIN_VALUE;
for (int i = 0; i < list.size(); i++) {
int ai = list.get(i);
if (last > ai) return false;
last = ai;
}
return true;
}
int distance(int x1, int y1, int x2, int y2) {
return (int) Math.sqrt((x2 - x1) * (x2 - x1) + (y2 - y1) * (y2 - y1));
}
boolean isSquare(int n) {
int sqrt = (int) Math.sqrt(n);
if (sqrt * sqrt == n) return true; else return false;
}
int gcd(int a, int b) {
if (a == 0) return b;
return gcd(b % a, a);
}
int lcm(int a, int b) {
return (a / gcd(a, b)) * b;
}
void reverseAArray(int[] arr, int s, int e) {
while (s < e) {
int temp = arr[e];
arr[e] = arr[s];
arr[s] = temp;
s++;
e--;
}
}
String sort(String inputString) {
char tempArray[] = inputString.toCharArray();
Arrays.sort(tempArray);
return new String(tempArray);
}
int ceil(int a, int b) {
return (a + b - 1) / b;
}
String reverseAString(String s) {
char[] ch = s.toCharArray();
String reverse = "";
for (int i = s.length() - 1; i >= 0; i--) {
reverse += ch[i];
}
return reverse;
}
public static void main(String[] $) {
try (Practice o = new Practice()) {
o.main();
o.flush();
}
}
void main() {
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 ans = 0;
for (int i = 1; i < n - 1; i++) {
if (arr[i] > arr[i - 1] && arr[i] > arr[i + 1]) {
ans++;
if (i >= n - 3) {
arr[i + 1] = arr[i];
} else {
arr[i + 1] = Math.max(arr[i], arr[i + 2]);
}
}
}
println(ans);
for (int i = 0; i < n; i++) {
print(arr[i] + " ");
}
println();
}
}
}
| Java | ["5\n3\n2 1 2\n4\n1 2 3 1\n5\n1 2 1 2 1\n9\n1 2 1 3 2 3 1 2 1\n9\n2 1 3 1 3 1 3 1 3"] | 2 seconds | ["0\n2 1 2\n1\n1 3 3 1\n1\n1 2 2 2 1\n2\n1 2 3 3 2 3 3 2 1\n2\n2 1 3 3 3 1 1 1 3"] | NoteIn the first example, the array contains no local maximum, so we don't need to perform operations.In the second example, we can change $$$a_2$$$ to $$$3$$$, then the array don't have local maximums. | Java 11 | standard input | [
"greedy"
] | 255abf8750541f54d8ff0f1f80e1f8e7 | Each test contains multiple test cases. The first line will contain a single integer $$$t$$$ $$$(1 \leq t \leq 10000)$$$ — the number of test cases. Then $$$t$$$ test cases follow. The first line of each test case contains a single integer $$$n$$$ $$$(2 \leq n \leq 2 \cdot 10^5)$$$ — the size of the array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots ,a_n$$$ $$$(1 \leq a_i \leq 10^9)$$$, the elements of array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, first output a line containing a single integer $$$m$$$ — minimum number of operations required. Then ouput a line consist of $$$n$$$ integers — the resulting array after the operations. Note that this array should differ in exactly $$$m$$$ elements from the initial array. If there are multiple answers, print any. | standard output | |
PASSED | 3d2497cacc4afff8bc68a2f1a184b622 | train_108.jsonl | 1645367700 | You are given an array $$$a$$$ of size $$$n$$$. Each element in this array is an integer between $$$1$$$ and $$$10^9$$$.You can perform several operations to this array. During an operation, you can replace an element in the array with any integer between $$$1$$$ and $$$10^9$$$. Output the minimum number of operations needed such that the resulting array doesn't contain any local maximums, and the resulting array after the operations.An element $$$a_i$$$ is a local maximum if it is strictly larger than both of its neighbors (that is, $$$a_i > a_{i - 1}$$$ and $$$a_i > a_{i + 1}$$$). Since $$$a_1$$$ and $$$a_n$$$ have only one neighbor each, they will never be a local maximum. | 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();
sc.nextLine();
while(t != 0)
{
t--;
int n = sc.nextInt();
sc.nextLine();
int[] a = new int[n];
for(int i=0; i<n; i++)
{
a[i] = sc.nextInt();
}
sc.nextLine();
int ans = 0;
for(int i=1; i<n-1; i++)
{
if(a[i]>a[i-1] && a[i]>a[i+1]) {
if(i+2>=n) a[i+1] = a[i];
else a[i+1] = Math.max(a[i+2], a[i]);
ans++;
}
}
int m = 0, i = 1;
// while(i<n-1)
// {
// if(a[i]>a[i-1] && a[i]>a[i+1])
// {
// m++;
// if(i+3<n)
// {
// if(a[i+2]>a[i+1] && a[i+2]>a[i+3])
// {
// if(a[i]>a[i+2])
// {
// ans[i+1] = a[i];
// i += 3;
// }
// else
// {
// ans[i+1] = a[i+2];
// i += 3;
// }
// }
// else
// {
// ans[i] = a[i+1];
// i += 3;
// }
// }
// else
// {
// ans[i-1] = a[i];
// i += 3;
// }
// }
// else
// {
// i++;
// }
// }
System.out.println(ans);
for(i=0; i<n; i++)
{
System.out.print(a[i]+" ");
}
System.out.println("");
}
}
} | Java | ["5\n3\n2 1 2\n4\n1 2 3 1\n5\n1 2 1 2 1\n9\n1 2 1 3 2 3 1 2 1\n9\n2 1 3 1 3 1 3 1 3"] | 2 seconds | ["0\n2 1 2\n1\n1 3 3 1\n1\n1 2 2 2 1\n2\n1 2 3 3 2 3 3 2 1\n2\n2 1 3 3 3 1 1 1 3"] | NoteIn the first example, the array contains no local maximum, so we don't need to perform operations.In the second example, we can change $$$a_2$$$ to $$$3$$$, then the array don't have local maximums. | Java 11 | standard input | [
"greedy"
] | 255abf8750541f54d8ff0f1f80e1f8e7 | Each test contains multiple test cases. The first line will contain a single integer $$$t$$$ $$$(1 \leq t \leq 10000)$$$ — the number of test cases. Then $$$t$$$ test cases follow. The first line of each test case contains a single integer $$$n$$$ $$$(2 \leq n \leq 2 \cdot 10^5)$$$ — the size of the array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots ,a_n$$$ $$$(1 \leq a_i \leq 10^9)$$$, the elements of array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, first output a line containing a single integer $$$m$$$ — minimum number of operations required. Then ouput a line consist of $$$n$$$ integers — the resulting array after the operations. Note that this array should differ in exactly $$$m$$$ elements from the initial array. If there are multiple answers, print any. | standard output | |
PASSED | 34ca3729f29e2012269a29a168dd0645 | train_108.jsonl | 1645367700 | You are given an array $$$a$$$ of size $$$n$$$. Each element in this array is an integer between $$$1$$$ and $$$10^9$$$.You can perform several operations to this array. During an operation, you can replace an element in the array with any integer between $$$1$$$ and $$$10^9$$$. Output the minimum number of operations needed such that the resulting array doesn't contain any local maximums, and the resulting array after the operations.An element $$$a_i$$$ is a local maximum if it is strictly larger than both of its neighbors (that is, $$$a_i > a_{i - 1}$$$ and $$$a_i > a_{i + 1}$$$). Since $$$a_1$$$ and $$$a_n$$$ have only one neighbor each, they will never be a local maximum. | 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));
int t = Integer.parseInt(br.readLine());
BufferedWriter output = new BufferedWriter(new OutputStreamWriter(System.out));
StringTokenizer st;
while (t-- > 0) {
int n = Integer.parseInt(br.readLine());
st = new StringTokenizer(br.readLine());
int arr[] = new int[n];
int count[] = new int[n];
for (int i = 0; i < n; i++) {
int e = Integer.parseInt(st.nextToken());
arr[i] = e;
}
for (int i = 1; i < n - 1; i++) {
if (arr[i] > arr[i - 1] && arr[i] > arr[i + 1]) {
count[i - 1]++;
count[i + 1]++;
}
}
int res = 0;
for (int i = 0; i < n; i++) {
if (count[i] == 2) {
res++;
arr[i] = Math.max(arr[i + 1], arr[i-1]);
count[i - 2]--;
count[i + 2]--;
count[i] = 0;
i += 2;
continue;
}
}
for (int i = 0; i < n; i++) {
if (count[i] == 1) {
res++;
if (i != n - 1) {
arr[i] = arr[i + 1];
count[i + 2]--;
}
else {
arr[i] = arr[i - 1];
count[i - 2]--;
}
}
}
output.write(res + "\n");
for (int i : arr)
output.write(i + " ");
output.write("\n");
// int k = Integer.parseInt(st.nextToken());
// char arr[] = br.readLine().toCharArray();
// output.write();
// int n = Integer.parseInt(st.nextToken());
// HashMap<Character, Integer> map = new HashMap<Character, Integer>();
// if
// output.write("YES\n");
// else
// output.write("NO\n");
// long a = Long.parseLong(st.nextToken());
// long b = Long.parseLong(st.nextToken());
// if(flag == 1)
// output.write("NO\n");
// else
// output.write("YES\n" + x + " " + y + " " + z + "\n");
// output.write(n+ "\n");
}
output.flush();
}
} | Java | ["5\n3\n2 1 2\n4\n1 2 3 1\n5\n1 2 1 2 1\n9\n1 2 1 3 2 3 1 2 1\n9\n2 1 3 1 3 1 3 1 3"] | 2 seconds | ["0\n2 1 2\n1\n1 3 3 1\n1\n1 2 2 2 1\n2\n1 2 3 3 2 3 3 2 1\n2\n2 1 3 3 3 1 1 1 3"] | NoteIn the first example, the array contains no local maximum, so we don't need to perform operations.In the second example, we can change $$$a_2$$$ to $$$3$$$, then the array don't have local maximums. | Java 11 | standard input | [
"greedy"
] | 255abf8750541f54d8ff0f1f80e1f8e7 | Each test contains multiple test cases. The first line will contain a single integer $$$t$$$ $$$(1 \leq t \leq 10000)$$$ — the number of test cases. Then $$$t$$$ test cases follow. The first line of each test case contains a single integer $$$n$$$ $$$(2 \leq n \leq 2 \cdot 10^5)$$$ — the size of the array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots ,a_n$$$ $$$(1 \leq a_i \leq 10^9)$$$, the elements of array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, first output a line containing a single integer $$$m$$$ — minimum number of operations required. Then ouput a line consist of $$$n$$$ integers — the resulting array after the operations. Note that this array should differ in exactly $$$m$$$ elements from the initial array. If there are multiple answers, print any. | standard output | |
PASSED | 09a70698d67395742b5918336fed97ab | train_108.jsonl | 1645367700 | You are given an array $$$a$$$ of size $$$n$$$. Each element in this array is an integer between $$$1$$$ and $$$10^9$$$.You can perform several operations to this array. During an operation, you can replace an element in the array with any integer between $$$1$$$ and $$$10^9$$$. Output the minimum number of operations needed such that the resulting array doesn't contain any local maximums, and the resulting array after the operations.An element $$$a_i$$$ is a local maximum if it is strictly larger than both of its neighbors (that is, $$$a_i > a_{i - 1}$$$ and $$$a_i > a_{i + 1}$$$). Since $$$a_1$$$ and $$$a_n$$$ have only one neighbor each, they will never be a local maximum. | 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;
while(t-->0)
{
int n = sc.nextInt();
long arr[] = sc.readArrayLong(n);
long ans = 0;
long maxi = 0;
for(int i=0;i<n;i++)
{
if(i == 0 || i == n - 1){
continue;
}
if(arr[i] > arr[i - 1] && arr[i] > arr[i + 1]){
long val1 = arr[i];
long val2 = (i + 2 < n) ? arr[i + 2] : Long.MIN_VALUE;
arr[i + 1] = Math.max(val1 , val2);
ans++;
}
}
System.out.println(ans);
for(long x:arr){
System.out.print(x+" ");
}
System.out.println();
}
}
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 void reverse_sorted(int[] arr)
{
ArrayList<Integer> list = new ArrayList<>();
for(int i=0;i<arr.length;i++)
{
list.add(arr[i]);
}
Collections.sort(list , Collections.reverseOrder());
for(int i=0;i<arr.length;i++)
{
arr[i] = list.get(i);
}
}
static int LowerBound(int a[], int x) { // x is the target value or key
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(ArrayList<Integer> list, int x) {// x is the key or target value
int l=-1,r=list.size();
while(l+1<r) {
int m=(l+r)>>>1;
if(list.get(m)<=x) l=m;
else r=m;
}
return l+1;
}
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 class Queue_Pair implements Comparable<Queue_Pair> {
int first , second;
public Queue_Pair(int first, int second) {
this.first=first;
this.second=second;
}
public int compareTo(Queue_Pair o) {
return Integer.compare(o.first, first);
}
}
static void leftRotate(int arr[], int d, int n)
{
for (int i = 0; i < d; i++)
leftRotatebyOne(arr, n);
}
static void leftRotatebyOne(int arr[], int n)
{
int i, temp;
temp = arr[0];
for (i = 0; i < n - 1; i++)
arr[i] = arr[i + 1];
arr[n-1] = temp;
}
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 boolean palindrome_array(char arr[], int n)
{
// Initialise flag to zero.
int flag = 0;
// Loop till array size n/2.
for (int i = 0; i <= n / 2 && n != 0; i++) {
// Check if first and last element are different
// Then set flag to 1.
if (arr[i] != arr[n - i - 1]) {
flag = 1;
break;
}
}
// If flag is set then print Not Palindrome
// else print Palindrome.
if (flag == 1)
return false;
else
return true;
}
static boolean allElementsEqual(int[] arr,int n)
{
int z=0;
for(int i=0;i<n-1;i++)
{
if(arr[i]==arr[i+1])
{
z++;
}
}
if(z==n-1)
{
return true;
}
else
{
return false;
}
}
static boolean allElementsDistinct(int[] arr,int n)
{
int z=0;
for(int i=0;i<n-1;i++)
{
if(arr[i]!=arr[i+1])
{
z++;
}
}
if(z==n-1)
{
return true;
}
else
{
return false;
}
}
public static void reverse(int[] 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
int 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;
}
}
public static void reverse_Long(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 boolean isSorted(int[] a)
{
for (int i = 0; i < a.length - 1; i++)
{
if (a[i] > a[i + 1]) {
return false;
}
}
return true;
}
static boolean isReverseSorted(int[] a)
{
for (int i = 0; i < a.length - 1; i++)
{
if (a[i] < a[i + 1]) {
return false;
}
}
return true;
}
static int[] rearrangeEvenAndOdd(int arr[], int n)
{
ArrayList<Integer> list = new ArrayList<>();
for(int i=0;i<n;i++)
{
if(arr[i]%2==0)
{
list.add(arr[i]);
}
}
for(int i=0;i<n;i++)
{
if(arr[i]%2!=0)
{
list.add(arr[i]);
}
}
int len = list.size();
int[] array = list.stream().mapToInt(i->i).toArray();
return array;
}
static long[] rearrangeEvenAndOddLong(long arr[], int n)
{
ArrayList<Long> list = new ArrayList<>();
for(int i=0;i<n;i++)
{
if(arr[i]%2==0)
{
list.add(arr[i]);
}
}
for(int i=0;i<n;i++)
{
if(arr[i]%2!=0)
{
list.add(arr[i]);
}
}
int len = list.size();
long[] array = list.stream().mapToLong(i->i).toArray();
return array;
}
static boolean isPrime(long n)
{
// Check if number is less than
// equal to 1
if (n <= 1)
return false;
// Check if number is 2
else if (n == 2)
return true;
// Check if n is a multiple of 2
else if (n % 2 == 0)
return false;
// If not, then just check the odds
for (long i = 3; i <= Math.sqrt(n); i += 2)
{
if (n % i == 0)
return false;
}
return true;
}
static long getSum(long n)
{
long sum = 0;
while (n != 0)
{
sum = sum + n % 10;
n = n/10;
}
return sum;
}
static int gcd(int a, int b)
{
if (b == 0)
return a;
return gcd(b, a % b);
}
static long gcdLong(long a, long b)
{
if (b == 0)
return a;
return gcdLong(b, a % b);
}
static void swap(int i, int j)
{
int temp = i;
i = j;
j = temp;
}
static int countDigit(int n)
{
return (int)Math.floor(Math.log10(n) + 1);
}
}
class Pair
{
int first;
int second;
Pair(int first , int second)
{
this.first = first;
this.second = second;
}
}
| Java | ["5\n3\n2 1 2\n4\n1 2 3 1\n5\n1 2 1 2 1\n9\n1 2 1 3 2 3 1 2 1\n9\n2 1 3 1 3 1 3 1 3"] | 2 seconds | ["0\n2 1 2\n1\n1 3 3 1\n1\n1 2 2 2 1\n2\n1 2 3 3 2 3 3 2 1\n2\n2 1 3 3 3 1 1 1 3"] | NoteIn the first example, the array contains no local maximum, so we don't need to perform operations.In the second example, we can change $$$a_2$$$ to $$$3$$$, then the array don't have local maximums. | Java 11 | standard input | [
"greedy"
] | 255abf8750541f54d8ff0f1f80e1f8e7 | Each test contains multiple test cases. The first line will contain a single integer $$$t$$$ $$$(1 \leq t \leq 10000)$$$ — the number of test cases. Then $$$t$$$ test cases follow. The first line of each test case contains a single integer $$$n$$$ $$$(2 \leq n \leq 2 \cdot 10^5)$$$ — the size of the array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots ,a_n$$$ $$$(1 \leq a_i \leq 10^9)$$$, the elements of array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, first output a line containing a single integer $$$m$$$ — minimum number of operations required. Then ouput a line consist of $$$n$$$ integers — the resulting array after the operations. Note that this array should differ in exactly $$$m$$$ elements from the initial array. If there are multiple answers, print any. | standard output | |
PASSED | 255a48badd8fe30bf1045da39b48648c | train_108.jsonl | 1645367700 | You are given an array $$$a$$$ of size $$$n$$$. Each element in this array is an integer between $$$1$$$ and $$$10^9$$$.You can perform several operations to this array. During an operation, you can replace an element in the array with any integer between $$$1$$$ and $$$10^9$$$. Output the minimum number of operations needed such that the resulting array doesn't contain any local maximums, and the resulting array after the operations.An element $$$a_i$$$ is a local maximum if it is strictly larger than both of its neighbors (that is, $$$a_i > a_{i - 1}$$$ and $$$a_i > a_{i + 1}$$$). Since $$$a_1$$$ and $$$a_n$$$ have only one neighbor each, they will never be a local maximum. | 256 megabytes | import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.Arrays;
import java.util.StringTokenizer;
public class B_Avoid_Local_Maximums {
static void solve(int n, long[] arr, long[] temp) {
Arrays.sort(temp);
long c = 0;
for (int i = 1; i < n - 1; i++) {
if (arr[i] > arr[i - 1] && arr[i] > arr[i + 1]) {
if (i + 2 < n) {
arr[i + 1] = Math.max(arr[i], arr[i + 2]);
} else {
arr[i + 1] = arr[i];
}
c++;
}
}
System.out.println(c);
for (int i = 0; i < n; i++) {
System.out.print(arr[i] + " ");
}
}
public static void main(String[] args) {
try {
FastReader s = new FastReader();
int t = s.nextInt();
while (t-- > 0) {
int n = s.nextInt();
long[] arr = new long[n];
long[] temp = new long[n];
for (int i = 0; i < n; i++) {
arr[i] = s.nextLong();
temp[i] = arr[i];
}
solve(n, arr, temp);
System.out.println();
}
/*
* for(int i=0;i<t;i++)
* {
* System.out.println(arr[i]);
* }
*/
} catch (Exception e) {
return;
}
}
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 | ["5\n3\n2 1 2\n4\n1 2 3 1\n5\n1 2 1 2 1\n9\n1 2 1 3 2 3 1 2 1\n9\n2 1 3 1 3 1 3 1 3"] | 2 seconds | ["0\n2 1 2\n1\n1 3 3 1\n1\n1 2 2 2 1\n2\n1 2 3 3 2 3 3 2 1\n2\n2 1 3 3 3 1 1 1 3"] | NoteIn the first example, the array contains no local maximum, so we don't need to perform operations.In the second example, we can change $$$a_2$$$ to $$$3$$$, then the array don't have local maximums. | Java 11 | standard input | [
"greedy"
] | 255abf8750541f54d8ff0f1f80e1f8e7 | Each test contains multiple test cases. The first line will contain a single integer $$$t$$$ $$$(1 \leq t \leq 10000)$$$ — the number of test cases. Then $$$t$$$ test cases follow. The first line of each test case contains a single integer $$$n$$$ $$$(2 \leq n \leq 2 \cdot 10^5)$$$ — the size of the array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots ,a_n$$$ $$$(1 \leq a_i \leq 10^9)$$$, the elements of array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, first output a line containing a single integer $$$m$$$ — minimum number of operations required. Then ouput a line consist of $$$n$$$ integers — the resulting array after the operations. Note that this array should differ in exactly $$$m$$$ elements from the initial array. If there are multiple answers, print any. | standard output | |
PASSED | 95c2ef55c8bb439391870f84d9e948f7 | train_108.jsonl | 1645367700 | You are given an array $$$a$$$ of size $$$n$$$. Each element in this array is an integer between $$$1$$$ and $$$10^9$$$.You can perform several operations to this array. During an operation, you can replace an element in the array with any integer between $$$1$$$ and $$$10^9$$$. Output the minimum number of operations needed such that the resulting array doesn't contain any local maximums, and the resulting array after the operations.An element $$$a_i$$$ is a local maximum if it is strictly larger than both of its neighbors (that is, $$$a_i > a_{i - 1}$$$ and $$$a_i > a_{i + 1}$$$). Since $$$a_1$$$ and $$$a_n$$$ have only one neighbor each, they will never be a local maximum. | 256 megabytes | import java.util.Scanner;
public class problemB {
public static void main(String[] args) {
Scanner sh = new Scanner(System.in);
int t = sh.nextInt();
while(t>0){
int n = sh.nextInt();
int a[] = new int[n];
for(int i = 0; i<n; i++){
a[i] = sh.nextInt();
}
int idx = 1;
int count = 0;
if(n !=2){
while(idx<n-1){
if(a[idx] > a[idx-1] && a[idx] > a[idx+1]){
if(idx+2<n-1 && a[idx+2] > a[idx+1] && a[idx+2] > a[idx+3]){
a[idx+1] = Math.max(a[idx], a[idx+2]); count++;
idx = idx+4;
}else{
a[idx] = Math.max(a[idx-1],a[idx+1]); count++;
idx = idx+2;
}
}else{
idx++;
}
}}
System.out.println(count);
for(int i = 0; i<n; i++){
System.out.print(a[i]+" ");
}
System.out.println();
t--;}
}
}
| Java | ["5\n3\n2 1 2\n4\n1 2 3 1\n5\n1 2 1 2 1\n9\n1 2 1 3 2 3 1 2 1\n9\n2 1 3 1 3 1 3 1 3"] | 2 seconds | ["0\n2 1 2\n1\n1 3 3 1\n1\n1 2 2 2 1\n2\n1 2 3 3 2 3 3 2 1\n2\n2 1 3 3 3 1 1 1 3"] | NoteIn the first example, the array contains no local maximum, so we don't need to perform operations.In the second example, we can change $$$a_2$$$ to $$$3$$$, then the array don't have local maximums. | Java 11 | standard input | [
"greedy"
] | 255abf8750541f54d8ff0f1f80e1f8e7 | Each test contains multiple test cases. The first line will contain a single integer $$$t$$$ $$$(1 \leq t \leq 10000)$$$ — the number of test cases. Then $$$t$$$ test cases follow. The first line of each test case contains a single integer $$$n$$$ $$$(2 \leq n \leq 2 \cdot 10^5)$$$ — the size of the array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots ,a_n$$$ $$$(1 \leq a_i \leq 10^9)$$$, the elements of array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, first output a line containing a single integer $$$m$$$ — minimum number of operations required. Then ouput a line consist of $$$n$$$ integers — the resulting array after the operations. Note that this array should differ in exactly $$$m$$$ elements from the initial array. If there are multiple answers, print any. | standard output | |
PASSED | df476805d129c99d09bcb7a84c2ebcad | train_108.jsonl | 1645367700 | You are given an array $$$a$$$ of size $$$n$$$. Each element in this array is an integer between $$$1$$$ and $$$10^9$$$.You can perform several operations to this array. During an operation, you can replace an element in the array with any integer between $$$1$$$ and $$$10^9$$$. Output the minimum number of operations needed such that the resulting array doesn't contain any local maximums, and the resulting array after the operations.An element $$$a_i$$$ is a local maximum if it is strictly larger than both of its neighbors (that is, $$$a_i > a_{i - 1}$$$ and $$$a_i > a_{i + 1}$$$). Since $$$a_1$$$ and $$$a_n$$$ have only one neighbor each, they will never be a local maximum. | 256 megabytes | //Utilities
import java.io.*;
import java.util.*;
public class a {
static int t;
static int n;
static int[] a;
static int res;
public static void main(String[] args) throws IOException {
t = in.iscan();
while (t-- > 0) {
n = in.iscan(); a = new int[n];
for (int i = 0; i < n; i++) {
a[i] = in.iscan();
}
res = 0;
for (int i = 1; i < n-1; i++) {
if (a[i-1] < a[i] && a[i] > a[i+1]) {
res++;
if (i + 2 < n) a[i+1] = Math.max(a[i], a[i+2]);
else a[i+1] = a[i];
i += 2;
}
}
out.println(res);
for (int i = 0; i < n; i++) {
out.print(a[i] + " ");
}
out.println();
}
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 | ["5\n3\n2 1 2\n4\n1 2 3 1\n5\n1 2 1 2 1\n9\n1 2 1 3 2 3 1 2 1\n9\n2 1 3 1 3 1 3 1 3"] | 2 seconds | ["0\n2 1 2\n1\n1 3 3 1\n1\n1 2 2 2 1\n2\n1 2 3 3 2 3 3 2 1\n2\n2 1 3 3 3 1 1 1 3"] | NoteIn the first example, the array contains no local maximum, so we don't need to perform operations.In the second example, we can change $$$a_2$$$ to $$$3$$$, then the array don't have local maximums. | Java 11 | standard input | [
"greedy"
] | 255abf8750541f54d8ff0f1f80e1f8e7 | Each test contains multiple test cases. The first line will contain a single integer $$$t$$$ $$$(1 \leq t \leq 10000)$$$ — the number of test cases. Then $$$t$$$ test cases follow. The first line of each test case contains a single integer $$$n$$$ $$$(2 \leq n \leq 2 \cdot 10^5)$$$ — the size of the array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots ,a_n$$$ $$$(1 \leq a_i \leq 10^9)$$$, the elements of array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, first output a line containing a single integer $$$m$$$ — minimum number of operations required. Then ouput a line consist of $$$n$$$ integers — the resulting array after the operations. Note that this array should differ in exactly $$$m$$$ elements from the initial array. If there are multiple answers, print any. | standard output | |
PASSED | 349489952b867abd651de76cc4604358 | train_108.jsonl | 1645367700 | You are given an array $$$a$$$ of size $$$n$$$. Each element in this array is an integer between $$$1$$$ and $$$10^9$$$.You can perform several operations to this array. During an operation, you can replace an element in the array with any integer between $$$1$$$ and $$$10^9$$$. Output the minimum number of operations needed such that the resulting array doesn't contain any local maximums, and the resulting array after the operations.An element $$$a_i$$$ is a local maximum if it is strictly larger than both of its neighbors (that is, $$$a_i > a_{i - 1}$$$ and $$$a_i > a_{i + 1}$$$). Since $$$a_1$$$ and $$$a_n$$$ have only one neighbor each, they will never be a local maximum. | 256 megabytes | import java.util.*;
public class Main{
public static void main(String[]args){
Scanner at = new Scanner(System.in);
int t = at.nextInt();
while(t-->0){
int n = at.nextInt();
int[]arr = new int[n];
for(int i = 0;i<n;i++){
arr[i] = at.nextInt();
}
int count = 0;
for(int i = 1;i<n-1;i++){
if(arr[i] > arr[i-1] && arr[i] > arr[i+1]){
// if(i == n-2){
// arr[i+1] = arr[i];
// }
arr[i+1] = Math.max(arr[i],(i+2) < n ? arr[i+2]:arr[i]);
count++;
}
}
System.out.println(count);
for(int i = 0;i<n;i++){
System.out.print(arr[i]+" ");
}
System.out.println();
}
}
} | Java | ["5\n3\n2 1 2\n4\n1 2 3 1\n5\n1 2 1 2 1\n9\n1 2 1 3 2 3 1 2 1\n9\n2 1 3 1 3 1 3 1 3"] | 2 seconds | ["0\n2 1 2\n1\n1 3 3 1\n1\n1 2 2 2 1\n2\n1 2 3 3 2 3 3 2 1\n2\n2 1 3 3 3 1 1 1 3"] | NoteIn the first example, the array contains no local maximum, so we don't need to perform operations.In the second example, we can change $$$a_2$$$ to $$$3$$$, then the array don't have local maximums. | Java 11 | standard input | [
"greedy"
] | 255abf8750541f54d8ff0f1f80e1f8e7 | Each test contains multiple test cases. The first line will contain a single integer $$$t$$$ $$$(1 \leq t \leq 10000)$$$ — the number of test cases. Then $$$t$$$ test cases follow. The first line of each test case contains a single integer $$$n$$$ $$$(2 \leq n \leq 2 \cdot 10^5)$$$ — the size of the array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots ,a_n$$$ $$$(1 \leq a_i \leq 10^9)$$$, the elements of array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, first output a line containing a single integer $$$m$$$ — minimum number of operations required. Then ouput a line consist of $$$n$$$ integers — the resulting array after the operations. Note that this array should differ in exactly $$$m$$$ elements from the initial array. If there are multiple answers, print any. | standard output | |
PASSED | 2e45d2aabb95334930aa67e30ce8ed4a | train_108.jsonl | 1645367700 | You are given an array $$$a$$$ of size $$$n$$$. Each element in this array is an integer between $$$1$$$ and $$$10^9$$$.You can perform several operations to this array. During an operation, you can replace an element in the array with any integer between $$$1$$$ and $$$10^9$$$. Output the minimum number of operations needed such that the resulting array doesn't contain any local maximums, and the resulting array after the operations.An element $$$a_i$$$ is a local maximum if it is strictly larger than both of its neighbors (that is, $$$a_i > a_{i - 1}$$$ and $$$a_i > a_{i + 1}$$$). Since $$$a_1$$$ and $$$a_n$$$ have only one neighbor each, they will never be a local maximum. | 256 megabytes | import javax.swing.*;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.*;
public class Main extends PrintWriter {
static BufferedReader s = new BufferedReader(new InputStreamReader(System.in));Main() { super(System.out); }public static void main(String[] args) throws IOException{ Main d1=new Main ();d1.main();d1.flush(); }
void main() throws IOException {
StringBuilder sb = new StringBuilder();
int t = 1;
t = i(s()[0]);
while (t-- > 0) {
int n = i(s()[0]);
int[] a=new int[n]; arri(a, n);
int count = 0;
for(int i = 1 ; i < n - 1; i++){
if(a[i] > a[i - 1] && a[i] > a[i + 1]){
count++;
if(i + 2 < n)
a[i + 1] = Math.max(a[i], a[i + 2]);
else
a[i + 1] = a[i];
}
}
sb.append(count + "\n");
for(int i = 0; i < n; i++){
sb.append(a[i] + " ");
}sb.append("\n");
}
System.out.println(sb);
}
static String[] s() throws IOException { return s.readLine().trim().split("\\s+"); }
static int i(String ss) {return Integer.parseInt(ss); }
static long l(String ss) {return Long.parseLong(ss); }
public void arr(long[] a,int n) throws IOException {String[] s2=s();for(int i=0;i<n;i++){ a[i]=l(s2[i]); }}
public void arri(int[] a,int n) throws IOException {String[] s2=s();for(int i=0;i<n;i++){ a[i]=(i(s2[i])); }}
}class Pair{
int start, end;
public Pair(long a, long b){
this.start = (int)a ; this.end = (int)b;
}
}
| Java | ["5\n3\n2 1 2\n4\n1 2 3 1\n5\n1 2 1 2 1\n9\n1 2 1 3 2 3 1 2 1\n9\n2 1 3 1 3 1 3 1 3"] | 2 seconds | ["0\n2 1 2\n1\n1 3 3 1\n1\n1 2 2 2 1\n2\n1 2 3 3 2 3 3 2 1\n2\n2 1 3 3 3 1 1 1 3"] | NoteIn the first example, the array contains no local maximum, so we don't need to perform operations.In the second example, we can change $$$a_2$$$ to $$$3$$$, then the array don't have local maximums. | Java 11 | standard input | [
"greedy"
] | 255abf8750541f54d8ff0f1f80e1f8e7 | Each test contains multiple test cases. The first line will contain a single integer $$$t$$$ $$$(1 \leq t \leq 10000)$$$ — the number of test cases. Then $$$t$$$ test cases follow. The first line of each test case contains a single integer $$$n$$$ $$$(2 \leq n \leq 2 \cdot 10^5)$$$ — the size of the array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots ,a_n$$$ $$$(1 \leq a_i \leq 10^9)$$$, the elements of array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, first output a line containing a single integer $$$m$$$ — minimum number of operations required. Then ouput a line consist of $$$n$$$ integers — the resulting array after the operations. Note that this array should differ in exactly $$$m$$$ elements from the initial array. If there are multiple answers, print any. | standard output | |
PASSED | bb3ff22d5d99d306b0ef27988a83d921 | train_108.jsonl | 1645367700 | You are given an array $$$a$$$ of size $$$n$$$. Each element in this array is an integer between $$$1$$$ and $$$10^9$$$.You can perform several operations to this array. During an operation, you can replace an element in the array with any integer between $$$1$$$ and $$$10^9$$$. Output the minimum number of operations needed such that the resulting array doesn't contain any local maximums, and the resulting array after the operations.An element $$$a_i$$$ is a local maximum if it is strictly larger than both of its neighbors (that is, $$$a_i > a_{i - 1}$$$ and $$$a_i > a_{i + 1}$$$). Since $$$a_1$$$ and $$$a_n$$$ have only one neighbor each, they will never be a local maximum. | 256 megabytes |
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.*;
public class Main {
static class Pair{
String s;
int index;
public Pair(String s, int index) {
this.s = s;
this.index = index;
}
@Override
public boolean equals(Object obj) {
Pair tmp = (Pair) obj;
return s.equals(tmp.s);
}
}
static ArrayList<Pair> set =new ArrayList<>();
static String input;
static String ans = null;
static void check( int num ){
if (input.length()<=2||ans!=null)return;
StringBuilder tmp = new StringBuilder(input);
// System.out.print(tmp+" ");
for (int j = 0; j < input.length(); j++) {
tmp.deleteCharAt(j);
// System.out.println(tmp+" "+input.charAt(j));
if (num==1) set.add(new Pair(tmp.toString(),j));
boolean isPalindrome = checkPalindrome(tmp);
if (isPalindrome){
if (ans==null) ans= input.charAt(j)+"";
else {
ans=input.charAt(j)+"";
}
return;
}
// System.out.print(tmp+" ");
tmp.insert(j,input.charAt(j));
// System.out.println(tmp);
}
}
private static boolean checkPalindrome(StringBuilder tmp) {
StringBuilder v = new StringBuilder(tmp);
// System.out.println(tmp);
return v.toString().equals(v.reverse().toString());
}
public static void main(String[] args) throws IOException {
Scanner in = new Scanner(System.in);
try (PrintWriter or = new PrintWriter(System.out)) {
int t = in.nextInt();
outer:
while (t-->0){
int n = in.nextInt();
int[] a = new int[n];
for (int i = 0; i < n; i++) {
a[i]=in.nextInt();
}
int ans=0;
for (int i = 1; i <n-1 ; i++) {
if (a[i]>a[i-1]&&a[i]>a[i+1]){
if (i+2<n )a[i+1]=Math.max(a[i],a[i+2]);
else a[i+1]=a[i];
++ans;
}
}
or.println(ans);
for (int item:a) or.print(item+" ");
or.println();
}
}
}
}
| Java | ["5\n3\n2 1 2\n4\n1 2 3 1\n5\n1 2 1 2 1\n9\n1 2 1 3 2 3 1 2 1\n9\n2 1 3 1 3 1 3 1 3"] | 2 seconds | ["0\n2 1 2\n1\n1 3 3 1\n1\n1 2 2 2 1\n2\n1 2 3 3 2 3 3 2 1\n2\n2 1 3 3 3 1 1 1 3"] | NoteIn the first example, the array contains no local maximum, so we don't need to perform operations.In the second example, we can change $$$a_2$$$ to $$$3$$$, then the array don't have local maximums. | Java 11 | standard input | [
"greedy"
] | 255abf8750541f54d8ff0f1f80e1f8e7 | Each test contains multiple test cases. The first line will contain a single integer $$$t$$$ $$$(1 \leq t \leq 10000)$$$ — the number of test cases. Then $$$t$$$ test cases follow. The first line of each test case contains a single integer $$$n$$$ $$$(2 \leq n \leq 2 \cdot 10^5)$$$ — the size of the array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots ,a_n$$$ $$$(1 \leq a_i \leq 10^9)$$$, the elements of array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, first output a line containing a single integer $$$m$$$ — minimum number of operations required. Then ouput a line consist of $$$n$$$ integers — the resulting array after the operations. Note that this array should differ in exactly $$$m$$$ elements from the initial array. If there are multiple answers, print any. | standard output | |
PASSED | 25386cd18b70c22732e6c3a165c31fdb | train_108.jsonl | 1645367700 | You are given an array $$$a$$$ of size $$$n$$$. Each element in this array is an integer between $$$1$$$ and $$$10^9$$$.You can perform several operations to this array. During an operation, you can replace an element in the array with any integer between $$$1$$$ and $$$10^9$$$. Output the minimum number of operations needed such that the resulting array doesn't contain any local maximums, and the resulting array after the operations.An element $$$a_i$$$ is a local maximum if it is strictly larger than both of its neighbors (that is, $$$a_i > a_{i - 1}$$$ and $$$a_i > a_{i + 1}$$$). Since $$$a_1$$$ and $$$a_n$$$ have only one neighbor each, they will never be a local maximum. | 256 megabytes | import java.util.*;
import java.io.*;
public class B {
public void prayGod() throws IOException {
int t = nextInt();
while (t-- > 0) {
int n = nextInt();
int[] a = nextIntArray(n);
ArrayList<Integer> localMaxima = new ArrayList<>();
for (int i = 1; i < n - 1; i++) {
if (a[i] > a[i - 1] && a[i] > a[i + 1]) {
localMaxima.add(i);
}
}
if (localMaxima.size() == 0) {
out.println(0);
for (int v : a) {
out.printf("%d ", v);
}
out.println();
continue;
}
if (localMaxima.size() == 1) {
int idx = localMaxima.get(0);
a[idx] = Math.max(a[idx - 1], a[idx + 1]);
out.println(1);
for (int v : a) {
out.printf("%d ", v);
}
out.println();
continue;
}
int ret = 0, ptr = 0;
for (int i = 1; i < localMaxima.size(); i++) {
int prev = localMaxima.get(i - 1), curr = localMaxima.get(i);
if (curr - prev != 2) {
for (int j = ptr; j < i; j += 2) {
int pos = localMaxima.get(j);
if (j == i - 1)
a[pos] = Math.max(a[pos - 1], a[pos + 1]);
else
a[pos + 1] = Math.max(a[pos], a[pos + 2]);
ret++;
}
ptr = i;
}
}
for (int j = ptr; j < localMaxima.size(); j += 2) {
int pos = localMaxima.get(j);
if (j == localMaxima.size() - 1)
a[pos] = Math.max(a[pos - 1], a[pos + 1]);
else
a[pos + 1] = Math.max(a[pos], a[pos + 2]);
ret++;
}
out.println(ret);
for (int v : a) {
out.printf("%d ", v);
}
out.println();
}
}
public int gcd(int x, int y) {
if (y == 0)
return x;
return gcd(y, x % y);
}
public long binpow(int a, int b) {
if (b < 0)
return 0;
long ret = 1, curr = a;
while (b > 0) {
if (b % 2 == 1)
ret = (ret * curr) % mod;
b /= 2;
curr = (curr * curr) % mod;
}
return ret;
}
public void printVerdict(boolean verdict) {
if (verdict)
out.println(VERDICT_YES);
else
out.println(VERDICT_NO);
}
static final String VERDICT_YES = "YES";
static final String VERDICT_NO = "NO";
static final boolean RUN_TIMING = true;
static final boolean AUTOFLUSH = false;
static final boolean FILE_INPUT = false;
static final boolean FILE_OUTPUT = false;
static int iinf = 0x3f3f3f3f;
static long inf = (long) 1e18 + 10;
static long mod = (long) 1e9 + 7;
static char[] inputBuffer = new char[1 << 20];
static PushbackReader in = new PushbackReader(new BufferedReader(new InputStreamReader(System.in)), 1 << 20);
static PrintWriter out = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)), AUTOFLUSH);
// int data-type
public int nextInt() throws IOException {
return Integer.parseInt(next());
}
public int[] nextIntArray(int n) throws IOException {
int[] arr = new int[n];
for (int i = 0; i < n; i++)
arr[i] = nextInt();
return arr;
}
public void sort(int[] a) {
shuffle(a);
Arrays.sort(a);
}
public static void printArray(int[] arr) {
for (int i = 0; i < arr.length; i++)
out.print(arr[i] + " ");
out.println();
}
// long data-type
public long nextLong() throws IOException {
return Long.parseLong(next());
}
public long[] nextLongArray(int n) throws IOException {
long[] arr = new long[n];
for (int i = 0; i < n; i++)
arr[i] = nextLong();
return arr;
}
public static void printArray(long[] arr) {
for (int i = 0; i < arr.length; i++)
out.print(arr[i] + " ");
out.println();
}
public void sort(long[] a) {
shuffle(a);
Arrays.sort(a);
}
// double data-type
public double nextDouble() throws IOException {
return Double.parseDouble(next());
}
public double[] nextDoubleArray(int n) throws IOException {
double[] arr = new double[n];
for (int i = 0; i < n; i++)
arr[i] = nextDouble();
return arr;
}
public static void printArray(double[] arr) {
for (int i = 0; i < arr.length; i++)
out.print(arr[i] + " ");
out.println();
}
// Generic type
public <T> void sort(T[] a) {
shuffle(a);
Arrays.sort(a);
}
public static <T> void printArray(T[] arr) {
for (int i = 0; i < arr.length; i++)
out.print(arr[i] + " ");
out.println();
}
public String next() throws IOException {
int len = 0;
int c;
do {
c = in.read();
} while (Character.isWhitespace(c) && c != -1);
if (c == -1) {
throw new NoSuchElementException("Reached EOF");
}
do {
inputBuffer[len] = (char) c;
len++;
c = in.read();
} while (!Character.isWhitespace(c) && c != -1);
while (c != '\n' && Character.isWhitespace(c) && c != -1) {
c = in.read();
}
if (c != -1 && c != '\n') {
in.unread(c);
}
return new String(inputBuffer, 0, len);
}
public String nextLine() throws IOException {
int len = 0;
int c;
while ((c = in.read()) != '\n' && c != -1) {
if (c == '\r') {
continue;
}
inputBuffer[len] = (char) c;
len++;
}
return new String(inputBuffer, 0, len);
}
public boolean hasNext() throws IOException {
String line = nextLine();
if (line.isEmpty()) {
return false;
}
in.unread('\n');
in.unread(line.toCharArray());
return true;
}
public void shuffle(int[] arr) {
int n = arr.length;
for (int i = 0; i < n; i++) {
int j = (int) (Math.random() * (n - i));
int temp = arr[i];
arr[i] = arr[j];
arr[j] = temp;
}
}
public void shuffle(long[] arr) {
int n = arr.length;
for (int i = 0; i < n; i++) {
int j = (int) (Math.random() * (n - i));
long temp = arr[i];
arr[i] = arr[j];
arr[j] = temp;
}
}
public void shuffle(Object[] arr) {
int n = arr.length;
for (int i = 0; i < n; i++) {
int j = (int) (Math.random() * (n - i));
Object temp = arr[i];
arr[i] = arr[j];
arr[j] = temp;
}
}
public static void main(String[] args) throws IOException {
if (FILE_INPUT)
in = new PushbackReader(new BufferedReader(new FileReader(new File("output.txt"))), 1 << 20);
if (FILE_OUTPUT)
out = new PrintWriter(new FileWriter(new File("output.txt")));
long time = 0;
time -= System.nanoTime();
new B().prayGod();
time += System.nanoTime();
if (RUN_TIMING)
System.err.printf("%.3f ms%n", time / 1000000.0);
out.flush();
in.close();
}
} | Java | ["5\n3\n2 1 2\n4\n1 2 3 1\n5\n1 2 1 2 1\n9\n1 2 1 3 2 3 1 2 1\n9\n2 1 3 1 3 1 3 1 3"] | 2 seconds | ["0\n2 1 2\n1\n1 3 3 1\n1\n1 2 2 2 1\n2\n1 2 3 3 2 3 3 2 1\n2\n2 1 3 3 3 1 1 1 3"] | NoteIn the first example, the array contains no local maximum, so we don't need to perform operations.In the second example, we can change $$$a_2$$$ to $$$3$$$, then the array don't have local maximums. | Java 11 | standard input | [
"greedy"
] | 255abf8750541f54d8ff0f1f80e1f8e7 | Each test contains multiple test cases. The first line will contain a single integer $$$t$$$ $$$(1 \leq t \leq 10000)$$$ — the number of test cases. Then $$$t$$$ test cases follow. The first line of each test case contains a single integer $$$n$$$ $$$(2 \leq n \leq 2 \cdot 10^5)$$$ — the size of the array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots ,a_n$$$ $$$(1 \leq a_i \leq 10^9)$$$, the elements of array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, first output a line containing a single integer $$$m$$$ — minimum number of operations required. Then ouput a line consist of $$$n$$$ integers — the resulting array after the operations. Note that this array should differ in exactly $$$m$$$ elements from the initial array. If there are multiple answers, print any. | standard output | |
PASSED | 2c1169165cd35adc3a20e90c126decd4 | train_108.jsonl | 1645367700 | You are given an array $$$a$$$ of size $$$n$$$. Each element in this array is an integer between $$$1$$$ and $$$10^9$$$.You can perform several operations to this array. During an operation, you can replace an element in the array with any integer between $$$1$$$ and $$$10^9$$$. Output the minimum number of operations needed such that the resulting array doesn't contain any local maximums, and the resulting array after the operations.An element $$$a_i$$$ is a local maximum if it is strictly larger than both of its neighbors (that is, $$$a_i > a_{i - 1}$$$ and $$$a_i > a_{i + 1}$$$). Since $$$a_1$$$ and $$$a_n$$$ have only one neighbor each, they will never be a local maximum. | 256 megabytes | import java.io.*;
import java.util.*;
public class Main {
static FastScanner sc = new FastScanner();
static PrintWriter out = new PrintWriter(new BufferedOutputStream(System.out));
public static void main(String[] args) {
int t = sc.nextInt();
int max = 2_000_00;
while(t-- > 0) {
int n = sc.nextInt();
int[] a = sc.readArray(n);
int cnt = 0;
for(int i = 1; i < n - 1; i++){
if (a[i] > a[i + 1] && a[i] > a[i - 1]) {
if(i +2 < n && a[i+2] >= a[i]){
a[i+1] = a[i+2];
}else{
a[i+1] = a[i];
}
cnt++;
i++;
}
}
out.println(cnt);
for (int num : a) {
out.print(num + " ");
}
out.println();
}
out.close();
}
private 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();
}
boolean hasNext() {
return st.hasMoreTokens();
}
char[] readCharArray(int n) {
char[] arr = new char[n];
try {
br.read(arr);
br.readLine();
} catch (IOException e) {
e.printStackTrace();
}
return arr;
}
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[] readLongArray(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());
}
}
}
| Java | ["5\n3\n2 1 2\n4\n1 2 3 1\n5\n1 2 1 2 1\n9\n1 2 1 3 2 3 1 2 1\n9\n2 1 3 1 3 1 3 1 3"] | 2 seconds | ["0\n2 1 2\n1\n1 3 3 1\n1\n1 2 2 2 1\n2\n1 2 3 3 2 3 3 2 1\n2\n2 1 3 3 3 1 1 1 3"] | NoteIn the first example, the array contains no local maximum, so we don't need to perform operations.In the second example, we can change $$$a_2$$$ to $$$3$$$, then the array don't have local maximums. | Java 11 | standard input | [
"greedy"
] | 255abf8750541f54d8ff0f1f80e1f8e7 | Each test contains multiple test cases. The first line will contain a single integer $$$t$$$ $$$(1 \leq t \leq 10000)$$$ — the number of test cases. Then $$$t$$$ test cases follow. The first line of each test case contains a single integer $$$n$$$ $$$(2 \leq n \leq 2 \cdot 10^5)$$$ — the size of the array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots ,a_n$$$ $$$(1 \leq a_i \leq 10^9)$$$, the elements of array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, first output a line containing a single integer $$$m$$$ — minimum number of operations required. Then ouput a line consist of $$$n$$$ integers — the resulting array after the operations. Note that this array should differ in exactly $$$m$$$ elements from the initial array. If there are multiple answers, print any. | standard output | |
PASSED | 9d66ee0f7e6b72724d669f3bcbfa3fd0 | train_108.jsonl | 1645367700 | You are given an array $$$a$$$ of size $$$n$$$. Each element in this array is an integer between $$$1$$$ and $$$10^9$$$.You can perform several operations to this array. During an operation, you can replace an element in the array with any integer between $$$1$$$ and $$$10^9$$$. Output the minimum number of operations needed such that the resulting array doesn't contain any local maximums, and the resulting array after the operations.An element $$$a_i$$$ is a local maximum if it is strictly larger than both of its neighbors (that is, $$$a_i > a_{i - 1}$$$ and $$$a_i > a_{i + 1}$$$). Since $$$a_1$$$ and $$$a_n$$$ have only one neighbor each, they will never be a local maximum. | 256 megabytes | import java.util.Arrays;
import java.util.List;
import java.util.Scanner;
public class Main {
public static void main(String[] args) {
Scanner scan= new Scanner(System.in);
int T=scan.nextInt();
for (int i = 0; i < T; i++) {
int n=scan.nextInt();
int[] li=new int[n];
for (int j = 0; j < n; j++) {
li[j]=scan.nextInt();
}
int c=0;
for (int j = 0; j < n; j++) {
if(j!=0 && j!=n-1 && (li[j]> li[j-1]) && (li[j]>li[j+1])){
if(j+2<n){
li[j+1]=Math.max(li[j],li[j+2]);
}else{
li[j+1]=li[j];
}
c++;
}
}
System.out.println(c+"\n"+Arrays.toString(li).replace('[',' ').replace(']',' ').replace(',',' '));
}
}
} | Java | ["5\n3\n2 1 2\n4\n1 2 3 1\n5\n1 2 1 2 1\n9\n1 2 1 3 2 3 1 2 1\n9\n2 1 3 1 3 1 3 1 3"] | 2 seconds | ["0\n2 1 2\n1\n1 3 3 1\n1\n1 2 2 2 1\n2\n1 2 3 3 2 3 3 2 1\n2\n2 1 3 3 3 1 1 1 3"] | NoteIn the first example, the array contains no local maximum, so we don't need to perform operations.In the second example, we can change $$$a_2$$$ to $$$3$$$, then the array don't have local maximums. | Java 11 | standard input | [
"greedy"
] | 255abf8750541f54d8ff0f1f80e1f8e7 | Each test contains multiple test cases. The first line will contain a single integer $$$t$$$ $$$(1 \leq t \leq 10000)$$$ — the number of test cases. Then $$$t$$$ test cases follow. The first line of each test case contains a single integer $$$n$$$ $$$(2 \leq n \leq 2 \cdot 10^5)$$$ — the size of the array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots ,a_n$$$ $$$(1 \leq a_i \leq 10^9)$$$, the elements of array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, first output a line containing a single integer $$$m$$$ — minimum number of operations required. Then ouput a line consist of $$$n$$$ integers — the resulting array after the operations. Note that this array should differ in exactly $$$m$$$ elements from the initial array. If there are multiple answers, print any. | standard output | |
PASSED | 4e59422eab83521ce4b224b4c2e73946 | train_108.jsonl | 1645367700 | You are given an array $$$a$$$ of size $$$n$$$. Each element in this array is an integer between $$$1$$$ and $$$10^9$$$.You can perform several operations to this array. During an operation, you can replace an element in the array with any integer between $$$1$$$ and $$$10^9$$$. Output the minimum number of operations needed such that the resulting array doesn't contain any local maximums, and the resulting array after the operations.An element $$$a_i$$$ is a local maximum if it is strictly larger than both of its neighbors (that is, $$$a_i > a_{i - 1}$$$ and $$$a_i > a_{i + 1}$$$). Since $$$a_1$$$ and $$$a_n$$$ have only one neighbor each, they will never be a local maximum. | 256 megabytes | import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.*;
import static java.lang.Math.*;
import static java.lang.Math.ceil;
import static java.util.Arrays.sort;
public class Round11 {
public static void main(String[] args) {
FastReader fastreader = new FastReader();
PrintWriter out = new PrintWriter(System.out);
int t = fastreader.nextInt();
while (t-- > 0) {
int n = fastreader.nextInt();
int[] a = new int[n];
for (int i = 0; i < n; i++) a[i] = fastreader.nextInt();
int copy[] = copy(a);
int m = 0;
for (int i = 1; i < n - 1; i++) {
if (a[i] > a[i - 1] && a[i] > a[i + 1]) {
if (a[i - 1] != copy[i - 1]) {
a[i - 1] = a[i];
} else {
m++;
a[i + 1] = a[i];
}
}
}
StringBuilder ans = new StringBuilder();
for (int i = 0; i < n; i++) {
ans.append(a[i]).append(" ");
}
out.println(m);
out.println(ans);
}
out.close();
}
// constants
static final int IBIG = 1000000007;
static final int IMAX = 2147483647;
static final long LMAX = 9223372036854775807L;
static Random __r = new Random();
// math util
static int minof(int a, int b, int c) {
return min(a, min(b, c));
}
static int minof(int... x) {
if (x.length == 1) return x[0];
if (x.length == 2) return min(x[0], x[1]);
if (x.length == 3) return min(x[0], min(x[1], x[2]));
int min = x[0];
for (int i = 1; i < x.length; ++i) if (x[i] < min) min = x[i];
return min;
}
static long minof(long a, long b, long c) {
return min(a, min(b, c));
}
static long minof(long... x) {
if (x.length == 1) return x[0];
if (x.length == 2) return min(x[0], x[1]);
if (x.length == 3) return min(x[0], min(x[1], x[2]));
long min = x[0];
for (int i = 1; i < x.length; ++i) if (x[i] < min) min = x[i];
return min;
}
static int maxof(int a, int b, int c) {
return max(a, max(b, c));
}
static int maxof(int... x) {
if (x.length == 1) return x[0];
if (x.length == 2) return max(x[0], x[1]);
if (x.length == 3) return max(x[0], max(x[1], x[2]));
int max = x[0];
for (int i = 1; i < x.length; ++i) if (x[i] > max) max = x[i];
return max;
}
static long maxof(long a, long b, long c) {
return max(a, max(b, c));
}
static long maxof(long... x) {
if (x.length == 1) return x[0];
if (x.length == 2) return max(x[0], x[1]);
if (x.length == 3) return max(x[0], max(x[1], x[2]));
long max = x[0];
for (int i = 1; i < x.length; ++i) if (x[i] > max) max = x[i];
return max;
}
static int powi(int a, int b) {
if (a == 0) return 0;
int ans = 1;
while (b > 0) {
if ((b & 1) > 0) ans *= a;
a *= a;
b >>= 1;
}
return ans;
}
static long powl(long a, int b) {
if (a == 0) return 0;
long ans = 1;
while (b > 0) {
if ((b & 1) > 0) ans *= a;
a *= a;
b >>= 1;
}
return ans;
}
static int fli(double d) {
return (int) d;
}
static int cei(double d) {
return (int) ceil(d);
}
static long fll(double d) {
return (long) d;
}
static long cel(double d) {
return (long) ceil(d);
}
static int gcd(int a, int b) {
return b == 0 ? a : gcd(b, a % b);
}
static int lcm(int a, int b) {
return (a / gcd(a, b)) * b;
}
static long gcd(long a, long b) {
return b == 0 ? a : gcd(b, a % b);
}
static long lcm(long a, long b) {
return (a / gcd(a, b)) * b;
}
static int[] exgcd(int a, int b) {
if (b == 0) return new int[]{1, 0};
int[] y = exgcd(b, a % b);
return new int[]{y[1], y[0] - y[1] * (a / b)};
}
static long[] exgcd(long a, long b) {
if (b == 0) return new long[]{1, 0};
long[] y = exgcd(b, a % b);
return new long[]{y[1], y[0] - y[1] * (a / b)};
}
static int randInt(int min, int max) {
return __r.nextInt(max - min + 1) + min;
}
static long mix(long x) {
x += 0x9e3779b97f4a7c15L;
x = (x ^ (x >> 30)) * 0xbf58476d1ce4e5b9L;
x = (x ^ (x >> 27)) * 0x94d049bb133111ebL;
return x ^ (x >> 31);
}
public static boolean[] findPrimes(int limit) {
assert limit >= 2;
final boolean[] nonPrimes = new boolean[limit];
nonPrimes[0] = true;
nonPrimes[1] = true;
int sqrt = (int) Math.sqrt(limit);
for (int i = 2; i <= sqrt; i++) {
if (nonPrimes[i]) continue;
for (int j = i; j < limit; j += i) {
if (!nonPrimes[j] && i != j) nonPrimes[j] = true;
}
}
return nonPrimes;
}
// array util
static void reverse(int[] a) {
for (int i = 0, n = a.length, half = n / 2; i < half; ++i) {
int swap = a[i];
a[i] = a[n - i - 1];
a[n - i - 1] = swap;
}
}
static void reverse(long[] a) {
for (int i = 0, n = a.length, half = n / 2; i < half; ++i) {
long swap = a[i];
a[i] = a[n - i - 1];
a[n - i - 1] = swap;
}
}
static void reverse(double[] a) {
for (int i = 0, n = a.length, half = n / 2; i < half; ++i) {
double swap = a[i];
a[i] = a[n - i - 1];
a[n - i - 1] = swap;
}
}
static void reverse(char[] a) {
for (int i = 0, n = a.length, half = n / 2; i < half; ++i) {
char swap = a[i];
a[i] = a[n - i - 1];
a[n - i - 1] = swap;
}
}
static void shuffle(int[] a) {
int n = a.length - 1;
for (int i = 0; i < n; ++i) {
int ind = randInt(i, n);
int swap = a[i];
a[i] = a[ind];
a[ind] = swap;
}
}
static void shuffle(long[] a) {
int n = a.length - 1;
for (int i = 0; i < n; ++i) {
int ind = randInt(i, n);
long swap = a[i];
a[i] = a[ind];
a[ind] = swap;
}
}
static void shuffle(double[] a) {
int n = a.length - 1;
for (int i = 0; i < n; ++i) {
int ind = randInt(i, n);
double swap = a[i];
a[i] = a[ind];
a[ind] = swap;
}
}
static void rsort(int[] a) {
shuffle(a);
sort(a);
}
static void rsort(long[] a) {
shuffle(a);
sort(a);
}
static void rsort(double[] a) {
shuffle(a);
sort(a);
}
static int[] copy(int[] a) {
int[] ans = new int[a.length];
for (int i = 0; i < a.length; ++i) ans[i] = a[i];
return ans;
}
static long[] copy(long[] a) {
long[] ans = new long[a.length];
for (int i = 0; i < a.length; ++i) ans[i] = a[i];
return ans;
}
static double[] copy(double[] a) {
double[] ans = new double[a.length];
for (int i = 0; i < a.length; ++i) ans[i] = a[i];
return ans;
}
static char[] copy(char[] a) {
char[] ans = new char[a.length];
for (int i = 0; i < a.length; ++i) ans[i] = a[i];
return ans;
}
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 | ["5\n3\n2 1 2\n4\n1 2 3 1\n5\n1 2 1 2 1\n9\n1 2 1 3 2 3 1 2 1\n9\n2 1 3 1 3 1 3 1 3"] | 2 seconds | ["0\n2 1 2\n1\n1 3 3 1\n1\n1 2 2 2 1\n2\n1 2 3 3 2 3 3 2 1\n2\n2 1 3 3 3 1 1 1 3"] | NoteIn the first example, the array contains no local maximum, so we don't need to perform operations.In the second example, we can change $$$a_2$$$ to $$$3$$$, then the array don't have local maximums. | Java 11 | standard input | [
"greedy"
] | 255abf8750541f54d8ff0f1f80e1f8e7 | Each test contains multiple test cases. The first line will contain a single integer $$$t$$$ $$$(1 \leq t \leq 10000)$$$ — the number of test cases. Then $$$t$$$ test cases follow. The first line of each test case contains a single integer $$$n$$$ $$$(2 \leq n \leq 2 \cdot 10^5)$$$ — the size of the array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots ,a_n$$$ $$$(1 \leq a_i \leq 10^9)$$$, the elements of array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, first output a line containing a single integer $$$m$$$ — minimum number of operations required. Then ouput a line consist of $$$n$$$ integers — the resulting array after the operations. Note that this array should differ in exactly $$$m$$$ elements from the initial array. If there are multiple answers, print any. | standard output | |
PASSED | 40b8d2c5468219e9ac4d6895c932f4f3 | train_108.jsonl | 1645367700 | You are given an array $$$a$$$ of size $$$n$$$. Each element in this array is an integer between $$$1$$$ and $$$10^9$$$.You can perform several operations to this array. During an operation, you can replace an element in the array with any integer between $$$1$$$ and $$$10^9$$$. Output the minimum number of operations needed such that the resulting array doesn't contain any local maximums, and the resulting array after the operations.An element $$$a_i$$$ is a local maximum if it is strictly larger than both of its neighbors (that is, $$$a_i > a_{i - 1}$$$ and $$$a_i > a_{i + 1}$$$). Since $$$a_1$$$ and $$$a_n$$$ have only one neighbor each, they will never be a local maximum. | 256 megabytes | import java.util.Scanner;
public class B_Avoid_Local_Maximums {
public static void main(String[] args) {
Scanner scan = new Scanner(System.in);
int t = scan.nextInt();
while (t-->0){
int n = scan.nextInt();
int[] arr = new int[n];
int count=0;
for (int i = 0; i < n; i++)
arr[i] = scan.nextInt();
for (int i = 1; i < n-1; i++) {
if( arr[i] > arr[i-1] && arr[i]> arr[i+1] ){
if( i+2 < n ) {
arr[i + 1] = Math.max(arr[i], arr[i + 2]);
}
else {
arr[i+1]=arr[i];
}
count++;
}
}
System.out.println(count);
for(int x : arr){
System.out.print(x+" ");
}
System.out.println();
}
}
} | Java | ["5\n3\n2 1 2\n4\n1 2 3 1\n5\n1 2 1 2 1\n9\n1 2 1 3 2 3 1 2 1\n9\n2 1 3 1 3 1 3 1 3"] | 2 seconds | ["0\n2 1 2\n1\n1 3 3 1\n1\n1 2 2 2 1\n2\n1 2 3 3 2 3 3 2 1\n2\n2 1 3 3 3 1 1 1 3"] | NoteIn the first example, the array contains no local maximum, so we don't need to perform operations.In the second example, we can change $$$a_2$$$ to $$$3$$$, then the array don't have local maximums. | Java 11 | standard input | [
"greedy"
] | 255abf8750541f54d8ff0f1f80e1f8e7 | Each test contains multiple test cases. The first line will contain a single integer $$$t$$$ $$$(1 \leq t \leq 10000)$$$ — the number of test cases. Then $$$t$$$ test cases follow. The first line of each test case contains a single integer $$$n$$$ $$$(2 \leq n \leq 2 \cdot 10^5)$$$ — the size of the array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots ,a_n$$$ $$$(1 \leq a_i \leq 10^9)$$$, the elements of array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, first output a line containing a single integer $$$m$$$ — minimum number of operations required. Then ouput a line consist of $$$n$$$ integers — the resulting array after the operations. Note that this array should differ in exactly $$$m$$$ elements from the initial array. If there are multiple answers, print any. | standard output | |
PASSED | 130ee6f704a9acb094f41f0552895bdd | train_108.jsonl | 1645367700 | You are given an array $$$a$$$ of size $$$n$$$. Each element in this array is an integer between $$$1$$$ and $$$10^9$$$.You can perform several operations to this array. During an operation, you can replace an element in the array with any integer between $$$1$$$ and $$$10^9$$$. Output the minimum number of operations needed such that the resulting array doesn't contain any local maximums, and the resulting array after the operations.An element $$$a_i$$$ is a local maximum if it is strictly larger than both of its neighbors (that is, $$$a_i > a_{i - 1}$$$ and $$$a_i > a_{i + 1}$$$). Since $$$a_1$$$ and $$$a_n$$$ have only one neighbor each, they will never be a local maximum. | 256 megabytes | import java.util.Scanner;
public class Main {
public static void main (String[] args) {
Scanner s = new Scanner(System.in);
int z = s.nextInt();
while (z-->0) {
int a = s.nextInt();
int[] b = new int[a] ;
for (int i = 0; i <a ; i++){
b[i] = s.nextInt();
}
int count = 0;
for (int i = a-2; i >1 ; i--){
if (b[i-1] < b[i] && b[i] > b[i+1]){
b[i-1] = Math.max(b[i], b[i-2]);
count ++;
}
}
if(a>2) {
if (b[1] > b[2] && b[1] > b[0]) {
b[0] = b[1];
count++;
}
}
System.out.println(count);
for (int i = 0; i < a; i++) {
System.out.print(b[i] + " ");
}
System.out.println();
}
}
} | Java | ["5\n3\n2 1 2\n4\n1 2 3 1\n5\n1 2 1 2 1\n9\n1 2 1 3 2 3 1 2 1\n9\n2 1 3 1 3 1 3 1 3"] | 2 seconds | ["0\n2 1 2\n1\n1 3 3 1\n1\n1 2 2 2 1\n2\n1 2 3 3 2 3 3 2 1\n2\n2 1 3 3 3 1 1 1 3"] | NoteIn the first example, the array contains no local maximum, so we don't need to perform operations.In the second example, we can change $$$a_2$$$ to $$$3$$$, then the array don't have local maximums. | Java 11 | standard input | [
"greedy"
] | 255abf8750541f54d8ff0f1f80e1f8e7 | Each test contains multiple test cases. The first line will contain a single integer $$$t$$$ $$$(1 \leq t \leq 10000)$$$ — the number of test cases. Then $$$t$$$ test cases follow. The first line of each test case contains a single integer $$$n$$$ $$$(2 \leq n \leq 2 \cdot 10^5)$$$ — the size of the array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots ,a_n$$$ $$$(1 \leq a_i \leq 10^9)$$$, the elements of array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, first output a line containing a single integer $$$m$$$ — minimum number of operations required. Then ouput a line consist of $$$n$$$ integers — the resulting array after the operations. Note that this array should differ in exactly $$$m$$$ elements from the initial array. If there are multiple answers, print any. | standard output | |
PASSED | 8b93187f45d4b754da1001dfe850fb8a | train_108.jsonl | 1645367700 | You are given an array $$$a$$$ of size $$$n$$$. Each element in this array is an integer between $$$1$$$ and $$$10^9$$$.You can perform several operations to this array. During an operation, you can replace an element in the array with any integer between $$$1$$$ and $$$10^9$$$. Output the minimum number of operations needed such that the resulting array doesn't contain any local maximums, and the resulting array after the operations.An element $$$a_i$$$ is a local maximum if it is strictly larger than both of its neighbors (that is, $$$a_i > a_{i - 1}$$$ and $$$a_i > a_{i + 1}$$$). Since $$$a_1$$$ and $$$a_n$$$ have only one neighbor each, they will never be a local maximum. | 256 megabytes | //Some of the methods are copied from GeeksforGeeks Website
import java.util.*;
import java.lang.*;
import java.io.*;
public class Main
{
//static Scanner sc=new Scanner(System.in);
static Reader sc=new Reader();
// static FastReader sc=new FastReader(System.in);
static long mod = (long)(1e9)+ 7;
static int max_num=(int)1e5+5;
public static void main (String[] args) throws java.lang.Exception
{
try{
/*
Collections.sort(al,(a,b)->a.x-b.x);
Collections.sort(al,Collections.reverseOrder());
long n=sc.nextLong();
String s=sc.next();
char a[]=s.toCharArray();
StringBuilder sb=new StringBuilder();
map.put(a[i],map.getOrDefault(a[i],0)+1);
*/
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 op=0;
for(int i=1;i<n-2;i++)
{
if(a[i]>a[i-1] && a[i]>a[i+1])
{
op++;
a[i+1]=Math.max(a[i],a[i+2]);
}
}
if(n>=3 && a[n-2]>a[n-3] && a[n-2]>a[n-1])
{
op++;
a[n-1]=a[n-2];
}
out.println(op);
print(a);
}
out.flush();
out.close();
}
catch(Exception e)
{}
}
/*
...SOLUTION ENDS HERE...........SOLUTION ENDS HERE...
*/
static void flag(boolean flag)
{
out.println(flag ? "YES" : "NO");
out.flush();
}
/*
Map<Long,Long> map=new HashMap<>();
for(int i=0;i<n;i++)
{
if(!map.containsKey(a[i]))
map.put(a[i],1);
else
map.replace(a[i],map.get(a[i])+1);
}
Set<Map.Entry<Long,Long>> hmap=map.entrySet();
for(Map.Entry<Long,Long> data : hmap)
{
}
Iterator<Integer> itr = set.iterator();
while(itr.hasNext())
{
int val=itr.next();
}
*/
// static class Pair
// {
// int x,y;
// Pair(int x,int y)
// {
// this.x=x;
// this.y=y;
// }
// }
// Arrays.sort(p, new Comparator<Pair>()
// {
// @Override
// public int compare(Pair o1,Pair o2)
// {
// if(o1.x>o2.x) return 1;
// else if(o1.x==o2.x)
// {
// if(o1.y>o2.y) return 1;
// else return -1;
// }
// else return -1;
// }});
static void print(int a[])
{
int n=a.length;
for(int i=0;i<n;i++)
{
out.print(a[i]+" ");
}
out.println();
out.flush();
}
static void print(long a[])
{
int n=a.length;
for(int i=0;i<n;i++)
{
out.print(a[i]+" ");
}
out.println();
out.flush();
}
static void print_int(ArrayList<Integer> al)
{
int si=al.size();
for(int i=0;i<si;i++)
{
out.print(al.get(i)+" ");
}
out.println();
out.flush();
}
static void print_long(ArrayList<Long> al)
{
int si=al.size();
for(int i=0;i<si;i++)
{
out.print(al.get(i)+" ");
}
out.println();
out.flush();
}
static int LowerBound(int a[], int x)
{ // x is the target value or key
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)
{// x is the key or target value
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 void DFS(ArrayList<Integer> graph[],boolean[] visited, int u)
{
visited[u]=true;
int v=0;
for(int i=0;i<graph[u].size();i++)
{
v=graph[u].get(i);
if(!visited[v])
DFS(graph,visited,v);
}
}
static boolean[] prime(int num)
{
boolean[] bool = new boolean[num];
for (int i = 0; i< bool.length; i++) {
bool[i] = true;
}
for (int i = 2; i< Math.sqrt(num); i++) {
if(bool[i] == true) {
for(int j = (i*i); j<num; j = j+i) {
bool[j] = false;
}
}
}
if(num >= 0) {
bool[0] = false;
bool[1] = false;
}
return bool;
}
static long nCr(long a,long b,long mod)
{
return (((fact[(int)a] * modInverse(fact[(int)b],mod))%mod * modInverse(fact[(int)(a - b)],mod))%mod + mod)%mod;
}
static long fact[]=new long[max_num];
static void fact_fill()
{
fact[0]=1l;
for(int i=1;i<max_num;i++)
{
fact[i]=(fact[i-1]*(long)i);
if(fact[i]>=mod)
fact[i]%=mod;
}
}
static long modInverse(long a, long m)
{
return power(a, m - 2, m);
}
static long power(long x, long y, long m)
{
if (y == 0)
return 1;
long p = power(x, y / 2, m) % m;
p = (long)((p * (long)p) % m);
if (y % 2 == 0)
return p;
else
return (long)((x * (long)p) % m);
}
static long sum_array(int a[])
{
int n=a.length;
long sum=0;
for(int i=0;i<n;i++)
sum+=a[i];
return sum;
}
static long sum_array(long a[])
{
int n=a.length;
long sum=0;
for(int i=0;i<n;i++)
sum+=a[i];
return sum;
}
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 void reverse_array(int a[])
{
int n=a.length;
int i,t;
for (i = 0; i < n / 2; i++) {
t = a[i];
a[i] = a[n - i - 1];
a[n - i - 1] = t;
}
}
static void reverse_array(long a[])
{
int n=a.length;
int i; long t;
for (i = 0; i < n / 2; i++) {
t = a[i];
a[i] = a[n - i - 1];
a[n - i - 1] = t;
}
}
static long gcd(long a, long b)
{
if (a == 0)
return b;
return gcd(b%a, a);
}
static int gcd(int a, int b)
{
if (a == 0)
return b;
return gcd(b%a, a);
}
static class FastReader{
byte[] buf = new byte[2048];
int index, total;
InputStream in;
FastReader(InputStream is) {
in = is;
}
int scan() throws IOException {
if (index >= total) {
index = 0;
total = in.read(buf);
if (total <= 0) {
return -1;
}
}
return buf[index++];
}
String next() throws IOException {
int c;
for (c = scan(); c <= 32; c = scan());
StringBuilder sb = new StringBuilder();
for (; c > 32; c = scan()) {
sb.append((char) c);
}
return sb.toString();
}
int nextInt() throws IOException {
int c, val = 0;
for (c = scan(); c <= 32; c = scan());
boolean neg = c == '-';
if (c == '-' || c == '+') {
c = scan();
}
for (; c >= '0' && c <= '9'; c = scan()) {
val = (val << 3) + (val << 1) + (c & 15);
}
return neg ? -val : val;
}
long nextLong() throws IOException {
int c;
long val = 0;
for (c = scan(); c <= 32; c = scan());
boolean neg = c == '-';
if (c == '-' || c == '+') {
c = scan();
}
for (; c >= '0' && c <= '9'; c = scan()) {
val = (val << 3) + (val << 1) + (c & 15);
}
return neg ? -val : val;
}
}
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[64]; // line length
int cnt = 0, c;
while ((c = read()) != -1)
{
if (c == '\n')
break;
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 PrintWriter out=new PrintWriter(System.out);
static int int_max=Integer.MAX_VALUE;
static int int_min=Integer.MIN_VALUE;
static long long_max=Long.MAX_VALUE;
static long long_min=Long.MIN_VALUE;
}
// Thank You ! | Java | ["5\n3\n2 1 2\n4\n1 2 3 1\n5\n1 2 1 2 1\n9\n1 2 1 3 2 3 1 2 1\n9\n2 1 3 1 3 1 3 1 3"] | 2 seconds | ["0\n2 1 2\n1\n1 3 3 1\n1\n1 2 2 2 1\n2\n1 2 3 3 2 3 3 2 1\n2\n2 1 3 3 3 1 1 1 3"] | NoteIn the first example, the array contains no local maximum, so we don't need to perform operations.In the second example, we can change $$$a_2$$$ to $$$3$$$, then the array don't have local maximums. | Java 11 | standard input | [
"greedy"
] | 255abf8750541f54d8ff0f1f80e1f8e7 | Each test contains multiple test cases. The first line will contain a single integer $$$t$$$ $$$(1 \leq t \leq 10000)$$$ — the number of test cases. Then $$$t$$$ test cases follow. The first line of each test case contains a single integer $$$n$$$ $$$(2 \leq n \leq 2 \cdot 10^5)$$$ — the size of the array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots ,a_n$$$ $$$(1 \leq a_i \leq 10^9)$$$, the elements of array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, first output a line containing a single integer $$$m$$$ — minimum number of operations required. Then ouput a line consist of $$$n$$$ integers — the resulting array after the operations. Note that this array should differ in exactly $$$m$$$ elements from the initial array. If there are multiple answers, print any. | standard output | |
PASSED | 63ea1eb9659856a358b3bd99bc318a67 | train_108.jsonl | 1645367700 | You are given an array $$$a$$$ of size $$$n$$$. Each element in this array is an integer between $$$1$$$ and $$$10^9$$$.You can perform several operations to this array. During an operation, you can replace an element in the array with any integer between $$$1$$$ and $$$10^9$$$. Output the minimum number of operations needed such that the resulting array doesn't contain any local maximums, and the resulting array after the operations.An element $$$a_i$$$ is a local maximum if it is strictly larger than both of its neighbors (that is, $$$a_i > a_{i - 1}$$$ and $$$a_i > a_{i + 1}$$$). Since $$$a_1$$$ and $$$a_n$$$ have only one neighbor each, they will never be a local maximum. | 256 megabytes | import java.util.Scanner;
public class Main {
public static void main(String args[]) {
Scanner scan = new Scanner(System.in);
int testCases = Integer.valueOf(scan.nextLine());
while(testCases>0) {
int size = Integer.valueOf(scan.nextLine());
String raw = scan.nextLine();
String[] data = raw.split(" ");
int[] array = new int[size];
for(int i = 0 ; i<size ; i++) {
array[i] = Integer.valueOf(data[i]);
}
int count = 0;
for(int i = 0 ; i<size-2 ; i++) {
int first = array[i];
int second = array[i+1];
int third = array[i+2];
if(second > first && second>third) {
if(i+4 < size) {
int fourth = array[i+3];
int fifth = array[i+4];
if(fourth > fifth && fourth > third && fourth >= second) {
array[i+2] = array[i+3];
count++;
}
else {
array[i+2] = array[i+1];
count++;
}
}
else {
array[i+2] = array[i+1];
count++;
}
}
}
System.out.println(count);
for(int i:array) {
System.out.print(i+" ");
}
System.out.println();
testCases--;
}
}
}
| Java | ["5\n3\n2 1 2\n4\n1 2 3 1\n5\n1 2 1 2 1\n9\n1 2 1 3 2 3 1 2 1\n9\n2 1 3 1 3 1 3 1 3"] | 2 seconds | ["0\n2 1 2\n1\n1 3 3 1\n1\n1 2 2 2 1\n2\n1 2 3 3 2 3 3 2 1\n2\n2 1 3 3 3 1 1 1 3"] | NoteIn the first example, the array contains no local maximum, so we don't need to perform operations.In the second example, we can change $$$a_2$$$ to $$$3$$$, then the array don't have local maximums. | Java 11 | standard input | [
"greedy"
] | 255abf8750541f54d8ff0f1f80e1f8e7 | Each test contains multiple test cases. The first line will contain a single integer $$$t$$$ $$$(1 \leq t \leq 10000)$$$ — the number of test cases. Then $$$t$$$ test cases follow. The first line of each test case contains a single integer $$$n$$$ $$$(2 \leq n \leq 2 \cdot 10^5)$$$ — the size of the array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots ,a_n$$$ $$$(1 \leq a_i \leq 10^9)$$$, the elements of array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, first output a line containing a single integer $$$m$$$ — minimum number of operations required. Then ouput a line consist of $$$n$$$ integers — the resulting array after the operations. Note that this array should differ in exactly $$$m$$$ elements from the initial array. If there are multiple answers, print any. | standard output | |
PASSED | 24be61e8ae908659b1862408a7e56f4b | train_108.jsonl | 1645367700 | You are given an array $$$a$$$ of size $$$n$$$. Each element in this array is an integer between $$$1$$$ and $$$10^9$$$.You can perform several operations to this array. During an operation, you can replace an element in the array with any integer between $$$1$$$ and $$$10^9$$$. Output the minimum number of operations needed such that the resulting array doesn't contain any local maximums, and the resulting array after the operations.An element $$$a_i$$$ is a local maximum if it is strictly larger than both of its neighbors (that is, $$$a_i > a_{i - 1}$$$ and $$$a_i > a_{i + 1}$$$). Since $$$a_1$$$ and $$$a_n$$$ have only one neighbor each, they will never be a local maximum. | 256 megabytes | import java.util.*;
import java.io.*;
public class Main {
static StringBuilder sb;
static dsu dsu;
static long fact[];
static int mod = (int) (1e9 + 7);
static void solve() {
int n=i();
int[]arr=new int[n];
for(int i=0;i<n;i++)arr[i]=i();
int op=0;
for(int i=1;i<n-1;i++){
if(arr[i]>arr[i-1]&&arr[i]>arr[i+1]){
if(i+2<n){
if(arr[i+2]<=arr[i+1]){// sor
op++;
arr[i+1]=arr[i];
}
else {
arr[i+1]=arr[i];
arr[i+1]=Math.max(arr[i+2],arr[i]);
op++;
}
}
else{
op++;
arr[i+1]=1000000000;
}
}
}
sb.append(op+"\n");
for(int i=0;i<n;i++){
sb.append(arr[i]+" ");
}
sb.append("\n");
}
public static void main(String[] args) {
sb = new StringBuilder();
int test = i();
fact=new long[(int)1e6+10];
fact[0]=fact[1]=1;
for(int i=2;i<fact.length;i++)
{ fact[i]=((long)(i%mod)*(long)(fact[i-1]%mod))%mod; }
while (test-- > 0) {
solve();
}
System.out.println(sb);
}
//**************NCR%P******************
static long ncr(int n, int r) {
if (r > n)
return (long) 0;
long res = fact[n] % mod;
// System.out.println(res);
res = ((long) (res % mod) * (long) (p(fact[r], mod - 2) % mod)) % mod;
res = ((long) (res % mod) * (long) (p(fact[n - r], mod - 2) % mod)) % mod;
// System.out.println(res);
return res;
}
static long p(long x, long y)// POWER FXN //
{
if (y == 0)
return 1;
long res = 1;
while (y > 0) {
if (y % 2 == 1) {
res = (res * x) % mod;
y--;
}
x = (x * x) % mod;
y = y / 2;
}
return res;
}
//**************END******************
// *************Disjoint set
// union*********//
static class dsu {
int parent[];
dsu(int n) {
parent = new int[n];
for (int i = 0; i < n; i++)
parent[i] = i;
}
int find(int a) {
if (parent[a] ==a)
return a;
else {
int x = find(parent[a]);
parent[a] = x;
return x;
}
}
void merge(int a, int b) {
a = find(a);
b = find(b);
if (a == b)
return;
parent[b] = a;
}
}
//**************PRIME FACTORIZE **********************************//
static TreeMap<Integer, Integer> prime(long n) {
TreeMap<Integer, Integer> h = new TreeMap<>();
long num = n;
for (int i = 2; i <= Math.sqrt(num); i++) {
if (n % i == 0) {
int nt = 0;
while (n % i == 0) {
n = n / i;
nt++;
}
h.put(i, nt);
}
}
if (n != 1)
h.put((int) n, 1);
return h;
}
//****CLASS PAIR ************************************************
static class Pair implements Comparable<Pair> {
int x;
long y;
Pair(int x, long y) {
this.x = x;
this.y = y;
}
public int compareTo(Pair o) {
return (int) (this.y - o.y);
}
}
//****CLASS PAIR **************************************************
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 class OutputWriter {
private final PrintWriter writer;
public OutputWriter(OutputStream outputStream) {
writer = new PrintWriter(new BufferedWriter(new OutputStreamWriter(outputStream)));
}
public OutputWriter(Writer writer) {
this.writer = new PrintWriter(writer);
}
public void print(Object... objects) {
for (int i = 0; i < objects.length; i++) {
if (i != 0)
writer.print(' ');
writer.print(objects[i]);
}
}
public void printLine(Object... objects) {
print(objects);
writer.println();
}
public void close() {
writer.close();
}
public void flush() {
writer.flush();
}
}
static InputReader in = new InputReader(System.in);
static OutputWriter out = new OutputWriter(System.out);
public static long[] sort(long[] a2) {
int n = a2.length;
ArrayList<Long> l = new ArrayList<>();
for (long i : a2)
l.add(i);
Collections.sort(l);
for (int i = 0; i < l.size(); i++)
a2[i] = l.get(i);
return a2;
}
public static char[] sort(char[] a2) {
int n = a2.length;
ArrayList<Character> l = new ArrayList<>();
for (char i : a2)
l.add(i);
Collections.sort(l);
for (int i = 0; i < l.size(); i++)
a2[i] = l.get(i);
return a2;
}
public static long pow(long x, long y) {
long res = 1;
while (y > 0) {
if (y % 2 != 0) {
res = (res * x);// % modulus;
y--;
}
x = (x * x);// % modulus;
y = y / 2;
}
return res;
}
//GCD___+++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
public static long gcd(long x, long y) {
if (x == 0)
return y;
else
return gcd(y % x, x);
}
// ******LOWEST COMMON MULTIPLE
// *********************************************
public static long lcm(long x, long y) {
return (x * (y / gcd(x, y)));
}
//INPUT PATTERN********************************************************
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[] readArrayi(int n) {
int A[] = new int[n];
for (int i = 0; i < n; i++) {
A[i] = i();
}
return A;
}
public static long[] readArray(long n) {
long A[] = new long[(int) n];
for (int i = 0; i < n; i++) {
A[i] = l();
}
return A;
}
} | Java | ["5\n3\n2 1 2\n4\n1 2 3 1\n5\n1 2 1 2 1\n9\n1 2 1 3 2 3 1 2 1\n9\n2 1 3 1 3 1 3 1 3"] | 2 seconds | ["0\n2 1 2\n1\n1 3 3 1\n1\n1 2 2 2 1\n2\n1 2 3 3 2 3 3 2 1\n2\n2 1 3 3 3 1 1 1 3"] | NoteIn the first example, the array contains no local maximum, so we don't need to perform operations.In the second example, we can change $$$a_2$$$ to $$$3$$$, then the array don't have local maximums. | Java 11 | standard input | [
"greedy"
] | 255abf8750541f54d8ff0f1f80e1f8e7 | Each test contains multiple test cases. The first line will contain a single integer $$$t$$$ $$$(1 \leq t \leq 10000)$$$ — the number of test cases. Then $$$t$$$ test cases follow. The first line of each test case contains a single integer $$$n$$$ $$$(2 \leq n \leq 2 \cdot 10^5)$$$ — the size of the array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots ,a_n$$$ $$$(1 \leq a_i \leq 10^9)$$$, the elements of array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, first output a line containing a single integer $$$m$$$ — minimum number of operations required. Then ouput a line consist of $$$n$$$ integers — the resulting array after the operations. Note that this array should differ in exactly $$$m$$$ elements from the initial array. If there are multiple answers, print any. | standard output | |
PASSED | af93a97a1956fede57df50fa70c17c50 | train_108.jsonl | 1645367700 | You are given an array $$$a$$$ of size $$$n$$$. Each element in this array is an integer between $$$1$$$ and $$$10^9$$$.You can perform several operations to this array. During an operation, you can replace an element in the array with any integer between $$$1$$$ and $$$10^9$$$. Output the minimum number of operations needed such that the resulting array doesn't contain any local maximums, and the resulting array after the operations.An element $$$a_i$$$ is a local maximum if it is strictly larger than both of its neighbors (that is, $$$a_i > a_{i - 1}$$$ and $$$a_i > a_{i + 1}$$$). Since $$$a_1$$$ and $$$a_n$$$ have only one neighbor each, they will never be a local maximum. | 256 megabytes | import java.util.*;
import java.io.*;
import java.security.KeyStore.Entry;
public class Main{
static final Random random=new Random();
static long mod=1000000007L;
static HashMap<String,Integer>map=new HashMap<>();
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;
}
int[] readIntArray(int n){
int[] res=new int[n];
for(int i=0;i<n;i++)res[i]=nextInt();
return res;
}
}
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();
}
}
static int ans=0;
public static void main(String[] args) {
try {
FastReader in=new FastReader();
FastWriter out = new FastWriter();
int testCases=in.nextInt();
//int testCases=1;
while(testCases-- > 0){
solve(in);
}
out.close();
} catch (Exception e) {
System.out.println(e);
return;
}
}
public static void solve( FastReader in){
int n=in.nextInt();
//String s=in.next();
//String t=in.next();
//long y=in.nextInt();
//long n=in.nextLong();
//int q=in.nextInt();
//int m=in.nextInt();
//long k=in.nextLong();
StringBuilder res=new StringBuilder();
int[] arr=in.readIntArray(n);
int ans=0;
for(int i=1;i<=n-2;i++){
if(arr[i-1]<arr[i] && arr[i]>arr[i+1]){
arr[i+1]=Math.max(arr[i],i==n-2?1:arr[i+2]);
ans++;
}
}
res.append(""+ans+"\n");
for(int i=0;i<=n-1;i++){
res.append(""+arr[i]+" ");
}
System.out.println(res.toString());
}
static void dfs1(Map<String,String> hp,int i,int n,String r){
if(i>n){
ans+=ch(hp,n,r);
return;
}
for(char x='a';x<='f';x++){
dfs1(hp,i+1,n,r+x);
}
}
static int ch(Map<String,String> hp,int n,String r){
for(int i=0;i<n-1;i++){
String x=r.substring(0,2);
if(!hp.containsKey(x))return 0;
r=hp.get(x)+r.substring(2);
}
if(r.equals("a"))return 1;
return 0;
}
static int gcd(int a,int b){
if(b==0){
return a;
}
return gcd(b,a%b);
}
static void sort(int[] arr)
{
ArrayList<Integer> ls = new ArrayList<Integer>();
for(int x: arr)
ls.add(x);
Collections.sort(ls);
for(int i=0; i < arr.length; i++)
arr[i] = ls.get(i);
}
static void reversesort(int[] arr)
{
ArrayList<Integer> ls = new ArrayList<Integer>();
for(int x: arr)
ls.add(x);
Collections.sort(ls);
Collections.reverse(ls);
for(int i=0; i < arr.length; i++)
arr[i] = ls.get(i);
}
static int lcm(int a, int b)
{
return (a / gcd(a, b)) * b;
}
static void debug(String x){
System.out.println(x);
}
static < E > void print(E res)
{
System.out.println(res);
}
static String rString(String s){
StringBuilder sb=new StringBuilder();
sb.append(s);
return sb.reverse().toString();
}
} | Java | ["5\n3\n2 1 2\n4\n1 2 3 1\n5\n1 2 1 2 1\n9\n1 2 1 3 2 3 1 2 1\n9\n2 1 3 1 3 1 3 1 3"] | 2 seconds | ["0\n2 1 2\n1\n1 3 3 1\n1\n1 2 2 2 1\n2\n1 2 3 3 2 3 3 2 1\n2\n2 1 3 3 3 1 1 1 3"] | NoteIn the first example, the array contains no local maximum, so we don't need to perform operations.In the second example, we can change $$$a_2$$$ to $$$3$$$, then the array don't have local maximums. | Java 11 | standard input | [
"greedy"
] | 255abf8750541f54d8ff0f1f80e1f8e7 | Each test contains multiple test cases. The first line will contain a single integer $$$t$$$ $$$(1 \leq t \leq 10000)$$$ — the number of test cases. Then $$$t$$$ test cases follow. The first line of each test case contains a single integer $$$n$$$ $$$(2 \leq n \leq 2 \cdot 10^5)$$$ — the size of the array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots ,a_n$$$ $$$(1 \leq a_i \leq 10^9)$$$, the elements of array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, first output a line containing a single integer $$$m$$$ — minimum number of operations required. Then ouput a line consist of $$$n$$$ integers — the resulting array after the operations. Note that this array should differ in exactly $$$m$$$ elements from the initial array. If there are multiple answers, print any. | standard output | |
PASSED | b6f0eabe3201bae77d260978d50fb73f | train_108.jsonl | 1645367700 | You are given an array $$$a$$$ of size $$$n$$$. Each element in this array is an integer between $$$1$$$ and $$$10^9$$$.You can perform several operations to this array. During an operation, you can replace an element in the array with any integer between $$$1$$$ and $$$10^9$$$. Output the minimum number of operations needed such that the resulting array doesn't contain any local maximums, and the resulting array after the operations.An element $$$a_i$$$ is a local maximum if it is strictly larger than both of its neighbors (that is, $$$a_i > a_{i - 1}$$$ and $$$a_i > a_{i + 1}$$$). Since $$$a_1$$$ and $$$a_n$$$ have only one neighbor each, they will never be a local maximum. | 256 megabytes | import java.util.*;
public class codeforce {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int a = sc.nextInt();
while (a-- > 0) {
int len = sc.nextInt();
long arr[] = new long[len];
for (int i = 0; i < len; i++) {
arr[i] = sc.nextLong();
}
int x = 0;
for (int i=1;i<len-1;i++){
if(arr[i]>arr[i-1] && arr[i]>arr[i+1]){
// System.out.println("local maxima");
if(local(arr,i+2)){
arr[i+1] = Math.max(arr[i], arr[i+2]);
x++;
}else{
x++;
arr[i] = Math.max(arr[i-1], arr[i+1]);
}
}
}
System.out.println(x);
for(int i=0;i<len;i++){
if(i!=len-1) {
System.out.print(arr[i] + " ");
}else{
System.out.println(arr[i]);
}
}
}
}
public static boolean local(long arr[],int i){
if(i < arr.length-1) {
if (arr[i] > arr[i - 1] && arr[i] > arr[i + 1]) {
return true;
}
}
return false;
}
} | Java | ["5\n3\n2 1 2\n4\n1 2 3 1\n5\n1 2 1 2 1\n9\n1 2 1 3 2 3 1 2 1\n9\n2 1 3 1 3 1 3 1 3"] | 2 seconds | ["0\n2 1 2\n1\n1 3 3 1\n1\n1 2 2 2 1\n2\n1 2 3 3 2 3 3 2 1\n2\n2 1 3 3 3 1 1 1 3"] | NoteIn the first example, the array contains no local maximum, so we don't need to perform operations.In the second example, we can change $$$a_2$$$ to $$$3$$$, then the array don't have local maximums. | Java 11 | standard input | [
"greedy"
] | 255abf8750541f54d8ff0f1f80e1f8e7 | Each test contains multiple test cases. The first line will contain a single integer $$$t$$$ $$$(1 \leq t \leq 10000)$$$ — the number of test cases. Then $$$t$$$ test cases follow. The first line of each test case contains a single integer $$$n$$$ $$$(2 \leq n \leq 2 \cdot 10^5)$$$ — the size of the array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots ,a_n$$$ $$$(1 \leq a_i \leq 10^9)$$$, the elements of array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, first output a line containing a single integer $$$m$$$ — minimum number of operations required. Then ouput a line consist of $$$n$$$ integers — the resulting array after the operations. Note that this array should differ in exactly $$$m$$$ elements from the initial array. If there are multiple answers, print any. | standard output | |
PASSED | f52c80fffeb63bbdf037f08bb2236fe1 | train_108.jsonl | 1645367700 | You are given an array $$$a$$$ of size $$$n$$$. Each element in this array is an integer between $$$1$$$ and $$$10^9$$$.You can perform several operations to this array. During an operation, you can replace an element in the array with any integer between $$$1$$$ and $$$10^9$$$. Output the minimum number of operations needed such that the resulting array doesn't contain any local maximums, and the resulting array after the operations.An element $$$a_i$$$ is a local maximum if it is strictly larger than both of its neighbors (that is, $$$a_i > a_{i - 1}$$$ and $$$a_i > a_{i + 1}$$$). Since $$$a_1$$$ and $$$a_n$$$ have only one neighbor each, they will never be a local maximum. | 256 megabytes | //package MyPackage;
import java.util.*;
import java.io.*;
public class codec{
// static class Pair implements Comparable<Pair>
// {
// int a;
// int b;
// Pair(int a, int b)
// {
// this.a = a;
// this.b = b;
// }
// public int compareTo(Pair o)
// {
// return this.a - o.a;
// }
// }
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();
}
}
// private static boolean isPal(String s)
// {
// int l = 0, r = s.length() - 1;
// while(l < r)
// {
// if(s.charAt(l++) != s.charAt(r--))
// return false;
// }
// return true;
// }
// public static void swap(int[][] arr, int i, int j, int a, int b) {
//// arr[i] = (arr[i] + arr[j]) - (arr[j] = arr[i]);
// int t = arr[i][j];
// arr[i][j] = arr[a][b];
// arr[a][b] = t;
// }
// static int[] pre;
//
// private static long func(int i, int b[], int c[], int k, long dp[][])
// {
//// if(k < 0) return Long.MIN_VALUE;
//
// if(k == 0 || i >= b.length)
// return 0;
//
// if(dp[i][k] != -1)
// return dp[i][k];
//
// if(k - pre[b[i]] < 0)
// {
// long nahi = func(i + 1, b, c, k, dp);
// return nahi;
// }
// else
// {
// long pick = c[i] + func(i + 1, b, c, k - pre[b[i]], dp);
// long notpick = func(i + 1, b, c, k, dp);
//
// return dp[i][k] = Math.max(pick, notpick);
// }
// }
public static void main(String[] args) {
try {
FastReader in=new FastReader();
FastWriter out = new FastWriter();
StringBuilder sb = new StringBuilder();
int testCases=in.nextInt();
while(testCases-- > 0){
// write code here
int n = in.nextInt();
int a[] = new int[n];
for(int i = 0; i < n; i++)
{
a[i] = in.nextInt();
}
// for(int i = 1; i < n; i++)
// {
// int or = a[i] | a[i - 1];
// int sum = a[i] + a[i - 1];
// if(i + 1 < n && sum >= a[i + 1])
// {
// a[i] = 0;
// a[i - 1] = or - a[i + 1];
// i++;
// }
// else
// {
// a[i] = or - 1;
// a[i - 1] = 1;
// }
// }
//
// int s = 0;
// for(int i: a)
// s += i;
//
// sb.append(s + "\n");
int c = 0;
for(int i = 1; i < n - 1;)
{
if(a[i] > a[i - 1] && a[i] > a[i + 1])
{
if(i + 2 < n && a[i] < a[i + 2])
{
a[i + 1] = a[i + 2];
i += 3;
}
else
{
a[i + 1] = a[i];
i += 2;
}
c++;
}
else
i++;
}
sb.append(c + "\n");
for(int i: a)
sb.append(i + " ");
sb.append("\n");
}
out.println(sb);
out.close();
}
catch (Exception e) {
return;
}
}
} | Java | ["5\n3\n2 1 2\n4\n1 2 3 1\n5\n1 2 1 2 1\n9\n1 2 1 3 2 3 1 2 1\n9\n2 1 3 1 3 1 3 1 3"] | 2 seconds | ["0\n2 1 2\n1\n1 3 3 1\n1\n1 2 2 2 1\n2\n1 2 3 3 2 3 3 2 1\n2\n2 1 3 3 3 1 1 1 3"] | NoteIn the first example, the array contains no local maximum, so we don't need to perform operations.In the second example, we can change $$$a_2$$$ to $$$3$$$, then the array don't have local maximums. | Java 11 | standard input | [
"greedy"
] | 255abf8750541f54d8ff0f1f80e1f8e7 | Each test contains multiple test cases. The first line will contain a single integer $$$t$$$ $$$(1 \leq t \leq 10000)$$$ — the number of test cases. Then $$$t$$$ test cases follow. The first line of each test case contains a single integer $$$n$$$ $$$(2 \leq n \leq 2 \cdot 10^5)$$$ — the size of the array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots ,a_n$$$ $$$(1 \leq a_i \leq 10^9)$$$, the elements of array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, first output a line containing a single integer $$$m$$$ — minimum number of operations required. Then ouput a line consist of $$$n$$$ integers — the resulting array after the operations. Note that this array should differ in exactly $$$m$$$ elements from the initial array. If there are multiple answers, print any. | standard output | |
PASSED | c44e6f0ca50baf2d6b248ff2b76102d3 | train_108.jsonl | 1645367700 | You are given an array $$$a$$$ of size $$$n$$$. Each element in this array is an integer between $$$1$$$ and $$$10^9$$$.You can perform several operations to this array. During an operation, you can replace an element in the array with any integer between $$$1$$$ and $$$10^9$$$. Output the minimum number of operations needed such that the resulting array doesn't contain any local maximums, and the resulting array after the operations.An element $$$a_i$$$ is a local maximum if it is strictly larger than both of its neighbors (that is, $$$a_i > a_{i - 1}$$$ and $$$a_i > a_{i + 1}$$$). Since $$$a_1$$$ and $$$a_n$$$ have only one neighbor each, they will never be a local maximum. | 256 megabytes | import java.util.*;
import java.lang.*;
import java.io.*;
public class Main
{
public static void main (String[] args) throws java.lang.Exception
{
Reader scan = new Reader();
int tc = scan.nextInt();
for (int i = 0; i < tc; i++){
int n = scan.nextInt();
ArrayList <Long> arr = new ArrayList <Long>();
for (int j = 0; j < n; j++){
long k = scan.nextLong();
arr.add(k);
}
solve(arr);
}
}
public static void solve(ArrayList <Long> arr){
long maximums = 0;
if (arr.size()==2){
System.out.println(0);
printarr(arr);
return;
}
for (int i = 1; i < arr.size()-1; i++){
if (arr.get(i) > arr.get(i-1) && arr.get(i) > arr.get(i+1)){
maximums++;
long first = arr.get(i);
long second = 0;
if (i < arr.size()-2)
second = arr.get(i+2);
else
second = -1;
arr.set(i+1, Math.max(first, second));
}
}
System.out.println(maximums);
printarr(arr);
}
public static long gcd(long a, long b){
if (b == 0)
return a;
return gcd(b, a % b);
}
public static boolean isPrime(long n){
if (n == 1)
return false;
if (n == 2 || n == 3)
return true;
if (n % 2 == 0 || n % 3 == 0)
return false;
for (long i = 5; i*i <= n; i += 6){
if (n % i == 0 || n % (i+2) == 0)
return false;
}
return true;
}
public static boolean isPerfectSquare(long x){
long root = (long)Math.sqrt(x);
return ((root *root) == x);
}
public static long modPow(long x, long n){
long mod = 1000000007;
long result = 1;
while (n > 0){
if (n % 2 != 0)
result = (result % mod) * (x % mod) % mod;
x *= x % mod;
n /= 2;
}
return result;
}
public static void printarr(String[]arr){
for (int i = 0; i < arr.length; i++){
System.out.print(arr[i] + " ");
}
System.out.println();
}
public static void printarr(ArrayList <Long> arr){
for (int i = 0; i < arr.size(); i++){
System.out.print(arr.get(i) + " ");
}
System.out.println();
}
static class Reader {
final private int BUFFER_SIZE = 1 << 64;
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[1000000]; // 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();
}
}
} | Java | ["5\n3\n2 1 2\n4\n1 2 3 1\n5\n1 2 1 2 1\n9\n1 2 1 3 2 3 1 2 1\n9\n2 1 3 1 3 1 3 1 3"] | 2 seconds | ["0\n2 1 2\n1\n1 3 3 1\n1\n1 2 2 2 1\n2\n1 2 3 3 2 3 3 2 1\n2\n2 1 3 3 3 1 1 1 3"] | NoteIn the first example, the array contains no local maximum, so we don't need to perform operations.In the second example, we can change $$$a_2$$$ to $$$3$$$, then the array don't have local maximums. | Java 11 | standard input | [
"greedy"
] | 255abf8750541f54d8ff0f1f80e1f8e7 | Each test contains multiple test cases. The first line will contain a single integer $$$t$$$ $$$(1 \leq t \leq 10000)$$$ — the number of test cases. Then $$$t$$$ test cases follow. The first line of each test case contains a single integer $$$n$$$ $$$(2 \leq n \leq 2 \cdot 10^5)$$$ — the size of the array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots ,a_n$$$ $$$(1 \leq a_i \leq 10^9)$$$, the elements of array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, first output a line containing a single integer $$$m$$$ — minimum number of operations required. Then ouput a line consist of $$$n$$$ integers — the resulting array after the operations. Note that this array should differ in exactly $$$m$$$ elements from the initial array. If there are multiple answers, print any. | standard output | |
PASSED | de1f03a6c82973a3d45722289c34ad50 | train_108.jsonl | 1645367700 | You are given an array $$$a$$$ of size $$$n$$$. Each element in this array is an integer between $$$1$$$ and $$$10^9$$$.You can perform several operations to this array. During an operation, you can replace an element in the array with any integer between $$$1$$$ and $$$10^9$$$. Output the minimum number of operations needed such that the resulting array doesn't contain any local maximums, and the resulting array after the operations.An element $$$a_i$$$ is a local maximum if it is strictly larger than both of its neighbors (that is, $$$a_i > a_{i - 1}$$$ and $$$a_i > a_{i + 1}$$$). Since $$$a_1$$$ and $$$a_n$$$ have only one neighbor each, they will never be a local maximum. | 256 megabytes | import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.Scanner;
import java.util.StringTokenizer;
import java.util.*;
public class Main {
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 s = new FastReader();
int t = s.nextInt();
while (t-- > 0){
int n = s.nextInt();
int[] ar = new int[n];
for (int i = 0; i < n; i++){
ar[i] = s.nextInt();
}
int cnt = 0;
for (int i = 1; i < n-1; i++){
if (ar[i] > ar[i-1] && ar[i] > ar[i+1]) {
cnt++;
if (i+2 < n-1 && ar[i+2] > ar[i+1]){
if (ar[i+2] >= ar[i]){
ar[i+1] = ar[i+2];
} else {
ar[i+1] = ar[i];
}
} else {
ar[i+1] = ar[i];
}
}
}
System.out.println(cnt);
for (int i = 0; i < n; i++){
System.out.print(ar[i] + " ");
}
System.out.println();
}
}
}
| Java | ["5\n3\n2 1 2\n4\n1 2 3 1\n5\n1 2 1 2 1\n9\n1 2 1 3 2 3 1 2 1\n9\n2 1 3 1 3 1 3 1 3"] | 2 seconds | ["0\n2 1 2\n1\n1 3 3 1\n1\n1 2 2 2 1\n2\n1 2 3 3 2 3 3 2 1\n2\n2 1 3 3 3 1 1 1 3"] | NoteIn the first example, the array contains no local maximum, so we don't need to perform operations.In the second example, we can change $$$a_2$$$ to $$$3$$$, then the array don't have local maximums. | Java 11 | standard input | [
"greedy"
] | 255abf8750541f54d8ff0f1f80e1f8e7 | Each test contains multiple test cases. The first line will contain a single integer $$$t$$$ $$$(1 \leq t \leq 10000)$$$ — the number of test cases. Then $$$t$$$ test cases follow. The first line of each test case contains a single integer $$$n$$$ $$$(2 \leq n \leq 2 \cdot 10^5)$$$ — the size of the array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots ,a_n$$$ $$$(1 \leq a_i \leq 10^9)$$$, the elements of array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, first output a line containing a single integer $$$m$$$ — minimum number of operations required. Then ouput a line consist of $$$n$$$ integers — the resulting array after the operations. Note that this array should differ in exactly $$$m$$$ elements from the initial array. If there are multiple answers, print any. | standard output | |
PASSED | 4cdf4ec305ed8bfaf179bc4445742683 | train_108.jsonl | 1645367700 | You are given an array $$$a$$$ of size $$$n$$$. Each element in this array is an integer between $$$1$$$ and $$$10^9$$$.You can perform several operations to this array. During an operation, you can replace an element in the array with any integer between $$$1$$$ and $$$10^9$$$. Output the minimum number of operations needed such that the resulting array doesn't contain any local maximums, and the resulting array after the operations.An element $$$a_i$$$ is a local maximum if it is strictly larger than both of its neighbors (that is, $$$a_i > a_{i - 1}$$$ and $$$a_i > a_{i + 1}$$$). Since $$$a_1$$$ and $$$a_n$$$ have only one neighbor each, they will never be a local maximum. | 256 megabytes |
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.Scanner;
import java.util.StringTokenizer;
public class problemB {
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 []agrs) {
FastReader fastReader=new FastReader();
int t=fastReader.nextInt();
StringBuffer stringBuffer=new StringBuffer();
for(int i=0;i<t;i++){
int len=fastReader.nextInt();
long[] arr=new long[len];
for(int j=0;j<len;j++){
arr[j] = fastReader.nextLong();
}
long steps = 0;
for(int j=1;j<len-1;j++){
long prev=arr[j-1];
long next=arr[j+1];
if(arr[j]>prev && arr[j]>next){
if(j+2<len && arr[j+1]<arr[j+2]){
arr[j+1] = Math.max(arr[j+2],arr[j]);
steps +=1;
}
else{
arr[j]= Math.max(next,prev);
steps +=1;
}
}
}
stringBuffer.append(steps).append("\n");
for(int j=0;j<len;j++){
stringBuffer.append(arr[j]).append(" ");
}
stringBuffer.append("\n");
}
System.out.println(stringBuffer.toString());
}
} | Java | ["5\n3\n2 1 2\n4\n1 2 3 1\n5\n1 2 1 2 1\n9\n1 2 1 3 2 3 1 2 1\n9\n2 1 3 1 3 1 3 1 3"] | 2 seconds | ["0\n2 1 2\n1\n1 3 3 1\n1\n1 2 2 2 1\n2\n1 2 3 3 2 3 3 2 1\n2\n2 1 3 3 3 1 1 1 3"] | NoteIn the first example, the array contains no local maximum, so we don't need to perform operations.In the second example, we can change $$$a_2$$$ to $$$3$$$, then the array don't have local maximums. | Java 11 | standard input | [
"greedy"
] | 255abf8750541f54d8ff0f1f80e1f8e7 | Each test contains multiple test cases. The first line will contain a single integer $$$t$$$ $$$(1 \leq t \leq 10000)$$$ — the number of test cases. Then $$$t$$$ test cases follow. The first line of each test case contains a single integer $$$n$$$ $$$(2 \leq n \leq 2 \cdot 10^5)$$$ — the size of the array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots ,a_n$$$ $$$(1 \leq a_i \leq 10^9)$$$, the elements of array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, first output a line containing a single integer $$$m$$$ — minimum number of operations required. Then ouput a line consist of $$$n$$$ integers — the resulting array after the operations. Note that this array should differ in exactly $$$m$$$ elements from the initial array. If there are multiple answers, print any. | standard output | |
PASSED | 4942ca24e915f44fd24039d57e082440 | train_108.jsonl | 1645367700 | You are given an array $$$a$$$ of size $$$n$$$. Each element in this array is an integer between $$$1$$$ and $$$10^9$$$.You can perform several operations to this array. During an operation, you can replace an element in the array with any integer between $$$1$$$ and $$$10^9$$$. Output the minimum number of operations needed such that the resulting array doesn't contain any local maximums, and the resulting array after the operations.An element $$$a_i$$$ is a local maximum if it is strictly larger than both of its neighbors (that is, $$$a_i > a_{i - 1}$$$ and $$$a_i > a_{i + 1}$$$). Since $$$a_1$$$ and $$$a_n$$$ have only one neighbor each, they will never be a local maximum. | 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 res = cal(val, pow/2, mod);
long ret = (res*res)%mod;
if(pow%2 == 0) return ret;
return (val*ret)%mod;
}
static long gcd(long a, long b) { return b == 0 ? a:gcd(b, a%b); }
}
static int mod = 998244353;
static long inf = (long) Long.MAX_VALUE;
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);
int count = 0;
for(int i = 1; i < n-1; i++) {
if(a[i] > a[i+1] && a[i] > a[i-1]) {
if(i+2 < n) a[i+1] = Math.max(a[i], a[i+2]);
else a[i+1] = a[i];
count++;
}
}
ret.append(count + "\n");
for(int aa:a) ret.append(aa + " ");
ret.append("\n");
}
System.out.println(ret);
}
}
| Java | ["5\n3\n2 1 2\n4\n1 2 3 1\n5\n1 2 1 2 1\n9\n1 2 1 3 2 3 1 2 1\n9\n2 1 3 1 3 1 3 1 3"] | 2 seconds | ["0\n2 1 2\n1\n1 3 3 1\n1\n1 2 2 2 1\n2\n1 2 3 3 2 3 3 2 1\n2\n2 1 3 3 3 1 1 1 3"] | NoteIn the first example, the array contains no local maximum, so we don't need to perform operations.In the second example, we can change $$$a_2$$$ to $$$3$$$, then the array don't have local maximums. | Java 11 | standard input | [
"greedy"
] | 255abf8750541f54d8ff0f1f80e1f8e7 | Each test contains multiple test cases. The first line will contain a single integer $$$t$$$ $$$(1 \leq t \leq 10000)$$$ — the number of test cases. Then $$$t$$$ test cases follow. The first line of each test case contains a single integer $$$n$$$ $$$(2 \leq n \leq 2 \cdot 10^5)$$$ — the size of the array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots ,a_n$$$ $$$(1 \leq a_i \leq 10^9)$$$, the elements of array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, first output a line containing a single integer $$$m$$$ — minimum number of operations required. Then ouput a line consist of $$$n$$$ integers — the resulting array after the operations. Note that this array should differ in exactly $$$m$$$ elements from the initial array. If there are multiple answers, print any. | standard output | |
PASSED | 33713e02ff34a81cab8f3f2d1f3af1cf | train_108.jsonl | 1645367700 | You are given an array $$$a$$$ of size $$$n$$$. Each element in this array is an integer between $$$1$$$ and $$$10^9$$$.You can perform several operations to this array. During an operation, you can replace an element in the array with any integer between $$$1$$$ and $$$10^9$$$. Output the minimum number of operations needed such that the resulting array doesn't contain any local maximums, and the resulting array after the operations.An element $$$a_i$$$ is a local maximum if it is strictly larger than both of its neighbors (that is, $$$a_i > a_{i - 1}$$$ and $$$a_i > a_{i + 1}$$$). Since $$$a_1$$$ and $$$a_n$$$ have only one neighbor each, they will never be a local maximum. | 256 megabytes | import java.util.*;
public class Main {
public static void main(String args[]) {
Scanner sc = new Scanner(System.in);
int t = sc.nextInt();
while(t>0){
t--;
int n = sc.nextInt();
int[] arr =new int[n];
int max=0;
for(int i=0;i<n;i++){
arr[i]=sc.nextInt();
if(arr[i]>max)
max=arr[i];
}
int count=0;
for(int i=1;i<n-1;i++){
if(arr[i]>arr[i-1]&&arr[i]>arr[i+1]){
if(i+1==n-1)
arr[i-1]=arr[i];
else if(i+2<n)
if(arr[i]<arr[i+2])
arr[i+1]=arr[i+2];
else
arr[i+1]=arr[i];
count++;
}
}
System.out.println(count);
for(int i :arr){
System.out.print(i+" ");
}
System.out.println();
}
}
} | Java | ["5\n3\n2 1 2\n4\n1 2 3 1\n5\n1 2 1 2 1\n9\n1 2 1 3 2 3 1 2 1\n9\n2 1 3 1 3 1 3 1 3"] | 2 seconds | ["0\n2 1 2\n1\n1 3 3 1\n1\n1 2 2 2 1\n2\n1 2 3 3 2 3 3 2 1\n2\n2 1 3 3 3 1 1 1 3"] | NoteIn the first example, the array contains no local maximum, so we don't need to perform operations.In the second example, we can change $$$a_2$$$ to $$$3$$$, then the array don't have local maximums. | Java 11 | standard input | [
"greedy"
] | 255abf8750541f54d8ff0f1f80e1f8e7 | Each test contains multiple test cases. The first line will contain a single integer $$$t$$$ $$$(1 \leq t \leq 10000)$$$ — the number of test cases. Then $$$t$$$ test cases follow. The first line of each test case contains a single integer $$$n$$$ $$$(2 \leq n \leq 2 \cdot 10^5)$$$ — the size of the array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots ,a_n$$$ $$$(1 \leq a_i \leq 10^9)$$$, the elements of array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, first output a line containing a single integer $$$m$$$ — minimum number of operations required. Then ouput a line consist of $$$n$$$ integers — the resulting array after the operations. Note that this array should differ in exactly $$$m$$$ elements from the initial array. If there are multiple answers, print any. | standard output | |
PASSED | 01894925844de5cbd0dd2b47ced1c75b | train_108.jsonl | 1645367700 | You are given an array $$$a$$$ of size $$$n$$$. Each element in this array is an integer between $$$1$$$ and $$$10^9$$$.You can perform several operations to this array. During an operation, you can replace an element in the array with any integer between $$$1$$$ and $$$10^9$$$. Output the minimum number of operations needed such that the resulting array doesn't contain any local maximums, and the resulting array after the operations.An element $$$a_i$$$ is a local maximum if it is strictly larger than both of its neighbors (that is, $$$a_i > a_{i - 1}$$$ and $$$a_i > a_{i + 1}$$$). Since $$$a_1$$$ and $$$a_n$$$ have only one neighbor each, they will never be a local maximum. | 256 megabytes | import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.ArrayList;
public class cf772b {
public static void main(String[] args) throws IOException {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
int t = Integer.parseInt(br.readLine());
for (int i = 0; i < t; ++i) {
int num = Integer.parseInt(br.readLine());
String[] strings = br.readLine().split(" ");
ArrayList<Integer> list = new ArrayList<>();
ArrayList<Boolean> bList = new ArrayList<>();
for (int j = 0; j < num; ++j) {
list.add(Integer.parseInt(strings[j]));
bList.add(false);
}
int m = 0;
for (int j = 1; j < num - 1; ++j) {
if (isLocMax(list, j)) {
bList.set(j, true);
}
}
for (int j = 1; j < num - 2; ++j) {
if (bList.get(j) == true) {
if (bList.get(j + 2) == true) {
list.set(j + 1, Math.max(list.get(j), list.get(j + 2)));
bList.set(j + 2, false);
} else {
list.set(j + 1, list.get(j));
}
m++;
}
}
if (isLocMax(list, num - 2)) {
m++;
list.set(num - 1, list.get(num - 2));
}
System.out.println(m);
for (int j = 0; j < num; ++j) {
System.out.print(list.get(j));
System.out.print(" ");
}
System.out.println("");
}
}
public static boolean isLocMax(ArrayList<Integer> list, int loc) {
if (loc == 0 || loc == list.size() - 1) {
return false;
}
if (list.get(loc) > list.get(loc - 1) && list.get(loc) > list.get(loc + 1)) {
return true;
}
return false;
}
}
| Java | ["5\n3\n2 1 2\n4\n1 2 3 1\n5\n1 2 1 2 1\n9\n1 2 1 3 2 3 1 2 1\n9\n2 1 3 1 3 1 3 1 3"] | 2 seconds | ["0\n2 1 2\n1\n1 3 3 1\n1\n1 2 2 2 1\n2\n1 2 3 3 2 3 3 2 1\n2\n2 1 3 3 3 1 1 1 3"] | NoteIn the first example, the array contains no local maximum, so we don't need to perform operations.In the second example, we can change $$$a_2$$$ to $$$3$$$, then the array don't have local maximums. | Java 11 | standard input | [
"greedy"
] | 255abf8750541f54d8ff0f1f80e1f8e7 | Each test contains multiple test cases. The first line will contain a single integer $$$t$$$ $$$(1 \leq t \leq 10000)$$$ — the number of test cases. Then $$$t$$$ test cases follow. The first line of each test case contains a single integer $$$n$$$ $$$(2 \leq n \leq 2 \cdot 10^5)$$$ — the size of the array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots ,a_n$$$ $$$(1 \leq a_i \leq 10^9)$$$, the elements of array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, first output a line containing a single integer $$$m$$$ — minimum number of operations required. Then ouput a line consist of $$$n$$$ integers — the resulting array after the operations. Note that this array should differ in exactly $$$m$$$ elements from the initial array. If there are multiple answers, print any. | standard output | |
PASSED | 7d4378d4d0ab4fd0f4864f8925be443a | train_108.jsonl | 1645367700 | You are given an array $$$a$$$ of size $$$n$$$. Each element in this array is an integer between $$$1$$$ and $$$10^9$$$.You can perform several operations to this array. During an operation, you can replace an element in the array with any integer between $$$1$$$ and $$$10^9$$$. Output the minimum number of operations needed such that the resulting array doesn't contain any local maximums, and the resulting array after the operations.An element $$$a_i$$$ is a local maximum if it is strictly larger than both of its neighbors (that is, $$$a_i > a_{i - 1}$$$ and $$$a_i > a_{i + 1}$$$). Since $$$a_1$$$ and $$$a_n$$$ have only one neighbor each, they will never be a local maximum. | 256 megabytes | import java.util.*;
public class Main {
public static void main(String[] args) {
Scanner scn=new Scanner(System.in);
int tc=scn.nextInt();
while(tc-->0){
int n=scn.nextInt();
int arr[]=new int[n];
for(int i=0;i<n;i++){
arr[i]=scn.nextInt();
}
int count=0;
boolean flag=false;
for(int i=1;i<n-1;i++){
if(i!=n-2 && arr[i]>arr[i-1] && arr[i]>arr[i+1]){
if(i==n-4)
flag=true;
if(arr[i]==arr[i+2]){
arr[i+1]=arr[i];
}
else if(arr[i]>arr[i+2]){
arr[i+1]=arr[i];
}
else if(arr[i]<arr[i+2]){
arr[i+1]=arr[i+2];
}
//System.out.println(i+" ind");
count++;
}
if(i==n-2 && arr[i]>arr[i+1] && arr[i]>arr[i-1]){
if(flag==true){
arr[i-1]=arr[i];
}
else{
count++;
arr[i-1]=arr[i];
}
}
}
System.out.println(count);
for(int i=0;i<n;i++){
System.out.print(arr[i]+" ");
}
System.out.println();
}
}
} | Java | ["5\n3\n2 1 2\n4\n1 2 3 1\n5\n1 2 1 2 1\n9\n1 2 1 3 2 3 1 2 1\n9\n2 1 3 1 3 1 3 1 3"] | 2 seconds | ["0\n2 1 2\n1\n1 3 3 1\n1\n1 2 2 2 1\n2\n1 2 3 3 2 3 3 2 1\n2\n2 1 3 3 3 1 1 1 3"] | NoteIn the first example, the array contains no local maximum, so we don't need to perform operations.In the second example, we can change $$$a_2$$$ to $$$3$$$, then the array don't have local maximums. | Java 11 | standard input | [
"greedy"
] | 255abf8750541f54d8ff0f1f80e1f8e7 | Each test contains multiple test cases. The first line will contain a single integer $$$t$$$ $$$(1 \leq t \leq 10000)$$$ — the number of test cases. Then $$$t$$$ test cases follow. The first line of each test case contains a single integer $$$n$$$ $$$(2 \leq n \leq 2 \cdot 10^5)$$$ — the size of the array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots ,a_n$$$ $$$(1 \leq a_i \leq 10^9)$$$, the elements of array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 800 | For each test case, first output a line containing a single integer $$$m$$$ — minimum number of operations required. Then ouput a line consist of $$$n$$$ integers — the resulting array after the operations. Note that this array should differ in exactly $$$m$$$ elements from the initial array. If there are multiple answers, print any. | standard output |
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