title stringlengths 1 100 | titleSlug stringlengths 3 77 | Java int64 0 1 | Python3 int64 1 1 | content stringlengths 28 44.4k | voteCount int64 0 3.67k | question_content stringlengths 65 5k | question_hints stringclasses 970
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find-the-minimum-number-of-fibonacci-numbers-whose-sum-is-k | find-the-minimum-number-of-fibonacci-numbers-whose-sum-is-k | 0 | 1 | # Code\n```\nclass Solution:\n def findMinFibonacciNumbers(self, k: int) -> int:\n self.count = 0\n def fib(k):\n l = [0,1]\n for i in range(1,k+1):\n if l[-2]+l[-1]<k:\n l.append(l[-1] + l[-2])\n elif l[-2]+l[-1]==k:\n ... | 0 | A delivery company wants to build a new service center in a new city. The company knows the positions of all the customers in this city on a 2D-Map and wants to build the new center in a position such that **the sum of the euclidean distances to all customers is minimum**.
Given an array `positions` where `positions[i... | Generate all Fibonacci numbers up to the limit (they are few). Use greedy solution, taking at every time the greatest Fibonacci number which is smaller than or equal to the current number. Subtract this Fibonacci number from the current number and repeat again the process. |
Python Solution using Greedy approach | find-the-minimum-number-of-fibonacci-numbers-whose-sum-is-k | 0 | 1 | # Approach\nInitially we will create a pre calculated fibonacci sequence and later on using greedy approach we will count the valid additions needed for achieving the value of `k`.\n\n# Complexity\n- Time complexity:\nThe number of iterations in this loop is proportional to the length of the arr list, which is $$O(n)$$... | 0 | Given an integer `k`, _return the minimum number of Fibonacci numbers whose sum is equal to_ `k`. The same Fibonacci number can be used multiple times.
The Fibonacci numbers are defined as:
* `F1 = 1`
* `F2 = 1`
* `Fn = Fn-1 + Fn-2` for `n > 2.`
It is guaranteed that for the given constraints we can always fin... | Use BFS. BFS on (x,y,r) x,y is coordinate, r is remain number of obstacles you can remove. |
Python Solution using Greedy approach | find-the-minimum-number-of-fibonacci-numbers-whose-sum-is-k | 0 | 1 | # Approach\nInitially we will create a pre calculated fibonacci sequence and later on using greedy approach we will count the valid additions needed for achieving the value of `k`.\n\n# Complexity\n- Time complexity:\nThe number of iterations in this loop is proportional to the length of the arr list, which is $$O(n)$$... | 0 | A delivery company wants to build a new service center in a new city. The company knows the positions of all the customers in this city on a 2D-Map and wants to build the new center in a position such that **the sum of the euclidean distances to all customers is minimum**.
Given an array `positions` where `positions[i... | Generate all Fibonacci numbers up to the limit (they are few). Use greedy solution, taking at every time the greatest Fibonacci number which is smaller than or equal to the current number. Subtract this Fibonacci number from the current number and repeat again the process. |
Python solution !!!98% Beat . | find-the-minimum-number-of-fibonacci-numbers-whose-sum-is-k | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\nFist find fibonacci numbers less than k ,append them in a list .List must be in increasing order.\n\n**Find the integer numbers whose sumation is equal to k**\nCheck t... | 0 | Given an integer `k`, _return the minimum number of Fibonacci numbers whose sum is equal to_ `k`. The same Fibonacci number can be used multiple times.
The Fibonacci numbers are defined as:
* `F1 = 1`
* `F2 = 1`
* `Fn = Fn-1 + Fn-2` for `n > 2.`
It is guaranteed that for the given constraints we can always fin... | Use BFS. BFS on (x,y,r) x,y is coordinate, r is remain number of obstacles you can remove. |
Python solution !!!98% Beat . | find-the-minimum-number-of-fibonacci-numbers-whose-sum-is-k | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\nFist find fibonacci numbers less than k ,append them in a list .List must be in increasing order.\n\n**Find the integer numbers whose sumation is equal to k**\nCheck t... | 0 | A delivery company wants to build a new service center in a new city. The company knows the positions of all the customers in this city on a 2D-Map and wants to build the new center in a position such that **the sum of the euclidean distances to all customers is minimum**.
Given an array `positions` where `positions[i... | Generate all Fibonacci numbers up to the limit (they are few). Use greedy solution, taking at every time the greatest Fibonacci number which is smaller than or equal to the current number. Subtract this Fibonacci number from the current number and repeat again the process. |
A proof of the correctness for greedy, and comparison to other problems | find-the-minimum-number-of-fibonacci-numbers-whose-sum-is-k | 0 | 1 | # Approach\nThere are tons of good solutions posted by others, but I want to explain why greedy is correct for this problem, and compare to another one.\nFirst consider a variation of this question. Given a number K, choose a minimal amount of numbers amoung a *given list* instead of *Fibonacci sequence*. For example:\... | 0 | Given an integer `k`, _return the minimum number of Fibonacci numbers whose sum is equal to_ `k`. The same Fibonacci number can be used multiple times.
The Fibonacci numbers are defined as:
* `F1 = 1`
* `F2 = 1`
* `Fn = Fn-1 + Fn-2` for `n > 2.`
It is guaranteed that for the given constraints we can always fin... | Use BFS. BFS on (x,y,r) x,y is coordinate, r is remain number of obstacles you can remove. |
A proof of the correctness for greedy, and comparison to other problems | find-the-minimum-number-of-fibonacci-numbers-whose-sum-is-k | 0 | 1 | # Approach\nThere are tons of good solutions posted by others, but I want to explain why greedy is correct for this problem, and compare to another one.\nFirst consider a variation of this question. Given a number K, choose a minimal amount of numbers amoung a *given list* instead of *Fibonacci sequence*. For example:\... | 0 | A delivery company wants to build a new service center in a new city. The company knows the positions of all the customers in this city on a 2D-Map and wants to build the new center in a position such that **the sum of the euclidean distances to all customers is minimum**.
Given an array `positions` where `positions[i... | Generate all Fibonacci numbers up to the limit (they are few). Use greedy solution, taking at every time the greatest Fibonacci number which is smaller than or equal to the current number. Subtract this Fibonacci number from the current number and repeat again the process. |
Python Simple Solution | find-the-minimum-number-of-fibonacci-numbers-whose-sum-is-k | 0 | 1 | # Approach\n<!-- Describe your approach to solving the problem. -->\nBecuase there is always an answer, we will just subtract the biggest Fibonacci number possible until we get k down to zero. We create a helper function to subtract the maximum number, then simply count how many times we do this.\n\nIn the helper funci... | 0 | Given an integer `k`, _return the minimum number of Fibonacci numbers whose sum is equal to_ `k`. The same Fibonacci number can be used multiple times.
The Fibonacci numbers are defined as:
* `F1 = 1`
* `F2 = 1`
* `Fn = Fn-1 + Fn-2` for `n > 2.`
It is guaranteed that for the given constraints we can always fin... | Use BFS. BFS on (x,y,r) x,y is coordinate, r is remain number of obstacles you can remove. |
Python Simple Solution | find-the-minimum-number-of-fibonacci-numbers-whose-sum-is-k | 0 | 1 | # Approach\n<!-- Describe your approach to solving the problem. -->\nBecuase there is always an answer, we will just subtract the biggest Fibonacci number possible until we get k down to zero. We create a helper function to subtract the maximum number, then simply count how many times we do this.\n\nIn the helper funci... | 0 | A delivery company wants to build a new service center in a new city. The company knows the positions of all the customers in this city on a 2D-Map and wants to build the new center in a position such that **the sum of the euclidean distances to all customers is minimum**.
Given an array `positions` where `positions[i... | Generate all Fibonacci numbers up to the limit (they are few). Use greedy solution, taking at every time the greatest Fibonacci number which is smaller than or equal to the current number. Subtract this Fibonacci number from the current number and repeat again the process. |
Python sol. with utube video | find-the-minimum-number-of-fibonacci-numbers-whose-sum-is-k | 0 | 1 | # Code\n```\n# https://www.youtube.com/watch?v=w3zD7kyJE8U\nclass Solution:\n def findMinFibonacciNumbers(self, k: int) -> int:\n fib=[1,1]\n a=b=1\n for i in range(3,100):\n c=a+b\n if c>10**9:\n break\n a=b\n b=c\n fib.appen... | 0 | Given an integer `k`, _return the minimum number of Fibonacci numbers whose sum is equal to_ `k`. The same Fibonacci number can be used multiple times.
The Fibonacci numbers are defined as:
* `F1 = 1`
* `F2 = 1`
* `Fn = Fn-1 + Fn-2` for `n > 2.`
It is guaranteed that for the given constraints we can always fin... | Use BFS. BFS on (x,y,r) x,y is coordinate, r is remain number of obstacles you can remove. |
Python sol. with utube video | find-the-minimum-number-of-fibonacci-numbers-whose-sum-is-k | 0 | 1 | # Code\n```\n# https://www.youtube.com/watch?v=w3zD7kyJE8U\nclass Solution:\n def findMinFibonacciNumbers(self, k: int) -> int:\n fib=[1,1]\n a=b=1\n for i in range(3,100):\n c=a+b\n if c>10**9:\n break\n a=b\n b=c\n fib.appen... | 0 | A delivery company wants to build a new service center in a new city. The company knows the positions of all the customers in this city on a 2D-Map and wants to build the new center in a position such that **the sum of the euclidean distances to all customers is minimum**.
Given an array `positions` where `positions[i... | Generate all Fibonacci numbers up to the limit (they are few). Use greedy solution, taking at every time the greatest Fibonacci number which is smaller than or equal to the current number. Subtract this Fibonacci number from the current number and repeat again the process. |
Simple solution with Backtracking in Python3 / TypeScript | the-k-th-lexicographical-string-of-all-happy-strings-of-length-n | 0 | 1 | # Intuition\nHere\'s a brief explanation of a problem:\n- there\'re `n` and `k` integers\n- our goal is to get `k` - th permutation of a **happy string** with length `n`\n\nA **happy string** consists with letters `abc`, that\'s formed as permutation.\nThe neighbours **can\'t be equal** such as `s[i] != s[i+1] (or s[i-... | 2 | A **happy string** is a string that:
* consists only of letters of the set `['a', 'b', 'c']`.
* `s[i] != s[i + 1]` for all values of `i` from `1` to `s.length - 1` (string is 1-indexed).
For example, strings **"abc ", "ac ", "b "** and **"abcbabcbcb "** are all happy strings and strings **"aa ", "baa "** and **"a... | null |
Python 3 || 9 lines, binary map w/example || T/M: 99% / 87% | the-k-th-lexicographical-string-of-all-happy-strings-of-length-n | 0 | 1 | ```\nclass Solution:\n def getHappyString(self, n: int, k: int) -> str:\n\n d = {\'a\':\'bc\',\'b\':\'ac\',\'c\':\'ab\'} # Example: n = 5, k = 20\n\n div,r = divmod(k-1,2**(n-1)) # div,r = divmod(19,16) = 1,3\n\n if div > 2: return \'\'\n prev = ans = \'... | 4 | A **happy string** is a string that:
* consists only of letters of the set `['a', 'b', 'c']`.
* `s[i] != s[i + 1]` for all values of `i` from `1` to `s.length - 1` (string is 1-indexed).
For example, strings **"abc ", "ac ", "b "** and **"abcbabcbcb "** are all happy strings and strings **"aa ", "baa "** and **"a... | null |
Python3 O(n) solution using math with clear explanation | the-k-th-lexicographical-string-of-all-happy-strings-of-length-n | 0 | 1 | let\'s consider n=3 and k=9\nThe lexicographical order of the strings are ["aba", "abc", "aca", "acb", "bab", "bac", "bca", "bcb", "cab", "cac", "cba", "cbc"]\n\nWe can observe that each element a,b,c has repeated 2**(n-1) times as the first character which is 4 in our case they are ["aba", "abc", "aca", "acb"], ["bab... | 15 | A **happy string** is a string that:
* consists only of letters of the set `['a', 'b', 'c']`.
* `s[i] != s[i + 1]` for all values of `i` from `1` to `s.length - 1` (string is 1-indexed).
For example, strings **"abc ", "ac ", "b "** and **"abcbabcbcb "** are all happy strings and strings **"aa ", "baa "** and **"a... | null |
1st char to last | the-k-th-lexicographical-string-of-all-happy-strings-of-length-n | 0 | 1 | # Intuition\nYou could just generate all the cases by using python itertools.product, but that would take too long.\n\nIt seems you should be able to more or less go straight to the kth item.\n\n# Approach\nFirst figure out how many happy lists there are\n\nDraw on a whiteboard n=3 get a feel for it\n\n\n# Complexity\n... | 0 | A **happy string** is a string that:
* consists only of letters of the set `['a', 'b', 'c']`.
* `s[i] != s[i + 1]` for all values of `i` from `1` to `s.length - 1` (string is 1-indexed).
For example, strings **"abc ", "ac ", "b "** and **"abcbabcbcb "** are all happy strings and strings **"aa ", "baa "** and **"a... | null |
Easy to understand Python3 solution | the-k-th-lexicographical-string-of-all-happy-strings-of-length-n | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ --... | 0 | A **happy string** is a string that:
* consists only of letters of the set `['a', 'b', 'c']`.
* `s[i] != s[i + 1]` for all values of `i` from `1` to `s.length - 1` (string is 1-indexed).
For example, strings **"abc ", "ac ", "b "** and **"abcbabcbcb "** are all happy strings and strings **"aa ", "baa "** and **"a... | null |
Concise solution on python3 | the-k-th-lexicographical-string-of-all-happy-strings-of-length-n | 0 | 1 | \n# Code\n```\nclass Solution:\n def getHappyString(self, n: int, k: int) -> str:\n \n cnt = 0\n ans = ""\n def go(res):\n if len(res) == n:\n nonlocal cnt, ans\n cnt += 1\n if cnt == k:\n ans = res\n ... | 0 | A **happy string** is a string that:
* consists only of letters of the set `['a', 'b', 'c']`.
* `s[i] != s[i + 1]` for all values of `i` from `1` to `s.length - 1` (string is 1-indexed).
For example, strings **"abc ", "ac ", "b "** and **"abcbabcbcb "** are all happy strings and strings **"aa ", "baa "** and **"a... | null |
Clean partition dp solution | restore-the-array | 0 | 1 | # Code\n```\nclass Solution:\n def numberOfArrays(self, s: str, k: int) -> int:\n\n n = len(s)\n mod = 10**9+7\n\n @lru_cache(None)\n def rec(i=0):\n\n if i==n :\n return 1\n \n if s[i]=="0":\n return 0\n \n ... | 2 | A program was supposed to print an array of integers. The program forgot to print whitespaces and the array is printed as a string of digits `s` and all we know is that all integers in the array were in the range `[1, k]` and there are no leading zeros in the array.
Given the string `s` and the integer `k`, return _th... | Try sorting nums. If nums is consecutive and sorted in ascending order, then nums[i] + 1 = nums[i + 1] for every i in the range 0 ≤ i < nums.length - 1. |
Image Explanation🏆- [Easiest & Concise] - C++/Java/Python | restore-the-array | 1 | 1 | # Video Solution (`Aryan Mittal`) - Link in LeetCode Profile\n`Restore The Array` by `Aryan Mittal`\n\n\n\n# Approach & Intution\n. Then your subproblem becomes the rest of the string.)\n\n# Approach\n<!-- Descri... | 1 | A program was supposed to print an array of integers. The program forgot to print whitespaces and the array is printed as a string of digits `s` and all we know is that all integers in the array were in the range `[1, k]` and there are no leading zeros in the array.
Given the string `s` and the integer `k`, return _th... | Try sorting nums. If nums is consecutive and sorted in ascending order, then nums[i] + 1 = nums[i + 1] for every i in the range 0 ≤ i < nums.length - 1. |
python3 , TopDown + Bottom Up | restore-the-array | 0 | 1 | # Intuition\nWe need to explore all the number in range starting with single digit possible number , 2 digits possible numbers and so on.\n\n\n# Code [TopDown]\n```\nclass Solution:\n def numberOfArrays(self, s: str, k: int) -> int:\n n = len(s)\n MOD = 1000000007\n dp = [-1]*n\n def coun... | 1 | A program was supposed to print an array of integers. The program forgot to print whitespaces and the array is printed as a string of digits `s` and all we know is that all integers in the array were in the range `[1, k]` and there are no leading zeros in the array.
Given the string `s` and the integer `k`, return _th... | Try sorting nums. If nums is consecutive and sorted in ascending order, then nums[i] + 1 = nums[i + 1] for every i in the range 0 ≤ i < nums.length - 1. |
✨ Linear time and near constant space | Details explanation | S: 100% M: 100% ✨ | restore-the-array | 0 | 1 | \n\n# Approach\nSimilar to Solution 3 of the Editorial but without the second loop.\n\n# Complexity\nn = len(s)\n- Time complexity: $$O(n)$$\n- Space complexity: $$O(log_{10}{k})$$ ~ $$O(1)$$\n<!-- Add your... | 1 | A program was supposed to print an array of integers. The program forgot to print whitespaces and the array is printed as a string of digits `s` and all we know is that all integers in the array were in the range `[1, k]` and there are no leading zeros in the array.
Given the string `s` and the integer `k`, return _th... | Try sorting nums. If nums is consecutive and sorted in ascending order, then nums[i] + 1 = nums[i + 1] for every i in the range 0 ≤ i < nums.length - 1. |
python3 Solution | restore-the-array | 0 | 1 | \n```\nclass Solution:\n def numberOfArrays(self, s: str, k: int) -> int:\n n=len(s)\n mod=10**9+7\n dp=[0]*(n+1)\n dp[-1]=1\n for i in range(n-1,-1,-1):\n if s[i]==\'0\':\n continue\n\n num=0\n j=i\n while j<n and int(s[i:... | 1 | A program was supposed to print an array of integers. The program forgot to print whitespaces and the array is printed as a string of digits `s` and all we know is that all integers in the array were in the range `[1, k]` and there are no leading zeros in the array.
Given the string `s` and the integer `k`, return _th... | Try sorting nums. If nums is consecutive and sorted in ascending order, then nums[i] + 1 = nums[i + 1] for every i in the range 0 ≤ i < nums.length - 1. |
Python (Simple DP) | restore-the-array | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ --... | 2 | A program was supposed to print an array of integers. The program forgot to print whitespaces and the array is printed as a string of digits `s` and all we know is that all integers in the array were in the range `[1, k]` and there are no leading zeros in the array.
Given the string `s` and the integer `k`, return _th... | Try sorting nums. If nums is consecutive and sorted in ascending order, then nums[i] + 1 = nums[i + 1] for every i in the range 0 ≤ i < nums.length - 1. |
[Python] Clean solutions with explanation. O(N) Time and Space. | reformat-the-string | 0 | 1 | First we get lists of letters and digits seperately. \nThen we append the bigger list first. In this problem, i use flag to keep track of which one to append, this will make the code cleaner.\n**1st Solution**\n```python\nclass Solution:\n def reformat(self, s: str) -> str:\n letters = [c for c in s if c.isal... | 16 | You are given an alphanumeric string `s`. (**Alphanumeric string** is a string consisting of lowercase English letters and digits).
You have to find a permutation of the string where no letter is followed by another letter and no digit is followed by another digit. That is, no two adjacent characters have the same typ... | null |
Python3 Solution with separated lists | reformat-the-string | 0 | 1 | # Intuition\nSeparate list into letters and digits. If the length of the lists differ by more than 1, then return empty string (cannot create alternating new string)\n\nCreate flag to determine whether to start with digits or letters first (whichever list is longer)\n\n# Code\n```\nclass Solution:\n def getLettersAn... | 0 | You are given an alphanumeric string `s`. (**Alphanumeric string** is a string consisting of lowercase English letters and digits).
You have to find a permutation of the string where no letter is followed by another letter and no digit is followed by another digit. That is, no two adjacent characters have the same typ... | null |
Python solution | reformat-the-string | 0 | 1 | ```\nclass Solution:\n def reformat(self, s: str) -> str:\n alphabet: str = "abcdefghijklmnopqrstuvwxyz"\n digits: str = "1234567890"\n permutation: str = ""\n\n letters: list[str] = [letter for letter in s if letter in alphabet]\n numbers: list[str] = [number for number in s if nu... | 0 | You are given an alphanumeric string `s`. (**Alphanumeric string** is a string consisting of lowercase English letters and digits).
You have to find a permutation of the string where no letter is followed by another letter and no digit is followed by another digit. That is, no two adjacent characters have the same typ... | null |
beats 94% uisng zip | reformat-the-string | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ --... | 0 | You are given an alphanumeric string `s`. (**Alphanumeric string** is a string consisting of lowercase English letters and digits).
You have to find a permutation of the string where no letter is followed by another letter and no digit is followed by another digit. That is, no two adjacent characters have the same typ... | null |
Python solution easy to understand | reformat-the-string | 0 | 1 | \n\n# Code\n```\nclass Solution:\n def reformat(self, s: str) -> str:\n nums=[i for i in s if i.isnumeric()]\n s=[i for i in s if i.isalpha()]\n if len(nums)!=len(s) and len(nums)-1!=len(s) and len(nums)+1!=len(s):\n return ""\n else:\n if len(nums)==len(s):\n ... | 0 | You are given an alphanumeric string `s`. (**Alphanumeric string** is a string consisting of lowercase English letters and digits).
You have to find a permutation of the string where no letter is followed by another letter and no digit is followed by another digit. That is, no two adjacent characters have the same typ... | null |
Alternating Characters and Digits in a String: Detailed Explanation Included | reformat-the-string | 0 | 1 | # Approach & Intuition:\n\nWhen given a string that contains both alphabetic and numeric characters, the goal is to reformat the string so that no two adjacent characters are of the same type and to alternate between letters and digits as much as possible. The core challenges here are determining the right order and en... | 0 | You are given an alphanumeric string `s`. (**Alphanumeric string** is a string consisting of lowercase English letters and digits).
You have to find a permutation of the string where no letter is followed by another letter and no digit is followed by another digit. That is, no two adjacent characters have the same typ... | null |
Len of list comprehension Python solution | reformat-the-string | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ --... | 0 | You are given an alphanumeric string `s`. (**Alphanumeric string** is a string consisting of lowercase English letters and digits).
You have to find a permutation of the string where no letter is followed by another letter and no digit is followed by another digit. That is, no two adjacent characters have the same typ... | null |
reformat string | reformat-the-string | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\nO(n + m + k), where n is the length of the string s and m is the length of the shorter list among alpha and nums, and k is the leng... | 0 | You are given an alphanumeric string `s`. (**Alphanumeric string** is a string consisting of lowercase English letters and digits).
You have to find a permutation of the string where no letter is followed by another letter and no digit is followed by another digit. That is, no two adjacent characters have the same typ... | null |
Interesting way to intentionally choose the array for consideration - Python3 | reformat-the-string | 0 | 1 | It just a smart way to overcome which array (digit or characters) is larger then we consider that array prior to the other by using index: -1, 0, 1. When applying absolute, we will always get 0 and 1!!\n# Code\n```\nclass Solution:\n def reformat(self, s: str) -> str:\n divide = [[], []]\n res = ""\n ... | 0 | You are given an alphanumeric string `s`. (**Alphanumeric string** is a string consisting of lowercase English letters and digits).
You have to find a permutation of the string where no letter is followed by another letter and no digit is followed by another digit. That is, no two adjacent characters have the same typ... | null |
Python3 Naive Solution | reformat-the-string | 0 | 1 | \n\n# Code\n```\nclass Solution:\n def reformat(self, s: str) -> str:\n \n digits=[]\n chrs=[]\n \n for ch in s:\n if ch.isdigit():\n digits.append(ch)\n else:\n chrs.append(ch)\n \n if abs(len(digits)-len(ch... | 0 | You are given an alphanumeric string `s`. (**Alphanumeric string** is a string consisting of lowercase English letters and digits).
You have to find a permutation of the string where no letter is followed by another letter and no digit is followed by another digit. That is, no two adjacent characters have the same typ... | null |
Easy Python Solution Using zip() || ✅✅✅ | reformat-the-string | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ --... | 0 | You are given an alphanumeric string `s`. (**Alphanumeric string** is a string consisting of lowercase English letters and digits).
You have to find a permutation of the string where no letter is followed by another letter and no digit is followed by another digit. That is, no two adjacent characters have the same typ... | null |
Solution in Python3. Beats 100% of users with Python3 | display-table-of-food-orders-in-a-restaurant | 0 | 1 | # Intuition\r\n<!-- Describe your first thoughts on how to solve this problem. -->\r\n1. We have the input as an array of arrays, where each item has three information to provide about the order : [[Customer_Name, Table_Number, Food_Item]]. \r\n2. The output should be in the form of an array of arrays which represents ... | 2 | Given the array `orders`, which represents the orders that customers have done in a restaurant. More specifically `orders[i]=[customerNamei,tableNumberi,foodItemi]` where `customerNamei` is the name of the customer, `tableNumberi` is the table customer sit at, and `foodItemi` is the item customer orders.
_Return the r... | null |
Python3 O(NlogN) solution with dictionaries (98.54% Runtime) | display-table-of-food-orders-in-a-restaurant | 0 | 1 | # Intuition\r\n<!-- Describe your first thoughts on how to solve this problem. -->\r\n\r\nUse a dictionary to store mappings of table -> menus\r\n\r\n# Approach\r\n<!-- Describe your approach to solving the... | 0 | Given the array `orders`, which represents the orders that customers have done in a restaurant. More specifically `orders[i]=[customerNamei,tableNumberi,foodItemi]` where `customerNamei` is the name of the customer, `tableNumberi` is the table customer sit at, and `foodItemi` is the item customer orders.
_Return the r... | null |
✅ Easy Python Solution | HashMap | Sorting | display-table-of-food-orders-in-a-restaurant | 0 | 1 | # Code\r\n```\r\nclass Solution:\r\n def displayTable(self, orders: List[List[str]]) -> List[List[str]]:\r\n hashMap = {}\r\n sortedFoodItems = []\r\n ans = []\r\n temp = [\'Table\']\r\n for order in orders:\r\n if order[2] not in sortedFoodItems:\r\n sort... | 0 | Given the array `orders`, which represents the orders that customers have done in a restaurant. More specifically `orders[i]=[customerNamei,tableNumberi,foodItemi]` where `customerNamei` is the name of the customer, `tableNumberi` is the table customer sit at, and `foodItemi` is the item customer orders.
_Return the r... | null |
Beginner friendly | display-table-of-food-orders-in-a-restaurant | 0 | 1 | # Intuition\r\n<!-- Describe your first thoughts on how to solve this problem. -->\r\n\r\n# Approach\r\n<!-- Describe your approach to solving the problem. -->\r\n\r\n# Complexity\r\n- Time complexity:\r\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\r\n\r\n- Space complexity:\r\n<!-- Add your space complexity ... | 0 | Given the array `orders`, which represents the orders that customers have done in a restaurant. More specifically `orders[i]=[customerNamei,tableNumberi,foodItemi]` where `customerNamei` is the name of the customer, `tableNumberi` is the table customer sit at, and `foodItemi` is the item customer orders.
_Return the r... | null |
Python Solution | display-table-of-food-orders-in-a-restaurant | 0 | 1 | # Code\r\n```\r\nclass Solution:\r\n def displayTable(self, orders: List[List[str]]) -> List[List[str]]:\r\n d=dict()\r\n s=set()\r\n for i in orders:\r\n if int(i[1]) in d.keys():\r\n d1=d.get(int(i[1]))\r\n if(i[2] in d1.keys()):\r\n ... | 0 | Given the array `orders`, which represents the orders that customers have done in a restaurant. More specifically `orders[i]=[customerNamei,tableNumberi,foodItemi]` where `customerNamei` is the name of the customer, `tableNumberi` is the table customer sit at, and `foodItemi` is the item customer orders.
_Return the r... | null |
Python 100%, very short code. Explanation line by line. | display-table-of-food-orders-in-a-restaurant | 0 | 1 | # Intuition\nIt\'s quite simple problem, if you are familiar with working with dataframes.\n\n# Approach\n1. Create sorted list of all items in menu\n2. Create dictionary of positions of all items in menu, so later when we go trough order, we will know where to put each item for each table.\n3. create dictionary where... | 0 | Given the array `orders`, which represents the orders that customers have done in a restaurant. More specifically `orders[i]=[customerNamei,tableNumberi,foodItemi]` where `customerNamei` is the name of the customer, `tableNumberi` is the table customer sit at, and `foodItemi` is the item customer orders.
_Return the r... | null |
Fast python3 solution using Hashmap and set | display-table-of-food-orders-in-a-restaurant | 0 | 1 | # Approach\r\nMy approach was to have a dictionary keep track of the number of each food served to each table. That is a dictionary with key, table number and value another dictionary with key, food and value number of orders.\r\n\r\nWe must however sort the orders list by table number so that we can compute the orders... | 0 | Given the array `orders`, which represents the orders that customers have done in a restaurant. More specifically `orders[i]=[customerNamei,tableNumberi,foodItemi]` where `customerNamei` is the name of the customer, `tableNumberi` is the table customer sit at, and `foodItemi` is the item customer orders.
_Return the r... | null |
python3 straight forward solution beats +90% | display-table-of-food-orders-in-a-restaurant | 0 | 1 | # Code\r\n```\r\nclass Solution:\r\n def displayTable(self, orders: List[List[str]]) -> List[List[str]]:\r\n hm = {}\r\n itms = set()\r\n foods = {}\r\n for i in orders:\r\n itms.add(i[2])\r\n\r\n itms = sorted(itms)\r\n\r\n for k,itm in enumerate(itms):\r\n ... | 0 | Given the array `orders`, which represents the orders that customers have done in a restaurant. More specifically `orders[i]=[customerNamei,tableNumberi,foodItemi]` where `customerNamei` is the name of the customer, `tableNumberi` is the table customer sit at, and `foodItemi` is the item customer orders.
_Return the r... | null |
Python Solution, hashmap O(N + TlogT + FlogF + T * F) | display-table-of-food-orders-in-a-restaurant | 0 | 1 | It\'s quit straightforward to use a hashmap to save meals that each table has.\nTime: O(N + TlogT + FlogF + T * F)\nN, T, F stands for amount of orders, tables and foods\nCost come from iterating orders and sorting tables and foods.\nAnd the last for takes T * F to dump the result.\nSpace: O(T * F)\n# Code\n```\nimport... | 0 | Given the array `orders`, which represents the orders that customers have done in a restaurant. More specifically `orders[i]=[customerNamei,tableNumberi,foodItemi]` where `customerNamei` is the name of the customer, `tableNumberi` is the table customer sit at, and `foodItemi` is the item customer orders.
_Return the r... | null |
Easy to understand Python3 solution | display-table-of-food-orders-in-a-restaurant | 0 | 1 | # Intuition\r\n<!-- Describe your first thoughts on how to solve this problem. -->\r\n\r\n# Approach\r\n<!-- Describe your approach to solving the problem. -->\r\n\r\n# Complexity\r\n- Time complexity:\r\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\r\n\r\n- Space complexity:\r\n<!-- Add your space complexity ... | 0 | Given the array `orders`, which represents the orders that customers have done in a restaurant. More specifically `orders[i]=[customerNamei,tableNumberi,foodItemi]` where `customerNamei` is the name of the customer, `tableNumberi` is the table customer sit at, and `foodItemi` is the item customer orders.
_Return the r... | null |
Elegant and short, perfect for interview | minimum-number-of-frogs-croaking | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nThe idea is somewhat similar to the famous problem of checking validity of braces. For example "(()())" is a valid sequence and "()(()" is invalid. The expected solution is to keep the number of open braces `num_open`. Increment the value... | 2 | You are given the string `croakOfFrogs`, which represents a combination of the string `"croak "` from different frogs, that is, multiple frogs can croak at the same time, so multiple `"croak "` are mixed.
_Return the minimum number of_ different _frogs to finish all the croaks in the given string._
A valid `"croak "`... | null |
Elegant and short, perfect for interview | minimum-number-of-frogs-croaking | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nThe idea is somewhat similar to the famous problem of checking validity of braces. For example "(()())" is a valid sequence and "()(()" is invalid. The expected solution is to keep the number of open braces `num_open`. Increment the value... | 2 | Given an array of integers `arr`, and three integers `a`, `b` and `c`. You need to find the number of good triplets.
A triplet `(arr[i], arr[j], arr[k])` is **good** if the following conditions are true:
* `0 <= i < j < k < arr.length`
* `|arr[i] - arr[j]| <= a`
* `|arr[j] - arr[k]| <= b`
* `|arr[i] - arr[k]|... | keep the frequency of all characters from "croak" using a hashmap. For each character in the given string, greedily match it to a possible "croak". |
Py O(n) Sol: Easy to Understand, Ask Doubt | minimum-number-of-frogs-croaking | 0 | 1 | ```\nclass Solution:\n def minNumberOfFrogs(self, croakOfFrogs: str) -> int:\n c = r = o = a = k = max_frog_croak = present_frog_croak = 0\n # need to know, at particular point,\n # what are the max frog are croaking,\n\n for i, v in enumerate(croakOfFrogs):\n if v == \'c\':\n ... | 30 | You are given the string `croakOfFrogs`, which represents a combination of the string `"croak "` from different frogs, that is, multiple frogs can croak at the same time, so multiple `"croak "` are mixed.
_Return the minimum number of_ different _frogs to finish all the croaks in the given string._
A valid `"croak "`... | null |
Py O(n) Sol: Easy to Understand, Ask Doubt | minimum-number-of-frogs-croaking | 0 | 1 | ```\nclass Solution:\n def minNumberOfFrogs(self, croakOfFrogs: str) -> int:\n c = r = o = a = k = max_frog_croak = present_frog_croak = 0\n # need to know, at particular point,\n # what are the max frog are croaking,\n\n for i, v in enumerate(croakOfFrogs):\n if v == \'c\':\n ... | 30 | Given an array of integers `arr`, and three integers `a`, `b` and `c`. You need to find the number of good triplets.
A triplet `(arr[i], arr[j], arr[k])` is **good** if the following conditions are true:
* `0 <= i < j < k < arr.length`
* `|arr[i] - arr[j]| <= a`
* `|arr[j] - arr[k]| <= b`
* `|arr[i] - arr[k]|... | keep the frequency of all characters from "croak" using a hashmap. For each character in the given string, greedily match it to a possible "croak". |
</> | minimum-number-of-frogs-croaking | 0 | 1 | \n# Code\n```\nclass Solution:\n def minNumberOfFrogs(self, croakOfFrogs: str) -> int:\n Frogs = 0\n sounds = [0] * 5\n d = {\n "c" : 0,\n "r" : 1,\n "o" : 2,\n "a" : 3,\n "k" : 4\n }\n\n for s in croakOfFrogs:\n sou... | 1 | You are given the string `croakOfFrogs`, which represents a combination of the string `"croak "` from different frogs, that is, multiple frogs can croak at the same time, so multiple `"croak "` are mixed.
_Return the minimum number of_ different _frogs to finish all the croaks in the given string._
A valid `"croak "`... | null |
</> | minimum-number-of-frogs-croaking | 0 | 1 | \n# Code\n```\nclass Solution:\n def minNumberOfFrogs(self, croakOfFrogs: str) -> int:\n Frogs = 0\n sounds = [0] * 5\n d = {\n "c" : 0,\n "r" : 1,\n "o" : 2,\n "a" : 3,\n "k" : 4\n }\n\n for s in croakOfFrogs:\n sou... | 1 | Given an array of integers `arr`, and three integers `a`, `b` and `c`. You need to find the number of good triplets.
A triplet `(arr[i], arr[j], arr[k])` is **good** if the following conditions are true:
* `0 <= i < j < k < arr.length`
* `|arr[i] - arr[j]| <= a`
* `|arr[j] - arr[k]| <= b`
* `|arr[i] - arr[k]|... | keep the frequency of all characters from "croak" using a hashmap. For each character in the given string, greedily match it to a possible "croak". |
Python 3 | Greedy, Simulation, Clean code | Explanantion | minimum-number-of-frogs-croaking | 0 | 1 | # Explanation\n- Need to find the maximum number of "croak" ongoing at some time point `t`. \n\t- e.g. `crcoakroak`\n\t\t- `0123456789` - time\n\t\t- `cr*oak****` - first \n\t\t- `**c***roak` - second\n\t- At time 2,3,4,5, there are 2 "croak" ongoing at the same time, thus there will be 2 frogs needed\n- To simulate th... | 6 | You are given the string `croakOfFrogs`, which represents a combination of the string `"croak "` from different frogs, that is, multiple frogs can croak at the same time, so multiple `"croak "` are mixed.
_Return the minimum number of_ different _frogs to finish all the croaks in the given string._
A valid `"croak "`... | null |
Python 3 | Greedy, Simulation, Clean code | Explanantion | minimum-number-of-frogs-croaking | 0 | 1 | # Explanation\n- Need to find the maximum number of "croak" ongoing at some time point `t`. \n\t- e.g. `crcoakroak`\n\t\t- `0123456789` - time\n\t\t- `cr*oak****` - first \n\t\t- `**c***roak` - second\n\t- At time 2,3,4,5, there are 2 "croak" ongoing at the same time, thus there will be 2 frogs needed\n- To simulate th... | 6 | Given an array of integers `arr`, and three integers `a`, `b` and `c`. You need to find the number of good triplets.
A triplet `(arr[i], arr[j], arr[k])` is **good** if the following conditions are true:
* `0 <= i < j < k < arr.length`
* `|arr[i] - arr[j]| <= a`
* `|arr[j] - arr[k]| <= b`
* `|arr[i] - arr[k]|... | keep the frequency of all characters from "croak" using a hashmap. For each character in the given string, greedily match it to a possible "croak". |
Monotonic sequence, quick | minimum-number-of-frogs-croaking | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nList record every word count, the list is decreasing sequence\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\ncount[\'c\'] - count[\'k\'] == active_forg\nif pre word less cur word is invalid\n\n# Complexity\n- Time... | 0 | You are given the string `croakOfFrogs`, which represents a combination of the string `"croak "` from different frogs, that is, multiple frogs can croak at the same time, so multiple `"croak "` are mixed.
_Return the minimum number of_ different _frogs to finish all the croaks in the given string._
A valid `"croak "`... | null |
Monotonic sequence, quick | minimum-number-of-frogs-croaking | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nList record every word count, the list is decreasing sequence\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\ncount[\'c\'] - count[\'k\'] == active_forg\nif pre word less cur word is invalid\n\n# Complexity\n- Time... | 0 | Given an array of integers `arr`, and three integers `a`, `b` and `c`. You need to find the number of good triplets.
A triplet `(arr[i], arr[j], arr[k])` is **good** if the following conditions are true:
* `0 <= i < j < k < arr.length`
* `|arr[i] - arr[j]| <= a`
* `|arr[j] - arr[k]| <= b`
* `|arr[i] - arr[k]|... | keep the frequency of all characters from "croak" using a hashmap. For each character in the given string, greedily match it to a possible "croak". |
Python3 - Easy to understand | minimum-number-of-frogs-croaking | 0 | 1 | # Approach\nThink of the problem as a state machine: frog A begins with "c" then transitions to "r" then transitions to "o" all the way to "k." If the input were to be always valid, we would only care about the maximum number of frogs concurrently croaking. In the context of the state machine, this is the number of fro... | 0 | You are given the string `croakOfFrogs`, which represents a combination of the string `"croak "` from different frogs, that is, multiple frogs can croak at the same time, so multiple `"croak "` are mixed.
_Return the minimum number of_ different _frogs to finish all the croaks in the given string._
A valid `"croak "`... | null |
Python3 - Easy to understand | minimum-number-of-frogs-croaking | 0 | 1 | # Approach\nThink of the problem as a state machine: frog A begins with "c" then transitions to "r" then transitions to "o" all the way to "k." If the input were to be always valid, we would only care about the maximum number of frogs concurrently croaking. In the context of the state machine, this is the number of fro... | 0 | Given an array of integers `arr`, and three integers `a`, `b` and `c`. You need to find the number of good triplets.
A triplet `(arr[i], arr[j], arr[k])` is **good** if the following conditions are true:
* `0 <= i < j < k < arr.length`
* `|arr[i] - arr[j]| <= a`
* `|arr[j] - arr[k]| <= b`
* `|arr[i] - arr[k]|... | keep the frequency of all characters from "croak" using a hashmap. For each character in the given string, greedily match it to a possible "croak". |
EASY PYTHON SOLUTION USING SIMPLE DP | build-array-where-you-can-find-the-maximum-exactly-k-comparisons | 0 | 1 | \n# Code\n```\nclass Solution:\n def dp(self,i,n,m,k,last,dct):\n if i==n:\n if k==0:\n return 1\n return 0\n if (i,k,last) in dct:\n return dct[(i,k,last)]\n val=0\n for j in range(1,m+1):\n if j<=last:\n val+=self... | 4 | You are given three integers `n`, `m` and `k`. Consider the following algorithm to find the maximum element of an array of positive integers:
You should build the array arr which has the following properties:
* `arr` has exactly `n` integers.
* `1 <= arr[i] <= m` where `(0 <= i < n)`.
* After applying the menti... | null |
EASY PYTHON SOLUTION USING SIMPLE DP | build-array-where-you-can-find-the-maximum-exactly-k-comparisons | 0 | 1 | \n# Code\n```\nclass Solution:\n def dp(self,i,n,m,k,last,dct):\n if i==n:\n if k==0:\n return 1\n return 0\n if (i,k,last) in dct:\n return dct[(i,k,last)]\n val=0\n for j in range(1,m+1):\n if j<=last:\n val+=self... | 4 | Given an integer array `arr` of **distinct** integers and an integer `k`.
A game will be played between the first two elements of the array (i.e. `arr[0]` and `arr[1]`). In each round of the game, we compare `arr[0]` with `arr[1]`, the larger integer wins and remains at position `0`, and the smaller integer moves to t... | Use dynamic programming approach. Build dp table where dp[a][b][c] is the number of ways you can start building the array starting from index a where the search_cost = c and the maximum used integer was b. Recursively, solve the small sub-problems first. Optimize your answer by stopping the search if you exceeded k cha... |
Python solution with optimize space and easy to understand | build-array-where-you-can-find-the-maximum-exactly-k-comparisons | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nThe problem involves finding the number of ways to build an array arr with certain constraints while using dynamic programming.\n# Approach\n<!-- Describe your approach to solving the problem. -->\nInitialize a 3D DP array dp where dp[i][... | 2 | You are given three integers `n`, `m` and `k`. Consider the following algorithm to find the maximum element of an array of positive integers:
You should build the array arr which has the following properties:
* `arr` has exactly `n` integers.
* `1 <= arr[i] <= m` where `(0 <= i < n)`.
* After applying the menti... | null |
Python solution with optimize space and easy to understand | build-array-where-you-can-find-the-maximum-exactly-k-comparisons | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nThe problem involves finding the number of ways to build an array arr with certain constraints while using dynamic programming.\n# Approach\n<!-- Describe your approach to solving the problem. -->\nInitialize a 3D DP array dp where dp[i][... | 2 | Given an integer array `arr` of **distinct** integers and an integer `k`.
A game will be played between the first two elements of the array (i.e. `arr[0]` and `arr[1]`). In each round of the game, we compare `arr[0]` with `arr[1]`, the larger integer wins and remains at position `0`, and the smaller integer moves to t... | Use dynamic programming approach. Build dp table where dp[a][b][c] is the number of ways you can start building the array starting from index a where the search_cost = c and the maximum used integer was b. Recursively, solve the small sub-problems first. Optimize your answer by stopping the search if you exceeded k cha... |
Simple Solution || Beginner Friendly || Easy to Understand ✅ | build-array-where-you-can-find-the-maximum-exactly-k-comparisons | 1 | 1 | # Approach\n<!-- Describe your approach to solving the problem. -->\n- We are given three integers `n, m, and k`.\n- We need to find the number of ways to build an array `arr` of `n` positive integers, each ranging from 1 to `m`, such that the "search_cost" of the array is equal to `k`.\n- "Search_cost" is defined as t... | 66 | You are given three integers `n`, `m` and `k`. Consider the following algorithm to find the maximum element of an array of positive integers:
You should build the array arr which has the following properties:
* `arr` has exactly `n` integers.
* `1 <= arr[i] <= m` where `(0 <= i < n)`.
* After applying the menti... | null |
Simple Solution || Beginner Friendly || Easy to Understand ✅ | build-array-where-you-can-find-the-maximum-exactly-k-comparisons | 1 | 1 | # Approach\n<!-- Describe your approach to solving the problem. -->\n- We are given three integers `n, m, and k`.\n- We need to find the number of ways to build an array `arr` of `n` positive integers, each ranging from 1 to `m`, such that the "search_cost" of the array is equal to `k`.\n- "Search_cost" is defined as t... | 66 | Given an integer array `arr` of **distinct** integers and an integer `k`.
A game will be played between the first two elements of the array (i.e. `arr[0]` and `arr[1]`). In each round of the game, we compare `arr[0]` with `arr[1]`, the larger integer wins and remains at position `0`, and the smaller integer moves to t... | Use dynamic programming approach. Build dp table where dp[a][b][c] is the number of ways you can start building the array starting from index a where the search_cost = c and the maximum used integer was b. Recursively, solve the small sub-problems first. Optimize your answer by stopping the search if you exceeded k cha... |
JAVA and C++ Faster | Fastest In Python | Dynamic Programming | O(n*k*m) | build-array-where-you-can-find-the-maximum-exactly-k-comparisons | 1 | 1 | \n\n# YouTube Video:\n[https://youtu.be/EwZ9yLxrYRg]()\n\n# Intuition:\n\n- To find the number of ways to construct the array `arr` satisfying the given conditions, we can use dynamic... | 13 | You are given three integers `n`, `m` and `k`. Consider the following algorithm to find the maximum element of an array of positive integers:
You should build the array arr which has the following properties:
* `arr` has exactly `n` integers.
* `1 <= arr[i] <= m` where `(0 <= i < n)`.
* After applying the menti... | null |
JAVA and C++ Faster | Fastest In Python | Dynamic Programming | O(n*k*m) | build-array-where-you-can-find-the-maximum-exactly-k-comparisons | 1 | 1 | \n\n# YouTube Video:\n[https://youtu.be/EwZ9yLxrYRg]()\n\n# Intuition:\n\n- To find the number of ways to construct the array `arr` satisfying the given conditions, we can use dynamic... | 13 | Given an integer array `arr` of **distinct** integers and an integer `k`.
A game will be played between the first two elements of the array (i.e. `arr[0]` and `arr[1]`). In each round of the game, we compare `arr[0]` with `arr[1]`, the larger integer wins and remains at position `0`, and the smaller integer moves to t... | Use dynamic programming approach. Build dp table where dp[a][b][c] is the number of ways you can start building the array starting from index a where the search_cost = c and the maximum used integer was b. Recursively, solve the small sub-problems first. Optimize your answer by stopping the search if you exceeded k cha... |
✅ 90.27% Dynamic Programming Optimized | build-array-where-you-can-find-the-maximum-exactly-k-comparisons | 1 | 1 | # Intuition\n\nUpon reading the problem, we\'re tasked with finding the number of arrays of length `n` with values in the range `[1, m]` such that the array has exactly `k` cost. The challenge is not only in understanding what constitutes this cost but also in devising an efficient algorithm to count these arrays give... | 75 | You are given three integers `n`, `m` and `k`. Consider the following algorithm to find the maximum element of an array of positive integers:
You should build the array arr which has the following properties:
* `arr` has exactly `n` integers.
* `1 <= arr[i] <= m` where `(0 <= i < n)`.
* After applying the menti... | null |
✅ 90.27% Dynamic Programming Optimized | build-array-where-you-can-find-the-maximum-exactly-k-comparisons | 1 | 1 | # Intuition\n\nUpon reading the problem, we\'re tasked with finding the number of arrays of length `n` with values in the range `[1, m]` such that the array has exactly `k` cost. The challenge is not only in understanding what constitutes this cost but also in devising an efficient algorithm to count these arrays give... | 75 | Given an integer array `arr` of **distinct** integers and an integer `k`.
A game will be played between the first two elements of the array (i.e. `arr[0]` and `arr[1]`). In each round of the game, we compare `arr[0]` with `arr[1]`, the larger integer wins and remains at position `0`, and the smaller integer moves to t... | Use dynamic programming approach. Build dp table where dp[a][b][c] is the number of ways you can start building the array starting from index a where the search_cost = c and the maximum used integer was b. Recursively, solve the small sub-problems first. Optimize your answer by stopping the search if you exceeded k cha... |
Python3 Solution | build-array-where-you-can-find-the-maximum-exactly-k-comparisons | 0 | 1 | \n```\nclass Solution:\n def numOfArrays(self, n: int, m: int, k: int) -> int:\n mod=10**9+7\n @lru_cache(None)\n def dp(i,h,k):\n if i==n and k==0:\n return 1\n if i==n or k<0:\n return 0\n return sum(dp(i+1,max(h,j),k-(j>h)) for j ... | 1 | You are given three integers `n`, `m` and `k`. Consider the following algorithm to find the maximum element of an array of positive integers:
You should build the array arr which has the following properties:
* `arr` has exactly `n` integers.
* `1 <= arr[i] <= m` where `(0 <= i < n)`.
* After applying the menti... | null |
Python3 Solution | build-array-where-you-can-find-the-maximum-exactly-k-comparisons | 0 | 1 | \n```\nclass Solution:\n def numOfArrays(self, n: int, m: int, k: int) -> int:\n mod=10**9+7\n @lru_cache(None)\n def dp(i,h,k):\n if i==n and k==0:\n return 1\n if i==n or k<0:\n return 0\n return sum(dp(i+1,max(h,j),k-(j>h)) for j ... | 1 | Given an integer array `arr` of **distinct** integers and an integer `k`.
A game will be played between the first two elements of the array (i.e. `arr[0]` and `arr[1]`). In each round of the game, we compare `arr[0]` with `arr[1]`, the larger integer wins and remains at position `0`, and the smaller integer moves to t... | Use dynamic programming approach. Build dp table where dp[a][b][c] is the number of ways you can start building the array starting from index a where the search_cost = c and the maximum used integer was b. Recursively, solve the small sub-problems first. Optimize your answer by stopping the search if you exceeded k cha... |
✅ Beats 92% || Dynamic Programming || Full Explanation || | build-array-where-you-can-find-the-maximum-exactly-k-comparisons | 1 | 1 | # Intuition\n1. dp[i][j][k] denotes the number of ways to build an array of length i, with the maximum element j and cost k.\n2. dp[i][j][k] = dp[i-1][j][k]*j + sum(dp[i-1][x][k-1]) for x in range(1,j)\n3. return sum(dp[n][i][k] for i in range(1,m+1))%(10**9+7)\n\n\n# Approach\n1. The maximum element in the array can b... | 1 | You are given three integers `n`, `m` and `k`. Consider the following algorithm to find the maximum element of an array of positive integers:
You should build the array arr which has the following properties:
* `arr` has exactly `n` integers.
* `1 <= arr[i] <= m` where `(0 <= i < n)`.
* After applying the menti... | null |
✅ Beats 92% || Dynamic Programming || Full Explanation || | build-array-where-you-can-find-the-maximum-exactly-k-comparisons | 1 | 1 | # Intuition\n1. dp[i][j][k] denotes the number of ways to build an array of length i, with the maximum element j and cost k.\n2. dp[i][j][k] = dp[i-1][j][k]*j + sum(dp[i-1][x][k-1]) for x in range(1,j)\n3. return sum(dp[n][i][k] for i in range(1,m+1))%(10**9+7)\n\n\n# Approach\n1. The maximum element in the array can b... | 1 | Given an integer array `arr` of **distinct** integers and an integer `k`.
A game will be played between the first two elements of the array (i.e. `arr[0]` and `arr[1]`). In each round of the game, we compare `arr[0]` with `arr[1]`, the larger integer wins and remains at position `0`, and the smaller integer moves to t... | Use dynamic programming approach. Build dp table where dp[a][b][c] is the number of ways you can start building the array starting from index a where the search_cost = c and the maximum used integer was b. Recursively, solve the small sub-problems first. Optimize your answer by stopping the search if you exceeded k cha... |
Python || Documented and Simple Memoization Solution | build-array-where-you-can-find-the-maximum-exactly-k-comparisons | 0 | 1 | ```\nfrom functools import cache\n\nMAX = 10 ** 9 + 7\n\n\nclass Solution:\n def numOfArrays(self, N: int, M: int, K: int) -> int:\n @cache\n def ways(n: int, m: int, k: int) -> int:\n """\n :param n:\n :param m:\n :param k:\n :return: number of wa... | 1 | You are given three integers `n`, `m` and `k`. Consider the following algorithm to find the maximum element of an array of positive integers:
You should build the array arr which has the following properties:
* `arr` has exactly `n` integers.
* `1 <= arr[i] <= m` where `(0 <= i < n)`.
* After applying the menti... | null |
Python || Documented and Simple Memoization Solution | build-array-where-you-can-find-the-maximum-exactly-k-comparisons | 0 | 1 | ```\nfrom functools import cache\n\nMAX = 10 ** 9 + 7\n\n\nclass Solution:\n def numOfArrays(self, N: int, M: int, K: int) -> int:\n @cache\n def ways(n: int, m: int, k: int) -> int:\n """\n :param n:\n :param m:\n :param k:\n :return: number of wa... | 1 | Given an integer array `arr` of **distinct** integers and an integer `k`.
A game will be played between the first two elements of the array (i.e. `arr[0]` and `arr[1]`). In each round of the game, we compare `arr[0]` with `arr[1]`, the larger integer wins and remains at position `0`, and the smaller integer moves to t... | Use dynamic programming approach. Build dp table where dp[a][b][c] is the number of ways you can start building the array starting from index a where the search_cost = c and the maximum used integer was b. Recursively, solve the small sub-problems first. Optimize your answer by stopping the search if you exceeded k cha... |
Just count, do sum and boomm | in O(n) time | python3 | maximum-score-after-splitting-a-string | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\ncount \'0\' and \'1\'. Do sum conditionally.\n# Approach\n<!-- Describe your approach to solving the problem. -->\nfirst of all count how many one have the input sitring and store the result. again loop thorugh the string and check how ma... | 1 | Given a string `s` of zeros and ones, _return the maximum score after splitting the string into two **non-empty** substrings_ (i.e. **left** substring and **right** substring).
The score after splitting a string is the number of **zeros** in the **left** substring plus the number of **ones** in the **right** substring... | If the smallest number in the possible-to-split array is V, then numbers V+1, V+2, ... V+k-1 must contain there as well. You can iteratively find k sets and remove them from array until it becomes empty. Failure to do so would mean that array is unsplittable. |
Just count, do sum and boomm | in O(n) time | python3 | maximum-score-after-splitting-a-string | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\ncount \'0\' and \'1\'. Do sum conditionally.\n# Approach\n<!-- Describe your approach to solving the problem. -->\nfirst of all count how many one have the input sitring and store the result. again loop thorugh the string and check how ma... | 1 | You are given two **sorted** arrays of distinct integers `nums1` and `nums2.`
A **valid path** is defined as follows:
* Choose array `nums1` or `nums2` to traverse (from index-0).
* Traverse the current array from left to right.
* If you are reading any value that is present in `nums1` and `nums2` you are allow... | Precompute a prefix sum of ones ('1'). Iterate from left to right counting the number of zeros ('0'), then use the precomputed prefix sum for counting ones ('1'). Update the answer. |
✅ PYTHON || Simple Beginner Solution | maximum-score-after-splitting-a-string | 0 | 1 | # Code\n```\nclass Solution:\n def maxScore(self, s: str) -> int:\n cnt = 0\n for i in range(1, len(s)):\n cnt = max(cnt, (s[:i].count(\'0\') + s[i:].count(\'1\')))\n return cnt\n\n``` | 1 | Given a string `s` of zeros and ones, _return the maximum score after splitting the string into two **non-empty** substrings_ (i.e. **left** substring and **right** substring).
The score after splitting a string is the number of **zeros** in the **left** substring plus the number of **ones** in the **right** substring... | If the smallest number in the possible-to-split array is V, then numbers V+1, V+2, ... V+k-1 must contain there as well. You can iteratively find k sets and remove them from array until it becomes empty. Failure to do so would mean that array is unsplittable. |
✅ PYTHON || Simple Beginner Solution | maximum-score-after-splitting-a-string | 0 | 1 | # Code\n```\nclass Solution:\n def maxScore(self, s: str) -> int:\n cnt = 0\n for i in range(1, len(s)):\n cnt = max(cnt, (s[:i].count(\'0\') + s[i:].count(\'1\')))\n return cnt\n\n``` | 1 | You are given two **sorted** arrays of distinct integers `nums1` and `nums2.`
A **valid path** is defined as follows:
* Choose array `nums1` or `nums2` to traverse (from index-0).
* Traverse the current array from left to right.
* If you are reading any value that is present in `nums1` and `nums2` you are allow... | Precompute a prefix sum of ones ('1'). Iterate from left to right counting the number of zeros ('0'), then use the precomputed prefix sum for counting ones ('1'). Update the answer. |
Python | Easy Solution✅ | maximum-score-after-splitting-a-string | 0 | 1 | ```\ndef maxScore(self, s: str) -> int:\n maxx = 0\n i=1\n while i <= len(s)-1:\n left = s[:i]\n right = s[i:]\n zero_count = left.count("0")\n one_count = right.count("1")\n maxx = max(maxx,(zero_count+one_count))\n i = i+1\n ... | 8 | Given a string `s` of zeros and ones, _return the maximum score after splitting the string into two **non-empty** substrings_ (i.e. **left** substring and **right** substring).
The score after splitting a string is the number of **zeros** in the **left** substring plus the number of **ones** in the **right** substring... | If the smallest number in the possible-to-split array is V, then numbers V+1, V+2, ... V+k-1 must contain there as well. You can iteratively find k sets and remove them from array until it becomes empty. Failure to do so would mean that array is unsplittable. |
Python | Easy Solution✅ | maximum-score-after-splitting-a-string | 0 | 1 | ```\ndef maxScore(self, s: str) -> int:\n maxx = 0\n i=1\n while i <= len(s)-1:\n left = s[:i]\n right = s[i:]\n zero_count = left.count("0")\n one_count = right.count("1")\n maxx = max(maxx,(zero_count+one_count))\n i = i+1\n ... | 8 | You are given two **sorted** arrays of distinct integers `nums1` and `nums2.`
A **valid path** is defined as follows:
* Choose array `nums1` or `nums2` to traverse (from index-0).
* Traverse the current array from left to right.
* If you are reading any value that is present in `nums1` and `nums2` you are allow... | Precompute a prefix sum of ones ('1'). Iterate from left to right counting the number of zeros ('0'), then use the precomputed prefix sum for counting ones ('1'). Update the answer. |
Linear time Python solution with explanation | maximum-score-after-splitting-a-string | 0 | 1 | ```\n# IDEA: on first pass count total number of 1s, on second pass keep track of how many 0s and 1s encountered so far\n# and calculate current score (subtract current 1s from total 1s and add current 0s), keep track of maximum\n# O(N) time, O(1) space\n#\nclass Solution:\n def maxScore(self, s: str) ->... | 7 | Given a string `s` of zeros and ones, _return the maximum score after splitting the string into two **non-empty** substrings_ (i.e. **left** substring and **right** substring).
The score after splitting a string is the number of **zeros** in the **left** substring plus the number of **ones** in the **right** substring... | If the smallest number in the possible-to-split array is V, then numbers V+1, V+2, ... V+k-1 must contain there as well. You can iteratively find k sets and remove them from array until it becomes empty. Failure to do so would mean that array is unsplittable. |
Linear time Python solution with explanation | maximum-score-after-splitting-a-string | 0 | 1 | ```\n# IDEA: on first pass count total number of 1s, on second pass keep track of how many 0s and 1s encountered so far\n# and calculate current score (subtract current 1s from total 1s and add current 0s), keep track of maximum\n# O(N) time, O(1) space\n#\nclass Solution:\n def maxScore(self, s: str) ->... | 7 | You are given two **sorted** arrays of distinct integers `nums1` and `nums2.`
A **valid path** is defined as follows:
* Choose array `nums1` or `nums2` to traverse (from index-0).
* Traverse the current array from left to right.
* If you are reading any value that is present in `nums1` and `nums2` you are allow... | Precompute a prefix sum of ones ('1'). Iterate from left to right counting the number of zeros ('0'), then use the precomputed prefix sum for counting ones ('1'). Update the answer. |
python one line solution | maximum-score-after-splitting-a-string | 0 | 1 | ```\nclass Solution:\n def maxScore(self, s: str) -> int:\n \n return max([s[:i].count(\'0\')+s[i:].count(\'1\') for i in range(1, len(s))])\n``` | 1 | Given a string `s` of zeros and ones, _return the maximum score after splitting the string into two **non-empty** substrings_ (i.e. **left** substring and **right** substring).
The score after splitting a string is the number of **zeros** in the **left** substring plus the number of **ones** in the **right** substring... | If the smallest number in the possible-to-split array is V, then numbers V+1, V+2, ... V+k-1 must contain there as well. You can iteratively find k sets and remove them from array until it becomes empty. Failure to do so would mean that array is unsplittable. |
python one line solution | maximum-score-after-splitting-a-string | 0 | 1 | ```\nclass Solution:\n def maxScore(self, s: str) -> int:\n \n return max([s[:i].count(\'0\')+s[i:].count(\'1\') for i in range(1, len(s))])\n``` | 1 | You are given two **sorted** arrays of distinct integers `nums1` and `nums2.`
A **valid path** is defined as follows:
* Choose array `nums1` or `nums2` to traverse (from index-0).
* Traverse the current array from left to right.
* If you are reading any value that is present in `nums1` and `nums2` you are allow... | Precompute a prefix sum of ones ('1'). Iterate from left to right counting the number of zeros ('0'), then use the precomputed prefix sum for counting ones ('1'). Update the answer. |
Python, two passes. Time: O(N) | maximum-score-after-splitting-a-string | 0 | 1 | ```\nclass Solution:\n def maxScore(self, s: str) -> int:\n score = 0 \n zeros = 0\n ones = s.count(\'1\') \n \n for i in range(len(s)-1):\n if s[i] == \'0\':\n zeros += 1\n else:\n ones -= 1\n score = max(score,... | 8 | Given a string `s` of zeros and ones, _return the maximum score after splitting the string into two **non-empty** substrings_ (i.e. **left** substring and **right** substring).
The score after splitting a string is the number of **zeros** in the **left** substring plus the number of **ones** in the **right** substring... | If the smallest number in the possible-to-split array is V, then numbers V+1, V+2, ... V+k-1 must contain there as well. You can iteratively find k sets and remove them from array until it becomes empty. Failure to do so would mean that array is unsplittable. |
Python, two passes. Time: O(N) | maximum-score-after-splitting-a-string | 0 | 1 | ```\nclass Solution:\n def maxScore(self, s: str) -> int:\n score = 0 \n zeros = 0\n ones = s.count(\'1\') \n \n for i in range(len(s)-1):\n if s[i] == \'0\':\n zeros += 1\n else:\n ones -= 1\n score = max(score,... | 8 | You are given two **sorted** arrays of distinct integers `nums1` and `nums2.`
A **valid path** is defined as follows:
* Choose array `nums1` or `nums2` to traverse (from index-0).
* Traverse the current array from left to right.
* If you are reading any value that is present in `nums1` and `nums2` you are allow... | Precompute a prefix sum of ones ('1'). Iterate from left to right counting the number of zeros ('0'), then use the precomputed prefix sum for counting ones ('1'). Update the answer. |
Beats 91.69% of users with Python3 | maximum-score-after-splitting-a-string | 0 | 1 | # Code\n```\nclass Solution:\n def maxScore(self, s: str) -> int:\n # straight forward approach\n # 41ms\n # Beats 65.17% of users with Python3\n """\n ret, n = 0, len(s)\n for i in range(1, n):\n a = s.count("0", 0, i)\n b = s.count("1", i, n)\n ... | 0 | Given a string `s` of zeros and ones, _return the maximum score after splitting the string into two **non-empty** substrings_ (i.e. **left** substring and **right** substring).
The score after splitting a string is the number of **zeros** in the **left** substring plus the number of **ones** in the **right** substring... | If the smallest number in the possible-to-split array is V, then numbers V+1, V+2, ... V+k-1 must contain there as well. You can iteratively find k sets and remove them from array until it becomes empty. Failure to do so would mean that array is unsplittable. |
Beats 91.69% of users with Python3 | maximum-score-after-splitting-a-string | 0 | 1 | # Code\n```\nclass Solution:\n def maxScore(self, s: str) -> int:\n # straight forward approach\n # 41ms\n # Beats 65.17% of users with Python3\n """\n ret, n = 0, len(s)\n for i in range(1, n):\n a = s.count("0", 0, i)\n b = s.count("1", i, n)\n ... | 0 | You are given two **sorted** arrays of distinct integers `nums1` and `nums2.`
A **valid path** is defined as follows:
* Choose array `nums1` or `nums2` to traverse (from index-0).
* Traverse the current array from left to right.
* If you are reading any value that is present in `nums1` and `nums2` you are allow... | Precompute a prefix sum of ones ('1'). Iterate from left to right counting the number of zeros ('0'), then use the precomputed prefix sum for counting ones ('1'). Update the answer. |
[Python | beats 95.96% in time | no need to count each time] | maximum-score-after-splitting-a-string | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nOnce you count the number of zeros in the left substring, and the number of ones in the right substring, you don\'t have to count them every time you make new substrings. All you need to do is check whether the new character moved from ri... | 0 | Given a string `s` of zeros and ones, _return the maximum score after splitting the string into two **non-empty** substrings_ (i.e. **left** substring and **right** substring).
The score after splitting a string is the number of **zeros** in the **left** substring plus the number of **ones** in the **right** substring... | If the smallest number in the possible-to-split array is V, then numbers V+1, V+2, ... V+k-1 must contain there as well. You can iteratively find k sets and remove them from array until it becomes empty. Failure to do so would mean that array is unsplittable. |
[Python | beats 95.96% in time | no need to count each time] | maximum-score-after-splitting-a-string | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nOnce you count the number of zeros in the left substring, and the number of ones in the right substring, you don\'t have to count them every time you make new substrings. All you need to do is check whether the new character moved from ri... | 0 | You are given two **sorted** arrays of distinct integers `nums1` and `nums2.`
A **valid path** is defined as follows:
* Choose array `nums1` or `nums2` to traverse (from index-0).
* Traverse the current array from left to right.
* If you are reading any value that is present in `nums1` and `nums2` you are allow... | Precompute a prefix sum of ones ('1'). Iterate from left to right counting the number of zeros ('0'), then use the precomputed prefix sum for counting ones ('1'). Update the answer. |
🐍🐍🐍 One line solution 🐍🐍🐍 | maximum-score-after-splitting-a-string | 0 | 1 | # Code\n```\nclass Solution:\n def maxScore(self, s: str) -> int:\n return max([s[:i].count(\'0\')+s[i:].count(\'1\') for i in range(1,len(s))])\n \n``` | 0 | Given a string `s` of zeros and ones, _return the maximum score after splitting the string into two **non-empty** substrings_ (i.e. **left** substring and **right** substring).
The score after splitting a string is the number of **zeros** in the **left** substring plus the number of **ones** in the **right** substring... | If the smallest number in the possible-to-split array is V, then numbers V+1, V+2, ... V+k-1 must contain there as well. You can iteratively find k sets and remove them from array until it becomes empty. Failure to do so would mean that array is unsplittable. |
🐍🐍🐍 One line solution 🐍🐍🐍 | maximum-score-after-splitting-a-string | 0 | 1 | # Code\n```\nclass Solution:\n def maxScore(self, s: str) -> int:\n return max([s[:i].count(\'0\')+s[i:].count(\'1\') for i in range(1,len(s))])\n \n``` | 0 | You are given two **sorted** arrays of distinct integers `nums1` and `nums2.`
A **valid path** is defined as follows:
* Choose array `nums1` or `nums2` to traverse (from index-0).
* Traverse the current array from left to right.
* If you are reading any value that is present in `nums1` and `nums2` you are allow... | Precompute a prefix sum of ones ('1'). Iterate from left to right counting the number of zeros ('0'), then use the precomputed prefix sum for counting ones ('1'). Update the answer. |
python simple solution | maximum-score-after-splitting-a-string | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ --... | 0 | Given a string `s` of zeros and ones, _return the maximum score after splitting the string into two **non-empty** substrings_ (i.e. **left** substring and **right** substring).
The score after splitting a string is the number of **zeros** in the **left** substring plus the number of **ones** in the **right** substring... | If the smallest number in the possible-to-split array is V, then numbers V+1, V+2, ... V+k-1 must contain there as well. You can iteratively find k sets and remove them from array until it becomes empty. Failure to do so would mean that array is unsplittable. |
python simple solution | maximum-score-after-splitting-a-string | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ --... | 0 | You are given two **sorted** arrays of distinct integers `nums1` and `nums2.`
A **valid path** is defined as follows:
* Choose array `nums1` or `nums2` to traverse (from index-0).
* Traverse the current array from left to right.
* If you are reading any value that is present in `nums1` and `nums2` you are allow... | Precompute a prefix sum of ones ('1'). Iterate from left to right counting the number of zeros ('0'), then use the precomputed prefix sum for counting ones ('1'). Update the answer. |
easy solution | maximum-score-after-splitting-a-string | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ --... | 0 | Given a string `s` of zeros and ones, _return the maximum score after splitting the string into two **non-empty** substrings_ (i.e. **left** substring and **right** substring).
The score after splitting a string is the number of **zeros** in the **left** substring plus the number of **ones** in the **right** substring... | If the smallest number in the possible-to-split array is V, then numbers V+1, V+2, ... V+k-1 must contain there as well. You can iteratively find k sets and remove them from array until it becomes empty. Failure to do so would mean that array is unsplittable. |
easy solution | maximum-score-after-splitting-a-string | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ --... | 0 | You are given two **sorted** arrays of distinct integers `nums1` and `nums2.`
A **valid path** is defined as follows:
* Choose array `nums1` or `nums2` to traverse (from index-0).
* Traverse the current array from left to right.
* If you are reading any value that is present in `nums1` and `nums2` you are allow... | Precompute a prefix sum of ones ('1'). Iterate from left to right counting the number of zeros ('0'), then use the precomputed prefix sum for counting ones ('1'). Update the answer. |
[Python3] O(n) - Clean and Simple Sliding Window Solution | maximum-points-you-can-obtain-from-cards | 0 | 1 | **Implementation**\n\n1. Calculate the sum of all cards and store it in total.\n\n2. Calculate the sum of the initial subarray of size `n - k` and store it in `subarray_sum`.\n\n3. Initialize `min_sum` to `subarray_sum`. This variable will track the overall minimum subarray sum we\'ve seen so far. After iterating over... | 42 | There are several cards **arranged in a row**, and each card has an associated number of points. The points are given in the integer array `cardPoints`.
In one step, you can take one card from the beginning or from the end of the row. You have to take exactly `k` cards.
Your score is the sum of the points of the card... | Check out the constraints, (maxSize <=26). This means you can explore all substrings in O(n * 26). Find the Maximum Number of Occurrences of a Substring with bruteforce. |
Python Solution | Sliding Window Approach | maximum-points-you-can-obtain-from-cards | 0 | 1 | ```\nclass Solution:\n def maxScore(self, cardPoints: List[int], k: int) -> int:\n if len(cardPoints) <= k:\n return sum(cardPoints)\n\n points = cardPoints[len(cardPoints) - k:] + cardPoints[:k]\n temp_sum = sum(points[:k])\n largest = temp_sum\n for i in range(len(poin... | 1 | There are several cards **arranged in a row**, and each card has an associated number of points. The points are given in the integer array `cardPoints`.
In one step, you can take one card from the beginning or from the end of the row. You have to take exactly `k` cards.
Your score is the sum of the points of the card... | Check out the constraints, (maxSize <=26). This means you can explore all substrings in O(n * 26). Find the Maximum Number of Occurrences of a Substring with bruteforce. |
Recursive + DP Solution | maximum-points-you-can-obtain-from-cards | 0 | 1 | **Recursive Solution :-**\n\n```\nclass Solution:\n def maxScore(self, cardPoints: List[int], k: int) -> int:\n n = len(cardPoints)\n if k == 1:\n return max(cardPoints[0], cardPoints[-1])\n \n maximumScore = max(cardPoints[0] + self.maxScore(cardPoints[1:], k - 1), cardPoints[... | 6 | There are several cards **arranged in a row**, and each card has an associated number of points. The points are given in the integer array `cardPoints`.
In one step, you can take one card from the beginning or from the end of the row. You have to take exactly `k` cards.
Your score is the sum of the points of the card... | Check out the constraints, (maxSize <=26). This means you can explore all substrings in O(n * 26). Find the Maximum Number of Occurrences of a Substring with bruteforce. |
📌 Convert into maximum sum subarray of length k using sliding window | maximum-points-you-can-obtain-from-cards | 0 | 1 | ```\nclass Solution:\n def maxScore(self, cardPoints: List[int], k: int) -> int:\n length = len(cardPoints)\n s = sum(cardPoints[-k:])\n maximum = s\n j = length-k\n i=0\n while i!=j and j<length:\n s+=cardPoints[i]\n s-=cardPoints[j]\n i+=1\... | 6 | There are several cards **arranged in a row**, and each card has an associated number of points. The points are given in the integer array `cardPoints`.
In one step, you can take one card from the beginning or from the end of the row. You have to take exactly `k` cards.
Your score is the sum of the points of the card... | Check out the constraints, (maxSize <=26). This means you can explore all substrings in O(n * 26). Find the Maximum Number of Occurrences of a Substring with bruteforce. |
Python/JS/Java/Go/C++ O( k ) sliding window. [w/ Visualization] | maximum-points-you-can-obtain-from-cards | 1 | 1 | O( k ) sliding window solution\n\n---\n\n**Diagram & Visualization**:\n\n\n\n---\n\n**Implementation**:\n\nPython\n\n```\nclass Solution:\n def maxScore(self, cardPoints: List[int], k: int) -> int:\n \n size = len(cardPoints... | 53 | There are several cards **arranged in a row**, and each card has an associated number of points. The points are given in the integer array `cardPoints`.
In one step, you can take one card from the beginning or from the end of the row. You have to take exactly `k` cards.
Your score is the sum of the points of the card... | Check out the constraints, (maxSize <=26). This means you can explore all substrings in O(n * 26). Find the Maximum Number of Occurrences of a Substring with bruteforce. |
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