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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Easy Solution || A Sorting Approach || Python || Beats 100% | widest-vertical-area-between-two-points-containing-no-points | 0 | 1 | # BEATS\n\n\n\n# Intuition\nThe problem involves finding the widest vertical area between two points without including any points within the area. Since the goal is to maximize the width, it\'s essential to... | 1 | Given `n` `points` on a 2D plane where `points[i] = [xi, yi]`, Return _the **widest vertical area** between two points such that no points are inside the area._
A **vertical area** is an area of fixed-width extending infinitely along the y-axis (i.e., infinite height). The **widest vertical area** is the one with the ... | Use dynamic programming. The state of the DP can be the current index and the remaining characters to delete. Having a prefix sum for each character can help you determine for a certain character c in some specific range, how many characters you need to delete to merge all occurrences of c in that range. |
Easy Solution || A Sorting Approach || Python || Beats 100% | widest-vertical-area-between-two-points-containing-no-points | 0 | 1 | # BEATS\n\n\n\n# Intuition\nThe problem involves finding the widest vertical area between two points without including any points within the area. Since the goal is to maximize the width, it\'s essential to... | 1 | You are working in a ball factory where you have `n` balls numbered from `lowLimit` up to `highLimit` **inclusive** (i.e., `n == highLimit - lowLimit + 1`), and an infinite number of boxes numbered from `1` to `infinity`.
Your job at this factory is to put each ball in the box with a number equal to the sum of digits ... | Try sorting the points Think is the y-axis of a point relevant |
97% TC and 95% SC easy python solution with explanation | count-substrings-that-differ-by-one-character | 0 | 1 | Believe me, this ques is like must to have in the mind, as this gonna help you solve many such questions.\n1. We have to calculate all substrings which differ by just a character.\n2. So, maintain 2 states, first one will be for the count of substrings which is present in "t" string as well(all characters equal). The s... | 6 | Given two strings `s` and `t`, find the number of ways you can choose a non-empty substring of `s` and replace a **single character** by a different character such that the resulting substring is a substring of `t`. In other words, find the number of substrings in `s` that differ from some substring in `t` by **exactly... | The problem can be reworded as, giving a set of points on a 2d-plane, return the geometric median. Loop over each triplet of points (positions[i], positions[j], positions[k]) where i < j < k, get the centre of the circle which goes throw the 3 points, check if all other points lie in this circle. |
97% TC and 95% SC easy python solution with explanation | count-substrings-that-differ-by-one-character | 0 | 1 | Believe me, this ques is like must to have in the mind, as this gonna help you solve many such questions.\n1. We have to calculate all substrings which differ by just a character.\n2. So, maintain 2 states, first one will be for the count of substrings which is present in "t" string as well(all characters equal). The s... | 6 | There is an integer array `nums` that consists of `n` **unique** elements, but you have forgotten it. However, you do remember every pair of adjacent elements in `nums`.
You are given a 2D integer array `adjacentPairs` of size `n - 1` where each `adjacentPairs[i] = [ui, vi]` indicates that the elements `ui` and `vi` a... | Take every substring of s, change a character, and see how many substrings of t match that substring. Use a Trie to store all substrings of t as a dictionary. |
Python (Simple DP) | count-substrings-that-differ-by-one-character | 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. $$O(n)$$ --... | 1 | Given two strings `s` and `t`, find the number of ways you can choose a non-empty substring of `s` and replace a **single character** by a different character such that the resulting substring is a substring of `t`. In other words, find the number of substrings in `s` that differ from some substring in `t` by **exactly... | The problem can be reworded as, giving a set of points on a 2d-plane, return the geometric median. Loop over each triplet of points (positions[i], positions[j], positions[k]) where i < j < k, get the centre of the circle which goes throw the 3 points, check if all other points lie in this circle. |
Python (Simple DP) | count-substrings-that-differ-by-one-character | 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. $$O(n)$$ --... | 1 | There is an integer array `nums` that consists of `n` **unique** elements, but you have forgotten it. However, you do remember every pair of adjacent elements in `nums`.
You are given a 2D integer array `adjacentPairs` of size `n - 1` where each `adjacentPairs[i] = [ui, vi]` indicates that the elements `ui` and `vi` a... | Take every substring of s, change a character, and see how many substrings of t match that substring. Use a Trie to store all substrings of t as a dictionary. |
Python3 solution with explanations | count-substrings-that-differ-by-one-character | 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. -->\nxayy, xxbyy differs with two prefixing chars, and two suffixing chars\ntotal number of substrings that differs with one char is 9:\nwe could pick 0,1,2 x from left, an... | 0 | Given two strings `s` and `t`, find the number of ways you can choose a non-empty substring of `s` and replace a **single character** by a different character such that the resulting substring is a substring of `t`. In other words, find the number of substrings in `s` that differ from some substring in `t` by **exactly... | The problem can be reworded as, giving a set of points on a 2d-plane, return the geometric median. Loop over each triplet of points (positions[i], positions[j], positions[k]) where i < j < k, get the centre of the circle which goes throw the 3 points, check if all other points lie in this circle. |
Python3 solution with explanations | count-substrings-that-differ-by-one-character | 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. -->\nxayy, xxbyy differs with two prefixing chars, and two suffixing chars\ntotal number of substrings that differs with one char is 9:\nwe could pick 0,1,2 x from left, an... | 0 | There is an integer array `nums` that consists of `n` **unique** elements, but you have forgotten it. However, you do remember every pair of adjacent elements in `nums`.
You are given a 2D integer array `adjacentPairs` of size `n - 1` where each `adjacentPairs[i] = [ui, vi]` indicates that the elements `ui` and `vi` a... | Take every substring of s, change a character, and see how many substrings of t match that substring. Use a Trie to store all substrings of t as a dictionary. |
Python3 Dynamic Programming O(n^2) Runtime + Space | count-substrings-that-differ-by-one-character | 0 | 1 | # Code\n```\nclass Solution:\n\n def countSubstrings(self, s: str, t: str) -> int:\n\n m, n = len(s), len(t)\n \n # O(n^2) space\n left = [[0] * (n+1) for _ in range(m+1)]\n right = [[0] * (n+1) for _ in range(m+1)]\n\n for i in range(m):\n for j in range(n):\n ... | 0 | Given two strings `s` and `t`, find the number of ways you can choose a non-empty substring of `s` and replace a **single character** by a different character such that the resulting substring is a substring of `t`. In other words, find the number of substrings in `s` that differ from some substring in `t` by **exactly... | The problem can be reworded as, giving a set of points on a 2d-plane, return the geometric median. Loop over each triplet of points (positions[i], positions[j], positions[k]) where i < j < k, get the centre of the circle which goes throw the 3 points, check if all other points lie in this circle. |
Python3 Dynamic Programming O(n^2) Runtime + Space | count-substrings-that-differ-by-one-character | 0 | 1 | # Code\n```\nclass Solution:\n\n def countSubstrings(self, s: str, t: str) -> int:\n\n m, n = len(s), len(t)\n \n # O(n^2) space\n left = [[0] * (n+1) for _ in range(m+1)]\n right = [[0] * (n+1) for _ in range(m+1)]\n\n for i in range(m):\n for j in range(n):\n ... | 0 | There is an integer array `nums` that consists of `n` **unique** elements, but you have forgotten it. However, you do remember every pair of adjacent elements in `nums`.
You are given a 2D integer array `adjacentPairs` of size `n - 1` where each `adjacentPairs[i] = [ui, vi]` indicates that the elements `ui` and `vi` a... | Take every substring of s, change a character, and see how many substrings of t match that substring. Use a Trie to store all substrings of t as a dictionary. |
Python 3 || 7 lines using transpose of target || T/M: 84% / 79% | number-of-ways-to-form-a-target-string-given-a-dictionary | 0 | 1 | ```\nclass Solution:\n def numWays(self, words: List[str], target: str) -> int:\n \n m, n = len(words[0]),len(target)\n ans = [1]+ [0]*n\n words = list(map(Counter,zip(*map(list,words))))\n\n for word in words:\n for i in reversed(range(n)):\n ans[i+1] += ans[i]... | 3 | You are given a list of strings of the **same length** `words` and a string `target`.
Your task is to form `target` using the given `words` under the following rules:
* `target` should be formed from left to right.
* To form the `ith` character (**0-indexed**) of `target`, you can choose the `kth` character of th... | null |
Python 3 || 7 lines using transpose of target || T/M: 84% / 79% | number-of-ways-to-form-a-target-string-given-a-dictionary | 0 | 1 | ```\nclass Solution:\n def numWays(self, words: List[str], target: str) -> int:\n \n m, n = len(words[0]),len(target)\n ans = [1]+ [0]*n\n words = list(map(Counter,zip(*map(list,words))))\n\n for word in words:\n for i in reversed(range(n)):\n ans[i+1] += ans[i]... | 3 | You are given a **(0-indexed)** array of positive integers `candiesCount` where `candiesCount[i]` represents the number of candies of the `ith` type you have. You are also given a 2D array `queries` where `queries[i] = [favoriteTypei, favoriteDayi, dailyCapi]`.
You play a game with the following rules:
* You start ... | For each index i, store the frequency of each character in the ith row. Use dynamic programing to calculate the number of ways to get the target string using the frequency array, |
python 3 - top down dp | number-of-ways-to-form-a-target-string-given-a-dictionary | 0 | 1 | # Intuition\nThis question\'s constraints are quite harsh imo.\n\nThere are 2 state variables:\n1.index of target\n2.index of word in words list\n\nYou can\'t save indexes of all characters in each word in a defaultdict(list) and then do binary search, which will cost you log(n) time and get TLE. This was my first tria... | 1 | You are given a list of strings of the **same length** `words` and a string `target`.
Your task is to form `target` using the given `words` under the following rules:
* `target` should be formed from left to right.
* To form the `ith` character (**0-indexed**) of `target`, you can choose the `kth` character of th... | null |
python 3 - top down dp | number-of-ways-to-form-a-target-string-given-a-dictionary | 0 | 1 | # Intuition\nThis question\'s constraints are quite harsh imo.\n\nThere are 2 state variables:\n1.index of target\n2.index of word in words list\n\nYou can\'t save indexes of all characters in each word in a defaultdict(list) and then do binary search, which will cost you log(n) time and get TLE. This was my first tria... | 1 | You are given a **(0-indexed)** array of positive integers `candiesCount` where `candiesCount[i]` represents the number of candies of the `ith` type you have. You are also given a 2D array `queries` where `queries[i] = [favoriteTypei, favoriteDayi, dailyCapi]`.
You play a game with the following rules:
* You start ... | For each index i, store the frequency of each character in the ith row. Use dynamic programing to calculate the number of ways to get the target string using the frequency array, |
Python3 Solution | number-of-ways-to-form-a-target-string-given-a-dictionary | 0 | 1 | \n```\nclass Solution:\n def numWays(self, words: List[str], target: str) -> int:\n n=len(words[0])\n m=len(target)\n mod=10**9+7\n dp=[0]*(m+1)\n dp[0]=1\n count=[[0]*26 for _ in range(n)]\n for i in range(n):\n for word in words:\n count[i]... | 1 | You are given a list of strings of the **same length** `words` and a string `target`.
Your task is to form `target` using the given `words` under the following rules:
* `target` should be formed from left to right.
* To form the `ith` character (**0-indexed**) of `target`, you can choose the `kth` character of th... | null |
Python3 Solution | number-of-ways-to-form-a-target-string-given-a-dictionary | 0 | 1 | \n```\nclass Solution:\n def numWays(self, words: List[str], target: str) -> int:\n n=len(words[0])\n m=len(target)\n mod=10**9+7\n dp=[0]*(m+1)\n dp[0]=1\n count=[[0]*26 for _ in range(n)]\n for i in range(n):\n for word in words:\n count[i]... | 1 | You are given a **(0-indexed)** array of positive integers `candiesCount` where `candiesCount[i]` represents the number of candies of the `ith` type you have. You are also given a 2D array `queries` where `queries[i] = [favoriteTypei, favoriteDayi, dailyCapi]`.
You play a game with the following rules:
* You start ... | For each index i, store the frequency of each character in the ith row. Use dynamic programing to calculate the number of ways to get the target string using the frequency array, |
Python DP (Memoization) | number-of-ways-to-form-a-target-string-given-a-dictionary | 0 | 1 | ```\nclass word_index:\n def __init__(self):\n self.char_set = set()\n self.frequencies = defaultdict(int)\n def add_char(self, char):\n self.char_set.add(char)\n self.frequencies[char] += 1\n\nclass Solution(object):\n def numWays(self, words, target):\n # seperate each word... | 1 | You are given a list of strings of the **same length** `words` and a string `target`.
Your task is to form `target` using the given `words` under the following rules:
* `target` should be formed from left to right.
* To form the `ith` character (**0-indexed**) of `target`, you can choose the `kth` character of th... | null |
Python DP (Memoization) | number-of-ways-to-form-a-target-string-given-a-dictionary | 0 | 1 | ```\nclass word_index:\n def __init__(self):\n self.char_set = set()\n self.frequencies = defaultdict(int)\n def add_char(self, char):\n self.char_set.add(char)\n self.frequencies[char] += 1\n\nclass Solution(object):\n def numWays(self, words, target):\n # seperate each word... | 1 | You are given a **(0-indexed)** array of positive integers `candiesCount` where `candiesCount[i]` represents the number of candies of the `ith` type you have. You are also given a 2D array `queries` where `queries[i] = [favoriteTypei, favoriteDayi, dailyCapi]`.
You play a game with the following rules:
* You start ... | For each index i, store the frequency of each character in the ith row. Use dynamic programing to calculate the number of ways to get the target string using the frequency array, |
Python DP | number-of-ways-to-form-a-target-string-given-a-dictionary | 0 | 1 | ```\nclass Solution:\n def numWays(self, words: List[str], target: str) -> int:\n n, m = len(words[0]), len(target)\n maps = [Counter() for _ in range(n)]\n for word in words:\n for i, c in enumerate(word):\n maps[i][c] += 1\n @cache\n def dp(i, j):\n ... | 1 | You are given a list of strings of the **same length** `words` and a string `target`.
Your task is to form `target` using the given `words` under the following rules:
* `target` should be formed from left to right.
* To form the `ith` character (**0-indexed**) of `target`, you can choose the `kth` character of th... | null |
Python DP | number-of-ways-to-form-a-target-string-given-a-dictionary | 0 | 1 | ```\nclass Solution:\n def numWays(self, words: List[str], target: str) -> int:\n n, m = len(words[0]), len(target)\n maps = [Counter() for _ in range(n)]\n for word in words:\n for i, c in enumerate(word):\n maps[i][c] += 1\n @cache\n def dp(i, j):\n ... | 1 | You are given a **(0-indexed)** array of positive integers `candiesCount` where `candiesCount[i]` represents the number of candies of the `ith` type you have. You are also given a 2D array `queries` where `queries[i] = [favoriteTypei, favoriteDayi, dailyCapi]`.
You play a game with the following rules:
* You start ... | For each index i, store the frequency of each character in the ith row. Use dynamic programing to calculate the number of ways to get the target string using the frequency array, |
Image Explanation🏆- [DP - Complete Intuition] - C++/Java/Python | number-of-ways-to-form-a-target-string-given-a-dictionary | 1 | 1 | # Video Solution (`Aryan Mittal`) - Link in LeetCode Profile\n`Number of Ways to Form a Target String Given a Dictionary` by `Aryan Mittal`\n\n\n\n# Approach & Intution\n of `target`, you can choose the `kth` character of th... | null |
Image Explanation🏆- [DP - Complete Intuition] - C++/Java/Python | number-of-ways-to-form-a-target-string-given-a-dictionary | 1 | 1 | # Video Solution (`Aryan Mittal`) - Link in LeetCode Profile\n`Number of Ways to Form a Target String Given a Dictionary` by `Aryan Mittal`\n\n\n\n# Approach & Intution\n** array of positive integers `candiesCount` where `candiesCount[i]` represents the number of candies of the `ith` type you have. You are also given a 2D array `queries` where `queries[i] = [favoriteTypei, favoriteDayi, dailyCapi]`.
You play a game with the following rules:
* You start ... | For each index i, store the frequency of each character in the ith row. Use dynamic programing to calculate the number of ways to get the target string using the frequency array, |
C++/Python Solution | check-array-formation-through-concatenation | 0 | 1 | ### C++\n**Runtime: 8 ms, faster than 58.49% of C++ online submissions for Check Array Formation Through Concatenation.\nMemory Usage: 10.2 MB, less than 76.10% of C++ online submissions for Check Array Formation Through Concatenation.**\n```\nclass Solution {\npublic:\n bool canFormArray(vector<int>& arr, vector<ve... | 1 | You are given an array of **distinct** integers `arr` and an array of integer arrays `pieces`, where the integers in `pieces` are **distinct**. Your goal is to form `arr` by concatenating the arrays in `pieces` **in any order**. However, you are **not** allowed to reorder the integers in each array `pieces[i]`.
Return... | Try to solve it by keeping for each file chunk, the users who have this chunk. Try to solve it by keeping all the users in the system with their owned chunks, and when you request a chunk, check all users for it. |
C++/Python Solution | check-array-formation-through-concatenation | 0 | 1 | ### C++\n**Runtime: 8 ms, faster than 58.49% of C++ online submissions for Check Array Formation Through Concatenation.\nMemory Usage: 10.2 MB, less than 76.10% of C++ online submissions for Check Array Formation Through Concatenation.**\n```\nclass Solution {\npublic:\n bool canFormArray(vector<int>& arr, vector<ve... | 1 | You are given an integer array `nums` where the `ith` bag contains `nums[i]` balls. You are also given an integer `maxOperations`.
You can perform the following operation at most `maxOperations` times:
* Take any bag of balls and divide it into two new bags with a **positive** number of balls.
* For example, ... | Note that the distinct part means that every position in the array belongs to only one piece Note that you can get the piece every position belongs to naively |
"Python" easy explanation blackboard | check-array-formation-through-concatenation | 0 | 1 | * **Simple and easy python3 solution**\n* **The approch is well explained in the below image**\n\n\n\n\n```\nclass Solution:\n def canFormArray(self, arr: List[int], pieces: List[List[int]]) -> bool:\n ... | 29 | You are given an array of **distinct** integers `arr` and an array of integer arrays `pieces`, where the integers in `pieces` are **distinct**. Your goal is to form `arr` by concatenating the arrays in `pieces` **in any order**. However, you are **not** allowed to reorder the integers in each array `pieces[i]`.
Return... | Try to solve it by keeping for each file chunk, the users who have this chunk. Try to solve it by keeping all the users in the system with their owned chunks, and when you request a chunk, check all users for it. |
"Python" easy explanation blackboard | check-array-formation-through-concatenation | 0 | 1 | * **Simple and easy python3 solution**\n* **The approch is well explained in the below image**\n\n\n\n\n```\nclass Solution:\n def canFormArray(self, arr: List[int], pieces: List[List[int]]) -> bool:\n ... | 29 | You are given an integer array `nums` where the `ith` bag contains `nums[i]` balls. You are also given an integer `maxOperations`.
You can perform the following operation at most `maxOperations` times:
* Take any bag of balls and divide it into two new bags with a **positive** number of balls.
* For example, ... | Note that the distinct part means that every position in the array belongs to only one piece Note that you can get the piece every position belongs to naively |
simple code made using dict and beat 90% | check-array-formation-through-concatenation | 0 | 1 | \n\n# Code\n```\nclass Solution:\n def canFormArray(self, arr: List[int], pieces: List[List[int]]) -> bool:\n \n ## create a dictionary\n # key: head number of piece\n # value: all number of single piece\n mapping = { piece[0]: piece for piece in pieces }\n \n result ... | 1 | You are given an array of **distinct** integers `arr` and an array of integer arrays `pieces`, where the integers in `pieces` are **distinct**. Your goal is to form `arr` by concatenating the arrays in `pieces` **in any order**. However, you are **not** allowed to reorder the integers in each array `pieces[i]`.
Return... | Try to solve it by keeping for each file chunk, the users who have this chunk. Try to solve it by keeping all the users in the system with their owned chunks, and when you request a chunk, check all users for it. |
simple code made using dict and beat 90% | check-array-formation-through-concatenation | 0 | 1 | \n\n# Code\n```\nclass Solution:\n def canFormArray(self, arr: List[int], pieces: List[List[int]]) -> bool:\n \n ## create a dictionary\n # key: head number of piece\n # value: all number of single piece\n mapping = { piece[0]: piece for piece in pieces }\n \n result ... | 1 | You are given an integer array `nums` where the `ith` bag contains `nums[i]` balls. You are also given an integer `maxOperations`.
You can perform the following operation at most `maxOperations` times:
* Take any bag of balls and divide it into two new bags with a **positive** number of balls.
* For example, ... | Note that the distinct part means that every position in the array belongs to only one piece Note that you can get the piece every position belongs to naively |
Python O(n) by dictionary [w/ Comment] | check-array-formation-through-concatenation | 0 | 1 | Python O(n) by dictionary\n\n---\n\n**Implementation**:\n\n```\nclass Solution:\n def canFormArray(self, arr: List[int], pieces: List[List[int]]) -> bool:\n \n ## dictionary\n # key: head number of piece\n # value: all number of single piece\n mapping = { piece[0]: piece for piece ... | 15 | You are given an array of **distinct** integers `arr` and an array of integer arrays `pieces`, where the integers in `pieces` are **distinct**. Your goal is to form `arr` by concatenating the arrays in `pieces` **in any order**. However, you are **not** allowed to reorder the integers in each array `pieces[i]`.
Return... | Try to solve it by keeping for each file chunk, the users who have this chunk. Try to solve it by keeping all the users in the system with their owned chunks, and when you request a chunk, check all users for it. |
Python O(n) by dictionary [w/ Comment] | check-array-formation-through-concatenation | 0 | 1 | Python O(n) by dictionary\n\n---\n\n**Implementation**:\n\n```\nclass Solution:\n def canFormArray(self, arr: List[int], pieces: List[List[int]]) -> bool:\n \n ## dictionary\n # key: head number of piece\n # value: all number of single piece\n mapping = { piece[0]: piece for piece ... | 15 | You are given an integer array `nums` where the `ith` bag contains `nums[i]` balls. You are also given an integer `maxOperations`.
You can perform the following operation at most `maxOperations` times:
* Take any bag of balls and divide it into two new bags with a **positive** number of balls.
* For example, ... | Note that the distinct part means that every position in the array belongs to only one piece Note that you can get the piece every position belongs to naively |
[Python3] 2-line O(N) | check-array-formation-through-concatenation | 0 | 1 | Algo\nWe can leverage on the fact that the integers in `pieces` are distinct and define a mapping `mp` to map from the first element of piece to piece. Then, we could linearly scan `arr` and check if what\'s in arr `arr[i:i+len(mp[x])]` is the same as the one in piece `mp[x]`. \n\nImplementation\n```\nclass Solution:\n... | 11 | You are given an array of **distinct** integers `arr` and an array of integer arrays `pieces`, where the integers in `pieces` are **distinct**. Your goal is to form `arr` by concatenating the arrays in `pieces` **in any order**. However, you are **not** allowed to reorder the integers in each array `pieces[i]`.
Return... | Try to solve it by keeping for each file chunk, the users who have this chunk. Try to solve it by keeping all the users in the system with their owned chunks, and when you request a chunk, check all users for it. |
[Python3] 2-line O(N) | check-array-formation-through-concatenation | 0 | 1 | Algo\nWe can leverage on the fact that the integers in `pieces` are distinct and define a mapping `mp` to map from the first element of piece to piece. Then, we could linearly scan `arr` and check if what\'s in arr `arr[i:i+len(mp[x])]` is the same as the one in piece `mp[x]`. \n\nImplementation\n```\nclass Solution:\n... | 11 | You are given an integer array `nums` where the `ith` bag contains `nums[i]` balls. You are also given an integer `maxOperations`.
You can perform the following operation at most `maxOperations` times:
* Take any bag of balls and divide it into two new bags with a **positive** number of balls.
* For example, ... | Note that the distinct part means that every position in the array belongs to only one piece Note that you can get the piece every position belongs to naively |
check-array-formation-through-concatenation | check-array-formation-through-concatenation | 0 | 1 | # Code\n```\nclass Solution:\n def canFormArray(self, arr: List[int], pieces: List[List[int]]) -> bool:\n d = {}\n l = []\n for i in pieces:\n if i[0] not in d:\n d[i[0]]= i\n for i in arr:\n if d.get(i) is not None:\n l+= d[i]\n ... | 0 | You are given an array of **distinct** integers `arr` and an array of integer arrays `pieces`, where the integers in `pieces` are **distinct**. Your goal is to form `arr` by concatenating the arrays in `pieces` **in any order**. However, you are **not** allowed to reorder the integers in each array `pieces[i]`.
Return... | Try to solve it by keeping for each file chunk, the users who have this chunk. Try to solve it by keeping all the users in the system with their owned chunks, and when you request a chunk, check all users for it. |
check-array-formation-through-concatenation | check-array-formation-through-concatenation | 0 | 1 | # Code\n```\nclass Solution:\n def canFormArray(self, arr: List[int], pieces: List[List[int]]) -> bool:\n d = {}\n l = []\n for i in pieces:\n if i[0] not in d:\n d[i[0]]= i\n for i in arr:\n if d.get(i) is not None:\n l+= d[i]\n ... | 0 | You are given an integer array `nums` where the `ith` bag contains `nums[i]` balls. You are also given an integer `maxOperations`.
You can perform the following operation at most `maxOperations` times:
* Take any bag of balls and divide it into two new bags with a **positive** number of balls.
* For example, ... | Note that the distinct part means that every position in the array belongs to only one piece Note that you can get the piece every position belongs to naively |
My Solution :) | check-array-formation-through-concatenation | 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)\n\n- Space complexity:\nO(n)\n# Code\n```\nclass Solution:\n def canFormArray(self, arr: List[int], pieces: List[List[int]]... | 0 | You are given an array of **distinct** integers `arr` and an array of integer arrays `pieces`, where the integers in `pieces` are **distinct**. Your goal is to form `arr` by concatenating the arrays in `pieces` **in any order**. However, you are **not** allowed to reorder the integers in each array `pieces[i]`.
Return... | Try to solve it by keeping for each file chunk, the users who have this chunk. Try to solve it by keeping all the users in the system with their owned chunks, and when you request a chunk, check all users for it. |
My Solution :) | check-array-formation-through-concatenation | 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)\n\n- Space complexity:\nO(n)\n# Code\n```\nclass Solution:\n def canFormArray(self, arr: List[int], pieces: List[List[int]]... | 0 | You are given an integer array `nums` where the `ith` bag contains `nums[i]` balls. You are also given an integer `maxOperations`.
You can perform the following operation at most `maxOperations` times:
* Take any bag of balls and divide it into two new bags with a **positive** number of balls.
* For example, ... | Note that the distinct part means that every position in the array belongs to only one piece Note that you can get the piece every position belongs to naively |
Python solution | check-array-formation-through-concatenation | 0 | 1 | ```\nclass Solution:\n def canFormArray(self, arr: List[int], pieces: List[List[int]]) -> bool:\n is_concatenated, index, _len = True, 0, len(pieces)\n\n while pieces:\n if pieces[0][0] not in arr: return False\n start = arr.index(pieces[0][0])\n if arr[start:len(pieces... | 0 | You are given an array of **distinct** integers `arr` and an array of integer arrays `pieces`, where the integers in `pieces` are **distinct**. Your goal is to form `arr` by concatenating the arrays in `pieces` **in any order**. However, you are **not** allowed to reorder the integers in each array `pieces[i]`.
Return... | Try to solve it by keeping for each file chunk, the users who have this chunk. Try to solve it by keeping all the users in the system with their owned chunks, and when you request a chunk, check all users for it. |
Python solution | check-array-formation-through-concatenation | 0 | 1 | ```\nclass Solution:\n def canFormArray(self, arr: List[int], pieces: List[List[int]]) -> bool:\n is_concatenated, index, _len = True, 0, len(pieces)\n\n while pieces:\n if pieces[0][0] not in arr: return False\n start = arr.index(pieces[0][0])\n if arr[start:len(pieces... | 0 | You are given an integer array `nums` where the `ith` bag contains `nums[i]` balls. You are also given an integer `maxOperations`.
You can perform the following operation at most `maxOperations` times:
* Take any bag of balls and divide it into two new bags with a **positive** number of balls.
* For example, ... | Note that the distinct part means that every position in the array belongs to only one piece Note that you can get the piece every position belongs to naively |
One line not readable solution beats 94.12 % of users with Python 3 | check-array-formation-through-concatenation | 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 array of **distinct** integers `arr` and an array of integer arrays `pieces`, where the integers in `pieces` are **distinct**. Your goal is to form `arr` by concatenating the arrays in `pieces` **in any order**. However, you are **not** allowed to reorder the integers in each array `pieces[i]`.
Return... | Try to solve it by keeping for each file chunk, the users who have this chunk. Try to solve it by keeping all the users in the system with their owned chunks, and when you request a chunk, check all users for it. |
One line not readable solution beats 94.12 % of users with Python 3 | check-array-formation-through-concatenation | 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 integer array `nums` where the `ith` bag contains `nums[i]` balls. You are also given an integer `maxOperations`.
You can perform the following operation at most `maxOperations` times:
* Take any bag of balls and divide it into two new bags with a **positive** number of balls.
* For example, ... | Note that the distinct part means that every position in the array belongs to only one piece Note that you can get the piece every position belongs to naively |
Python O(n) by dictionary | check-array-formation-through-concatenation | 0 | 1 | # Intuition\nPython O(n) by dictionary\n\n# Approach\nPython O(n) by dictionary\n# Complexity\n- Time complexity:\n$$O(n)$$ \n\n- Space complexity:\n $$O(n)$$ \n\n# Code\n```\nclass Solution:\n def canFormArray(self, arr: List[int], pieces: List[List[int]]) -> bool:\n d = {}\n l = []\n for i in ... | 0 | You are given an array of **distinct** integers `arr` and an array of integer arrays `pieces`, where the integers in `pieces` are **distinct**. Your goal is to form `arr` by concatenating the arrays in `pieces` **in any order**. However, you are **not** allowed to reorder the integers in each array `pieces[i]`.
Return... | Try to solve it by keeping for each file chunk, the users who have this chunk. Try to solve it by keeping all the users in the system with their owned chunks, and when you request a chunk, check all users for it. |
Python O(n) by dictionary | check-array-formation-through-concatenation | 0 | 1 | # Intuition\nPython O(n) by dictionary\n\n# Approach\nPython O(n) by dictionary\n# Complexity\n- Time complexity:\n$$O(n)$$ \n\n- Space complexity:\n $$O(n)$$ \n\n# Code\n```\nclass Solution:\n def canFormArray(self, arr: List[int], pieces: List[List[int]]) -> bool:\n d = {}\n l = []\n for i in ... | 0 | You are given an integer array `nums` where the `ith` bag contains `nums[i]` balls. You are also given an integer `maxOperations`.
You can perform the following operation at most `maxOperations` times:
* Take any bag of balls and divide it into two new bags with a **positive** number of balls.
* For example, ... | Note that the distinct part means that every position in the array belongs to only one piece Note that you can get the piece every position belongs to naively |
Beginner-friendly || Simple solution with Dynamic Programming in Python3 / TypeScript | count-sorted-vowel-strings | 0 | 1 | # Intuition\nLet\'s briefly explain what the problem is:\n- there\'s an integer `n`\n- our goal is to find **how many** distinct and **lexicographically** correct strings exist, which are created from **vowels**\n\nThe approach is straightforward: if we represent the final string as a **tree with nodes**, this helps to... | 1 | Given an integer `n`, return _the number of strings of length_ `n` _that consist only of vowels (_`a`_,_ `e`_,_ `i`_,_ `o`_,_ `u`_) and are **lexicographically sorted**._
A string `s` is **lexicographically sorted** if for all valid `i`, `s[i]` is the same as or comes before `s[i+1]` in the alphabet.
**Example 1:**
... | null |
Beginner-friendly || Simple solution with Dynamic Programming in Python3 / TypeScript | count-sorted-vowel-strings | 0 | 1 | # Intuition\nLet\'s briefly explain what the problem is:\n- there\'s an integer `n`\n- our goal is to find **how many** distinct and **lexicographically** correct strings exist, which are created from **vowels**\n\nThe approach is straightforward: if we represent the final string as a **tree with nodes**, this helps to... | 1 | You are given an undirected graph. You are given an integer `n` which is the number of nodes in the graph and an array `edges`, where each `edges[i] = [ui, vi]` indicates that there is an undirected edge between `ui` and `vi`.
A **connected trio** is a set of **three** nodes where there is an edge between **every** pa... | For each character, its possible values will depend on the value of its previous character, because it needs to be not smaller than it. Think backtracking. Build a recursive function count(n, last_character) that counts the number of valid strings of length n and whose first characters are not less than last_character.... |
Python Best Solution | count-sorted-vowel-strings | 0 | 1 | \n# Code\n```\nclass Solution:\n def countVowelStrings(self, n: int) -> int:\n mul=1\n j=1\n for i in range(5,5+n):\n mul=(mul*i)//j\n j+=1\n return mul\n \n``` | 11 | Given an integer `n`, return _the number of strings of length_ `n` _that consist only of vowels (_`a`_,_ `e`_,_ `i`_,_ `o`_,_ `u`_) and are **lexicographically sorted**._
A string `s` is **lexicographically sorted** if for all valid `i`, `s[i]` is the same as or comes before `s[i+1]` in the alphabet.
**Example 1:**
... | null |
Python Best Solution | count-sorted-vowel-strings | 0 | 1 | \n# Code\n```\nclass Solution:\n def countVowelStrings(self, n: int) -> int:\n mul=1\n j=1\n for i in range(5,5+n):\n mul=(mul*i)//j\n j+=1\n return mul\n \n``` | 11 | You are given an undirected graph. You are given an integer `n` which is the number of nodes in the graph and an array `edges`, where each `edges[i] = [ui, vi]` indicates that there is an undirected edge between `ui` and `vi`.
A **connected trio** is a set of **three** nodes where there is an edge between **every** pa... | For each character, its possible values will depend on the value of its previous character, because it needs to be not smaller than it. Think backtracking. Build a recursive function count(n, last_character) that counts the number of valid strings of length n and whose first characters are not less than last_character.... |
✅🔥DP & 5 Variables Approach - C++/Java/Python | count-sorted-vowel-strings | 1 | 1 | # Intuition\n\n- Here we can observe a pattern to this problem.\n\n```\n\t\t a e i o u\n n=1 1 1 1 1 1 /a, e, i, o, u\n n=2 5 4 3 2 1 /a-> aa,ae,ai,ao,au | e-> ee,ei,eo,eu | i-> ii,io,iu | o-> oo,ou | u-> uu\n n=3 15 10 6 3 1\n```\n\n- If we observe from last there will be only 1 element... | 183 | Given an integer `n`, return _the number of strings of length_ `n` _that consist only of vowels (_`a`_,_ `e`_,_ `i`_,_ `o`_,_ `u`_) and are **lexicographically sorted**._
A string `s` is **lexicographically sorted** if for all valid `i`, `s[i]` is the same as or comes before `s[i+1]` in the alphabet.
**Example 1:**
... | null |
✅🔥DP & 5 Variables Approach - C++/Java/Python | count-sorted-vowel-strings | 1 | 1 | # Intuition\n\n- Here we can observe a pattern to this problem.\n\n```\n\t\t a e i o u\n n=1 1 1 1 1 1 /a, e, i, o, u\n n=2 5 4 3 2 1 /a-> aa,ae,ai,ao,au | e-> ee,ei,eo,eu | i-> ii,io,iu | o-> oo,ou | u-> uu\n n=3 15 10 6 3 1\n```\n\n- If we observe from last there will be only 1 element... | 183 | You are given an undirected graph. You are given an integer `n` which is the number of nodes in the graph and an array `edges`, where each `edges[i] = [ui, vi]` indicates that there is an undirected edge between `ui` and `vi`.
A **connected trio** is a set of **three** nodes where there is an edge between **every** pa... | For each character, its possible values will depend on the value of its previous character, because it needs to be not smaller than it. Think backtracking. Build a recursive function count(n, last_character) that counts the number of valid strings of length n and whose first characters are not less than last_character.... |
✅ Python 4 approaches (DP, Maths) | count-sorted-vowel-strings | 0 | 1 | 1. ##### DP Tabulation\n```\nclass Solution:\n def countVowelStrings(self, n: int) -> int: \n dp = [[0] * 6 for _ in range(n+1)]\n for i in range(1, 6):\n dp[1][i] = i\n \n for i in range(2, n+1):\n dp[i][1]=1\n for j in range(2, 6):\n ... | 50 | Given an integer `n`, return _the number of strings of length_ `n` _that consist only of vowels (_`a`_,_ `e`_,_ `i`_,_ `o`_,_ `u`_) and are **lexicographically sorted**._
A string `s` is **lexicographically sorted** if for all valid `i`, `s[i]` is the same as or comes before `s[i+1]` in the alphabet.
**Example 1:**
... | null |
✅ Python 4 approaches (DP, Maths) | count-sorted-vowel-strings | 0 | 1 | 1. ##### DP Tabulation\n```\nclass Solution:\n def countVowelStrings(self, n: int) -> int: \n dp = [[0] * 6 for _ in range(n+1)]\n for i in range(1, 6):\n dp[1][i] = i\n \n for i in range(2, n+1):\n dp[i][1]=1\n for j in range(2, 6):\n ... | 50 | You are given an undirected graph. You are given an integer `n` which is the number of nodes in the graph and an array `edges`, where each `edges[i] = [ui, vi]` indicates that there is an undirected edge between `ui` and `vi`.
A **connected trio** is a set of **three** nodes where there is an edge between **every** pa... | For each character, its possible values will depend on the value of its previous character, because it needs to be not smaller than it. Think backtracking. Build a recursive function count(n, last_character) that counts the number of valid strings of length n and whose first characters are not less than last_character.... |
Dynamic Programming - Python [100%] [Explanation + code] | count-sorted-vowel-strings | 0 | 1 | In this problem, the pattern I observed was prepending \'a\' to all the strings of length n-1 does not affect any order. Similarly, \'e\' can be prepended to strings starting with \'e\' and greater vowels and so on.\n\nSo we have our subproblem.\n\n**How do we fill the DP Table?**\nLets, take an example of n = 3\n\n![i... | 103 | Given an integer `n`, return _the number of strings of length_ `n` _that consist only of vowels (_`a`_,_ `e`_,_ `i`_,_ `o`_,_ `u`_) and are **lexicographically sorted**._
A string `s` is **lexicographically sorted** if for all valid `i`, `s[i]` is the same as or comes before `s[i+1]` in the alphabet.
**Example 1:**
... | null |
Dynamic Programming - Python [100%] [Explanation + code] | count-sorted-vowel-strings | 0 | 1 | In this problem, the pattern I observed was prepending \'a\' to all the strings of length n-1 does not affect any order. Similarly, \'e\' can be prepended to strings starting with \'e\' and greater vowels and so on.\n\nSo we have our subproblem.\n\n**How do we fill the DP Table?**\nLets, take an example of n = 3\n\n![i... | 103 | You are given an undirected graph. You are given an integer `n` which is the number of nodes in the graph and an array `edges`, where each `edges[i] = [ui, vi]` indicates that there is an undirected edge between `ui` and `vi`.
A **connected trio** is a set of **three** nodes where there is an edge between **every** pa... | For each character, its possible values will depend on the value of its previous character, because it needs to be not smaller than it. Think backtracking. Build a recursive function count(n, last_character) that counts the number of valid strings of length n and whose first characters are not less than last_character.... |
Easy method for beginners to understand | count-sorted-vowel-strings | 0 | 1 | # Intuition\nWe need to just calculate the permutation for given value n (Eg n=3\n then result = 5*(5+1)*(5+2)//(1*2*3)= 35)\n\n# Code\n```\nclass Solution:\n def countVowelStrings(self, n: int) -> int:\n mul=1\n j=1\n for i in range(5,5+n):\n mul=(mul*i)//j\n j+=1\n ... | 1 | Given an integer `n`, return _the number of strings of length_ `n` _that consist only of vowels (_`a`_,_ `e`_,_ `i`_,_ `o`_,_ `u`_) and are **lexicographically sorted**._
A string `s` is **lexicographically sorted** if for all valid `i`, `s[i]` is the same as or comes before `s[i+1]` in the alphabet.
**Example 1:**
... | null |
Easy method for beginners to understand | count-sorted-vowel-strings | 0 | 1 | # Intuition\nWe need to just calculate the permutation for given value n (Eg n=3\n then result = 5*(5+1)*(5+2)//(1*2*3)= 35)\n\n# Code\n```\nclass Solution:\n def countVowelStrings(self, n: int) -> int:\n mul=1\n j=1\n for i in range(5,5+n):\n mul=(mul*i)//j\n j+=1\n ... | 1 | You are given an undirected graph. You are given an integer `n` which is the number of nodes in the graph and an array `edges`, where each `edges[i] = [ui, vi]` indicates that there is an undirected edge between `ui` and `vi`.
A **connected trio** is a set of **three** nodes where there is an edge between **every** pa... | For each character, its possible values will depend on the value of its previous character, because it needs to be not smaller than it. Think backtracking. Build a recursive function count(n, last_character) that counts the number of valid strings of length n and whose first characters are not less than last_character.... |
3 different solutions (backtracking, dynamic programming, math) | count-sorted-vowel-strings | 0 | 1 | \n\n**Backtracking solution:**\n**Time complexity:** `O(n^5)`\nWe try all the possibilities where adding the new vowel doesn\'t break the lexicographic order. For example, if the last character of our actual possibility is \'e\', we can\'t add an \'a\' after it.\n```\ndef count(n, last=\'\'):\n if n == 0:\n r... | 30 | Given an integer `n`, return _the number of strings of length_ `n` _that consist only of vowels (_`a`_,_ `e`_,_ `i`_,_ `o`_,_ `u`_) and are **lexicographically sorted**._
A string `s` is **lexicographically sorted** if for all valid `i`, `s[i]` is the same as or comes before `s[i+1]` in the alphabet.
**Example 1:**
... | null |
3 different solutions (backtracking, dynamic programming, math) | count-sorted-vowel-strings | 0 | 1 | \n\n**Backtracking solution:**\n**Time complexity:** `O(n^5)`\nWe try all the possibilities where adding the new vowel doesn\'t break the lexicographic order. For example, if the last character of our actual possibility is \'e\', we can\'t add an \'a\' after it.\n```\ndef count(n, last=\'\'):\n if n == 0:\n r... | 30 | You are given an undirected graph. You are given an integer `n` which is the number of nodes in the graph and an array `edges`, where each `edges[i] = [ui, vi]` indicates that there is an undirected edge between `ui` and `vi`.
A **connected trio** is a set of **three** nodes where there is an edge between **every** pa... | For each character, its possible values will depend on the value of its previous character, because it needs to be not smaller than it. Think backtracking. Build a recursive function count(n, last_character) that counts the number of valid strings of length n and whose first characters are not less than last_character.... |
Python | Min Heap | With Explanation | Easy to Understand | furthest-building-you-can-reach | 0 | 1 | ```\nclass Solution:\n def furthestBuilding(self, heights: List[int], bricks: int, ladders: int) -> int:\n # prepare: use a min heap to store each difference(climb) between two contiguous buildings\n # strategy: use the ladders for the longest climbs and the bricks for the shortest climbs\n \n ... | 36 | You are given an integer array `heights` representing the heights of buildings, some `bricks`, and some `ladders`.
You start your journey from building `0` and move to the next building by possibly using bricks or ladders.
While moving from building `i` to building `i+1` (**0-indexed**),
* If the current building'... | Simulate the process until there are not enough empty bottles for even one full bottle of water. |
Python | Min Heap | With Explanation | Easy to Understand | furthest-building-you-can-reach | 0 | 1 | ```\nclass Solution:\n def furthestBuilding(self, heights: List[int], bricks: int, ladders: int) -> int:\n # prepare: use a min heap to store each difference(climb) between two contiguous buildings\n # strategy: use the ladders for the longest climbs and the bricks for the shortest climbs\n \n ... | 36 | There are `n` buildings in a line. You are given an integer array `heights` of size `n` that represents the heights of the buildings in the line.
The ocean is to the right of the buildings. A building has an ocean view if the building can see the ocean without obstructions. Formally, a building has an ocean view if al... | Assume the problem is to check whether you can reach the last building or not. You'll have to do a set of jumps, and choose for each one whether to do it using a ladder or bricks. It's always optimal to use ladders in the largest jumps. Iterate on the buildings, maintaining the largest r jumps and the sum of the remain... |
Python 3 || 13 lines, iterative & recursive versions , w/example || T/M: 94% / 81% | kth-smallest-instructions | 0 | 1 | The iterative version:\n```\nclass Solution:\n def kthSmallestPath(self, destination: List[int], k: int) -> str:\n\n (v, h), ans = destination, \'\' # Example: destination = [3, 4] k = 16 \n\n while h and v: # h v k c Ans \n ... | 3 | Bob is standing at cell `(0, 0)`, and he wants to reach `destination`: `(row, column)`. He can only travel **right** and **down**. You are going to help Bob by providing **instructions** for him to reach `destination`.
The **instructions** are represented as a string, where each character is either:
* `'H'`, meanin... | Start traversing the tree and each node should return a vector to its parent node. The vector should be of length 26 and have the count of all the labels in the sub-tree of this node. |
Python 3 || 13 lines, iterative & recursive versions , w/example || T/M: 94% / 81% | kth-smallest-instructions | 0 | 1 | The iterative version:\n```\nclass Solution:\n def kthSmallestPath(self, destination: List[int], k: int) -> str:\n\n (v, h), ans = destination, \'\' # Example: destination = [3, 4] k = 16 \n\n while h and v: # h v k c Ans \n ... | 3 | You are controlling a robot that is located somewhere in a room. The room is modeled as an `m x n` binary grid where `0` represents a wall and `1` represents an empty slot.
The robot starts at an unknown location in the room that is guaranteed to be empty, and you do not have access to the grid, but you can move the r... | There are nCr(row + column, row) possible instructions to reach (row, column). Try building the instructions one step at a time. How many instructions start with "H", and how does this compare with k? |
Solution | kth-smallest-instructions | 1 | 1 | ```C++ []\nclass Solution {\npublic:\n int getCount(int h, int v)\n {\n if(h == 0 || v == 0) return 1;\n long long res = 1;\n if(h < v) swap(h, v);\n for(int i = h + 1; i <= (h+v); i++) res *= i;\n for(int i = 1; i <= v; i++) res /= i;\n return (int)res;\n }\n strin... | 1 | Bob is standing at cell `(0, 0)`, and he wants to reach `destination`: `(row, column)`. He can only travel **right** and **down**. You are going to help Bob by providing **instructions** for him to reach `destination`.
The **instructions** are represented as a string, where each character is either:
* `'H'`, meanin... | Start traversing the tree and each node should return a vector to its parent node. The vector should be of length 26 and have the count of all the labels in the sub-tree of this node. |
Solution | kth-smallest-instructions | 1 | 1 | ```C++ []\nclass Solution {\npublic:\n int getCount(int h, int v)\n {\n if(h == 0 || v == 0) return 1;\n long long res = 1;\n if(h < v) swap(h, v);\n for(int i = h + 1; i <= (h+v); i++) res *= i;\n for(int i = 1; i <= v; i++) res /= i;\n return (int)res;\n }\n strin... | 1 | You are controlling a robot that is located somewhere in a room. The room is modeled as an `m x n` binary grid where `0` represents a wall and `1` represents an empty slot.
The robot starts at an unknown location in the room that is guaranteed to be empty, and you do not have access to the grid, but you can move the r... | There are nCr(row + column, row) possible instructions to reach (row, column). Try building the instructions one step at a time. How many instructions start with "H", and how does this compare with k? |
[Python] Simple & Fast Solution, explained | kth-smallest-instructions | 0 | 1 | **Observations**\n1. We know for certain number of **\'H\'** and **\'V\'**, the total number of instructions are **comb(h + v, v)**\n2. They should be sorted **lexicographically**, which means: the **order** of instructions with first character of **\'H\'** are always smaller than instructions with first character of *... | 17 | Bob is standing at cell `(0, 0)`, and he wants to reach `destination`: `(row, column)`. He can only travel **right** and **down**. You are going to help Bob by providing **instructions** for him to reach `destination`.
The **instructions** are represented as a string, where each character is either:
* `'H'`, meanin... | Start traversing the tree and each node should return a vector to its parent node. The vector should be of length 26 and have the count of all the labels in the sub-tree of this node. |
[Python] Simple & Fast Solution, explained | kth-smallest-instructions | 0 | 1 | **Observations**\n1. We know for certain number of **\'H\'** and **\'V\'**, the total number of instructions are **comb(h + v, v)**\n2. They should be sorted **lexicographically**, which means: the **order** of instructions with first character of **\'H\'** are always smaller than instructions with first character of *... | 17 | You are controlling a robot that is located somewhere in a room. The room is modeled as an `m x n` binary grid where `0` represents a wall and `1` represents an empty slot.
The robot starts at an unknown location in the room that is guaranteed to be empty, and you do not have access to the grid, but you can move the r... | There are nCr(row + column, row) possible instructions to reach (row, column). Try building the instructions one step at a time. How many instructions start with "H", and how does this compare with k? |
✔ Python3 Solution | Combination | kth-smallest-instructions | 0 | 1 | # Solution - 1\n```Python\nclass Solution:\n def kthSmallestPath(self, D, K):\n K -= 1\n V, H = D\n ans = \'\'\n while K and H and V:\n C = comb(H + V - 1, H - 1)\n if C <= K:\n ans, V, K = ans + \'V\', V - 1, K - C\n else:\n ... | 2 | Bob is standing at cell `(0, 0)`, and he wants to reach `destination`: `(row, column)`. He can only travel **right** and **down**. You are going to help Bob by providing **instructions** for him to reach `destination`.
The **instructions** are represented as a string, where each character is either:
* `'H'`, meanin... | Start traversing the tree and each node should return a vector to its parent node. The vector should be of length 26 and have the count of all the labels in the sub-tree of this node. |
✔ Python3 Solution | Combination | kth-smallest-instructions | 0 | 1 | # Solution - 1\n```Python\nclass Solution:\n def kthSmallestPath(self, D, K):\n K -= 1\n V, H = D\n ans = \'\'\n while K and H and V:\n C = comb(H + V - 1, H - 1)\n if C <= K:\n ans, V, K = ans + \'V\', V - 1, K - C\n else:\n ... | 2 | You are controlling a robot that is located somewhere in a room. The room is modeled as an `m x n` binary grid where `0` represents a wall and `1` represents an empty slot.
The robot starts at an unknown location in the room that is guaranteed to be empty, and you do not have access to the grid, but you can move the r... | There are nCr(row + column, row) possible instructions to reach (row, column). Try building the instructions one step at a time. How many instructions start with "H", and how does this compare with k? |
[Python3] greedy | kth-smallest-instructions | 0 | 1 | Algo \nGiven `m` H and `n` V to place, we check if it is possible to place H. If so, decrement `n`; otherwise decrement `m`. Since there are `comb(m+n-1, n-1)` instructions starting with H, the condition to check for placing H is `comb(m+n-1, n-1) >= k`. \n\nImplementation \n```\nclass Solution:\n def kthSmallestPat... | 5 | Bob is standing at cell `(0, 0)`, and he wants to reach `destination`: `(row, column)`. He can only travel **right** and **down**. You are going to help Bob by providing **instructions** for him to reach `destination`.
The **instructions** are represented as a string, where each character is either:
* `'H'`, meanin... | Start traversing the tree and each node should return a vector to its parent node. The vector should be of length 26 and have the count of all the labels in the sub-tree of this node. |
[Python3] greedy | kth-smallest-instructions | 0 | 1 | Algo \nGiven `m` H and `n` V to place, we check if it is possible to place H. If so, decrement `n`; otherwise decrement `m`. Since there are `comb(m+n-1, n-1)` instructions starting with H, the condition to check for placing H is `comb(m+n-1, n-1) >= k`. \n\nImplementation \n```\nclass Solution:\n def kthSmallestPat... | 5 | You are controlling a robot that is located somewhere in a room. The room is modeled as an `m x n` binary grid where `0` represents a wall and `1` represents an empty slot.
The robot starts at an unknown location in the room that is guaranteed to be empty, and you do not have access to the grid, but you can move the r... | There are nCr(row + column, row) possible instructions to reach (row, column). Try building the instructions one step at a time. How many instructions start with "H", and how does this compare with k? |
Recursive Combinatorics 99% runtime | kth-smallest-instructions | 0 | 1 | # Intuition\nWe might be tempted to brute force enumerate all strings then sort and get the top K. But, a quick calculation shows that if we naively enumerate even the top K strings, it is too much (nCr)\n\nThis motivates us to find a more guided approach towards finding the best string. The key insight is found by pla... | 0 | Bob is standing at cell `(0, 0)`, and he wants to reach `destination`: `(row, column)`. He can only travel **right** and **down**. You are going to help Bob by providing **instructions** for him to reach `destination`.
The **instructions** are represented as a string, where each character is either:
* `'H'`, meanin... | Start traversing the tree and each node should return a vector to its parent node. The vector should be of length 26 and have the count of all the labels in the sub-tree of this node. |
Recursive Combinatorics 99% runtime | kth-smallest-instructions | 0 | 1 | # Intuition\nWe might be tempted to brute force enumerate all strings then sort and get the top K. But, a quick calculation shows that if we naively enumerate even the top K strings, it is too much (nCr)\n\nThis motivates us to find a more guided approach towards finding the best string. The key insight is found by pla... | 0 | You are controlling a robot that is located somewhere in a room. The room is modeled as an `m x n` binary grid where `0` represents a wall and `1` represents an empty slot.
The robot starts at an unknown location in the room that is guaranteed to be empty, and you do not have access to the grid, but you can move the r... | There are nCr(row + column, row) possible instructions to reach (row, column). Try building the instructions one step at a time. How many instructions start with "H", and how does this compare with k? |
Python Combinatorial solution. Beats 95%. 10 lines | kth-smallest-instructions | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nPython combinatorial search. Beats 97%. \n\n# Approach\n<!-- Describe your approach to solving the problem. -->\nRecursive combinatorial search\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\nO(... | 0 | Bob is standing at cell `(0, 0)`, and he wants to reach `destination`: `(row, column)`. He can only travel **right** and **down**. You are going to help Bob by providing **instructions** for him to reach `destination`.
The **instructions** are represented as a string, where each character is either:
* `'H'`, meanin... | Start traversing the tree and each node should return a vector to its parent node. The vector should be of length 26 and have the count of all the labels in the sub-tree of this node. |
Python Combinatorial solution. Beats 95%. 10 lines | kth-smallest-instructions | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nPython combinatorial search. Beats 97%. \n\n# Approach\n<!-- Describe your approach to solving the problem. -->\nRecursive combinatorial search\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\nO(... | 0 | You are controlling a robot that is located somewhere in a room. The room is modeled as an `m x n` binary grid where `0` represents a wall and `1` represents an empty slot.
The robot starts at an unknown location in the room that is guaranteed to be empty, and you do not have access to the grid, but you can move the r... | There are nCr(row + column, row) possible instructions to reach (row, column). Try building the instructions one step at a time. How many instructions start with "H", and how does this compare with k? |
Good combinatorics problem | Commented and Explained | kth-smallest-instructions | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nOne of the best ways to solve a search space sometimes is to DFS. This is one of those times thanks to the fact that Bob is picky. Because Bob is picky, we can use that to prune the dfs space as we go along, thus providing a way to quickl... | 0 | Bob is standing at cell `(0, 0)`, and he wants to reach `destination`: `(row, column)`. He can only travel **right** and **down**. You are going to help Bob by providing **instructions** for him to reach `destination`.
The **instructions** are represented as a string, where each character is either:
* `'H'`, meanin... | Start traversing the tree and each node should return a vector to its parent node. The vector should be of length 26 and have the count of all the labels in the sub-tree of this node. |
Good combinatorics problem | Commented and Explained | kth-smallest-instructions | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nOne of the best ways to solve a search space sometimes is to DFS. This is one of those times thanks to the fact that Bob is picky. Because Bob is picky, we can use that to prune the dfs space as we go along, thus providing a way to quickl... | 0 | You are controlling a robot that is located somewhere in a room. The room is modeled as an `m x n` binary grid where `0` represents a wall and `1` represents an empty slot.
The robot starts at an unknown location in the room that is guaranteed to be empty, and you do not have access to the grid, but you can move the r... | There are nCr(row + column, row) possible instructions to reach (row, column). Try building the instructions one step at a time. How many instructions start with "H", and how does this compare with k? |
Is k small enough to insert symbol H at the end of the string? | kth-smallest-instructions | 0 | 1 | # Intuition\nWe iteratively add symbols from the the beggining of the string to its end. What we can do is to think of when the first symbol is "H" and when it is \'V\'. In fact if k is "too large" then it\'s \'V\'. In fact too large means it\'s larger then (n choose h-1) where n is the length of the string and h is th... | 0 | Bob is standing at cell `(0, 0)`, and he wants to reach `destination`: `(row, column)`. He can only travel **right** and **down**. You are going to help Bob by providing **instructions** for him to reach `destination`.
The **instructions** are represented as a string, where each character is either:
* `'H'`, meanin... | Start traversing the tree and each node should return a vector to its parent node. The vector should be of length 26 and have the count of all the labels in the sub-tree of this node. |
Is k small enough to insert symbol H at the end of the string? | kth-smallest-instructions | 0 | 1 | # Intuition\nWe iteratively add symbols from the the beggining of the string to its end. What we can do is to think of when the first symbol is "H" and when it is \'V\'. In fact if k is "too large" then it\'s \'V\'. In fact too large means it\'s larger then (n choose h-1) where n is the length of the string and h is th... | 0 | You are controlling a robot that is located somewhere in a room. The room is modeled as an `m x n` binary grid where `0` represents a wall and `1` represents an empty slot.
The robot starts at an unknown location in the room that is guaranteed to be empty, and you do not have access to the grid, but you can move the r... | There are nCr(row + column, row) possible instructions to reach (row, column). Try building the instructions one step at a time. How many instructions start with "H", and how does this compare with k? |
Tracing Path from 2D array | kth-smallest-instructions | 0 | 1 | ### Thought Process 1:Greedy TLE\n`Why not store all possible paths and return the k-1 th path`\n```\nclass Solution:\n def kthSmallestPath(self, arr: List[int], k: int) -> str:\n row = arr[0]\n col = arr[1]\n self.path = []\n def f(i,j,s):\n if i == row and j == col:\n ... | 0 | Bob is standing at cell `(0, 0)`, and he wants to reach `destination`: `(row, column)`. He can only travel **right** and **down**. You are going to help Bob by providing **instructions** for him to reach `destination`.
The **instructions** are represented as a string, where each character is either:
* `'H'`, meanin... | Start traversing the tree and each node should return a vector to its parent node. The vector should be of length 26 and have the count of all the labels in the sub-tree of this node. |
Tracing Path from 2D array | kth-smallest-instructions | 0 | 1 | ### Thought Process 1:Greedy TLE\n`Why not store all possible paths and return the k-1 th path`\n```\nclass Solution:\n def kthSmallestPath(self, arr: List[int], k: int) -> str:\n row = arr[0]\n col = arr[1]\n self.path = []\n def f(i,j,s):\n if i == row and j == col:\n ... | 0 | You are controlling a robot that is located somewhere in a room. The room is modeled as an `m x n` binary grid where `0` represents a wall and `1` represents an empty slot.
The robot starts at an unknown location in the room that is guaranteed to be empty, and you do not have access to the grid, but you can move the r... | There are nCr(row + column, row) possible instructions to reach (row, column). Try building the instructions one step at a time. How many instructions start with "H", and how does this compare with k? |
Python (Simple DP) | kth-smallest-instructions | 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 | Bob is standing at cell `(0, 0)`, and he wants to reach `destination`: `(row, column)`. He can only travel **right** and **down**. You are going to help Bob by providing **instructions** for him to reach `destination`.
The **instructions** are represented as a string, where each character is either:
* `'H'`, meanin... | Start traversing the tree and each node should return a vector to its parent node. The vector should be of length 26 and have the count of all the labels in the sub-tree of this node. |
Python (Simple DP) | kth-smallest-instructions | 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 controlling a robot that is located somewhere in a room. The room is modeled as an `m x n` binary grid where `0` represents a wall and `1` represents an empty slot.
The robot starts at an unknown location in the room that is guaranteed to be empty, and you do not have access to the grid, but you can move the r... | There are nCr(row + column, row) possible instructions to reach (row, column). Try building the instructions one step at a time. How many instructions start with "H", and how does this compare with k? |
get-maximum-in-generated-array | get-maximum-in-generated-array | 0 | 1 | # Code\n```\nclass Solution:\n def getMaximumGenerated(self, n: int) -> int:\n nums = []\n nums.append(0)\n nums.append(1)\n for i in range(2,n+1):\n if i%2==0:\n nums.append(nums[i//2])\n else:\n nums.append(nums[i//2] + nums[(i//2) + 1... | 1 | You are given an integer `n`. A **0-indexed** integer array `nums` of length `n + 1` is generated in the following way:
* `nums[0] = 0`
* `nums[1] = 1`
* `nums[2 * i] = nums[i]` when `2 <= 2 * i <= n`
* `nums[2 * i + 1] = nums[i] + nums[i + 1]` when `2 <= 2 * i + 1 <= n`
Return _the **maximum** integer in the... | Keep track of how many positive numbers are missing as you scan the array. |
get-maximum-in-generated-array | get-maximum-in-generated-array | 0 | 1 | # Code\n```\nclass Solution:\n def getMaximumGenerated(self, n: int) -> int:\n nums = []\n nums.append(0)\n nums.append(1)\n for i in range(2,n+1):\n if i%2==0:\n nums.append(nums[i//2])\n else:\n nums.append(nums[i//2] + nums[(i//2) + 1... | 1 | You have `n` boxes. You are given a binary string `boxes` of length `n`, where `boxes[i]` is `'0'` if the `ith` box is **empty**, and `'1'` if it contains **one** ball.
In one operation, you can move **one** ball from a box to an adjacent box. Box `i` is adjacent to box `j` if `abs(i - j) == 1`. Note that after doing ... | Try generating the array. Make sure not to fall in the base case of 0. |
bottom up approach easiest 9 lines code python | get-maximum-in-generated-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. $$O(n)$$ --... | 1 | You are given an integer `n`. A **0-indexed** integer array `nums` of length `n + 1` is generated in the following way:
* `nums[0] = 0`
* `nums[1] = 1`
* `nums[2 * i] = nums[i]` when `2 <= 2 * i <= n`
* `nums[2 * i + 1] = nums[i] + nums[i + 1]` when `2 <= 2 * i + 1 <= n`
Return _the **maximum** integer in the... | Keep track of how many positive numbers are missing as you scan the array. |
bottom up approach easiest 9 lines code python | get-maximum-in-generated-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. $$O(n)$$ --... | 1 | You have `n` boxes. You are given a binary string `boxes` of length `n`, where `boxes[i]` is `'0'` if the `ith` box is **empty**, and `'1'` if it contains **one** ball.
In one operation, you can move **one** ball from a box to an adjacent box. Box `i` is adjacent to box `j` if `abs(i - j) == 1`. Note that after doing ... | Try generating the array. Make sure not to fall in the base case of 0. |
Simplest code with runtime 45 ms | get-maximum-in-generated-array | 0 | 1 | \n\n# Code\n```\nclass Solution:\n def getMaximumGenerated(self, n: int) -> int:\n dp=[0,1]\n for i in range(2,n+1):\n dp += [dp[i//2] if i%2==0 else dp[i//2]+ dp[i//2+1]]\n return max(dp[:n+1])\n``` | 1 | You are given an integer `n`. A **0-indexed** integer array `nums` of length `n + 1` is generated in the following way:
* `nums[0] = 0`
* `nums[1] = 1`
* `nums[2 * i] = nums[i]` when `2 <= 2 * i <= n`
* `nums[2 * i + 1] = nums[i] + nums[i + 1]` when `2 <= 2 * i + 1 <= n`
Return _the **maximum** integer in the... | Keep track of how many positive numbers are missing as you scan the array. |
Simplest code with runtime 45 ms | get-maximum-in-generated-array | 0 | 1 | \n\n# Code\n```\nclass Solution:\n def getMaximumGenerated(self, n: int) -> int:\n dp=[0,1]\n for i in range(2,n+1):\n dp += [dp[i//2] if i%2==0 else dp[i//2]+ dp[i//2+1]]\n return max(dp[:n+1])\n``` | 1 | You have `n` boxes. You are given a binary string `boxes` of length `n`, where `boxes[i]` is `'0'` if the `ith` box is **empty**, and `'1'` if it contains **one** ball.
In one operation, you can move **one** ball from a box to an adjacent box. Box `i` is adjacent to box `j` if `abs(i - j) == 1`. Note that after doing ... | Try generating the array. Make sure not to fall in the base case of 0. |
✅Python3 || 23ms/Beats 97.74% | get-maximum-in-generated-array | 0 | 1 | \n\nPlease UPVOTE if it helps \u2764\uFE0F\uD83D\uDE0A\nThank You and Happy To Help You!!\n\n# Python3\n```\nclass Solution:\n def getMaximumGenerated(self, n: int) -> int:\n dp=[0,1]\n for... | 14 | You are given an integer `n`. A **0-indexed** integer array `nums` of length `n + 1` is generated in the following way:
* `nums[0] = 0`
* `nums[1] = 1`
* `nums[2 * i] = nums[i]` when `2 <= 2 * i <= n`
* `nums[2 * i + 1] = nums[i] + nums[i + 1]` when `2 <= 2 * i + 1 <= n`
Return _the **maximum** integer in the... | Keep track of how many positive numbers are missing as you scan the array. |
✅Python3 || 23ms/Beats 97.74% | get-maximum-in-generated-array | 0 | 1 | \n\nPlease UPVOTE if it helps \u2764\uFE0F\uD83D\uDE0A\nThank You and Happy To Help You!!\n\n# Python3\n```\nclass Solution:\n def getMaximumGenerated(self, n: int) -> int:\n dp=[0,1]\n for... | 14 | You have `n` boxes. You are given a binary string `boxes` of length `n`, where `boxes[i]` is `'0'` if the `ith` box is **empty**, and `'1'` if it contains **one** ball.
In one operation, you can move **one** ball from a box to an adjacent box. Box `i` is adjacent to box `j` if `abs(i - j) == 1`. Note that after doing ... | Try generating the array. Make sure not to fall in the base case of 0. |
Awesome Code--->Python3 | get-maximum-in-generated-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. $$O(n)$$ --... | 4 | You are given an integer `n`. A **0-indexed** integer array `nums` of length `n + 1` is generated in the following way:
* `nums[0] = 0`
* `nums[1] = 1`
* `nums[2 * i] = nums[i]` when `2 <= 2 * i <= n`
* `nums[2 * i + 1] = nums[i] + nums[i + 1]` when `2 <= 2 * i + 1 <= n`
Return _the **maximum** integer in the... | Keep track of how many positive numbers are missing as you scan the array. |
Awesome Code--->Python3 | get-maximum-in-generated-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. $$O(n)$$ --... | 4 | You have `n` boxes. You are given a binary string `boxes` of length `n`, where `boxes[i]` is `'0'` if the `ith` box is **empty**, and `'1'` if it contains **one** ball.
In one operation, you can move **one** ball from a box to an adjacent box. Box `i` is adjacent to box `j` if `abs(i - j) == 1`. Note that after doing ... | Try generating the array. Make sure not to fall in the base case of 0. |
Python simple solution | get-maximum-in-generated-array | 0 | 1 | **Python :**\n\n```\ndef getMaximumGenerated(self, n: int) -> int:\n\tif not n:\n\t\treturn n\n\n\tnums = [0] * (n + 1)\n\tnums[1] = 1\n\n\tfor i in range(2, n + 1): \n\t\tif i % 2:\n\t\t\tnums[i] = nums[i // 2] + nums[i // 2 + 1]\n\n\t\telse:\n\t\t\tnums[i] = nums[i // 2]\n\n\treturn max(nums)\n```\n\n**Like... | 8 | You are given an integer `n`. A **0-indexed** integer array `nums` of length `n + 1` is generated in the following way:
* `nums[0] = 0`
* `nums[1] = 1`
* `nums[2 * i] = nums[i]` when `2 <= 2 * i <= n`
* `nums[2 * i + 1] = nums[i] + nums[i + 1]` when `2 <= 2 * i + 1 <= n`
Return _the **maximum** integer in the... | Keep track of how many positive numbers are missing as you scan the array. |
Python simple solution | get-maximum-in-generated-array | 0 | 1 | **Python :**\n\n```\ndef getMaximumGenerated(self, n: int) -> int:\n\tif not n:\n\t\treturn n\n\n\tnums = [0] * (n + 1)\n\tnums[1] = 1\n\n\tfor i in range(2, n + 1): \n\t\tif i % 2:\n\t\t\tnums[i] = nums[i // 2] + nums[i // 2 + 1]\n\n\t\telse:\n\t\t\tnums[i] = nums[i // 2]\n\n\treturn max(nums)\n```\n\n**Like... | 8 | You have `n` boxes. You are given a binary string `boxes` of length `n`, where `boxes[i]` is `'0'` if the `ith` box is **empty**, and `'1'` if it contains **one** ball.
In one operation, you can move **one** ball from a box to an adjacent box. Box `i` is adjacent to box `j` if `abs(i - j) == 1`. Note that after doing ... | Try generating the array. Make sure not to fall in the base case of 0. |
Simple Python Solution | get-maximum-in-generated-array | 0 | 1 | **Upvote if the solution if helpful. Comment for any doubts or suggestions.**\n```\nclass Solution:\n def getMaximumGenerated(self, n: int) -> int:\n if n < 2:\n return n\n arr = [0]*(n+1)\n arr[1] = 1\n max_num = 0\n for i in range(2, n+1):\n if i%2 == 0:\n ... | 1 | You are given an integer `n`. A **0-indexed** integer array `nums` of length `n + 1` is generated in the following way:
* `nums[0] = 0`
* `nums[1] = 1`
* `nums[2 * i] = nums[i]` when `2 <= 2 * i <= n`
* `nums[2 * i + 1] = nums[i] + nums[i + 1]` when `2 <= 2 * i + 1 <= n`
Return _the **maximum** integer in the... | Keep track of how many positive numbers are missing as you scan the array. |
Simple Python Solution | get-maximum-in-generated-array | 0 | 1 | **Upvote if the solution if helpful. Comment for any doubts or suggestions.**\n```\nclass Solution:\n def getMaximumGenerated(self, n: int) -> int:\n if n < 2:\n return n\n arr = [0]*(n+1)\n arr[1] = 1\n max_num = 0\n for i in range(2, n+1):\n if i%2 == 0:\n ... | 1 | You have `n` boxes. You are given a binary string `boxes` of length `n`, where `boxes[i]` is `'0'` if the `ith` box is **empty**, and `'1'` if it contains **one** ball.
In one operation, you can move **one** ball from a box to an adjacent box. Box `i` is adjacent to box `j` if `abs(i - j) == 1`. Note that after doing ... | Try generating the array. Make sure not to fall in the base case of 0. |
DP | Python3 | get-maximum-in-generated-array | 0 | 1 | # Complexity\n- Time complexity: $$O(n)$$\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity: $$O(n)$$\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\nclass Solution:\n def getMaximumGenerated(self, n: int) -> int:\n dp = [0] * (n + 2)\n dp[1] = 1 \n ... | 1 | You are given an integer `n`. A **0-indexed** integer array `nums` of length `n + 1` is generated in the following way:
* `nums[0] = 0`
* `nums[1] = 1`
* `nums[2 * i] = nums[i]` when `2 <= 2 * i <= n`
* `nums[2 * i + 1] = nums[i] + nums[i + 1]` when `2 <= 2 * i + 1 <= n`
Return _the **maximum** integer in the... | Keep track of how many positive numbers are missing as you scan the array. |
DP | Python3 | get-maximum-in-generated-array | 0 | 1 | # Complexity\n- Time complexity: $$O(n)$$\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity: $$O(n)$$\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\nclass Solution:\n def getMaximumGenerated(self, n: int) -> int:\n dp = [0] * (n + 2)\n dp[1] = 1 \n ... | 1 | You have `n` boxes. You are given a binary string `boxes` of length `n`, where `boxes[i]` is `'0'` if the `ith` box is **empty**, and `'1'` if it contains **one** ball.
In one operation, you can move **one** ball from a box to an adjacent box. Box `i` is adjacent to box `j` if `abs(i - j) == 1`. Note that after doing ... | Try generating the array. Make sure not to fall in the base case of 0. |
Python generators are like space shuttles | get-maximum-in-generated-array | 0 | 1 | Constructive solutions were posted before, but did you know that generators allow you get rid of used array parts, like a space shuttle gets rid of spent fuel tanks?\n\n```\nfrom queue import deque\n\nclass Solution:\n def getMaximumGenerated(self, n: int) -> int:\n \n def arrgen(n):\n yield... | 12 | You are given an integer `n`. A **0-indexed** integer array `nums` of length `n + 1` is generated in the following way:
* `nums[0] = 0`
* `nums[1] = 1`
* `nums[2 * i] = nums[i]` when `2 <= 2 * i <= n`
* `nums[2 * i + 1] = nums[i] + nums[i + 1]` when `2 <= 2 * i + 1 <= n`
Return _the **maximum** integer in the... | Keep track of how many positive numbers are missing as you scan the array. |
Python generators are like space shuttles | get-maximum-in-generated-array | 0 | 1 | Constructive solutions were posted before, but did you know that generators allow you get rid of used array parts, like a space shuttle gets rid of spent fuel tanks?\n\n```\nfrom queue import deque\n\nclass Solution:\n def getMaximumGenerated(self, n: int) -> int:\n \n def arrgen(n):\n yield... | 12 | You have `n` boxes. You are given a binary string `boxes` of length `n`, where `boxes[i]` is `'0'` if the `ith` box is **empty**, and `'1'` if it contains **one** ball.
In one operation, you can move **one** ball from a box to an adjacent box. Box `i` is adjacent to box `j` if `abs(i - j) == 1`. Note that after doing ... | Try generating the array. Make sure not to fall in the base case of 0. |
Easy Python Solution O(n) with Explanation | get-maximum-in-generated-array | 0 | 1 | \tclass Solution:\n\t\tdef getMaximumGenerated(self, n: int) -> int:\n\t\t\tif n==0: return 0\n\t\t\tarr = [0, 1]\n\t\t\tfor i in range(2, n+1):\n\t\t\t\tif i%2==0:\n\t\t\t\t\tarr.append(arr[i//2])\n\t\t\t\telse:\n\t\t\t\t\tarr.append(arr[i//2] + arr[i//2 + 1])\n\t\t\treturn max(arr)\n\t\t\t\nThe Basic Idea is that giv... | 8 | You are given an integer `n`. A **0-indexed** integer array `nums` of length `n + 1` is generated in the following way:
* `nums[0] = 0`
* `nums[1] = 1`
* `nums[2 * i] = nums[i]` when `2 <= 2 * i <= n`
* `nums[2 * i + 1] = nums[i] + nums[i + 1]` when `2 <= 2 * i + 1 <= n`
Return _the **maximum** integer in the... | Keep track of how many positive numbers are missing as you scan the array. |
Easy Python Solution O(n) with Explanation | get-maximum-in-generated-array | 0 | 1 | \tclass Solution:\n\t\tdef getMaximumGenerated(self, n: int) -> int:\n\t\t\tif n==0: return 0\n\t\t\tarr = [0, 1]\n\t\t\tfor i in range(2, n+1):\n\t\t\t\tif i%2==0:\n\t\t\t\t\tarr.append(arr[i//2])\n\t\t\t\telse:\n\t\t\t\t\tarr.append(arr[i//2] + arr[i//2 + 1])\n\t\t\treturn max(arr)\n\t\t\t\nThe Basic Idea is that giv... | 8 | You have `n` boxes. You are given a binary string `boxes` of length `n`, where `boxes[i]` is `'0'` if the `ith` box is **empty**, and `'1'` if it contains **one** ball.
In one operation, you can move **one** ball from a box to an adjacent box. Box `i` is adjacent to box `j` if `abs(i - j) == 1`. Note that after doing ... | Try generating the array. Make sure not to fall in the base case of 0. |
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