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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Python3 [runtime faster than 90.82%, memory less than 93.18%] + testing | design-an-ordered-stream | 0 | 1 | ```\n"""\nexercise: 1656. Design an Ordered Stream\nhttps://leetcode.com/problems/design-an-ordered-stream/\ncompany: Bloomberg\n"""\n\n\n# Your OrderedStream object will be instantiated and called as such:\n# obj = OrderedStream(n)\n# param_1 = obj.insert(idKey,value)\n\nclass OrderedStream:\n\n def __init__(self, ... | 1 | There is a stream of `n` `(idKey, value)` pairs arriving in an **arbitrary** order, where `idKey` is an integer between `1` and `n` and `value` is a string. No two pairs have the same `id`.
Design a stream that returns the values in **increasing order of their IDs** by returning a **chunk** (list) of values after each... | Notice that the constraints are small enough for a brute force solution to pass. Loop through all triplets, and count the ones that are good. |
Python3 [runtime faster than 90.82%, memory less than 93.18%] + testing | design-an-ordered-stream | 0 | 1 | ```\n"""\nexercise: 1656. Design an Ordered Stream\nhttps://leetcode.com/problems/design-an-ordered-stream/\ncompany: Bloomberg\n"""\n\n\n# Your OrderedStream object will be instantiated and called as such:\n# obj = OrderedStream(n)\n# param_1 = obj.insert(idKey,value)\n\nclass OrderedStream:\n\n def __init__(self, ... | 1 | You are given two arrays of integers `nums1` and `nums2`, possibly of different lengths. The values in the arrays are between `1` and `6`, inclusive.
In one operation, you can change any integer's value in **any** of the arrays to **any** value between `1` and `6`, inclusive.
Return _the minimum number of operations ... | Maintain the next id that should be outputted. Maintain the ids that were inserted in the stream. Per each insert, make a loop where you check if the id that has the turn has been inserted, and if so increment the id that has the turn and continue the loop, else break. |
Python3 Solution + Better Description Because This One Is Really Bad | design-an-ordered-stream | 0 | 1 | The description as of June 13th, 2022 mentions nothing a ptr. There is a ptr that should be initialized at 1. If an item is inserted with an idKey above the ptr, return nothing. If an item is inserted that matches the ptr, return the largest chunk of contiguous values above the ptr. The ptr should update to the index i... | 11 | There is a stream of `n` `(idKey, value)` pairs arriving in an **arbitrary** order, where `idKey` is an integer between `1` and `n` and `value` is a string. No two pairs have the same `id`.
Design a stream that returns the values in **increasing order of their IDs** by returning a **chunk** (list) of values after each... | Notice that the constraints are small enough for a brute force solution to pass. Loop through all triplets, and count the ones that are good. |
Python3 Solution + Better Description Because This One Is Really Bad | design-an-ordered-stream | 0 | 1 | The description as of June 13th, 2022 mentions nothing a ptr. There is a ptr that should be initialized at 1. If an item is inserted with an idKey above the ptr, return nothing. If an item is inserted that matches the ptr, return the largest chunk of contiguous values above the ptr. The ptr should update to the index i... | 11 | You are given two arrays of integers `nums1` and `nums2`, possibly of different lengths. The values in the arrays are between `1` and `6`, inclusive.
In one operation, you can change any integer's value in **any** of the arrays to **any** value between `1` and `6`, inclusive.
Return _the minimum number of operations ... | Maintain the next id that should be outputted. Maintain the ids that were inserted in the stream. Per each insert, make a loop where you check if the id that has the turn has been inserted, and if so increment the id that has the turn and continue the loop, else break. |
python 3 || simple solution | design-an-ordered-stream | 0 | 1 | ```\nclass OrderedStream:\n\n def __init__(self, n: int):\n self.stream = [\'\'] * (n + 1)\n self.i = 0\n\n def insert(self, idKey: int, value: str) -> List[str]:\n self.stream[idKey - 1] = value\n res = []\n\n while self.stream[self.i]:\n res.append(self.stream[self.... | 2 | There is a stream of `n` `(idKey, value)` pairs arriving in an **arbitrary** order, where `idKey` is an integer between `1` and `n` and `value` is a string. No two pairs have the same `id`.
Design a stream that returns the values in **increasing order of their IDs** by returning a **chunk** (list) of values after each... | Notice that the constraints are small enough for a brute force solution to pass. Loop through all triplets, and count the ones that are good. |
python 3 || simple solution | design-an-ordered-stream | 0 | 1 | ```\nclass OrderedStream:\n\n def __init__(self, n: int):\n self.stream = [\'\'] * (n + 1)\n self.i = 0\n\n def insert(self, idKey: int, value: str) -> List[str]:\n self.stream[idKey - 1] = value\n res = []\n\n while self.stream[self.i]:\n res.append(self.stream[self.... | 2 | You are given two arrays of integers `nums1` and `nums2`, possibly of different lengths. The values in the arrays are between `1` and `6`, inclusive.
In one operation, you can change any integer's value in **any** of the arrays to **any** value between `1` and `6`, inclusive.
Return _the minimum number of operations ... | Maintain the next id that should be outputted. Maintain the ids that were inserted in the stream. Per each insert, make a loop where you check if the id that has the turn has been inserted, and if so increment the id that has the turn and continue the loop, else break. |
Python using dict, 212ms 14.6MB | design-an-ordered-stream | 0 | 1 | ```\nclass OrderedStream:\n\n def __init__(self, n: int):\n self.seen = {}\n self.ptr = 1\n\n def insert(self, id: int, value: str) -> List[str]:\n seen, ptr = self.seen, self.ptr\n \n seen[id] = value\n result = []\n while ptr in seen:\n result.append(s... | 16 | There is a stream of `n` `(idKey, value)` pairs arriving in an **arbitrary** order, where `idKey` is an integer between `1` and `n` and `value` is a string. No two pairs have the same `id`.
Design a stream that returns the values in **increasing order of their IDs** by returning a **chunk** (list) of values after each... | Notice that the constraints are small enough for a brute force solution to pass. Loop through all triplets, and count the ones that are good. |
Python using dict, 212ms 14.6MB | design-an-ordered-stream | 0 | 1 | ```\nclass OrderedStream:\n\n def __init__(self, n: int):\n self.seen = {}\n self.ptr = 1\n\n def insert(self, id: int, value: str) -> List[str]:\n seen, ptr = self.seen, self.ptr\n \n seen[id] = value\n result = []\n while ptr in seen:\n result.append(s... | 16 | You are given two arrays of integers `nums1` and `nums2`, possibly of different lengths. The values in the arrays are between `1` and `6`, inclusive.
In one operation, you can change any integer's value in **any** of the arrays to **any** value between `1` and `6`, inclusive.
Return _the minimum number of operations ... | Maintain the next id that should be outputted. Maintain the ids that were inserted in the stream. Per each insert, make a loop where you check if the id that has the turn has been inserted, and if so increment the id that has the turn and continue the loop, else break. |
Python Commented beats 90% | Iteration, Codesplitting | design-an-ordered-stream | 0 | 1 | class OrderedStream:\n\n def __init__(self, n: int):\n self.arr = [None]*n # initialize the array for size n\n self.ptr = 0 # remeber where the stream pointer is at\n def get_chunk(self):\n chunk = []\n \n # Construct the next chunk by checking bounds, and ensuring value is not None\n ... | 2 | There is a stream of `n` `(idKey, value)` pairs arriving in an **arbitrary** order, where `idKey` is an integer between `1` and `n` and `value` is a string. No two pairs have the same `id`.
Design a stream that returns the values in **increasing order of their IDs** by returning a **chunk** (list) of values after each... | Notice that the constraints are small enough for a brute force solution to pass. Loop through all triplets, and count the ones that are good. |
Python Commented beats 90% | Iteration, Codesplitting | design-an-ordered-stream | 0 | 1 | class OrderedStream:\n\n def __init__(self, n: int):\n self.arr = [None]*n # initialize the array for size n\n self.ptr = 0 # remeber where the stream pointer is at\n def get_chunk(self):\n chunk = []\n \n # Construct the next chunk by checking bounds, and ensuring value is not None\n ... | 2 | You are given two arrays of integers `nums1` and `nums2`, possibly of different lengths. The values in the arrays are between `1` and `6`, inclusive.
In one operation, you can change any integer's value in **any** of the arrays to **any** value between `1` and `6`, inclusive.
Return _the minimum number of operations ... | Maintain the next id that should be outputted. Maintain the ids that were inserted in the stream. Per each insert, make a loop where you check if the id that has the turn has been inserted, and if so increment the id that has the turn and continue the loop, else break. |
Beats 74.34%of users with Python3 | determine-if-two-strings-are-close | 0 | 1 | \n\n# Approach\n- The approach taken in the code is to use the Counter class from the collections module to count the frequency of characters in both input strings.\n- The code first checks if the lengths of the two strings are different, and if so, returns False because strings of different lengths cannot be close.\n-... | 2 | Two strings are considered **close** if you can attain one from the other using the following operations:
* Operation 1: Swap any two **existing** characters.
* For example, `abcde -> aecdb`
* Operation 2: Transform **every** occurrence of one **existing** character into another **existing** character, and d... | If k ≥ arr.length return the max element of the array. If k < arr.length simulate the game until a number wins k consecutive games. |
Three line, simple & fast solution | determine-if-two-strings-are-close | 0 | 1 | # Intuition\nTwo strings are close if they share the same characters and character frequencies, regardless of order.\n \n\n# Approach\n- **Character Set Comparison**: Ensure both strings contain identical characters.\n- **Frequency Comparison**: Check if the character frequencies match across both strings.\n\n# Com... | 3 | Two strings are considered **close** if you can attain one from the other using the following operations:
* Operation 1: Swap any two **existing** characters.
* For example, `abcde -> aecdb`
* Operation 2: Transform **every** occurrence of one **existing** character into another **existing** character, and d... | If k ≥ arr.length return the max element of the array. If k < arr.length simulate the game until a number wins k consecutive games. |
✅ 96.51% Sliding Window | minimum-operations-to-reduce-x-to-zero | 1 | 1 | # Comprehensive Guide to Solving "Minimum Operations to Reduce X to Zero"\n\n## Introduction & Problem Statement\n\nGiven an integer array `nums` and an integer `x`, the task is to find the minimum number of operations to reduce `x` to exactly 0 by removing either the leftmost or rightmost element from the array `nums`... | 140 | You are given an integer array `nums` and an integer `x`. In one operation, you can either remove the leftmost or the rightmost element from the array `nums` and subtract its value from `x`. Note that this **modifies** the array for future operations.
Return _the **minimum number** of operations to reduce_ `x` _to **e... | For each row of the grid calculate the most right 1 in the grid in the array maxRight. To check if there exist answer, sort maxRight and check if maxRight[i] ≤ i for all possible i's. If there exist an answer, simulate the swaps. |
✅ 96.51% Sliding Window | minimum-operations-to-reduce-x-to-zero | 1 | 1 | # Comprehensive Guide to Solving "Minimum Operations to Reduce X to Zero"\n\n## Introduction & Problem Statement\n\nGiven an integer array `nums` and an integer `x`, the task is to find the minimum number of operations to reduce `x` to exactly 0 by removing either the leftmost or rightmost element from the array `nums`... | 140 | There are `n` cars traveling at different speeds in the same direction along a one-lane road. You are given an array `cars` of length `n`, where `cars[i] = [positioni, speedi]` represents:
* `positioni` is the distance between the `ith` car and the beginning of the road in meters. It is guaranteed that `positioni < ... | Think in reverse; instead of finding the minimum prefix + suffix, find the maximum subarray. Finding the maximum subarray is standard and can be done greedily. |
🚀95.97% || Two Pointers - Sliding Window || Commented Code🚀 | minimum-operations-to-reduce-x-to-zero | 1 | 1 | # Porblem Description\nGiven an array of integers, `nums`, and an integer `x`. Each element in `nums` can be subtracted from x. The **goal** is to reduce `x` to exactly `0` using a **minimum** number of operations.\n\nIn each **operation**, you can choose to **remove** either the `leftmost` or the `rightmost` element f... | 60 | You are given an integer array `nums` and an integer `x`. In one operation, you can either remove the leftmost or the rightmost element from the array `nums` and subtract its value from `x`. Note that this **modifies** the array for future operations.
Return _the **minimum number** of operations to reduce_ `x` _to **e... | For each row of the grid calculate the most right 1 in the grid in the array maxRight. To check if there exist answer, sort maxRight and check if maxRight[i] ≤ i for all possible i's. If there exist an answer, simulate the swaps. |
🚀95.97% || Two Pointers - Sliding Window || Commented Code🚀 | minimum-operations-to-reduce-x-to-zero | 1 | 1 | # Porblem Description\nGiven an array of integers, `nums`, and an integer `x`. Each element in `nums` can be subtracted from x. The **goal** is to reduce `x` to exactly `0` using a **minimum** number of operations.\n\nIn each **operation**, you can choose to **remove** either the `leftmost` or the `rightmost` element f... | 60 | There are `n` cars traveling at different speeds in the same direction along a one-lane road. You are given an array `cars` of length `n`, where `cars[i] = [positioni, speedi]` represents:
* `positioni` is the distance between the `ith` car and the beginning of the road in meters. It is guaranteed that `positioni < ... | Think in reverse; instead of finding the minimum prefix + suffix, find the maximum subarray. Finding the maximum subarray is standard and can be done greedily. |
Simple Python Solution | minimum-operations-to-reduce-x-to-zero | 0 | 1 | # Complexity\n- Time complexity: **O(N)**\n- Space complexity: **O(N)**\n\n# Code\n```\nclass Solution:\n def minOperations(self, nums: List[int], x: int) -> int:\n n = len(nums)\n totalSum = sum(nums)\n if totalSum == x:\n return n\n target = totalSum - x\n hashmap = { ... | 1 | You are given an integer array `nums` and an integer `x`. In one operation, you can either remove the leftmost or the rightmost element from the array `nums` and subtract its value from `x`. Note that this **modifies** the array for future operations.
Return _the **minimum number** of operations to reduce_ `x` _to **e... | For each row of the grid calculate the most right 1 in the grid in the array maxRight. To check if there exist answer, sort maxRight and check if maxRight[i] ≤ i for all possible i's. If there exist an answer, simulate the swaps. |
Simple Python Solution | minimum-operations-to-reduce-x-to-zero | 0 | 1 | # Complexity\n- Time complexity: **O(N)**\n- Space complexity: **O(N)**\n\n# Code\n```\nclass Solution:\n def minOperations(self, nums: List[int], x: int) -> int:\n n = len(nums)\n totalSum = sum(nums)\n if totalSum == x:\n return n\n target = totalSum - x\n hashmap = { ... | 1 | There are `n` cars traveling at different speeds in the same direction along a one-lane road. You are given an array `cars` of length `n`, where `cars[i] = [positioni, speedi]` represents:
* `positioni` is the distance between the `ith` car and the beginning of the road in meters. It is guaranteed that `positioni < ... | Think in reverse; instead of finding the minimum prefix + suffix, find the maximum subarray. Finding the maximum subarray is standard and can be done greedily. |
Python 3 | Beats 92.74% Time | Beats 100% Memory | Sliding Window Solution | minimum-operations-to-reduce-x-to-zero | 0 | 1 | # Intuition\n\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- Ti... | 1 | You are given an integer array `nums` and an integer `x`. In one operation, you can either remove the leftmost or the rightmost element from the array `nums` and subtract its value from `x`. Note that this **modifies** the array for future operations.
Return _the **minimum number** of operations to reduce_ `x` _to **e... | For each row of the grid calculate the most right 1 in the grid in the array maxRight. To check if there exist answer, sort maxRight and check if maxRight[i] ≤ i for all possible i's. If there exist an answer, simulate the swaps. |
Python 3 | Beats 92.74% Time | Beats 100% Memory | Sliding Window Solution | minimum-operations-to-reduce-x-to-zero | 0 | 1 | # Intuition\n\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- Ti... | 1 | There are `n` cars traveling at different speeds in the same direction along a one-lane road. You are given an array `cars` of length `n`, where `cars[i] = [positioni, speedi]` represents:
* `positioni` is the distance between the `ith` car and the beginning of the road in meters. It is guaranteed that `positioni < ... | Think in reverse; instead of finding the minimum prefix + suffix, find the maximum subarray. Finding the maximum subarray is standard and can be done greedily. |
Maximum length subarray sum equal K | minimum-operations-to-reduce-x-to-zero | 0 | 1 | # Approach\nIf we have to find ***X as sum from starting and ending of array***, \nSo we basically need to find - \n ***maximum length of subarray*** with `sum = total-x`\n\n# Complexity\n- Time complexity: O(n)\n\n- Space complexity: O(1)\n\n# Code\n```\nclass Solution:\n def minOperations(self, nums: List[int], x:... | 1 | You are given an integer array `nums` and an integer `x`. In one operation, you can either remove the leftmost or the rightmost element from the array `nums` and subtract its value from `x`. Note that this **modifies** the array for future operations.
Return _the **minimum number** of operations to reduce_ `x` _to **e... | For each row of the grid calculate the most right 1 in the grid in the array maxRight. To check if there exist answer, sort maxRight and check if maxRight[i] ≤ i for all possible i's. If there exist an answer, simulate the swaps. |
Maximum length subarray sum equal K | minimum-operations-to-reduce-x-to-zero | 0 | 1 | # Approach\nIf we have to find ***X as sum from starting and ending of array***, \nSo we basically need to find - \n ***maximum length of subarray*** with `sum = total-x`\n\n# Complexity\n- Time complexity: O(n)\n\n- Space complexity: O(1)\n\n# Code\n```\nclass Solution:\n def minOperations(self, nums: List[int], x:... | 1 | There are `n` cars traveling at different speeds in the same direction along a one-lane road. You are given an array `cars` of length `n`, where `cars[i] = [positioni, speedi]` represents:
* `positioni` is the distance between the `ith` car and the beginning of the road in meters. It is guaranteed that `positioni < ... | Think in reverse; instead of finding the minimum prefix + suffix, find the maximum subarray. Finding the maximum subarray is standard and can be done greedily. |
How we think about a solution - O(n) time, O(1) space - Python, JavaScript, Java, C++ | minimum-operations-to-reduce-x-to-zero | 1 | 1 | Welcome to my article. Before starting the article, why do I have to get multiple downvotes for a miniute every day? That is obviously deliberate downvotes. I can tell who is doing it. Please show your respect to others! Thanks.\n\n# Intuition\nTry to find target number with sum(nums) - x which is `unnecessary numbers`... | 39 | You are given an integer array `nums` and an integer `x`. In one operation, you can either remove the leftmost or the rightmost element from the array `nums` and subtract its value from `x`. Note that this **modifies** the array for future operations.
Return _the **minimum number** of operations to reduce_ `x` _to **e... | For each row of the grid calculate the most right 1 in the grid in the array maxRight. To check if there exist answer, sort maxRight and check if maxRight[i] ≤ i for all possible i's. If there exist an answer, simulate the swaps. |
How we think about a solution - O(n) time, O(1) space - Python, JavaScript, Java, C++ | minimum-operations-to-reduce-x-to-zero | 1 | 1 | Welcome to my article. Before starting the article, why do I have to get multiple downvotes for a miniute every day? That is obviously deliberate downvotes. I can tell who is doing it. Please show your respect to others! Thanks.\n\n# Intuition\nTry to find target number with sum(nums) - x which is `unnecessary numbers`... | 39 | There are `n` cars traveling at different speeds in the same direction along a one-lane road. You are given an array `cars` of length `n`, where `cars[i] = [positioni, speedi]` represents:
* `positioni` is the distance between the `ith` car and the beginning of the road in meters. It is guaranteed that `positioni < ... | Think in reverse; instead of finding the minimum prefix + suffix, find the maximum subarray. Finding the maximum subarray is standard and can be done greedily. |
✅97.65%🔥Sum of Subarray Reduction🔥 | minimum-operations-to-reduce-x-to-zero | 1 | 1 | # Problem\n\n### This problem involves finding the minimum number of operations to reduce the sum of elements in the array nums to exactly zero, given that you can perform operations to remove elements from either the left or right end of the array.\n---\n# Solution\n\n##### **1.** Calculate the total sum of all elemen... | 11 | You are given an integer array `nums` and an integer `x`. In one operation, you can either remove the leftmost or the rightmost element from the array `nums` and subtract its value from `x`. Note that this **modifies** the array for future operations.
Return _the **minimum number** of operations to reduce_ `x` _to **e... | For each row of the grid calculate the most right 1 in the grid in the array maxRight. To check if there exist answer, sort maxRight and check if maxRight[i] ≤ i for all possible i's. If there exist an answer, simulate the swaps. |
✅97.65%🔥Sum of Subarray Reduction🔥 | minimum-operations-to-reduce-x-to-zero | 1 | 1 | # Problem\n\n### This problem involves finding the minimum number of operations to reduce the sum of elements in the array nums to exactly zero, given that you can perform operations to remove elements from either the left or right end of the array.\n---\n# Solution\n\n##### **1.** Calculate the total sum of all elemen... | 11 | There are `n` cars traveling at different speeds in the same direction along a one-lane road. You are given an array `cars` of length `n`, where `cars[i] = [positioni, speedi]` represents:
* `positioni` is the distance between the `ith` car and the beginning of the road in meters. It is guaranteed that `positioni < ... | Think in reverse; instead of finding the minimum prefix + suffix, find the maximum subarray. Finding the maximum subarray is standard and can be done greedily. |
🔥98.80% Acceptance📈 rate | Using Sliding Window🧭 | Explained in details✅ | minimum-operations-to-reduce-x-to-zero | 1 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nThe intuition behind this code is to find a subarray with the maximum possible sum (equal to targetSum) within the given nums array. The goal is to minimize the number of operations to reduce x to 0. By calculating the maximum subarray le... | 3 | You are given an integer array `nums` and an integer `x`. In one operation, you can either remove the leftmost or the rightmost element from the array `nums` and subtract its value from `x`. Note that this **modifies** the array for future operations.
Return _the **minimum number** of operations to reduce_ `x` _to **e... | For each row of the grid calculate the most right 1 in the grid in the array maxRight. To check if there exist answer, sort maxRight and check if maxRight[i] ≤ i for all possible i's. If there exist an answer, simulate the swaps. |
🔥98.80% Acceptance📈 rate | Using Sliding Window🧭 | Explained in details✅ | minimum-operations-to-reduce-x-to-zero | 1 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nThe intuition behind this code is to find a subarray with the maximum possible sum (equal to targetSum) within the given nums array. The goal is to minimize the number of operations to reduce x to 0. By calculating the maximum subarray le... | 3 | There are `n` cars traveling at different speeds in the same direction along a one-lane road. You are given an array `cars` of length `n`, where `cars[i] = [positioni, speedi]` represents:
* `positioni` is the distance between the `ith` car and the beginning of the road in meters. It is guaranteed that `positioni < ... | Think in reverse; instead of finding the minimum prefix + suffix, find the maximum subarray. Finding the maximum subarray is standard and can be done greedily. |
✅ [C++/Python] Simple Solution w/ Explanation | Sliding Window | minimum-operations-to-reduce-x-to-zero | 0 | 1 | Sometimes, converting a problem into some other familiar one helps a lot. This question is one of them.\nLet me state a different problem, and your task is to relate how solving this problem will help in solving the actual one.\n> Given an array containing integers, your task is to find the length of the longest subarr... | 203 | You are given an integer array `nums` and an integer `x`. In one operation, you can either remove the leftmost or the rightmost element from the array `nums` and subtract its value from `x`. Note that this **modifies** the array for future operations.
Return _the **minimum number** of operations to reduce_ `x` _to **e... | For each row of the grid calculate the most right 1 in the grid in the array maxRight. To check if there exist answer, sort maxRight and check if maxRight[i] ≤ i for all possible i's. If there exist an answer, simulate the swaps. |
✅ [C++/Python] Simple Solution w/ Explanation | Sliding Window | minimum-operations-to-reduce-x-to-zero | 0 | 1 | Sometimes, converting a problem into some other familiar one helps a lot. This question is one of them.\nLet me state a different problem, and your task is to relate how solving this problem will help in solving the actual one.\n> Given an array containing integers, your task is to find the length of the longest subarr... | 203 | There are `n` cars traveling at different speeds in the same direction along a one-lane road. You are given an array `cars` of length `n`, where `cars[i] = [positioni, speedi]` represents:
* `positioni` is the distance between the `ith` car and the beginning of the road in meters. It is guaranteed that `positioni < ... | Think in reverse; instead of finding the minimum prefix + suffix, find the maximum subarray. Finding the maximum subarray is standard and can be done greedily. |
[Python3] top-down dp | maximize-grid-happiness | 0 | 1 | \n```\nclass Solution:\n def getMaxGridHappiness(self, m: int, n: int, introvertsCount: int, extrovertsCount: int) -> int:\n \n @cache\n def fn(prev, i, j, intro, extro): \n """Return max grid happiness at (i, j)."""\n if i == m: return 0 # no more position\n if ... | 3 | You are given four integers, `m`, `n`, `introvertsCount`, and `extrovertsCount`. You have an `m x n` grid, and there are two types of people: introverts and extroverts. There are `introvertsCount` introverts and `extrovertsCount` extroverts.
You should decide how many people you want to live in the grid and assign eac... | Partition the array by common integers, and choose the path with larger sum with a DP technique. |
[Python3] top-down dp | maximize-grid-happiness | 0 | 1 | \n```\nclass Solution:\n def getMaxGridHappiness(self, m: int, n: int, introvertsCount: int, extrovertsCount: int) -> int:\n \n @cache\n def fn(prev, i, j, intro, extro): \n """Return max grid happiness at (i, j)."""\n if i == m: return 0 # no more position\n if ... | 3 | This is an **interactive problem**.
There is a robot in a hidden grid, and you are trying to get it from its starting cell to the target cell in this grid. The grid is of size `m x n`, and each cell in the grid is either empty or blocked. It is **guaranteed** that the starting cell and the target cell are different, a... | For each cell, it has 3 options, either it is empty, or contains an introvert, or an extrovert. You can do DP where you maintain the state of the previous row, the number of remaining introverts and extroverts, the current row and column, and try the 3 options for each cell. Assume that the previous columns in the curr... |
ez C/C++/Python string||0ms Beats 100% | check-if-two-string-arrays-are-equivalent | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nString concatenation vs pointers\n# Approach\n<!-- Describe your approach to solving the problem. -->\nC code is also provided which is rare. No string copy is used for this solution.\n\nPython code is 1-line code.\n# Complexity\n- Time c... | 7 | Given two string arrays `word1` and `word2`, return `true` _if the two arrays **represent** the same string, and_ `false` _otherwise._
A string is **represented** by an array if the array elements concatenated **in order** forms the string.
**Example 1:**
**Input:** word1 = \[ "ab ", "c "\], word2 = \[ "a ", "bc "... | Work backwards: try to go from nums to arr. You should try to divide by 2 as much as possible, but you can only divide by 2 if everything is even. |
ez C/C++/Python string||0ms Beats 100% | check-if-two-string-arrays-are-equivalent | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nString concatenation vs pointers\n# Approach\n<!-- Describe your approach to solving the problem. -->\nC code is also provided which is rare. No string copy is used for this solution.\n\nPython code is 1-line code.\n# Complexity\n- Time c... | 7 | The **beauty** of a string is the difference in frequencies between the most frequent and least frequent characters.
* For example, the beauty of `"abaacc "` is `3 - 1 = 2`.
Given a string `s`, return _the sum of **beauty** of all of its substrings._
**Example 1:**
**Input:** s = "aabcb "
**Output:** 5
**Explana... | Concatenate all strings in the first array into a single string in the given order, the same for the second array. Both arrays represent the same string if and only if the generated strings are the same. |
✅☑[C++/Java/Python/JavaScript] || One-Line Code || 3 Approaches || Beats 100% || EXPLAINED🔥 | check-if-two-string-arrays-are-equivalent | 1 | 1 | # PLEASE UPVOTE IF IT HELPED\n\n---\n\n\n# Approaches\n**(Also explained in the code)**\n\n#### ***Approach 1(Built in Function)***\n\n1. Use built-in function accumulate to add the two string vectors.\n2. Compare them, if equal then true else false.\n\n# Complexity\n- *Time complexity:*\n $$O(word1.size() + word2.s... | 3 | Given two string arrays `word1` and `word2`, return `true` _if the two arrays **represent** the same string, and_ `false` _otherwise._
A string is **represented** by an array if the array elements concatenated **in order** forms the string.
**Example 1:**
**Input:** word1 = \[ "ab ", "c "\], word2 = \[ "a ", "bc "... | Work backwards: try to go from nums to arr. You should try to divide by 2 as much as possible, but you can only divide by 2 if everything is even. |
✅☑[C++/Java/Python/JavaScript] || One-Line Code || 3 Approaches || Beats 100% || EXPLAINED🔥 | check-if-two-string-arrays-are-equivalent | 1 | 1 | # PLEASE UPVOTE IF IT HELPED\n\n---\n\n\n# Approaches\n**(Also explained in the code)**\n\n#### ***Approach 1(Built in Function)***\n\n1. Use built-in function accumulate to add the two string vectors.\n2. Compare them, if equal then true else false.\n\n# Complexity\n- *Time complexity:*\n $$O(word1.size() + word2.s... | 3 | The **beauty** of a string is the difference in frequencies between the most frequent and least frequent characters.
* For example, the beauty of `"abaacc "` is `3 - 1 = 2`.
Given a string `s`, return _the sum of **beauty** of all of its substrings._
**Example 1:**
**Input:** s = "aabcb "
**Output:** 5
**Explana... | Concatenate all strings in the first array into a single string in the given order, the same for the second array. Both arrays represent the same string if and only if the generated strings are the same. |
Simple Single Line Code. | check-if-two-string-arrays-are-equivalent | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ --... | 3 | Given two string arrays `word1` and `word2`, return `true` _if the two arrays **represent** the same string, and_ `false` _otherwise._
A string is **represented** by an array if the array elements concatenated **in order** forms the string.
**Example 1:**
**Input:** word1 = \[ "ab ", "c "\], word2 = \[ "a ", "bc "... | Work backwards: try to go from nums to arr. You should try to divide by 2 as much as possible, but you can only divide by 2 if everything is even. |
Simple Single Line Code. | check-if-two-string-arrays-are-equivalent | 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. $$O(n)$$ --... | 3 | The **beauty** of a string is the difference in frequencies between the most frequent and least frequent characters.
* For example, the beauty of `"abaacc "` is `3 - 1 = 2`.
Given a string `s`, return _the sum of **beauty** of all of its substrings._
**Example 1:**
**Input:** s = "aabcb "
**Output:** 5
**Explana... | Concatenate all strings in the first array into a single string in the given order, the same for the second array. Both arrays represent the same string if and only if the generated strings are the same. |
[Java/C++/Python]Beats 100% || Well explained || | check-if-two-string-arrays-are-equivalent | 1 | 1 | # Approach\n1. **StringBuilder Initialization:**\n - **Why StringBuilder?** StringBuilder is used for efficient string concatenation. It allows you to build strings dynamically, especially when dealing with a sequence of concatenations.\n2. **Concatenation Using For Loops:**\n - **Why For Loops?** For loops are ... | 2 | Given two string arrays `word1` and `word2`, return `true` _if the two arrays **represent** the same string, and_ `false` _otherwise._
A string is **represented** by an array if the array elements concatenated **in order** forms the string.
**Example 1:**
**Input:** word1 = \[ "ab ", "c "\], word2 = \[ "a ", "bc "... | Work backwards: try to go from nums to arr. You should try to divide by 2 as much as possible, but you can only divide by 2 if everything is even. |
[Java/C++/Python]Beats 100% || Well explained || | check-if-two-string-arrays-are-equivalent | 1 | 1 | # Approach\n1. **StringBuilder Initialization:**\n - **Why StringBuilder?** StringBuilder is used for efficient string concatenation. It allows you to build strings dynamically, especially when dealing with a sequence of concatenations.\n2. **Concatenation Using For Loops:**\n - **Why For Loops?** For loops are ... | 2 | The **beauty** of a string is the difference in frequencies between the most frequent and least frequent characters.
* For example, the beauty of `"abaacc "` is `3 - 1 = 2`.
Given a string `s`, return _the sum of **beauty** of all of its substrings._
**Example 1:**
**Input:** s = "aabcb "
**Output:** 5
**Explana... | Concatenate all strings in the first array into a single string in the given order, the same for the second array. Both arrays represent the same string if and only if the generated strings are the same. |
Rust/Python/Go oneliners and then a normal solution in py | check-if-two-string-arrays-are-equivalent | 0 | 1 | # Intuition\nIn all high-level languages you can just use some sort of join to create strings and check if they are the same. This gives you an easy 1 line solution wich runs in $O(n + m)$\n\n```Rust []\nimpl Solution {\n pub fn array_strings_are_equal(word1: Vec<String>, word2: Vec<String>) -> bool {\n return word... | 2 | Given two string arrays `word1` and `word2`, return `true` _if the two arrays **represent** the same string, and_ `false` _otherwise._
A string is **represented** by an array if the array elements concatenated **in order** forms the string.
**Example 1:**
**Input:** word1 = \[ "ab ", "c "\], word2 = \[ "a ", "bc "... | Work backwards: try to go from nums to arr. You should try to divide by 2 as much as possible, but you can only divide by 2 if everything is even. |
Rust/Python/Go oneliners and then a normal solution in py | check-if-two-string-arrays-are-equivalent | 0 | 1 | # Intuition\nIn all high-level languages you can just use some sort of join to create strings and check if they are the same. This gives you an easy 1 line solution wich runs in $O(n + m)$\n\n```Rust []\nimpl Solution {\n pub fn array_strings_are_equal(word1: Vec<String>, word2: Vec<String>) -> bool {\n return word... | 2 | The **beauty** of a string is the difference in frequencies between the most frequent and least frequent characters.
* For example, the beauty of `"abaacc "` is `3 - 1 = 2`.
Given a string `s`, return _the sum of **beauty** of all of its substrings._
**Example 1:**
**Input:** s = "aabcb "
**Output:** 5
**Explana... | Concatenate all strings in the first array into a single string in the given order, the same for the second array. Both arrays represent the same string if and only if the generated strings are the same. |
✅✅|| ONE LINE || EASY PEASY || Java, Python, C#, Rust, Ruby || 🔥🔥🔥 | check-if-two-string-arrays-are-equivalent | 1 | 1 | # Description\n\n## Intuition\nThe problem seems to be about comparing two lists of strings and checking if the concatenation of strings in each list results in equal strings. The provided code uses the `join` method to concatenate the strings in both lists and then compares the resulting strings.\n\n## Approach\nThe a... | 33 | Given two string arrays `word1` and `word2`, return `true` _if the two arrays **represent** the same string, and_ `false` _otherwise._
A string is **represented** by an array if the array elements concatenated **in order** forms the string.
**Example 1:**
**Input:** word1 = \[ "ab ", "c "\], word2 = \[ "a ", "bc "... | Work backwards: try to go from nums to arr. You should try to divide by 2 as much as possible, but you can only divide by 2 if everything is even. |
✅✅|| ONE LINE || EASY PEASY || Java, Python, C#, Rust, Ruby || 🔥🔥🔥 | check-if-two-string-arrays-are-equivalent | 1 | 1 | # Description\n\n## Intuition\nThe problem seems to be about comparing two lists of strings and checking if the concatenation of strings in each list results in equal strings. The provided code uses the `join` method to concatenate the strings in both lists and then compares the resulting strings.\n\n## Approach\nThe a... | 33 | The **beauty** of a string is the difference in frequencies between the most frequent and least frequent characters.
* For example, the beauty of `"abaacc "` is `3 - 1 = 2`.
Given a string `s`, return _the sum of **beauty** of all of its substrings._
**Example 1:**
**Input:** s = "aabcb "
**Output:** 5
**Explana... | Concatenate all strings in the first array into a single string in the given order, the same for the second array. Both arrays represent the same string if and only if the generated strings are the same. |
Python | Zip Chains, O(1) Space | check-if-two-string-arrays-are-equivalent | 0 | 1 | # Intuition\n\nThe focus of this problem is on developing a solution with $$O(1)$$ space usage. In most languages, this requires maintaining pointers manually, but not in Python (#blessed). \n\n# Approach\n\nThe following solution is my favorite:\n```py\nclass Solution:\n def arrayStringsAreEqual(self, word1: list[s... | 6 | Given two string arrays `word1` and `word2`, return `true` _if the two arrays **represent** the same string, and_ `false` _otherwise._
A string is **represented** by an array if the array elements concatenated **in order** forms the string.
**Example 1:**
**Input:** word1 = \[ "ab ", "c "\], word2 = \[ "a ", "bc "... | Work backwards: try to go from nums to arr. You should try to divide by 2 as much as possible, but you can only divide by 2 if everything is even. |
Python | Zip Chains, O(1) Space | check-if-two-string-arrays-are-equivalent | 0 | 1 | # Intuition\n\nThe focus of this problem is on developing a solution with $$O(1)$$ space usage. In most languages, this requires maintaining pointers manually, but not in Python (#blessed). \n\n# Approach\n\nThe following solution is my favorite:\n```py\nclass Solution:\n def arrayStringsAreEqual(self, word1: list[s... | 6 | The **beauty** of a string is the difference in frequencies between the most frequent and least frequent characters.
* For example, the beauty of `"abaacc "` is `3 - 1 = 2`.
Given a string `s`, return _the sum of **beauty** of all of its substrings._
**Example 1:**
**Input:** s = "aabcb "
**Output:** 5
**Explana... | Concatenate all strings in the first array into a single string in the given order, the same for the second array. Both arrays represent the same string if and only if the generated strings are the same. |
Easy Python Solution || One liner.......... || Beginner Friendly | check-if-two-string-arrays-are-equivalent | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nThe intuition behind this function is to check if two lists of strings, word1 and word2, would be equal when their elements are concatenated into single strings.\n# Approach\n<!-- Describe your approach to solving the problem. -->\n- The ... | 1 | Given two string arrays `word1` and `word2`, return `true` _if the two arrays **represent** the same string, and_ `false` _otherwise._
A string is **represented** by an array if the array elements concatenated **in order** forms the string.
**Example 1:**
**Input:** word1 = \[ "ab ", "c "\], word2 = \[ "a ", "bc "... | Work backwards: try to go from nums to arr. You should try to divide by 2 as much as possible, but you can only divide by 2 if everything is even. |
Easy Python Solution || One liner.......... || Beginner Friendly | check-if-two-string-arrays-are-equivalent | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nThe intuition behind this function is to check if two lists of strings, word1 and word2, would be equal when their elements are concatenated into single strings.\n# Approach\n<!-- Describe your approach to solving the problem. -->\n- The ... | 1 | The **beauty** of a string is the difference in frequencies between the most frequent and least frequent characters.
* For example, the beauty of `"abaacc "` is `3 - 1 = 2`.
Given a string `s`, return _the sum of **beauty** of all of its substrings._
**Example 1:**
**Input:** s = "aabcb "
**Output:** 5
**Explana... | Concatenate all strings in the first array into a single string in the given order, the same for the second array. Both arrays represent the same string if and only if the generated strings are the same. |
🔥 Easy solution | JAVA | Python 3 🔥| | check-if-two-string-arrays-are-equivalent | 1 | 1 | # Intuition\nCheck If Two String Arrays are Equivalent\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n**String.join:** \n 1. This method concatenates all the strings in the array into a single string without any delimiter. \n 2. It\'s used here to create two strings, joinedWor... | 1 | Given two string arrays `word1` and `word2`, return `true` _if the two arrays **represent** the same string, and_ `false` _otherwise._
A string is **represented** by an array if the array elements concatenated **in order** forms the string.
**Example 1:**
**Input:** word1 = \[ "ab ", "c "\], word2 = \[ "a ", "bc "... | Work backwards: try to go from nums to arr. You should try to divide by 2 as much as possible, but you can only divide by 2 if everything is even. |
🔥 Easy solution | JAVA | Python 3 🔥| | check-if-two-string-arrays-are-equivalent | 1 | 1 | # Intuition\nCheck If Two String Arrays are Equivalent\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n**String.join:** \n 1. This method concatenates all the strings in the array into a single string without any delimiter. \n 2. It\'s used here to create two strings, joinedWor... | 1 | The **beauty** of a string is the difference in frequencies between the most frequent and least frequent characters.
* For example, the beauty of `"abaacc "` is `3 - 1 = 2`.
Given a string `s`, return _the sum of **beauty** of all of its substrings._
**Example 1:**
**Input:** s = "aabcb "
**Output:** 5
**Explana... | Concatenate all strings in the first array into a single string in the given order, the same for the second array. Both arrays represent the same string if and only if the generated strings are the same. |
O(1) Space | no while loop | easy to understand | check-if-two-string-arrays-are-equivalent | 0 | 1 | # Complexity\n- Time complexity: $$O(N*K)$$\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity: $$O(1)$$\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\nclass Solution:\n def arrayStringsAreEqual(self, word1: List[str], word2: List[str]) -> bool:\n p2, i2 = 0,... | 1 | Given two string arrays `word1` and `word2`, return `true` _if the two arrays **represent** the same string, and_ `false` _otherwise._
A string is **represented** by an array if the array elements concatenated **in order** forms the string.
**Example 1:**
**Input:** word1 = \[ "ab ", "c "\], word2 = \[ "a ", "bc "... | Work backwards: try to go from nums to arr. You should try to divide by 2 as much as possible, but you can only divide by 2 if everything is even. |
O(1) Space | no while loop | easy to understand | check-if-two-string-arrays-are-equivalent | 0 | 1 | # Complexity\n- Time complexity: $$O(N*K)$$\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity: $$O(1)$$\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\nclass Solution:\n def arrayStringsAreEqual(self, word1: List[str], word2: List[str]) -> bool:\n p2, i2 = 0,... | 1 | The **beauty** of a string is the difference in frequencies between the most frequent and least frequent characters.
* For example, the beauty of `"abaacc "` is `3 - 1 = 2`.
Given a string `s`, return _the sum of **beauty** of all of its substrings._
**Example 1:**
**Input:** s = "aabcb "
**Output:** 5
**Explana... | Concatenate all strings in the first array into a single string in the given order, the same for the second array. Both arrays represent the same string if and only if the generated strings are the same. |
Simple one line python solution | check-if-two-string-arrays-are-equivalent | 0 | 1 | # Code\n```\nclass Solution:\n def arrayStringsAreEqual(self, word1: List[str], word2: List[str]) -> bool:\n return True if \'\'.join(word1)==\'\'.join(word2) else False\n``` | 1 | Given two string arrays `word1` and `word2`, return `true` _if the two arrays **represent** the same string, and_ `false` _otherwise._
A string is **represented** by an array if the array elements concatenated **in order** forms the string.
**Example 1:**
**Input:** word1 = \[ "ab ", "c "\], word2 = \[ "a ", "bc "... | Work backwards: try to go from nums to arr. You should try to divide by 2 as much as possible, but you can only divide by 2 if everything is even. |
Simple one line python solution | check-if-two-string-arrays-are-equivalent | 0 | 1 | # Code\n```\nclass Solution:\n def arrayStringsAreEqual(self, word1: List[str], word2: List[str]) -> bool:\n return True if \'\'.join(word1)==\'\'.join(word2) else False\n``` | 1 | The **beauty** of a string is the difference in frequencies between the most frequent and least frequent characters.
* For example, the beauty of `"abaacc "` is `3 - 1 = 2`.
Given a string `s`, return _the sum of **beauty** of all of its substrings._
**Example 1:**
**Input:** s = "aabcb "
**Output:** 5
**Explana... | Concatenate all strings in the first array into a single string in the given order, the same for the second array. Both arrays represent the same string if and only if the generated strings are the same. |
✔️ [Python3] GREEDY FILLING (🌸¬‿¬), Explained | smallest-string-with-a-given-numeric-value | 0 | 1 | **UPVOTE if you like (\uD83C\uDF38\u25E0\u203F\u25E0), If you have any question, feel free to ask.**\n\nSince we are forming the lexicographically smallest string, we just simply fill our result with `a`s ( o\u02D8\u25E1\u02D8o). But hold on, that result will not necessarily have the required score c(\uFF9F.\uFF9F*) U... | 64 | The **numeric value** of a **lowercase character** is defined as its position `(1-indexed)` in the alphabet, so the numeric value of `a` is `1`, the numeric value of `b` is `2`, the numeric value of `c` is `3`, and so on.
The **numeric value** of a **string** consisting of lowercase characters is defined as the sum of... | Keep track of the parent (previous position) to avoid considering an invalid path. Use DFS or BFS and keep track of visited cells to see if there is a cycle. |
✔️ [Python3] GREEDY FILLING (🌸¬‿¬), Explained | smallest-string-with-a-given-numeric-value | 0 | 1 | **UPVOTE if you like (\uD83C\uDF38\u25E0\u203F\u25E0), If you have any question, feel free to ask.**\n\nSince we are forming the lexicographically smallest string, we just simply fill our result with `a`s ( o\u02D8\u25E1\u02D8o). But hold on, that result will not necessarily have the required score c(\uFF9F.\uFF9F*) U... | 64 | You are given an undirected graph defined by an integer `n`, the number of nodes, and a 2D integer array `edges`, the edges in the graph, where `edges[i] = [ui, vi]` indicates that there is an **undirected** edge between `ui` and `vi`. You are also given an integer array `queries`.
Let `incident(a, b)` be defined as t... | Think greedily. If you build the string from the end to the beginning, it will always be optimal to put the highest possible character at the current index. |
[Java/Python 3] Two O(n) codes w/ brief explanation and analysis. | smallest-string-with-a-given-numeric-value | 1 | 1 | **Mehtod 1: Greedily reversely place values**\n\n1. Make sure each value of the n characters is at least `1`: initialized all as `\'a\'`;\n2. Put as more value at the end of the String as possible.\n```java\n public String getSmallestString(int n, int k) {\n k -= n;\n char[] ans = new char[n];\n ... | 63 | The **numeric value** of a **lowercase character** is defined as its position `(1-indexed)` in the alphabet, so the numeric value of `a` is `1`, the numeric value of `b` is `2`, the numeric value of `c` is `3`, and so on.
The **numeric value** of a **string** consisting of lowercase characters is defined as the sum of... | Keep track of the parent (previous position) to avoid considering an invalid path. Use DFS or BFS and keep track of visited cells to see if there is a cycle. |
[Java/Python 3] Two O(n) codes w/ brief explanation and analysis. | smallest-string-with-a-given-numeric-value | 1 | 1 | **Mehtod 1: Greedily reversely place values**\n\n1. Make sure each value of the n characters is at least `1`: initialized all as `\'a\'`;\n2. Put as more value at the end of the String as possible.\n```java\n public String getSmallestString(int n, int k) {\n k -= n;\n char[] ans = new char[n];\n ... | 63 | You are given an undirected graph defined by an integer `n`, the number of nodes, and a 2D integer array `edges`, the edges in the graph, where `edges[i] = [ui, vi]` indicates that there is an **undirected** edge between `ui` and `vi`. You are also given an integer array `queries`.
Let `incident(a, b)` be defined as t... | Think greedily. If you build the string from the end to the beginning, it will always be optimal to put the highest possible character at the current index. |
[Python] Simple solution - Greedy - O(N) [Explanation + Code + Comment] | smallest-string-with-a-given-numeric-value | 0 | 1 | The lexicographically smallest string of length n could be \'aaa\'.. of length n\nThe constraint is that the numeric value must be k.\nTo satisfy this constraint, we will greedily add either \'z\' or character form of (k - value) to the string.\n\nFor eg. \nn = 3, k = 32\nInitially, ans = \'aaa\', val = 3, i =2\nSince ... | 17 | The **numeric value** of a **lowercase character** is defined as its position `(1-indexed)` in the alphabet, so the numeric value of `a` is `1`, the numeric value of `b` is `2`, the numeric value of `c` is `3`, and so on.
The **numeric value** of a **string** consisting of lowercase characters is defined as the sum of... | Keep track of the parent (previous position) to avoid considering an invalid path. Use DFS or BFS and keep track of visited cells to see if there is a cycle. |
[Python] Simple solution - Greedy - O(N) [Explanation + Code + Comment] | smallest-string-with-a-given-numeric-value | 0 | 1 | The lexicographically smallest string of length n could be \'aaa\'.. of length n\nThe constraint is that the numeric value must be k.\nTo satisfy this constraint, we will greedily add either \'z\' or character form of (k - value) to the string.\n\nFor eg. \nn = 3, k = 32\nInitially, ans = \'aaa\', val = 3, i =2\nSince ... | 17 | You are given an undirected graph defined by an integer `n`, the number of nodes, and a 2D integer array `edges`, the edges in the graph, where `edges[i] = [ui, vi]` indicates that there is an **undirected** edge between `ui` and `vi`. You are also given an integer array `queries`.
Let `incident(a, b)` be defined as t... | Think greedily. If you build the string from the end to the beginning, it will always be optimal to put the highest possible character at the current index. |
✅ Python Solution | smallest-string-with-a-given-numeric-value | 0 | 1 | ```\nclass Solution:\n def getSmallestString(self, n: int, k: int) -> str:\n ans = \'\'\n while (n - 1) * 26 >= k:\n ans += \'a\'\n n -= 1; k -= 1\n ans += chr(ord(\'a\') + (k % 26 or 26) - 1)\n ans += \'z\' * (n - 1)\n return ans\n``` | 3 | The **numeric value** of a **lowercase character** is defined as its position `(1-indexed)` in the alphabet, so the numeric value of `a` is `1`, the numeric value of `b` is `2`, the numeric value of `c` is `3`, and so on.
The **numeric value** of a **string** consisting of lowercase characters is defined as the sum of... | Keep track of the parent (previous position) to avoid considering an invalid path. Use DFS or BFS and keep track of visited cells to see if there is a cycle. |
✅ Python Solution | smallest-string-with-a-given-numeric-value | 0 | 1 | ```\nclass Solution:\n def getSmallestString(self, n: int, k: int) -> str:\n ans = \'\'\n while (n - 1) * 26 >= k:\n ans += \'a\'\n n -= 1; k -= 1\n ans += chr(ord(\'a\') + (k % 26 or 26) - 1)\n ans += \'z\' * (n - 1)\n return ans\n``` | 3 | You are given an undirected graph defined by an integer `n`, the number of nodes, and a 2D integer array `edges`, the edges in the graph, where `edges[i] = [ui, vi]` indicates that there is an **undirected** edge between `ui` and `vi`. You are also given an integer array `queries`.
Let `incident(a, b)` be defined as t... | Think greedily. If you build the string from the end to the beginning, it will always be optimal to put the highest possible character at the current index. |
[Python3] O(n) time and O(1) space complexity solution | smallest-string-with-a-given-numeric-value | 0 | 1 | ``` python3\nclass Solution:\n def getSmallestString(self, n: int, k: int) -> str:\n result =[0]*n\n for position in range(n-1,-1,-1):\n add = min(k -position,26)\n result[position] = chr(ord("a")+add -1)\n k-=add\n \n return "".join(result)\n```\n\nCo... | 3 | The **numeric value** of a **lowercase character** is defined as its position `(1-indexed)` in the alphabet, so the numeric value of `a` is `1`, the numeric value of `b` is `2`, the numeric value of `c` is `3`, and so on.
The **numeric value** of a **string** consisting of lowercase characters is defined as the sum of... | Keep track of the parent (previous position) to avoid considering an invalid path. Use DFS or BFS and keep track of visited cells to see if there is a cycle. |
[Python3] O(n) time and O(1) space complexity solution | smallest-string-with-a-given-numeric-value | 0 | 1 | ``` python3\nclass Solution:\n def getSmallestString(self, n: int, k: int) -> str:\n result =[0]*n\n for position in range(n-1,-1,-1):\n add = min(k -position,26)\n result[position] = chr(ord("a")+add -1)\n k-=add\n \n return "".join(result)\n```\n\nCo... | 3 | You are given an undirected graph defined by an integer `n`, the number of nodes, and a 2D integer array `edges`, the edges in the graph, where `edges[i] = [ui, vi]` indicates that there is an **undirected** edge between `ui` and `vi`. You are also given an integer array `queries`.
Let `incident(a, b)` be defined as t... | Think greedily. If you build the string from the end to the beginning, it will always be optimal to put the highest possible character at the current index. |
Easy to understand with comments | smallest-string-with-a-given-numeric-value | 0 | 1 | we start from the back , keep adding the highest letter possible to the result , highest possible letter is k-(remaining spaces)(remaining as we subtract the value of k by the value of each letter added, the remaining spaces are n - current length of result)\n```\ndef getSmallestString(self, n: int, k: int) -> str:\n ... | 2 | The **numeric value** of a **lowercase character** is defined as its position `(1-indexed)` in the alphabet, so the numeric value of `a` is `1`, the numeric value of `b` is `2`, the numeric value of `c` is `3`, and so on.
The **numeric value** of a **string** consisting of lowercase characters is defined as the sum of... | Keep track of the parent (previous position) to avoid considering an invalid path. Use DFS or BFS and keep track of visited cells to see if there is a cycle. |
Easy to understand with comments | smallest-string-with-a-given-numeric-value | 0 | 1 | we start from the back , keep adding the highest letter possible to the result , highest possible letter is k-(remaining spaces)(remaining as we subtract the value of k by the value of each letter added, the remaining spaces are n - current length of result)\n```\ndef getSmallestString(self, n: int, k: int) -> str:\n ... | 2 | You are given an undirected graph defined by an integer `n`, the number of nodes, and a 2D integer array `edges`, the edges in the graph, where `edges[i] = [ui, vi]` indicates that there is an **undirected** edge between `ui` and `vi`. You are also given an integer array `queries`.
Let `incident(a, b)` be defined as t... | Think greedily. If you build the string from the end to the beginning, it will always be optimal to put the highest possible character at the current index. |
Easy python solution of O(n) TC | ways-to-make-a-fair-array | 0 | 1 | ```\ndef waysToMakeFair(self, nums: List[int]) -> int:\n\teven, odd = [0], [0]\n\tfor i in range(len(nums)):\n\t\tif(i % 2):\n\t\t\todd.append(odd[-1] + nums[i])\n\t\t\teven.append(even[-1])\n\t\telse:\n\t\t\teven.append(even[-1] + nums[i])\n\t\t\todd.append(odd[-1])\n\tans = 0\n\tfor i in range(1, len(nums)+1):\n\t\te... | 2 | You are given an integer array `nums`. You can choose **exactly one** index (**0-indexed**) and remove the element. Notice that the index of the elements may change after the removal.
For example, if `nums = [6,1,7,4,1]`:
* Choosing to remove index `1` results in `nums = [6,7,4,1]`.
* Choosing to remove index `2`... | null |
WEEB EXPLAINS PYTHON/C++ DP/PREFIX SUM SOLN | ways-to-make-a-fair-array | 0 | 1 | My initial sketch for the question:\n\t\n\nHere is my explanation for my messy sketch:\n\t\n\tCorrect... | 6 | You are given an integer array `nums`. You can choose **exactly one** index (**0-indexed**) and remove the element. Notice that the index of the elements may change after the removal.
For example, if `nums = [6,1,7,4,1]`:
* Choosing to remove index `1` results in `nums = [6,7,4,1]`.
* Choosing to remove index `2`... | null |
[Python3] O(N) - Simple code + Explanation | ways-to-make-a-fair-array | 0 | 1 | The key thing to notice is that when you remove an element at index i, the odd sum and even sum for elements less that i remains unchanged and the odd sum and even sum for elements greater than i switches.\n\nAfter removing element at index i, array is fair if odd_sum of elements greater than i + even_sum of elements l... | 7 | You are given an integer array `nums`. You can choose **exactly one** index (**0-indexed**) and remove the element. Notice that the index of the elements may change after the removal.
For example, if `nums = [6,1,7,4,1]`:
* Choosing to remove index `1` results in `nums = [6,7,4,1]`.
* Choosing to remove index `2`... | null |
✔ Python3 Solution | Clean & Concise | ways-to-make-a-fair-array | 0 | 1 | # Complexity\n- Time complexity: $$O(n)$$\n- Space complexity: $$O(1)$$\n\n# Code\n```Python\nclass Solution:\n def waysToMakeFair(self, A):\n ans, cur = 0, sum(A[::2]) - sum(A[1::2])\n for i in A:\n if i == cur: ans += 1\n cur = (i << 1) - cur\n return ans\n``` | 2 | You are given an integer array `nums`. You can choose **exactly one** index (**0-indexed**) and remove the element. Notice that the index of the elements may change after the removal.
For example, if `nums = [6,1,7,4,1]`:
* Choosing to remove index `1` results in `nums = [6,7,4,1]`.
* Choosing to remove index `2`... | null |
intuitive prefix sum approach explained | ways-to-make-a-fair-array | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nWe can see that when we remove an index, all elements AFTER that original index now swap parity (even or oddness). even becomes odd, odd becomes even.\nWe can use this to help us.\n\n# Approach\n<!-- Describe your approach to solving the ... | 0 | You are given an integer array `nums`. You can choose **exactly one** index (**0-indexed**) and remove the element. Notice that the index of the elements may change after the removal.
For example, if `nums = [6,1,7,4,1]`:
* Choosing to remove index `1` results in `nums = [6,7,4,1]`.
* Choosing to remove index `2`... | null |
Beats 99.63%, "Efficient Energy Management: The Task of Minimum Effort" | minimum-initial-energy-to-finish-tasks | 0 | 1 | # Intuition\n The idea is to sort the tasks based on the difference between the actual and minimum energy required.\n This way, we prioritize tasks with a larger energy difference, making it easier to manage energy.\n\n# Approach\n1. Sort the tasks based on the difference between actual and minimum energy.\n 2. Initial... | 0 | You are given an array `tasks` where `tasks[i] = [actuali, minimumi]`:
* `actuali` is the actual amount of energy you **spend to finish** the `ith` task.
* `minimumi` is the minimum amount of energy you **require to begin** the `ith` task.
For example, if the task is `[10, 12]` and your current energy is `11`, yo... | For the node i, calculate the height of each of its children and keep the first and second maximum heights (max1_i , max2_i). Check all nodes and return max( 2 + max1_i + max2_i ). |
Beats 99.63%, "Efficient Energy Management: The Task of Minimum Effort" | minimum-initial-energy-to-finish-tasks | 0 | 1 | # Intuition\n The idea is to sort the tasks based on the difference between the actual and minimum energy required.\n This way, we prioritize tasks with a larger energy difference, making it easier to manage energy.\n\n# Approach\n1. Sort the tasks based on the difference between actual and minimum energy.\n 2. Initial... | 0 | Given a binary string `s` **without leading zeros**, return `true` _if_ `s` _contains **at most one contiguous segment of ones**_. Otherwise, return `false`.
**Example 1:**
**Input:** s = "1001 "
**Output:** false
**Explanation:** The ones do not form a contiguous segment.
**Example 2:**
**Input:** s = "1... | We can easily figure that the f(x) : does x solve this array is monotonic so binary Search is doable Figure a sorting pattern |
Beats 95%, "Efficient Energy Management: The Task of Minimum Effort" | minimum-initial-energy-to-finish-tasks | 0 | 1 | # Intuition\n The idea is to sort the tasks based on the difference between the actual and minimum energy required.\n\n This way, we prioritize tasks with a larger energy difference, making it easier to manage energy.\n\n\n# Approach\n1. Sort the tasks based on the difference between actual and minimum energy.\n2. Init... | 0 | You are given an array `tasks` where `tasks[i] = [actuali, minimumi]`:
* `actuali` is the actual amount of energy you **spend to finish** the `ith` task.
* `minimumi` is the minimum amount of energy you **require to begin** the `ith` task.
For example, if the task is `[10, 12]` and your current energy is `11`, yo... | For the node i, calculate the height of each of its children and keep the first and second maximum heights (max1_i , max2_i). Check all nodes and return max( 2 + max1_i + max2_i ). |
Beats 95%, "Efficient Energy Management: The Task of Minimum Effort" | minimum-initial-energy-to-finish-tasks | 0 | 1 | # Intuition\n The idea is to sort the tasks based on the difference between the actual and minimum energy required.\n\n This way, we prioritize tasks with a larger energy difference, making it easier to manage energy.\n\n\n# Approach\n1. Sort the tasks based on the difference between actual and minimum energy.\n2. Init... | 0 | Given a binary string `s` **without leading zeros**, return `true` _if_ `s` _contains **at most one contiguous segment of ones**_. Otherwise, return `false`.
**Example 1:**
**Input:** s = "1001 "
**Output:** false
**Explanation:** The ones do not form a contiguous segment.
**Example 2:**
**Input:** s = "1... | We can easily figure that the f(x) : does x solve this array is monotonic so binary Search is doable Figure a sorting pattern |
sorting + binary search | minimum-initial-energy-to-finish-tasks | 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 `tasks` where `tasks[i] = [actuali, minimumi]`:
* `actuali` is the actual amount of energy you **spend to finish** the `ith` task.
* `minimumi` is the minimum amount of energy you **require to begin** the `ith` task.
For example, if the task is `[10, 12]` and your current energy is `11`, yo... | For the node i, calculate the height of each of its children and keep the first and second maximum heights (max1_i , max2_i). Check all nodes and return max( 2 + max1_i + max2_i ). |
sorting + binary search | minimum-initial-energy-to-finish-tasks | 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 binary string `s` **without leading zeros**, return `true` _if_ `s` _contains **at most one contiguous segment of ones**_. Otherwise, return `false`.
**Example 1:**
**Input:** s = "1001 "
**Output:** false
**Explanation:** The ones do not form a contiguous segment.
**Example 2:**
**Input:** s = "1... | We can easily figure that the f(x) : does x solve this array is monotonic so binary Search is doable Figure a sorting pattern |
Heap solution Python | minimum-initial-energy-to-finish-tasks | 0 | 1 | # Intuition\nThink about dont want to waste any resource as much as possible => start with a task that has minimum gap => least time waste. In this part, you can either sort by (start-end) or just put that into heap. The rest of the algorithm is explained below\n<!-- Describe your first thoughts on how to solve this pr... | 0 | You are given an array `tasks` where `tasks[i] = [actuali, minimumi]`:
* `actuali` is the actual amount of energy you **spend to finish** the `ith` task.
* `minimumi` is the minimum amount of energy you **require to begin** the `ith` task.
For example, if the task is `[10, 12]` and your current energy is `11`, yo... | For the node i, calculate the height of each of its children and keep the first and second maximum heights (max1_i , max2_i). Check all nodes and return max( 2 + max1_i + max2_i ). |
Heap solution Python | minimum-initial-energy-to-finish-tasks | 0 | 1 | # Intuition\nThink about dont want to waste any resource as much as possible => start with a task that has minimum gap => least time waste. In this part, you can either sort by (start-end) or just put that into heap. The rest of the algorithm is explained below\n<!-- Describe your first thoughts on how to solve this pr... | 0 | Given a binary string `s` **without leading zeros**, return `true` _if_ `s` _contains **at most one contiguous segment of ones**_. Otherwise, return `false`.
**Example 1:**
**Input:** s = "1001 "
**Output:** false
**Explanation:** The ones do not form a contiguous segment.
**Example 2:**
**Input:** s = "1... | We can easily figure that the f(x) : does x solve this array is monotonic so binary Search is doable Figure a sorting pattern |
python | minimum-initial-energy-to-finish-tasks | 0 | 1 | \n\n# Code\n```\nclass Solution:\n def minimumEffort(self, tasks: List[List[int]]) -> int:\n\n m = max(y for x, y in tasks)\n n = m + sum(x for x, y in tasks)\n res = float(\'inf\')\n\n # binary search between m, n\n # print(n, m)\n\n tasks.sort(key =lambda x: x[1]-x[0], rev... | 0 | You are given an array `tasks` where `tasks[i] = [actuali, minimumi]`:
* `actuali` is the actual amount of energy you **spend to finish** the `ith` task.
* `minimumi` is the minimum amount of energy you **require to begin** the `ith` task.
For example, if the task is `[10, 12]` and your current energy is `11`, yo... | For the node i, calculate the height of each of its children and keep the first and second maximum heights (max1_i , max2_i). Check all nodes and return max( 2 + max1_i + max2_i ). |
python | minimum-initial-energy-to-finish-tasks | 0 | 1 | \n\n# Code\n```\nclass Solution:\n def minimumEffort(self, tasks: List[List[int]]) -> int:\n\n m = max(y for x, y in tasks)\n n = m + sum(x for x, y in tasks)\n res = float(\'inf\')\n\n # binary search between m, n\n # print(n, m)\n\n tasks.sort(key =lambda x: x[1]-x[0], rev... | 0 | Given a binary string `s` **without leading zeros**, return `true` _if_ `s` _contains **at most one contiguous segment of ones**_. Otherwise, return `false`.
**Example 1:**
**Input:** s = "1001 "
**Output:** false
**Explanation:** The ones do not form a contiguous segment.
**Example 2:**
**Input:** s = "1... | We can easily figure that the f(x) : does x solve this array is monotonic so binary Search is doable Figure a sorting pattern |
[Python3] Simple Greedy Solution + Sorting | minimum-initial-energy-to-finish-tasks | 0 | 1 | # Intuition\nWe want to greedily choose the tasks - this should be clear since there exists an optimal ordering that will give us a minimal result, which is determined by some sorted property.\n\n# Approach\nAt first I tried sorting by required energy, but this doesn\'t work because the actual energy could be close to ... | 0 | You are given an array `tasks` where `tasks[i] = [actuali, minimumi]`:
* `actuali` is the actual amount of energy you **spend to finish** the `ith` task.
* `minimumi` is the minimum amount of energy you **require to begin** the `ith` task.
For example, if the task is `[10, 12]` and your current energy is `11`, yo... | For the node i, calculate the height of each of its children and keep the first and second maximum heights (max1_i , max2_i). Check all nodes and return max( 2 + max1_i + max2_i ). |
[Python3] Simple Greedy Solution + Sorting | minimum-initial-energy-to-finish-tasks | 0 | 1 | # Intuition\nWe want to greedily choose the tasks - this should be clear since there exists an optimal ordering that will give us a minimal result, which is determined by some sorted property.\n\n# Approach\nAt first I tried sorting by required energy, but this doesn\'t work because the actual energy could be close to ... | 0 | Given a binary string `s` **without leading zeros**, return `true` _if_ `s` _contains **at most one contiguous segment of ones**_. Otherwise, return `false`.
**Example 1:**
**Input:** s = "1001 "
**Output:** false
**Explanation:** The ones do not form a contiguous segment.
**Example 2:**
**Input:** s = "1... | We can easily figure that the f(x) : does x solve this array is monotonic so binary Search is doable Figure a sorting pattern |
Perfection | minimum-initial-energy-to-finish-tasks | 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 `tasks` where `tasks[i] = [actuali, minimumi]`:
* `actuali` is the actual amount of energy you **spend to finish** the `ith` task.
* `minimumi` is the minimum amount of energy you **require to begin** the `ith` task.
For example, if the task is `[10, 12]` and your current energy is `11`, yo... | For the node i, calculate the height of each of its children and keep the first and second maximum heights (max1_i , max2_i). Check all nodes and return max( 2 + max1_i + max2_i ). |
Perfection | minimum-initial-energy-to-finish-tasks | 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 binary string `s` **without leading zeros**, return `true` _if_ `s` _contains **at most one contiguous segment of ones**_. Otherwise, return `false`.
**Example 1:**
**Input:** s = "1001 "
**Output:** false
**Explanation:** The ones do not form a contiguous segment.
**Example 2:**
**Input:** s = "1... | We can easily figure that the f(x) : does x solve this array is monotonic so binary Search is doable Figure a sorting pattern |
Binary search solution | minimum-initial-energy-to-finish-tasks | 0 | 1 | # Code\n```\nclass Solution:\n def minimumEffort(self, tasks: List[List[int]]) -> int:\n def check(ans):\n for i in tasks:\n if ans >= i[1] :\n ans -= i[0]\n else :\n return False\n break\n return True... | 0 | You are given an array `tasks` where `tasks[i] = [actuali, minimumi]`:
* `actuali` is the actual amount of energy you **spend to finish** the `ith` task.
* `minimumi` is the minimum amount of energy you **require to begin** the `ith` task.
For example, if the task is `[10, 12]` and your current energy is `11`, yo... | For the node i, calculate the height of each of its children and keep the first and second maximum heights (max1_i , max2_i). Check all nodes and return max( 2 + max1_i + max2_i ). |
Binary search solution | minimum-initial-energy-to-finish-tasks | 0 | 1 | # Code\n```\nclass Solution:\n def minimumEffort(self, tasks: List[List[int]]) -> int:\n def check(ans):\n for i in tasks:\n if ans >= i[1] :\n ans -= i[0]\n else :\n return False\n break\n return True... | 0 | Given a binary string `s` **without leading zeros**, return `true` _if_ `s` _contains **at most one contiguous segment of ones**_. Otherwise, return `false`.
**Example 1:**
**Input:** s = "1001 "
**Output:** false
**Explanation:** The ones do not form a contiguous segment.
**Example 2:**
**Input:** s = "1... | We can easily figure that the f(x) : does x solve this array is monotonic so binary Search is doable Figure a sorting pattern |
Python (Simple Maths) | minimum-initial-energy-to-finish-tasks | 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 `tasks` where `tasks[i] = [actuali, minimumi]`:
* `actuali` is the actual amount of energy you **spend to finish** the `ith` task.
* `minimumi` is the minimum amount of energy you **require to begin** the `ith` task.
For example, if the task is `[10, 12]` and your current energy is `11`, yo... | For the node i, calculate the height of each of its children and keep the first and second maximum heights (max1_i , max2_i). Check all nodes and return max( 2 + max1_i + max2_i ). |
Python (Simple Maths) | minimum-initial-energy-to-finish-tasks | 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 binary string `s` **without leading zeros**, return `true` _if_ `s` _contains **at most one contiguous segment of ones**_. Otherwise, return `false`.
**Example 1:**
**Input:** s = "1001 "
**Output:** false
**Explanation:** The ones do not form a contiguous segment.
**Example 2:**
**Input:** s = "1... | We can easily figure that the f(x) : does x solve this array is monotonic so binary Search is doable Figure a sorting pattern |
Python3 solution using merge sort | minimum-initial-energy-to-finish-tasks | 0 | 1 | \n# Approach\n### 1. Sort the algorithm using merge sort so that the arrays inside the tasks array have descending differences between the actual energy and the minimum energy\n### 2. Add each of the differences between the minimum between the arrays to the first minimum\n\n# Complexity\n- Time complexity:\n$$O(nlog(n)... | 0 | You are given an array `tasks` where `tasks[i] = [actuali, minimumi]`:
* `actuali` is the actual amount of energy you **spend to finish** the `ith` task.
* `minimumi` is the minimum amount of energy you **require to begin** the `ith` task.
For example, if the task is `[10, 12]` and your current energy is `11`, yo... | For the node i, calculate the height of each of its children and keep the first and second maximum heights (max1_i , max2_i). Check all nodes and return max( 2 + max1_i + max2_i ). |
Python3 solution using merge sort | minimum-initial-energy-to-finish-tasks | 0 | 1 | \n# Approach\n### 1. Sort the algorithm using merge sort so that the arrays inside the tasks array have descending differences between the actual energy and the minimum energy\n### 2. Add each of the differences between the minimum between the arrays to the first minimum\n\n# Complexity\n- Time complexity:\n$$O(nlog(n)... | 0 | Given a binary string `s` **without leading zeros**, return `true` _if_ `s` _contains **at most one contiguous segment of ones**_. Otherwise, return `false`.
**Example 1:**
**Input:** s = "1001 "
**Output:** false
**Explanation:** The ones do not form a contiguous segment.
**Example 2:**
**Input:** s = "1... | We can easily figure that the f(x) : does x solve this array is monotonic so binary Search is doable Figure a sorting pattern |
O(n * log(n)) solution | minimum-initial-energy-to-finish-tasks | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nThe goal of this problem is to minimize the effort required to complete a given list of tasks. Each task has a minimum and maximum effort required to complete it. The solution should aim to find the minimum effort required to complete all... | 0 | You are given an array `tasks` where `tasks[i] = [actuali, minimumi]`:
* `actuali` is the actual amount of energy you **spend to finish** the `ith` task.
* `minimumi` is the minimum amount of energy you **require to begin** the `ith` task.
For example, if the task is `[10, 12]` and your current energy is `11`, yo... | For the node i, calculate the height of each of its children and keep the first and second maximum heights (max1_i , max2_i). Check all nodes and return max( 2 + max1_i + max2_i ). |
O(n * log(n)) solution | minimum-initial-energy-to-finish-tasks | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nThe goal of this problem is to minimize the effort required to complete a given list of tasks. Each task has a minimum and maximum effort required to complete it. The solution should aim to find the minimum effort required to complete all... | 0 | Given a binary string `s` **without leading zeros**, return `true` _if_ `s` _contains **at most one contiguous segment of ones**_. Otherwise, return `false`.
**Example 1:**
**Input:** s = "1001 "
**Output:** false
**Explanation:** The ones do not form a contiguous segment.
**Example 2:**
**Input:** s = "1... | We can easily figure that the f(x) : does x solve this array is monotonic so binary Search is doable Figure a sorting pattern |
Simplest code with runtime 40ms | maximum-repeating-substring | 0 | 1 | \n\n# Code\n```\nclass Solution:\n def maxRepeating(self, sequence: str, word: str) -> int:\n if len(sequence)<len(word):\n return 0\n k=1\n ans=0\n\n while k*word in sequence:\n ans+=1\n k+=1\n return ans \n``` | 1 | For a string `sequence`, a string `word` is **`k`\-repeating** if `word` concatenated `k` times is a substring of `sequence`. The `word`'s **maximum `k`\-repeating value** is the highest value `k` where `word` is `k`\-repeating in `sequence`. If `word` is not a substring of `sequence`, `word`'s maximum `k`\-repeating v... | Given the character counts, under what conditions can a palindrome be formed ? From left to right, use bitwise xor-operation to compute for any prefix the number of times modulo 2 of each digit. (mask ^= (1<<(s[i]-'0')). Expected complexity is O(n*A) where A is the alphabet (10). |
Simplest code with runtime 40ms | maximum-repeating-substring | 0 | 1 | \n\n# Code\n```\nclass Solution:\n def maxRepeating(self, sequence: str, word: str) -> int:\n if len(sequence)<len(word):\n return 0\n k=1\n ans=0\n\n while k*word in sequence:\n ans+=1\n k+=1\n return ans \n``` | 1 | You are given a 2D integer array `groups` of length `n`. You are also given an integer array `nums`.
You are asked if you can choose `n` **disjoint** subarrays from the array `nums` such that the `ith` subarray is equal to `groups[i]` (**0-indexed**), and if `i > 0`, the `(i-1)th` subarray appears **before** the `ith`... | The constraints are low enough for a brute force approach. Try every k value from 0 upwards until word is no longer k-repeating. |
Python, binary search on the length of the subsequence | maximum-repeating-substring | 0 | 1 | ```\nclass Solution:\n def maxRepeating(self, sequence: str, word: str) -> int:\n if word not in sequence:\n return 0\n\n left = 1\n right = len(sequence) // len(word)\n while left <= right:\n mid = (left + right) // 2\n if word * mid in sequence:\n ... | 23 | For a string `sequence`, a string `word` is **`k`\-repeating** if `word` concatenated `k` times is a substring of `sequence`. The `word`'s **maximum `k`\-repeating value** is the highest value `k` where `word` is `k`\-repeating in `sequence`. If `word` is not a substring of `sequence`, `word`'s maximum `k`\-repeating v... | Given the character counts, under what conditions can a palindrome be formed ? From left to right, use bitwise xor-operation to compute for any prefix the number of times modulo 2 of each digit. (mask ^= (1<<(s[i]-'0')). Expected complexity is O(n*A) where A is the alphabet (10). |
Python, binary search on the length of the subsequence | maximum-repeating-substring | 0 | 1 | ```\nclass Solution:\n def maxRepeating(self, sequence: str, word: str) -> int:\n if word not in sequence:\n return 0\n\n left = 1\n right = len(sequence) // len(word)\n while left <= right:\n mid = (left + right) // 2\n if word * mid in sequence:\n ... | 23 | You are given a 2D integer array `groups` of length `n`. You are also given an integer array `nums`.
You are asked if you can choose `n` **disjoint** subarrays from the array `nums` such that the `ith` subarray is equal to `groups[i]` (**0-indexed**), and if `i > 0`, the `(i-1)th` subarray appears **before** the `ith`... | The constraints are low enough for a brute force approach. Try every k value from 0 upwards until word is no longer k-repeating. |
python3 easy way | maximum-repeating-substring | 0 | 1 | ```\nclass Solution:\n def maxRepeating(self, sequence: str, word: str) -> int:\n if len(sequence) < len(word):\n return 0\n ans = 0\n k = 1\n while word*k in sequence:\n ans += 1\n k += 1\n return ans\n``` | 10 | For a string `sequence`, a string `word` is **`k`\-repeating** if `word` concatenated `k` times is a substring of `sequence`. The `word`'s **maximum `k`\-repeating value** is the highest value `k` where `word` is `k`\-repeating in `sequence`. If `word` is not a substring of `sequence`, `word`'s maximum `k`\-repeating v... | Given the character counts, under what conditions can a palindrome be formed ? From left to right, use bitwise xor-operation to compute for any prefix the number of times modulo 2 of each digit. (mask ^= (1<<(s[i]-'0')). Expected complexity is O(n*A) where A is the alphabet (10). |
python3 easy way | maximum-repeating-substring | 0 | 1 | ```\nclass Solution:\n def maxRepeating(self, sequence: str, word: str) -> int:\n if len(sequence) < len(word):\n return 0\n ans = 0\n k = 1\n while word*k in sequence:\n ans += 1\n k += 1\n return ans\n``` | 10 | You are given a 2D integer array `groups` of length `n`. You are also given an integer array `nums`.
You are asked if you can choose `n` **disjoint** subarrays from the array `nums` such that the `ith` subarray is equal to `groups[i]` (**0-indexed**), and if `i > 0`, the `(i-1)th` subarray appears **before** the `ith`... | The constraints are low enough for a brute force approach. Try every k value from 0 upwards until word is no longer k-repeating. |
Python3: O(N^2) solution intuitive | maximum-repeating-substring | 0 | 1 | O(n^2) still but found it to be more intuitive than the top solutions\n\n# Code\n```\nclass Solution:\n def maxRepeating(self, sequence: str, word: str) -> int:\n i = 0\n j = 0\n max_ct = 0\n\n if len(word) > len(sequence):\n return 0\n\n ct = 0\n while i < len(se... | 2 | For a string `sequence`, a string `word` is **`k`\-repeating** if `word` concatenated `k` times is a substring of `sequence`. The `word`'s **maximum `k`\-repeating value** is the highest value `k` where `word` is `k`\-repeating in `sequence`. If `word` is not a substring of `sequence`, `word`'s maximum `k`\-repeating v... | Given the character counts, under what conditions can a palindrome be formed ? From left to right, use bitwise xor-operation to compute for any prefix the number of times modulo 2 of each digit. (mask ^= (1<<(s[i]-'0')). Expected complexity is O(n*A) where A is the alphabet (10). |
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