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
values |
|---|---|---|---|---|---|---|---|
✅ 98.37% O(n) Final Winner | find-the-winner-of-an-array-game | 1 | 1 | # Intuition\nWhen first approaching this problem, we might think about simulating the entire process: comparing numbers, moving the smaller one to the end, and keeping track of the number of consecutive wins. However, a closer inspection reveals a couple of key insights. Firstly, if the game needs only one win (i.e., $... | 103 | 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. |
✅☑[C++/Java/Python/JavaScript] || 2 Approaches || EXPLAINED🔥 | find-the-winner-of-an-array-game | 1 | 1 | # PLEASE UPVOTE IF IT HELPED\n\n---\n\n\n# Approaches\n**(Also explained in the code)**\n\n#### ***Approach 1(With Queue)***\n1. **Initialization:**\n\n - The code initializes a variable `maxElement` to store the maximum element in the `arr`. It starts with the first element of the array.\n - A queue named `queue... | 2 | Given an integer array `arr` of **distinct** integers and an integer `k`.
A game will be played between the first two elements of the array (i.e. `arr[0]` and `arr[1]`). In each round of the game, we compare `arr[0]` with `arr[1]`, the larger integer wins and remains at position `0`, and the smaller integer moves to t... | Use dynamic programming approach. Build dp table where dp[a][b][c] is the number of ways you can start building the array starting from index a where the search_cost = c and the maximum used integer was b. Recursively, solve the small sub-problems first. Optimize your answer by stopping the search if you exceeded k cha... |
✅☑[C++/Java/Python/JavaScript] || 2 Approaches || EXPLAINED🔥 | find-the-winner-of-an-array-game | 1 | 1 | # PLEASE UPVOTE IF IT HELPED\n\n---\n\n\n# Approaches\n**(Also explained in the code)**\n\n#### ***Approach 1(With Queue)***\n1. **Initialization:**\n\n - The code initializes a variable `maxElement` to store the maximum element in the `arr`. It starts with the first element of the array.\n - A queue named `queue... | 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. |
Python3 Solution | find-the-winner-of-an-array-game | 0 | 1 | \n```\nclass Solution:\n def getWinner(self, arr: List[int], k: int) -> int:\n n=len(arr)\n cur=arr[0]\n count=0\n for i in range(1,n):\n if arr[i]>cur:\n cur=arr[i]\n count=0\n count+=1\n if count==k:\n break\n... | 2 | Given an integer array `arr` of **distinct** integers and an integer `k`.
A game will be played between the first two elements of the array (i.e. `arr[0]` and `arr[1]`). In each round of the game, we compare `arr[0]` with `arr[1]`, the larger integer wins and remains at position `0`, and the smaller integer moves to t... | Use dynamic programming approach. Build dp table where dp[a][b][c] is the number of ways you can start building the array starting from index a where the search_cost = c and the maximum used integer was b. Recursively, solve the small sub-problems first. Optimize your answer by stopping the search if you exceeded k cha... |
Python3 Solution | find-the-winner-of-an-array-game | 0 | 1 | \n```\nclass Solution:\n def getWinner(self, arr: List[int], k: int) -> int:\n n=len(arr)\n cur=arr[0]\n count=0\n for i in range(1,n):\n if arr[i]>cur:\n cur=arr[i]\n count=0\n count+=1\n if count==k:\n break\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. |
✅For Beginners✅ II ✅beats 100%✅ II just for loop & if-condition | find-the-winner-of-an-array-game | 1 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nThe code aims to determine the winner of a game between the first two elements of an integer array A. The game is played by comparing the values of A[0] and A[1] in each round. The larger of the two integers remains in position 0, and the... | 33 | Given an integer array `arr` of **distinct** integers and an integer `k`.
A game will be played between the first two elements of the array (i.e. `arr[0]` and `arr[1]`). In each round of the game, we compare `arr[0]` with `arr[1]`, the larger integer wins and remains at position `0`, and the smaller integer moves to t... | Use dynamic programming approach. Build dp table where dp[a][b][c] is the number of ways you can start building the array starting from index a where the search_cost = c and the maximum used integer was b. Recursively, solve the small sub-problems first. Optimize your answer by stopping the search if you exceeded k cha... |
✅For Beginners✅ II ✅beats 100%✅ II just for loop & if-condition | find-the-winner-of-an-array-game | 1 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nThe code aims to determine the winner of a game between the first two elements of an integer array A. The game is played by comparing the values of A[0] and A[1] in each round. The larger of the two integers remains in position 0, and the... | 33 | 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. |
Python 3 solution using array methods | find-the-winner-of-an-array-game | 0 | 1 | \n```\nclass Solution:\n def getWinner(self, arr: List[int], k: int) -> int:\n if k >= len(arr): return max(arr)\n\n resultIndex = 0\n maxInK = max(arr[resultIndex:resultIndex+k+1])\n while arr[resultIndex] < maxInK:\n resultIndex = arr.index(maxInK)\n maxInK = max(a... | 1 | Given an integer array `arr` of **distinct** integers and an integer `k`.
A game will be played between the first two elements of the array (i.e. `arr[0]` and `arr[1]`). In each round of the game, we compare `arr[0]` with `arr[1]`, the larger integer wins and remains at position `0`, and the smaller integer moves to t... | Use dynamic programming approach. Build dp table where dp[a][b][c] is the number of ways you can start building the array starting from index a where the search_cost = c and the maximum used integer was b. Recursively, solve the small sub-problems first. Optimize your answer by stopping the search if you exceeded k cha... |
Python 3 solution using array methods | find-the-winner-of-an-array-game | 0 | 1 | \n```\nclass Solution:\n def getWinner(self, arr: List[int], k: int) -> int:\n if k >= len(arr): return max(arr)\n\n resultIndex = 0\n maxInK = max(arr[resultIndex:resultIndex+k+1])\n while arr[resultIndex] < maxInK:\n resultIndex = arr.index(maxInK)\n maxInK = max(a... | 1 | 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. |
[Python3] bubble-ish sort | minimum-swaps-to-arrange-a-binary-grid | 0 | 1 | Algo \nBasic idea is bubble sort. Here, we transform each row into a number which the location of last index of 1. Then, we use a modified bubble sort algo to compute the `ans`. For each row, we find the first value which can be moved to this position via swaps, and update the `ans`. \n\nEdit: added comments to aid und... | 13 | Given an `n x n` binary `grid`, in one step you can choose two **adjacent rows** of the grid and swap them.
A grid is said to be **valid** if all the cells above the main diagonal are **zeros**.
Return _the minimum number of steps_ needed to make the grid valid, or **\-1** if the grid cannot be valid.
The main diago... | null |
[Python3] bubble-ish sort | minimum-swaps-to-arrange-a-binary-grid | 0 | 1 | Algo \nBasic idea is bubble sort. Here, we transform each row into a number which the location of last index of 1. Then, we use a modified bubble sort algo to compute the `ans`. For each row, we find the first value which can be moved to this position via swaps, and update the `ans`. \n\nEdit: added comments to aid und... | 13 | 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. |
Easy to understand Python Solution with Comments | minimum-swaps-to-arrange-a-binary-grid | 0 | 1 | # Complexity\n- Time complexity: $$O(n^2)$$\n<!-- Add your time complexity here, e.g. $$O(n^2)$$ -->\n\n- Space complexity: $$O(n)$$\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\nclass Solution:\n def minSwaps(self, grid: List[List[int]]) -> int:\n\n # Time : O(n^2)\n # Space... | 0 | Given an `n x n` binary `grid`, in one step you can choose two **adjacent rows** of the grid and swap them.
A grid is said to be **valid** if all the cells above the main diagonal are **zeros**.
Return _the minimum number of steps_ needed to make the grid valid, or **\-1** if the grid cannot be valid.
The main diago... | null |
Easy to understand Python Solution with Comments | minimum-swaps-to-arrange-a-binary-grid | 0 | 1 | # Complexity\n- Time complexity: $$O(n^2)$$\n<!-- Add your time complexity here, e.g. $$O(n^2)$$ -->\n\n- Space complexity: $$O(n)$$\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\nclass Solution:\n def minSwaps(self, grid: List[List[int]]) -> int:\n\n # Time : O(n^2)\n # Space... | 0 | 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 Greedy Solution 92% Faster | minimum-swaps-to-arrange-a-binary-grid | 0 | 1 | ```\nclass Solution:\n def minSwaps(self, grid: List[List[int]]) -> int:\n arr = []\n for row in grid:\n total = 0\n for i in range(len(row)-1,-1,-1):\n if row[i] == 1: break\n total += 1\n \n arr.append(total)\n\n left = ... | 0 | Given an `n x n` binary `grid`, in one step you can choose two **adjacent rows** of the grid and swap them.
A grid is said to be **valid** if all the cells above the main diagonal are **zeros**.
Return _the minimum number of steps_ needed to make the grid valid, or **\-1** if the grid cannot be valid.
The main diago... | null |
Python Greedy Solution 92% Faster | minimum-swaps-to-arrange-a-binary-grid | 0 | 1 | ```\nclass Solution:\n def minSwaps(self, grid: List[List[int]]) -> int:\n arr = []\n for row in grid:\n total = 0\n for i in range(len(row)-1,-1,-1):\n if row[i] == 1: break\n total += 1\n \n arr.append(total)\n\n left = ... | 0 | 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. |
Python3 | Easy Approach | minimum-swaps-to-arrange-a-binary-grid | 0 | 1 | # Code\n```\nclass Solution:\n def minSwaps(self, grid: List[List[int]]) -> int: \n n = len(grid)\n # count trailing zeros \n count = []\n for i in range(n):\n cnt = 0 \n for j in range(n -1 , -1 , -1 ):\n if grid[i][j] != 0 : \n bre... | 0 | Given an `n x n` binary `grid`, in one step you can choose two **adjacent rows** of the grid and swap them.
A grid is said to be **valid** if all the cells above the main diagonal are **zeros**.
Return _the minimum number of steps_ needed to make the grid valid, or **\-1** if the grid cannot be valid.
The main diago... | null |
Python3 | Easy Approach | minimum-swaps-to-arrange-a-binary-grid | 0 | 1 | # Code\n```\nclass Solution:\n def minSwaps(self, grid: List[List[int]]) -> int: \n n = len(grid)\n # count trailing zeros \n count = []\n for i in range(n):\n cnt = 0 \n for j in range(n -1 , -1 , -1 ):\n if grid[i][j] != 0 : \n bre... | 0 | 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] leverage syntactic sugar | minimum-swaps-to-arrange-a-binary-grid | 0 | 1 | # Approach\nThe same idea as others. Counting the tail zeros and bubble sort.\n\n# Complexity\nO(n^2)\n\n# Code\n```\nclass Solution:\n def minSwaps(self, grid: List[List[int]]) -> int:\n A = [sum(int(x == 0) for x in accumulate(row[::-1])) for row in grid]\n n = len(grid)\n\n res = 0\n f... | 0 | Given an `n x n` binary `grid`, in one step you can choose two **adjacent rows** of the grid and swap them.
A grid is said to be **valid** if all the cells above the main diagonal are **zeros**.
Return _the minimum number of steps_ needed to make the grid valid, or **\-1** if the grid cannot be valid.
The main diago... | null |
[python] leverage syntactic sugar | minimum-swaps-to-arrange-a-binary-grid | 0 | 1 | # Approach\nThe same idea as others. Counting the tail zeros and bubble sort.\n\n# Complexity\nO(n^2)\n\n# Code\n```\nclass Solution:\n def minSwaps(self, grid: List[List[int]]) -> int:\n A = [sum(int(x == 0) for x in accumulate(row[::-1])) for row in grid]\n n = len(grid)\n\n res = 0\n f... | 0 | 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] Very simple solution with explanation | get-the-maximum-score | 0 | 1 | ```py\nclass Solution:\n def maxSum(self, nums1: List[int], nums2: List[int]) -> int:\n ## RC ##\n ## APPROACH : GREEDY ##\n ## LOGIC ##\n ## 1. similar to merging 2 sorted arrays\n ## 2. Maintain sum for each array\n ## 3. when you find the same element in both arrays, only... | 6 | You are given two **sorted** arrays of distinct integers `nums1` and `nums2.`
A **valid path** is defined as follows:
* Choose array `nums1` or `nums2` to traverse (from index-0).
* Traverse the current array from left to right.
* If you are reading any value that is present in `nums1` and `nums2` you are allow... | Precompute a prefix sum of ones ('1'). Iterate from left to right counting the number of zeros ('0'), then use the precomputed prefix sum for counting ones ('1'). Update the answer. |
[Python] Very simple solution with explanation | get-the-maximum-score | 0 | 1 | ```py\nclass Solution:\n def maxSum(self, nums1: List[int], nums2: List[int]) -> int:\n ## RC ##\n ## APPROACH : GREEDY ##\n ## LOGIC ##\n ## 1. similar to merging 2 sorted arrays\n ## 2. Maintain sum for each array\n ## 3. when you find the same element in both arrays, only... | 6 | 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. |
prefix sum | get-the-maximum-score | 0 | 1 | # Code\n```\nclass Solution:\n def maxSum(self, nums1: List[int], nums2: List[int]) -> int:\n MOD = 10 ** 9 + 7\n num2idx1 = {num: i for i, num in enumerate(nums1)}\n num2idx2 = {num: i for i, num in enumerate(nums2)}\n pre1 = []\n cs = 0\n for num in nums1:\n cs ... | 0 | You are given two **sorted** arrays of distinct integers `nums1` and `nums2.`
A **valid path** is defined as follows:
* Choose array `nums1` or `nums2` to traverse (from index-0).
* Traverse the current array from left to right.
* If you are reading any value that is present in `nums1` and `nums2` you are allow... | Precompute a prefix sum of ones ('1'). Iterate from left to right counting the number of zeros ('0'), then use the precomputed prefix sum for counting ones ('1'). Update the answer. |
prefix sum | get-the-maximum-score | 0 | 1 | # Code\n```\nclass Solution:\n def maxSum(self, nums1: List[int], nums2: List[int]) -> int:\n MOD = 10 ** 9 + 7\n num2idx1 = {num: i for i, num in enumerate(nums1)}\n num2idx2 = {num: i for i, num in enumerate(nums2)}\n pre1 = []\n cs = 0\n for num in nums1:\n cs ... | 0 | 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. |
Python Solution using Two Pointer in O(n+m) and O(1) | get-the-maximum-score | 0 | 1 | # Code\n```\nclass Solution:\n def maxSum(self, nums1: List[int], nums2: List[int]) -> int:\n i=j=0\n ans=0\n mod=int(1e9+7)\n while(i<len(nums1) or j<len(nums2)):\n c1=c2=0\n while(i<len(nums1) and j<len(nums2) and nums1[i]!=nums2[j]):\n if(nums1[i]>n... | 0 | You are given two **sorted** arrays of distinct integers `nums1` and `nums2.`
A **valid path** is defined as follows:
* Choose array `nums1` or `nums2` to traverse (from index-0).
* Traverse the current array from left to right.
* If you are reading any value that is present in `nums1` and `nums2` you are allow... | Precompute a prefix sum of ones ('1'). Iterate from left to right counting the number of zeros ('0'), then use the precomputed prefix sum for counting ones ('1'). Update the answer. |
Python Solution using Two Pointer in O(n+m) and O(1) | get-the-maximum-score | 0 | 1 | # Code\n```\nclass Solution:\n def maxSum(self, nums1: List[int], nums2: List[int]) -> int:\n i=j=0\n ans=0\n mod=int(1e9+7)\n while(i<len(nums1) or j<len(nums2)):\n c1=c2=0\n while(i<len(nums1) and j<len(nums2) and nums1[i]!=nums2[j]):\n if(nums1[i]>n... | 0 | 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. |
1539. Kth Missing Positive Number | Simple Solution | kth-missing-positive-number | 0 | 1 | # Intuition\nWe are asked to find the missing smallest positive integer. We can use binary search to find the index of the smallest element which has k missing elements before it. If the element arr[middle] at index middle has p missing elements before it, we can get the number of missing elements before arr[middle+1] ... | 1 | Given an array `arr` of positive integers sorted in a **strictly increasing order**, and an integer `k`.
Return _the_ `kth` _**positive** integer that is **missing** from this array._
**Example 1:**
**Input:** arr = \[2,3,4,7,11\], k = 5
**Output:** 9
**Explanation:** The missing positive integers are \[1,5,6,8,9,10... | Notice that numbers with equal sums of row and column indexes belong to the same diagonal. Store them in tuples (sum, row, val), sort them, and then regroup the answer. |
1539. Kth Missing Positive Number | Simple Solution | kth-missing-positive-number | 0 | 1 | # Intuition\nWe are asked to find the missing smallest positive integer. We can use binary search to find the index of the smallest element which has k missing elements before it. If the element arr[middle] at index middle has p missing elements before it, we can get the number of missing elements before arr[middle+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. |
Easy Understanding Python Solution | kth-missing-positive-number | 0 | 1 | Beats 60% of other solutions.\n```\nclass Solution:\n def findKthPositive(self, arr: List[int], k: int) -> int:\n a= 0\n b=len(arr)-1\n while a <= b:\n m= a+(b-a)//2\n if arr[m]-m-1 < k:\n a = m+1\n else:\n b = m-1\n return a+... | 1 | Given an array `arr` of positive integers sorted in a **strictly increasing order**, and an integer `k`.
Return _the_ `kth` _**positive** integer that is **missing** from this array._
**Example 1:**
**Input:** arr = \[2,3,4,7,11\], k = 5
**Output:** 9
**Explanation:** The missing positive integers are \[1,5,6,8,9,10... | Notice that numbers with equal sums of row and column indexes belong to the same diagonal. Store them in tuples (sum, row, val), sort them, and then regroup the answer. |
Easy Understanding Python Solution | kth-missing-positive-number | 0 | 1 | Beats 60% of other solutions.\n```\nclass Solution:\n def findKthPositive(self, arr: List[int], k: int) -> int:\n a= 0\n b=len(arr)-1\n while a <= b:\n m= a+(b-a)//2\n if arr[m]-m-1 < k:\n a = m+1\n else:\n b = m-1\n return a+... | 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. |
✅Pyhhon3 easiest solution🔥🔥 | kth-missing-positive-number | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n**Binary search**\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n- setup counter to 0, and increment at first in while loop.\n- now find position of current counter value.\n- given array is sorted so we can find i... | 1 | Given an array `arr` of positive integers sorted in a **strictly increasing order**, and an integer `k`.
Return _the_ `kth` _**positive** integer that is **missing** from this array._
**Example 1:**
**Input:** arr = \[2,3,4,7,11\], k = 5
**Output:** 9
**Explanation:** The missing positive integers are \[1,5,6,8,9,10... | Notice that numbers with equal sums of row and column indexes belong to the same diagonal. Store them in tuples (sum, row, val), sort them, and then regroup the answer. |
✅Pyhhon3 easiest solution🔥🔥 | kth-missing-positive-number | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n**Binary search**\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n- setup counter to 0, and increment at first in while loop.\n- now find position of current counter value.\n- given array is sorted so we can find i... | 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. |
Awesome To Unveil Logic--->Python3 | kth-missing-positive-number | 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 an array `arr` of positive integers sorted in a **strictly increasing order**, and an integer `k`.
Return _the_ `kth` _**positive** integer that is **missing** from this array._
**Example 1:**
**Input:** arr = \[2,3,4,7,11\], k = 5
**Output:** 9
**Explanation:** The missing positive integers are \[1,5,6,8,9,10... | Notice that numbers with equal sums of row and column indexes belong to the same diagonal. Store them in tuples (sum, row, val), sort them, and then regroup the answer. |
Awesome To Unveil Logic--->Python3 | kth-missing-positive-number | 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. |
Python3 | kth-missing-positive-number | 0 | 1 | # Code\n```\nclass Solution:\n def findKthPositive(self, arr: List[int], k: int) -> int:\n a = [0] * len(arr)\n a[0] = arr[0] - 1\n\n for i in range(1, len(arr)):\n a[i] = a[i - 1] + (arr[i] - arr[i-1] - 1)\n \n idx = bisect_left(a, k)\n\n return arr[idx - 1] + (k... | 1 | Given an array `arr` of positive integers sorted in a **strictly increasing order**, and an integer `k`.
Return _the_ `kth` _**positive** integer that is **missing** from this array._
**Example 1:**
**Input:** arr = \[2,3,4,7,11\], k = 5
**Output:** 9
**Explanation:** The missing positive integers are \[1,5,6,8,9,10... | Notice that numbers with equal sums of row and column indexes belong to the same diagonal. Store them in tuples (sum, row, val), sort them, and then regroup the answer. |
Python3 | kth-missing-positive-number | 0 | 1 | # Code\n```\nclass Solution:\n def findKthPositive(self, arr: List[int], k: int) -> int:\n a = [0] * len(arr)\n a[0] = arr[0] - 1\n\n for i in range(1, len(arr)):\n a[i] = a[i - 1] + (arr[i] - arr[i-1] - 1)\n \n idx = bisect_left(a, k)\n\n return arr[idx - 1] + (k... | 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 Binary Search | 💣Beats 87% | 😎Understand able by NOOBS... | kth-missing-positive-number | 0 | 1 | # Intuition\n Binary search is the best way...\n Use of single WHILE loop!!!\n> Understand able by NOOBS!!!\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n Best approach by minimum time complaxity...\n Beats 87% as minimum!!!\n<!-- Describe your a... | 1 | Given an array `arr` of positive integers sorted in a **strictly increasing order**, and an integer `k`.
Return _the_ `kth` _**positive** integer that is **missing** from this array._
**Example 1:**
**Input:** arr = \[2,3,4,7,11\], k = 5
**Output:** 9
**Explanation:** The missing positive integers are \[1,5,6,8,9,10... | Notice that numbers with equal sums of row and column indexes belong to the same diagonal. Store them in tuples (sum, row, val), sort them, and then regroup the answer. |
🖋Simple Binary Search | 💣Beats 87% | 😎Understand able by NOOBS... | kth-missing-positive-number | 0 | 1 | # Intuition\n Binary search is the best way...\n Use of single WHILE loop!!!\n> Understand able by NOOBS!!!\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n Best approach by minimum time complaxity...\n Beats 87% as minimum!!!\n<!-- Describe your a... | 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 code for beginners😀 | kth-missing-positive-number | 0 | 1 | \n\n# Code\n```\nclass Solution:\n def findKthPositive(self, arr: List[int], k: int) -> int:\n i = 1\n while k>0:\n if i not in arr:\n k-=1\n i+=1\n return i-1\n \n``` | 1 | Given an array `arr` of positive integers sorted in a **strictly increasing order**, and an integer `k`.
Return _the_ `kth` _**positive** integer that is **missing** from this array._
**Example 1:**
**Input:** arr = \[2,3,4,7,11\], k = 5
**Output:** 9
**Explanation:** The missing positive integers are \[1,5,6,8,9,10... | Notice that numbers with equal sums of row and column indexes belong to the same diagonal. Store them in tuples (sum, row, val), sort them, and then regroup the answer. |
Simple 🐍python code for beginners😀 | kth-missing-positive-number | 0 | 1 | \n\n# Code\n```\nclass Solution:\n def findKthPositive(self, arr: List[int], k: int) -> int:\n i = 1\n while k>0:\n if i not in arr:\n k-=1\n i+=1\n return i-1\n \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. |
Python3 2 liner solution | kth-missing-positive-number | 0 | 1 | \n\n# Code\n```\nclass Solution:\n def findKthPositive(self, arr: List[int], k: int) -> int:\n ans=[i for i in range(1,max(arr)+k+1) if i not in arr]\n return ans[k-1]\n``` | 1 | Given an array `arr` of positive integers sorted in a **strictly increasing order**, and an integer `k`.
Return _the_ `kth` _**positive** integer that is **missing** from this array._
**Example 1:**
**Input:** arr = \[2,3,4,7,11\], k = 5
**Output:** 9
**Explanation:** The missing positive integers are \[1,5,6,8,9,10... | Notice that numbers with equal sums of row and column indexes belong to the same diagonal. Store them in tuples (sum, row, val), sort them, and then regroup the answer. |
Python3 2 liner solution | kth-missing-positive-number | 0 | 1 | \n\n# Code\n```\nclass Solution:\n def findKthPositive(self, arr: List[int], k: int) -> int:\n ans=[i for i in range(1,max(arr)+k+1) if i not in arr]\n return ans[k-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. |
Python| Fastest solution in python 100% | kth-missing-positive-number | 0 | 1 | # Code\n```\nclass Solution:\n def findKthPositive(self, arr: list[int], k: int) -> int:\n if k - (arr[-1] - len(arr)) > 0:\n return arr[-1] + k - (arr[-1] - len(arr))\n l, res = 1, []\n for i in arr:\n if i > l:\n res += [i for i in range(l,i)]\n ... | 1 | Given an array `arr` of positive integers sorted in a **strictly increasing order**, and an integer `k`.
Return _the_ `kth` _**positive** integer that is **missing** from this array._
**Example 1:**
**Input:** arr = \[2,3,4,7,11\], k = 5
**Output:** 9
**Explanation:** The missing positive integers are \[1,5,6,8,9,10... | Notice that numbers with equal sums of row and column indexes belong to the same diagonal. Store them in tuples (sum, row, val), sort them, and then regroup the answer. |
Python| Fastest solution in python 100% | kth-missing-positive-number | 0 | 1 | # Code\n```\nclass Solution:\n def findKthPositive(self, arr: list[int], k: int) -> int:\n if k - (arr[-1] - len(arr)) > 0:\n return arr[-1] + k - (arr[-1] - len(arr))\n l, res = 1, []\n for i in arr:\n if i > l:\n res += [i for i in range(l,i)]\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. |
✔ Python3 Solution | O(N) | can-convert-string-in-k-moves | 0 | 1 | # Complexity\n- Time complexity: $$O(n)$$\n- Space complexity: $$O(1)$$\n\n# Code\n```\nclass Solution:\n def canConvertString(self, s, t, k):\n if len(s) != len(t): return False\n dp = [-1] * 27\n for a, b in zip(s, t):\n n = ord(b) - ord(a)\n dp[n if n >= 0 else 26 + n] +... | 1 | Given two strings `s` and `t`, your goal is to convert `s` into `t` in `k` moves or less.
During the `ith` (`1 <= i <= k`) move you can:
* Choose any index `j` (1-indexed) from `s`, such that `1 <= j <= s.length` and `j` has not been chosen in any previous move, and shift the character at that index `i` times.
* ... | null |
✔ Python3 Solution | O(N) | can-convert-string-in-k-moves | 0 | 1 | # Complexity\n- Time complexity: $$O(n)$$\n- Space complexity: $$O(1)$$\n\n# Code\n```\nclass Solution:\n def canConvertString(self, s, t, k):\n if len(s) != len(t): return False\n dp = [-1] * 27\n for a, b in zip(s, t):\n n = ord(b) - ord(a)\n dp[n if n >= 0 else 26 + n] +... | 1 | A string `s` is called **good** if there are no two different characters in `s` that have the same **frequency**.
Given a string `s`, return _the **minimum** number of characters you need to delete to make_ `s` _**good**._
The **frequency** of a character in a string is the number of times it appears in the string. F... | Observe that shifting a letter x times has the same effect of shifting the letter x + 26 times. You need to check whether k is large enough to cover all shifts with the same remainder after modulo 26. |
📌📌 Detailed Explanation || Well-Coded || 92% faster 🐍 | can-convert-string-in-k-moves | 0 | 1 | ## IDEA :\n* First we need to calculate the difference of the conversion. for example conversion from `a to b` will have difference of 1.\n* Then we hold a dictiomary (hash map) diff to see how many times we want to convert each difference.\n\n* For example to go from `"aa" to "bb"` we want to go difference 1 for 2 tim... | 3 | Given two strings `s` and `t`, your goal is to convert `s` into `t` in `k` moves or less.
During the `ith` (`1 <= i <= k`) move you can:
* Choose any index `j` (1-indexed) from `s`, such that `1 <= j <= s.length` and `j` has not been chosen in any previous move, and shift the character at that index `i` times.
* ... | null |
📌📌 Detailed Explanation || Well-Coded || 92% faster 🐍 | can-convert-string-in-k-moves | 0 | 1 | ## IDEA :\n* First we need to calculate the difference of the conversion. for example conversion from `a to b` will have difference of 1.\n* Then we hold a dictiomary (hash map) diff to see how many times we want to convert each difference.\n\n* For example to go from `"aa" to "bb"` we want to go difference 1 for 2 tim... | 3 | A string `s` is called **good** if there are no two different characters in `s` that have the same **frequency**.
Given a string `s`, return _the **minimum** number of characters you need to delete to make_ `s` _**good**._
The **frequency** of a character in a string is the number of times it appears in the string. F... | Observe that shifting a letter x times has the same effect of shifting the letter x + 26 times. You need to check whether k is large enough to cover all shifts with the same remainder after modulo 26. |
Find the residue of modulo | can-convert-string-in-k-moves | 0 | 1 | # Approach\nif two strings have different length, can\'t convert s to t for all k.\nFind the modulo number of shifts by i and its upper bound for i = 1,2,...,25, meaning that do nothing for the residue\'s. \n\n# Complexity\n- Time complexity: O(n), n:length of s\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\... | 0 | Given two strings `s` and `t`, your goal is to convert `s` into `t` in `k` moves or less.
During the `ith` (`1 <= i <= k`) move you can:
* Choose any index `j` (1-indexed) from `s`, such that `1 <= j <= s.length` and `j` has not been chosen in any previous move, and shift the character at that index `i` times.
* ... | null |
Find the residue of modulo | can-convert-string-in-k-moves | 0 | 1 | # Approach\nif two strings have different length, can\'t convert s to t for all k.\nFind the modulo number of shifts by i and its upper bound for i = 1,2,...,25, meaning that do nothing for the residue\'s. \n\n# Complexity\n- Time complexity: O(n), n:length of s\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\... | 0 | A string `s` is called **good** if there are no two different characters in `s` that have the same **frequency**.
Given a string `s`, return _the **minimum** number of characters you need to delete to make_ `s` _**good**._
The **frequency** of a character in a string is the number of times it appears in the string. F... | Observe that shifting a letter x times has the same effect of shifting the letter x + 26 times. You need to check whether k is large enough to cover all shifts with the same remainder after modulo 26. |
[Python] 💡 Simple and Fast | Time O(n) | Space O(1) | minimum-insertions-to-balance-a-parentheses-string | 0 | 1 | # \uD83D\uDCA1 Idea\n\n* Record the number of \'(\' in an open bracket count int. \n* If there is a \'))\' or a \') \'then open bracket count -= 1\n* If there is a \'))\' or a \') and open bracket count = 0 an \'(\' must be inserted\n* If there is a single \')\' another \')\' must be inserted\n* At the end of the progr... | 87 | Given a parentheses string `s` containing only the characters `'('` and `')'`. A parentheses string is **balanced** if:
* Any left parenthesis `'('` must have a corresponding two consecutive right parenthesis `'))'`.
* Left parenthesis `'('` must go before the corresponding two consecutive right parenthesis `'))'`... | null |
[Python] 💡 Simple and Fast | Time O(n) | Space O(1) | minimum-insertions-to-balance-a-parentheses-string | 0 | 1 | # \uD83D\uDCA1 Idea\n\n* Record the number of \'(\' in an open bracket count int. \n* If there is a \'))\' or a \') \'then open bracket count -= 1\n* If there is a \'))\' or a \') and open bracket count = 0 an \'(\' must be inserted\n* If there is a single \')\' another \')\' must be inserted\n* At the end of the progr... | 87 | You have an `inventory` of different colored balls, and there is a customer that wants `orders` balls of **any** color.
The customer weirdly values the colored balls. Each colored ball's value is the number of balls **of that color** you currently have in your `inventory`. For example, if you own `6` yellow balls, the... | Use a stack to keep opening brackets. If you face single closing ')' add 1 to the answer and consider it as '))'. If you have '))' with empty stack, add 1 to the answer, If after finishing you have x opening remaining in the stack, add 2x to the answer. |
Python+ detailed explanation | minimum-insertions-to-balance-a-parentheses-string | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nWhile seeing parentheses problem, using stack is my intuition.\nres: record the number of times making insertion\nidx: pointer on s\nstk: store left parenthese\n\nIn this case, if we meet \'(\', just append left parenthese into stack.\nIf... | 1 | Given a parentheses string `s` containing only the characters `'('` and `')'`. A parentheses string is **balanced** if:
* Any left parenthesis `'('` must have a corresponding two consecutive right parenthesis `'))'`.
* Left parenthesis `'('` must go before the corresponding two consecutive right parenthesis `'))'`... | null |
Python+ detailed explanation | minimum-insertions-to-balance-a-parentheses-string | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nWhile seeing parentheses problem, using stack is my intuition.\nres: record the number of times making insertion\nidx: pointer on s\nstk: store left parenthese\n\nIn this case, if we meet \'(\', just append left parenthese into stack.\nIf... | 1 | You have an `inventory` of different colored balls, and there is a customer that wants `orders` balls of **any** color.
The customer weirdly values the colored balls. Each colored ball's value is the number of balls **of that color** you currently have in your `inventory`. For example, if you own `6` yellow balls, the... | Use a stack to keep opening brackets. If you face single closing ')' add 1 to the answer and consider it as '))'. If you have '))' with empty stack, add 1 to the answer, If after finishing you have x opening remaining in the stack, add 2x to the answer. |
[Python] Simple solution with detailed explanation | minimum-insertions-to-balance-a-parentheses-string | 0 | 1 | ```\nclass Solution:\n def minInsertions(self, s: str) -> int:\n ## RC ##\n ## APPROACH : STACK ##\n ## LOGIC ##\n ## 1. Only 3 conditions, open brace -> push to stack\n ## 2. 2 close braces -> pop from stack, if you donot have enough open braces before increment count(indicates on... | 23 | Given a parentheses string `s` containing only the characters `'('` and `')'`. A parentheses string is **balanced** if:
* Any left parenthesis `'('` must have a corresponding two consecutive right parenthesis `'))'`.
* Left parenthesis `'('` must go before the corresponding two consecutive right parenthesis `'))'`... | null |
[Python] Simple solution with detailed explanation | minimum-insertions-to-balance-a-parentheses-string | 0 | 1 | ```\nclass Solution:\n def minInsertions(self, s: str) -> int:\n ## RC ##\n ## APPROACH : STACK ##\n ## LOGIC ##\n ## 1. Only 3 conditions, open brace -> push to stack\n ## 2. 2 close braces -> pop from stack, if you donot have enough open braces before increment count(indicates on... | 23 | You have an `inventory` of different colored balls, and there is a customer that wants `orders` balls of **any** color.
The customer weirdly values the colored balls. Each colored ball's value is the number of balls **of that color** you currently have in your `inventory`. For example, if you own `6` yellow balls, the... | Use a stack to keep opening brackets. If you face single closing ')' add 1 to the answer and consider it as '))'. If you have '))' with empty stack, add 1 to the answer, If after finishing you have x opening remaining in the stack, add 2x to the answer. |
Fast and easy-understanding solution using python, deque | minimum-insertions-to-balance-a-parentheses-string | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nChange the string $s$ to a deque and then each time popleft the deque.\n\nIf the character $c$ is \'(\', then update left value to count the number of \'(\'.\n\nIf $c$ is \')\', we then check next character in deque. \nIf nothing in the d... | 0 | Given a parentheses string `s` containing only the characters `'('` and `')'`. A parentheses string is **balanced** if:
* Any left parenthesis `'('` must have a corresponding two consecutive right parenthesis `'))'`.
* Left parenthesis `'('` must go before the corresponding two consecutive right parenthesis `'))'`... | null |
Fast and easy-understanding solution using python, deque | minimum-insertions-to-balance-a-parentheses-string | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nChange the string $s$ to a deque and then each time popleft the deque.\n\nIf the character $c$ is \'(\', then update left value to count the number of \'(\'.\n\nIf $c$ is \')\', we then check next character in deque. \nIf nothing in the d... | 0 | You have an `inventory` of different colored balls, and there is a customer that wants `orders` balls of **any** color.
The customer weirdly values the colored balls. Each colored ball's value is the number of balls **of that color** you currently have in your `inventory`. For example, if you own `6` yellow balls, the... | Use a stack to keep opening brackets. If you face single closing ')' add 1 to the answer and consider it as '))'. If you have '))' with empty stack, add 1 to the answer, If after finishing you have x opening remaining in the stack, add 2x to the answer. |
python3, most intuitive to me, 45% | minimum-insertions-to-balance-a-parentheses-string | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nwe need to process the number of ) seen when 1. a when encountering a (, 2. when we reach the end of the array. \n\nThe logic looks most natural to me compared to other top answers\n# Approach\n<!-- Describe your approach to solving the p... | 0 | Given a parentheses string `s` containing only the characters `'('` and `')'`. A parentheses string is **balanced** if:
* Any left parenthesis `'('` must have a corresponding two consecutive right parenthesis `'))'`.
* Left parenthesis `'('` must go before the corresponding two consecutive right parenthesis `'))'`... | null |
python3, most intuitive to me, 45% | minimum-insertions-to-balance-a-parentheses-string | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nwe need to process the number of ) seen when 1. a when encountering a (, 2. when we reach the end of the array. \n\nThe logic looks most natural to me compared to other top answers\n# Approach\n<!-- Describe your approach to solving the p... | 0 | You have an `inventory` of different colored balls, and there is a customer that wants `orders` balls of **any** color.
The customer weirdly values the colored balls. Each colored ball's value is the number of balls **of that color** you currently have in your `inventory`. For example, if you own `6` yellow balls, the... | Use a stack to keep opening brackets. If you face single closing ')' add 1 to the answer and consider it as '))'. If you have '))' with empty stack, add 1 to the answer, If after finishing you have x opening remaining in the stack, add 2x to the answer. |
Python | 2 methods | stack or balance due | minimum-insertions-to-balance-a-parentheses-string | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n- stack\n- track balance due for `)`\n\n# Complexity\n- Time complexity: $$O(n)$$\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity: $$O(n)$$ and $$O(1)$$\n<!-- Add your space complexity here, e.g. $$O(n)$$ --... | 0 | Given a parentheses string `s` containing only the characters `'('` and `')'`. A parentheses string is **balanced** if:
* Any left parenthesis `'('` must have a corresponding two consecutive right parenthesis `'))'`.
* Left parenthesis `'('` must go before the corresponding two consecutive right parenthesis `'))'`... | null |
Python | 2 methods | stack or balance due | minimum-insertions-to-balance-a-parentheses-string | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n- stack\n- track balance due for `)`\n\n# Complexity\n- Time complexity: $$O(n)$$\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity: $$O(n)$$ and $$O(1)$$\n<!-- Add your space complexity here, e.g. $$O(n)$$ --... | 0 | You have an `inventory` of different colored balls, and there is a customer that wants `orders` balls of **any** color.
The customer weirdly values the colored balls. Each colored ball's value is the number of balls **of that color** you currently have in your `inventory`. For example, if you own `6` yellow balls, the... | Use a stack to keep opening brackets. If you face single closing ')' add 1 to the answer and consider it as '))'. If you have '))' with empty stack, add 1 to the answer, If after finishing you have x opening remaining in the stack, add 2x to the answer. |
Python3 O(1) Constant Storage Solution 🙌 | minimum-insertions-to-balance-a-parentheses-string | 0 | 1 | # Approach\n<!-- Describe your approach to solving the problem. -->\nKeep track of open parenthesis and insertions. Key observations:\n\n1. We encounter a (. In this case we increase the open count because we\'ll need to know when all are closed.\n2. We encounter a ). In this case we need to know:\na. Is it a single ) ... | 0 | Given a parentheses string `s` containing only the characters `'('` and `')'`. A parentheses string is **balanced** if:
* Any left parenthesis `'('` must have a corresponding two consecutive right parenthesis `'))'`.
* Left parenthesis `'('` must go before the corresponding two consecutive right parenthesis `'))'`... | null |
Python3 O(1) Constant Storage Solution 🙌 | minimum-insertions-to-balance-a-parentheses-string | 0 | 1 | # Approach\n<!-- Describe your approach to solving the problem. -->\nKeep track of open parenthesis and insertions. Key observations:\n\n1. We encounter a (. In this case we increase the open count because we\'ll need to know when all are closed.\n2. We encounter a ). In this case we need to know:\na. Is it a single ) ... | 0 | You have an `inventory` of different colored balls, and there is a customer that wants `orders` balls of **any** color.
The customer weirdly values the colored balls. Each colored ball's value is the number of balls **of that color** you currently have in your `inventory`. For example, if you own `6` yellow balls, the... | Use a stack to keep opening brackets. If you face single closing ')' add 1 to the answer and consider it as '))'. If you have '))' with empty stack, add 1 to the answer, If after finishing you have x opening remaining in the stack, add 2x to the answer. |
Python3 | Prefix xor | O(n) Solution | find-longest-awesome-substring | 0 | 1 | Time Complexity: O(n) -> O(512n) in worst case?\n\n```\nclass Solution:\n def longestAwesome(self, s: str) -> int:\n # li = [1, 2, 4, 8, 16, 32, 64, 128, 256, 512]\n li = [2**i for i in range(10)]\n # checker = {0, 1, 2, 4, 8, 16, 32, 64, 128, 256, 512}\n checker = set(li)\n check... | 1 | You are given a string `s`. An **awesome** substring is a non-empty substring of `s` such that we can make any number of swaps in order to make it a palindrome.
Return _the length of the maximum length **awesome substring** of_ `s`.
**Example 1:**
**Input:** s = "3242415 "
**Output:** 5
**Explanation:** "24241 " i... | Keep an array power where power[i] is the maximum power of the i-th character. The answer is max(power[i]). |
Python3 | Prefix xor | O(n) Solution | find-longest-awesome-substring | 0 | 1 | Time Complexity: O(n) -> O(512n) in worst case?\n\n```\nclass Solution:\n def longestAwesome(self, s: str) -> int:\n # li = [1, 2, 4, 8, 16, 32, 64, 128, 256, 512]\n li = [2**i for i in range(10)]\n # checker = {0, 1, 2, 4, 8, 16, 32, 64, 128, 256, 512}\n checker = set(li)\n check... | 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). |
100% | find-longest-awesome-substring | 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 a string `s`. An **awesome** substring is a non-empty substring of `s` such that we can make any number of swaps in order to make it a palindrome.
Return _the length of the maximum length **awesome substring** of_ `s`.
**Example 1:**
**Input:** s = "3242415 "
**Output:** 5
**Explanation:** "24241 " i... | Keep an array power where power[i] is the maximum power of the i-th character. The answer is max(power[i]). |
100% | find-longest-awesome-substring | 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 | 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 DP solution. | find-longest-awesome-substring | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nIn this problem, we are looking for the longest substring that is an awesome string. An awesome string is a string that has an even number of each digit from 0-9. To solve this problem, we can use dynamic programming to keep track of the ... | 0 | You are given a string `s`. An **awesome** substring is a non-empty substring of `s` such that we can make any number of swaps in order to make it a palindrome.
Return _the length of the maximum length **awesome substring** of_ `s`.
**Example 1:**
**Input:** s = "3242415 "
**Output:** 5
**Explanation:** "24241 " i... | Keep an array power where power[i] is the maximum power of the i-th character. The answer is max(power[i]). |
Python DP solution. | find-longest-awesome-substring | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nIn this problem, we are looking for the longest substring that is an awesome string. An awesome string is a string that has an even number of each digit from 0-9. To solve this problem, we can use dynamic programming to keep track of the ... | 0 | 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 Solution | Clean & Concise | O(n) | find-kth-bit-in-nth-binary-string | 0 | 1 | # Complexity\n- Time complexity: $$O(n)$$\n- Space complexity: $$O(1)$$\n\n# Code\n\n### Recursive Approach\n```\nclass Solution:\n def findKthBit(self, N, K, R = True):\n if K == 1: return \'0\' if R else \'1\'\n mid = (1 << (N - 1))\n if K < mid: return self.findKthBit(N - 1, K, R)\n if... | 1 | Given two positive integers `n` and `k`, the binary string `Sn` is formed as follows:
* `S1 = "0 "`
* `Si = Si - 1 + "1 " + reverse(invert(Si - 1))` for `i > 1`
Where `+` denotes the concatenation operation, `reverse(x)` returns the reversed string `x`, and `invert(x)` inverts all the bits in `x` (`0` changes to ... | Use dynamic programming to find the maximum digits to paint given a total cost. Build the largest number possible using this DP table. |
easy O(n) python solution beats 92% | find-kth-bit-in-nth-binary-string | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ --... | 1 | Given two positive integers `n` and `k`, the binary string `Sn` is formed as follows:
* `S1 = "0 "`
* `Si = Si - 1 + "1 " + reverse(invert(Si - 1))` for `i > 1`
Where `+` denotes the concatenation operation, `reverse(x)` returns the reversed string `x`, and `invert(x)` inverts all the bits in `x` (`0` changes to ... | Use dynamic programming to find the maximum digits to paint given a total cost. Build the largest number possible using this DP table. |
[Python3] 4-line recursive | find-kth-bit-in-nth-binary-string | 0 | 1 | The first digit is always "0" regardless of `n`. The middle digit at `2**(n-1)` is always `1`. If `k < 2**(n-1)`, one could reduce the problem to `n-1` which is half the size. Otherwise, focus on the other half by reversing the position and "0"/"1". \n\n```\nclass Solution:\n def findKthBit(self, n: int, k: int) -> ... | 12 | Given two positive integers `n` and `k`, the binary string `Sn` is formed as follows:
* `S1 = "0 "`
* `Si = Si - 1 + "1 " + reverse(invert(Si - 1))` for `i > 1`
Where `+` denotes the concatenation operation, `reverse(x)` returns the reversed string `x`, and `invert(x)` inverts all the bits in `x` (`0` changes to ... | Use dynamic programming to find the maximum digits to paint given a total cost. Build the largest number possible using this DP table. |
Python 4 Lines | Simple Solution | find-kth-bit-in-nth-binary-string | 0 | 1 | ```\nclass Solution:\n def findKthBit(self, n: int, k: int) -> str:\n i, s, hash_map = 1, \'0\', {\'1\': \'0\', \'0\': \'1\'}\n for i in range(1, n):\n s = s + \'1\' + \'\'.join((hash_map[i] for i in s))[::-1]\n return s[k-1] | 1 | Given two positive integers `n` and `k`, the binary string `Sn` is formed as follows:
* `S1 = "0 "`
* `Si = Si - 1 + "1 " + reverse(invert(Si - 1))` for `i > 1`
Where `+` denotes the concatenation operation, `reverse(x)` returns the reversed string `x`, and `invert(x)` inverts all the bits in `x` (`0` changes to ... | Use dynamic programming to find the maximum digits to paint given a total cost. Build the largest number possible using this DP table. |
Python 3 || 10 lines, prefix sum, w/example || T/M: 97% / 61% | maximum-number-of-non-overlapping-subarrays-with-sum-equals-target | 0 | 1 | ```\nclass Solution:\n def maxNonOverlapping(self, nums: List[int], target: int) -> int:\n\n d, ans = defaultdict(list), 0 # Example: nums = [2,2,5,2,5,9,4] target = 9 \n\n pref = enumerate(accumulate(nums, initial = 0)) # pref = [(0, 0),(1, 2),(2, 4),(3, 9),\n ... | 3 | Given an array `nums` and an integer `target`, return _the maximum number of **non-empty** **non-overlapping** subarrays such that the sum of values in each subarray is equal to_ `target`.
**Example 1:**
**Input:** nums = \[1,1,1,1,1\], target = 2
**Output:** 2
**Explanation:** There are 2 non-overlapping subarrays \... | null |
Python 3 || 10 lines, prefix sum, w/example || T/M: 97% / 61% | maximum-number-of-non-overlapping-subarrays-with-sum-equals-target | 0 | 1 | ```\nclass Solution:\n def maxNonOverlapping(self, nums: List[int], target: int) -> int:\n\n d, ans = defaultdict(list), 0 # Example: nums = [2,2,5,2,5,9,4] target = 9 \n\n pref = enumerate(accumulate(nums, initial = 0)) # pref = [(0, 0),(1, 2),(2, 4),(3, 9),\n ... | 3 | Given an integer array `instructions`, you are asked to create a sorted array from the elements in `instructions`. You start with an empty container `nums`. For each element from **left to right** in `instructions`, insert it into `nums`. The **cost** of each insertion is the **minimum** of the following:
* The numb... | Keep track of prefix sums to quickly look up what subarray that sums "target" can be formed at each step of scanning the input array. It can be proved that greedily forming valid subarrays as soon as one is found is optimal. |
[Python] Simple Solution - 2 Sum variant | maximum-number-of-non-overlapping-subarrays-with-sum-equals-target | 0 | 1 | ```\nclass Solution:\n def maxNonOverlapping(self, nums: List[int], target: int) -> int:\n ## RC ##\n ## APPROACH : HASHMAP - 2 SUM VARIANT ##\n lookup = {0 : -1}\n running_sum = 0\n count = 0\n for i in range(len(nums)):\n running_sum += nums[i]\n if ... | 17 | Given an array `nums` and an integer `target`, return _the maximum number of **non-empty** **non-overlapping** subarrays such that the sum of values in each subarray is equal to_ `target`.
**Example 1:**
**Input:** nums = \[1,1,1,1,1\], target = 2
**Output:** 2
**Explanation:** There are 2 non-overlapping subarrays \... | null |
[Python] Simple Solution - 2 Sum variant | maximum-number-of-non-overlapping-subarrays-with-sum-equals-target | 0 | 1 | ```\nclass Solution:\n def maxNonOverlapping(self, nums: List[int], target: int) -> int:\n ## RC ##\n ## APPROACH : HASHMAP - 2 SUM VARIANT ##\n lookup = {0 : -1}\n running_sum = 0\n count = 0\n for i in range(len(nums)):\n running_sum += nums[i]\n if ... | 17 | Given an integer array `instructions`, you are asked to create a sorted array from the elements in `instructions`. You start with an empty container `nums`. For each element from **left to right** in `instructions`, insert it into `nums`. The **cost** of each insertion is the **minimum** of the following:
* The numb... | Keep track of prefix sums to quickly look up what subarray that sums "target" can be formed at each step of scanning the input array. It can be proved that greedily forming valid subarrays as soon as one is found is optimal. |
python easy || beats 100% | maximum-number-of-non-overlapping-subarrays-with-sum-equals-target | 0 | 1 | ```\nclass Solution:\n def maxNonOverlapping(self, nums: List[int], target: int) -> int:\n f={}\n f[0]=1;s=0;ans=0\n for j in nums:\n s+=j\n if s-target in f:ans+=1;f={}\n f[s]=1\n return ans\n \n``` | 4 | Given an array `nums` and an integer `target`, return _the maximum number of **non-empty** **non-overlapping** subarrays such that the sum of values in each subarray is equal to_ `target`.
**Example 1:**
**Input:** nums = \[1,1,1,1,1\], target = 2
**Output:** 2
**Explanation:** There are 2 non-overlapping subarrays \... | null |
python easy || beats 100% | maximum-number-of-non-overlapping-subarrays-with-sum-equals-target | 0 | 1 | ```\nclass Solution:\n def maxNonOverlapping(self, nums: List[int], target: int) -> int:\n f={}\n f[0]=1;s=0;ans=0\n for j in nums:\n s+=j\n if s-target in f:ans+=1;f={}\n f[s]=1\n return ans\n \n``` | 4 | Given an integer array `instructions`, you are asked to create a sorted array from the elements in `instructions`. You start with an empty container `nums`. For each element from **left to right** in `instructions`, insert it into `nums`. The **cost** of each insertion is the **minimum** of the following:
* The numb... | Keep track of prefix sums to quickly look up what subarray that sums "target" can be formed at each step of scanning the input array. It can be proved that greedily forming valid subarrays as soon as one is found is optimal. |
Simple Python3 solution with sets and running sums (99%) | maximum-number-of-non-overlapping-subarrays-with-sum-equals-target | 0 | 1 | # Intuition\nWe want to greedily find the end index of the subarray, but don\'t care about the start index. As long as we end the subarray at the lowest possible index, we "leave room" for as many additional subarrays as possible.\n\nSince we\'re looking at subarrays, we are probably going to want some kind of running ... | 0 | Given an array `nums` and an integer `target`, return _the maximum number of **non-empty** **non-overlapping** subarrays such that the sum of values in each subarray is equal to_ `target`.
**Example 1:**
**Input:** nums = \[1,1,1,1,1\], target = 2
**Output:** 2
**Explanation:** There are 2 non-overlapping subarrays \... | null |
Simple Python3 solution with sets and running sums (99%) | maximum-number-of-non-overlapping-subarrays-with-sum-equals-target | 0 | 1 | # Intuition\nWe want to greedily find the end index of the subarray, but don\'t care about the start index. As long as we end the subarray at the lowest possible index, we "leave room" for as many additional subarrays as possible.\n\nSince we\'re looking at subarrays, we are probably going to want some kind of running ... | 0 | Given an integer array `instructions`, you are asked to create a sorted array from the elements in `instructions`. You start with an empty container `nums`. For each element from **left to right** in `instructions`, insert it into `nums`. The **cost** of each insertion is the **minimum** of the following:
* The numb... | Keep track of prefix sums to quickly look up what subarray that sums "target" can be formed at each step of scanning the input array. It can be proved that greedily forming valid subarrays as soon as one is found is optimal. |
Prefix sum of latest index with a pointer to the next index | maximum-number-of-non-overlapping-subarrays-with-sum-equals-target | 0 | 1 | # Intuition\nInital idea was to use sliding window, however, as the array consist of negative number that is not possible.\nHence, this hints to prefix sum.\nNow this is a typical prefix sum, find target. with just ignoring overlaps sub arrays. To do that we keep a start pointer for the next sub array that sums up to t... | 0 | Given an array `nums` and an integer `target`, return _the maximum number of **non-empty** **non-overlapping** subarrays such that the sum of values in each subarray is equal to_ `target`.
**Example 1:**
**Input:** nums = \[1,1,1,1,1\], target = 2
**Output:** 2
**Explanation:** There are 2 non-overlapping subarrays \... | null |
Prefix sum of latest index with a pointer to the next index | maximum-number-of-non-overlapping-subarrays-with-sum-equals-target | 0 | 1 | # Intuition\nInital idea was to use sliding window, however, as the array consist of negative number that is not possible.\nHence, this hints to prefix sum.\nNow this is a typical prefix sum, find target. with just ignoring overlaps sub arrays. To do that we keep a start pointer for the next sub array that sums up to t... | 0 | Given an integer array `instructions`, you are asked to create a sorted array from the elements in `instructions`. You start with an empty container `nums`. For each element from **left to right** in `instructions`, insert it into `nums`. The **cost** of each insertion is the **minimum** of the following:
* The numb... | Keep track of prefix sums to quickly look up what subarray that sums "target" can be formed at each step of scanning the input array. It can be proved that greedily forming valid subarrays as soon as one is found is optimal. |
Optimal solution O(N) | maximum-number-of-non-overlapping-subarrays-with-sum-equals-target | 0 | 1 | # Code\n```\nclass Solution:\n def maxNonOverlapping(self, nums: List[int], target: int) -> int:\n prefix = 0\n count = set([0])\n ans = 0\n\n for n in nums:\n prefix += n\n if prefix - target in count:\n ans += 1\n count = set()\n ... | 0 | Given an array `nums` and an integer `target`, return _the maximum number of **non-empty** **non-overlapping** subarrays such that the sum of values in each subarray is equal to_ `target`.
**Example 1:**
**Input:** nums = \[1,1,1,1,1\], target = 2
**Output:** 2
**Explanation:** There are 2 non-overlapping subarrays \... | null |
Optimal solution O(N) | maximum-number-of-non-overlapping-subarrays-with-sum-equals-target | 0 | 1 | # Code\n```\nclass Solution:\n def maxNonOverlapping(self, nums: List[int], target: int) -> int:\n prefix = 0\n count = set([0])\n ans = 0\n\n for n in nums:\n prefix += n\n if prefix - target in count:\n ans += 1\n count = set()\n ... | 0 | Given an integer array `instructions`, you are asked to create a sorted array from the elements in `instructions`. You start with an empty container `nums`. For each element from **left to right** in `instructions`, insert it into `nums`. The **cost** of each insertion is the **minimum** of the following:
* The numb... | Keep track of prefix sums to quickly look up what subarray that sums "target" can be formed at each step of scanning the input array. It can be proved that greedily forming valid subarrays as soon as one is found is optimal. |
Python thought process and debugging steps. Follow along | maximum-number-of-non-overlapping-subarrays-with-sum-equals-target | 0 | 1 | # Code\n```python []\nclass Solution:\n def maxNonOverlapping(self, nums: List[int], target: int) -> int:\n \'\'\'\n Initial thought: Keep two pointers and if the sum is equal to target, increase the count and set p1,p2 to p2+1\n If sum > target: increase p1, reduce s by nums[p1]\n And if... | 0 | Given an array `nums` and an integer `target`, return _the maximum number of **non-empty** **non-overlapping** subarrays such that the sum of values in each subarray is equal to_ `target`.
**Example 1:**
**Input:** nums = \[1,1,1,1,1\], target = 2
**Output:** 2
**Explanation:** There are 2 non-overlapping subarrays \... | null |
Python thought process and debugging steps. Follow along | maximum-number-of-non-overlapping-subarrays-with-sum-equals-target | 0 | 1 | # Code\n```python []\nclass Solution:\n def maxNonOverlapping(self, nums: List[int], target: int) -> int:\n \'\'\'\n Initial thought: Keep two pointers and if the sum is equal to target, increase the count and set p1,p2 to p2+1\n If sum > target: increase p1, reduce s by nums[p1]\n And if... | 0 | Given an integer array `instructions`, you are asked to create a sorted array from the elements in `instructions`. You start with an empty container `nums`. For each element from **left to right** in `instructions`, insert it into `nums`. The **cost** of each insertion is the **minimum** of the following:
* The numb... | Keep track of prefix sums to quickly look up what subarray that sums "target" can be formed at each step of scanning the input array. It can be proved that greedily forming valid subarrays as soon as one is found is optimal. |
Python 3 || 6 lines, dfs || T/M: 89% / 10% | minimum-cost-to-cut-a-stick | 0 | 1 | ```\nclass Solution:\n def minCost(self, n: int, cuts: List[int]) -> int:\n \n cuts = sorted(chain(cuts,[0,n]))\n \n @lru_cache(None)\n def dfs(l, r):\n length, M = cuts[r] - cuts[l], range(l+1, r)\n return min((dfs(l,i) + dfs(i,r) for i in M),\n ... | 5 | Given a wooden stick of length `n` units. The stick is labelled from `0` to `n`. For example, a stick of length **6** is labelled as follows:
Given an integer array `cuts` where `cuts[i]` denotes a position you should perform a cut at.
You should perform the cuts in order, you can change the order of the cuts as you ... | Start in any city and use the path to move to the next city. Eventually, you will reach a city with no path outgoing, this is the destination city. |
Python 3 || 6 lines, dfs || T/M: 89% / 10% | minimum-cost-to-cut-a-stick | 0 | 1 | ```\nclass Solution:\n def minCost(self, n: int, cuts: List[int]) -> int:\n \n cuts = sorted(chain(cuts,[0,n]))\n \n @lru_cache(None)\n def dfs(l, r):\n length, M = cuts[r] - cuts[l], range(l+1, r)\n return min((dfs(l,i) + dfs(i,r) for i in M),\n ... | 5 | You are given two linked lists: `list1` and `list2` of sizes `n` and `m` respectively.
Remove `list1`'s nodes from the `ath` node to the `bth` node, and put `list2` in their place.
The blue edges and nodes in the following figure indicate the result:
_Build the result list and return its head._
**Example 1:**
**In... | Build a dp array where dp[i][j] is the minimum cost to achieve all the cuts between i and j. When you try to get the minimum cost between i and j, try all possible cuts k between them, dp[i][j] = min(dp[i][k] + dp[k][j]) + (j - i) for all possible cuts k between them. |
Easy 2 line explanation for 2 liner code | Top down | minimum-cost-to-cut-a-stick | 0 | 1 | ```\n# Intuition\n1) Find cuts between range by binary search in sorted arr\n2) Loop k over cuts \n take min of (0,k) + (k,0) + length \n```\n\n\n\n\n# Code\n```\n\ndef minCost(self, n: int, cuts: List[int]) -> int:\n cuts.sort()\n\n @cache\n def help(i=0,j=n):\n x,y = bisect_right(cuts,i), bisect_r... | 2 | Given a wooden stick of length `n` units. The stick is labelled from `0` to `n`. For example, a stick of length **6** is labelled as follows:
Given an integer array `cuts` where `cuts[i]` denotes a position you should perform a cut at.
You should perform the cuts in order, you can change the order of the cuts as you ... | Start in any city and use the path to move to the next city. Eventually, you will reach a city with no path outgoing, this is the destination city. |
Easy 2 line explanation for 2 liner code | Top down | minimum-cost-to-cut-a-stick | 0 | 1 | ```\n# Intuition\n1) Find cuts between range by binary search in sorted arr\n2) Loop k over cuts \n take min of (0,k) + (k,0) + length \n```\n\n\n\n\n# Code\n```\n\ndef minCost(self, n: int, cuts: List[int]) -> int:\n cuts.sort()\n\n @cache\n def help(i=0,j=n):\n x,y = bisect_right(cuts,i), bisect_r... | 2 | You are given two linked lists: `list1` and `list2` of sizes `n` and `m` respectively.
Remove `list1`'s nodes from the `ath` node to the `bth` node, and put `list2` in their place.
The blue edges and nodes in the following figure indicate the result:
_Build the result list and return its head._
**Example 1:**
**In... | Build a dp array where dp[i][j] is the minimum cost to achieve all the cuts between i and j. When you try to get the minimum cost between i and j, try all possible cuts k between them, dp[i][j] = min(dp[i][k] + dp[k][j]) + (j - i) for all possible cuts k between them. |
Pretty beautiful simple and optimal python3 solution | minimum-cost-to-cut-a-stick | 0 | 1 | # Complexity\n- Time complexity: $$O(n ^ 3)$$\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity: $$O(n ^ 2)$$\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n``` python3 []\nfrom functools import cache\n\nclass Solution:\n def minCost(self, n: int, cuts: List[int]) -> in... | 1 | Given a wooden stick of length `n` units. The stick is labelled from `0` to `n`. For example, a stick of length **6** is labelled as follows:
Given an integer array `cuts` where `cuts[i]` denotes a position you should perform a cut at.
You should perform the cuts in order, you can change the order of the cuts as you ... | Start in any city and use the path to move to the next city. Eventually, you will reach a city with no path outgoing, this is the destination city. |
Pretty beautiful simple and optimal python3 solution | minimum-cost-to-cut-a-stick | 0 | 1 | # Complexity\n- Time complexity: $$O(n ^ 3)$$\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity: $$O(n ^ 2)$$\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n``` python3 []\nfrom functools import cache\n\nclass Solution:\n def minCost(self, n: int, cuts: List[int]) -> in... | 1 | You are given two linked lists: `list1` and `list2` of sizes `n` and `m` respectively.
Remove `list1`'s nodes from the `ath` node to the `bth` node, and put `list2` in their place.
The blue edges and nodes in the following figure indicate the result:
_Build the result list and return its head._
**Example 1:**
**In... | Build a dp array where dp[i][j] is the minimum cost to achieve all the cuts between i and j. When you try to get the minimum cost between i and j, try all possible cuts k between them, dp[i][j] = min(dp[i][k] + dp[k][j]) + (j - i) for all possible cuts k between them. |
"Minimum Cost to Cut a Stick"✔ | minimum-cost-to-cut-a-stick | 0 | 1 | # Intuition\nTo find the minimum total cost of the cuts, we can use dynamic programming. We can start by sorting the cuts in ascending order and then consider all possible subproblems.\n\n# Approach\n1. Sort the cuts array in ascending order.\n2. Create a new array called `cuts_arr` by adding 0 at the beginning and `n`... | 1 | Given a wooden stick of length `n` units. The stick is labelled from `0` to `n`. For example, a stick of length **6** is labelled as follows:
Given an integer array `cuts` where `cuts[i]` denotes a position you should perform a cut at.
You should perform the cuts in order, you can change the order of the cuts as you ... | Start in any city and use the path to move to the next city. Eventually, you will reach a city with no path outgoing, this is the destination city. |
"Minimum Cost to Cut a Stick"✔ | minimum-cost-to-cut-a-stick | 0 | 1 | # Intuition\nTo find the minimum total cost of the cuts, we can use dynamic programming. We can start by sorting the cuts in ascending order and then consider all possible subproblems.\n\n# Approach\n1. Sort the cuts array in ascending order.\n2. Create a new array called `cuts_arr` by adding 0 at the beginning and `n`... | 1 | You are given two linked lists: `list1` and `list2` of sizes `n` and `m` respectively.
Remove `list1`'s nodes from the `ath` node to the `bth` node, and put `list2` in their place.
The blue edges and nodes in the following figure indicate the result:
_Build the result list and return its head._
**Example 1:**
**In... | Build a dp array where dp[i][j] is the minimum cost to achieve all the cuts between i and j. When you try to get the minimum cost between i and j, try all possible cuts k between them, dp[i][j] = min(dp[i][k] + dp[k][j]) + (j - i) for all possible cuts k between them. |
Top down dynamic approach (8 lines) | minimum-cost-to-cut-a-stick | 0 | 1 | # Code\n```\nclass Solution:\n def minCost(self, n: int, cuts: List[int]) -> int:\n @cache\n def dp(leftRange, rightRange):\n ans = float(\'inf\')\n for cut in cuts:\n if leftRange < cut < rightRange:\n ans = min(ans, rightRange-leftRange + dp(lef... | 1 | Given a wooden stick of length `n` units. The stick is labelled from `0` to `n`. For example, a stick of length **6** is labelled as follows:
Given an integer array `cuts` where `cuts[i]` denotes a position you should perform a cut at.
You should perform the cuts in order, you can change the order of the cuts as you ... | Start in any city and use the path to move to the next city. Eventually, you will reach a city with no path outgoing, this is the destination city. |
Top down dynamic approach (8 lines) | minimum-cost-to-cut-a-stick | 0 | 1 | # Code\n```\nclass Solution:\n def minCost(self, n: int, cuts: List[int]) -> int:\n @cache\n def dp(leftRange, rightRange):\n ans = float(\'inf\')\n for cut in cuts:\n if leftRange < cut < rightRange:\n ans = min(ans, rightRange-leftRange + dp(lef... | 1 | You are given two linked lists: `list1` and `list2` of sizes `n` and `m` respectively.
Remove `list1`'s nodes from the `ath` node to the `bth` node, and put `list2` in their place.
The blue edges and nodes in the following figure indicate the result:
_Build the result list and return its head._
**Example 1:**
**In... | Build a dp array where dp[i][j] is the minimum cost to achieve all the cuts between i and j. When you try to get the minimum cost between i and j, try all possible cuts k between them, dp[i][j] = min(dp[i][k] + dp[k][j]) + (j - i) for all possible cuts k between them. |
Python Easy Solution | minimum-cost-to-cut-a-stick | 0 | 1 | \n\n# Code\n```\nimport sys\nclass Solution:\n def minCost(self, n: int, cuts: List[int]) -> int:\n d={}\n def solve(i,j):\n if (i,j) in d:\n return d[(i,j)]\n ans=sys.maxsize\n cost=j-i\n for cut in cuts:\n if cut<j and cut>i:\n... | 1 | Given a wooden stick of length `n` units. The stick is labelled from `0` to `n`. For example, a stick of length **6** is labelled as follows:
Given an integer array `cuts` where `cuts[i]` denotes a position you should perform a cut at.
You should perform the cuts in order, you can change the order of the cuts as you ... | Start in any city and use the path to move to the next city. Eventually, you will reach a city with no path outgoing, this is the destination city. |
Python Easy Solution | minimum-cost-to-cut-a-stick | 0 | 1 | \n\n# Code\n```\nimport sys\nclass Solution:\n def minCost(self, n: int, cuts: List[int]) -> int:\n d={}\n def solve(i,j):\n if (i,j) in d:\n return d[(i,j)]\n ans=sys.maxsize\n cost=j-i\n for cut in cuts:\n if cut<j and cut>i:\n... | 1 | You are given two linked lists: `list1` and `list2` of sizes `n` and `m` respectively.
Remove `list1`'s nodes from the `ath` node to the `bth` node, and put `list2` in their place.
The blue edges and nodes in the following figure indicate the result:
_Build the result list and return its head._
**Example 1:**
**In... | Build a dp array where dp[i][j] is the minimum cost to achieve all the cuts between i and j. When you try to get the minimum cost between i and j, try all possible cuts k between them, dp[i][j] = min(dp[i][k] + dp[k][j]) + (j - i) for all possible cuts k between them. |
Image Explanation🏆- [Recursion -> Memo(4 states - 2 states) -> Bottom Up] - C++/Java/Python | minimum-cost-to-cut-a-stick | 1 | 1 | \n\n# Video Solution (`Aryan Mittal`) - Link in LeetCode Profile\n`Minimum Cost to Cut a Stick` by `Aryan Mittal`\n\n\n\n# Approach & Intution\n![image.png](https://assets.leetcode.com/users/images/218b6ebe-f4... | 77 | Given a wooden stick of length `n` units. The stick is labelled from `0` to `n`. For example, a stick of length **6** is labelled as follows:
Given an integer array `cuts` where `cuts[i]` denotes a position you should perform a cut at.
You should perform the cuts in order, you can change the order of the cuts as you ... | Start in any city and use the path to move to the next city. Eventually, you will reach a city with no path outgoing, this is the destination city. |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.