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 |
|---|---|---|---|---|---|---|---|
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 | 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, Simple Solution | maximum-repeating-substring | 0 | 1 | We multiply the word and check occurence with `in`. \n```\nclass Solution:\n def maxRepeating(self, sequence: str, word: str) -> int:\n i = 0\n while word*(i+1) in sequence:\n i+=1\n return i\n``` | 5 | 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, Simple Solution | maximum-repeating-substring | 0 | 1 | We multiply the word and check occurence with `in`. \n```\nclass Solution:\n def maxRepeating(self, sequence: str, word: str) -> int:\n i = 0\n while word*(i+1) in sequence:\n i+=1\n return i\n``` | 5 | 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 Solution to the problem | maximum-repeating-substring | 0 | 1 | # Intuition\nThe intuition is to repeatedly concatenate the given word to itself and check if the resulting sequence is a substring of the given sequence. Keep track of the count until the condition is no longer satisfied.\n\n# Approach\nInitialize a count variable to 0.\nInitialize a variable current_sequence with 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). |
Python Solution to the problem | maximum-repeating-substring | 0 | 1 | # Intuition\nThe intuition is to repeatedly concatenate the given word to itself and check if the resulting sequence is a substring of the given sequence. Keep track of the count until the condition is no longer satisfied.\n\n# Approach\nInitialize a count variable to 0.\nInitialize a variable current_sequence with the... | 0 | 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 solution | maximum-repeating-substring | 0 | 1 | ```\nclass Solution:\n def maxRepeating(self, sequence: str, word: str) -> int:\n counter, new_one = 0, word\n\n while True:\n if new_one in sequence:\n counter += 1\n new_one += word\n else: break\n\n return counter\n # 7m\n\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 solution | maximum-repeating-substring | 0 | 1 | ```\nclass Solution:\n def maxRepeating(self, sequence: str, word: str) -> int:\n counter, new_one = 0, word\n\n while True:\n if new_one in sequence:\n counter += 1\n new_one += word\n else: break\n\n return counter\n # 7m\n\n``` | 0 | 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 Simple Solution!! | maximum-repeating-substring | 0 | 1 | # Code\n```\nclass Solution:\n def maxRepeating(self, sequence: str, word: str) -> int:\n\n k: int = 1\n while (word * k) in sequence:\n k += 1\n return k - 1\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 Simple Solution!! | maximum-repeating-substring | 0 | 1 | # Code\n```\nclass Solution:\n def maxRepeating(self, sequence: str, word: str) -> int:\n\n k: int = 1\n while (word * k) in sequence:\n k += 1\n return k - 1\n``` | 0 | 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. |
Solution with Linked List in Python3 / TypeScript | merge-in-between-linked-lists | 0 | 1 | # Intuition\nLet\'s briefly explain what the problem is:\n- there\'re **Linked Lists** `list1` and `list2`, integers `a` and `b`\n- our goal is to remove nodes at `list1` from `a` to `b` indexes and insert `list2` instead.\n\nThe algorithm is **quite simple**: store index `c` as current count of nodes, store nodes with... | 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. |
Solution with Linked List in Python3 / TypeScript | merge-in-between-linked-lists | 0 | 1 | # Intuition\nLet\'s briefly explain what the problem is:\n- there\'re **Linked Lists** `list1` and `list2`, integers `a` and `b`\n- our goal is to remove nodes at `list1` from `a` to `b` indexes and insert `list2` instead.\n\nThe algorithm is **quite simple**: store index `c` as current count of nodes, store nodes with... | 1 | You are given an integer matrix `isWater` of size `m x n` that represents a map of **land** and **water** cells.
* If `isWater[i][j] == 0`, cell `(i, j)` is a **land** cell.
* If `isWater[i][j] == 1`, cell `(i, j)` is a **water** cell.
You must assign each cell a height in a way that follows these rules:
* The... | Check which edges need to be changed. Let the next node of the (a-1)th node of list1 be the 0-th node in list 2. Let the next node of the last node of list2 be the (b+1)-th node in list 1. |
EFFICIENT MERGING WITH [O(n) && O(1)] COMPLEXITIES | merge-in-between-linked-lists | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nIntuition is a way more generalized, as it will be analogous to repair the damaged patch in a cloth or the rope.\nAnyone can do is to navigate to the damaged portions and then sew the new patch to the both ends of the older part.\n\n# App... | 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. |
EFFICIENT MERGING WITH [O(n) && O(1)] COMPLEXITIES | merge-in-between-linked-lists | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nIntuition is a way more generalized, as it will be analogous to repair the damaged patch in a cloth or the rope.\nAnyone can do is to navigate to the damaged portions and then sew the new patch to the both ends of the older part.\n\n# App... | 1 | You are given an integer matrix `isWater` of size `m x n` that represents a map of **land** and **water** cells.
* If `isWater[i][j] == 0`, cell `(i, j)` is a **land** cell.
* If `isWater[i][j] == 1`, cell `(i, j)` is a **water** cell.
You must assign each cell a height in a way that follows these rules:
* The... | Check which edges need to be changed. Let the next node of the (a-1)th node of list1 be the 0-th node in list 2. Let the next node of the last node of list2 be the (b+1)-th node in list 1. |
Easy | Python Solution | Loops | Arrays | merge-in-between-linked-lists | 0 | 1 | # Code\n```\n# Definition for singly-linked list.\n# class ListNode:\n# def __init__(self, val=0, next=None):\n# self.val = val\n# self.next = next\nclass Solution:\n def mergeInBetween(self, list1: ListNode, a: int, b: int, list2: ListNode) -> ListNode:\n res = []\n pos = 0\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. |
Easy | Python Solution | Loops | Arrays | merge-in-between-linked-lists | 0 | 1 | # Code\n```\n# Definition for singly-linked list.\n# class ListNode:\n# def __init__(self, val=0, next=None):\n# self.val = val\n# self.next = next\nclass Solution:\n def mergeInBetween(self, list1: ListNode, a: int, b: int, list2: ListNode) -> ListNode:\n res = []\n pos = 0\n ... | 1 | You are given an integer matrix `isWater` of size `m x n` that represents a map of **land** and **water** cells.
* If `isWater[i][j] == 0`, cell `(i, j)` is a **land** cell.
* If `isWater[i][j] == 1`, cell `(i, j)` is a **water** cell.
You must assign each cell a height in a way that follows these rules:
* The... | Check which edges need to be changed. Let the next node of the (a-1)th node of list1 be the 0-th node in list 2. Let the next node of the last node of list2 be the (b+1)-th node in list 1. |
Python3, easy to understand, | merge-in-between-linked-lists | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nWe will get to the node before \'a\' th index and will link the next pointer to list 2 and using another loop will get to the \'b\' th node.\n# Approach\n<!-- Describe your approach to solving the problem. -->\n1. We will get to the "a - ... | 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. |
Python3, easy to understand, | merge-in-between-linked-lists | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nWe will get to the node before \'a\' th index and will link the next pointer to list 2 and using another loop will get to the \'b\' th node.\n# Approach\n<!-- Describe your approach to solving the problem. -->\n1. We will get to the "a - ... | 2 | You are given an integer matrix `isWater` of size `m x n` that represents a map of **land** and **water** cells.
* If `isWater[i][j] == 0`, cell `(i, j)` is a **land** cell.
* If `isWater[i][j] == 1`, cell `(i, j)` is a **water** cell.
You must assign each cell a height in a way that follows these rules:
* The... | Check which edges need to be changed. Let the next node of the (a-1)th node of list1 be the 0-th node in list 2. Let the next node of the last node of list2 be the (b+1)-th node in list 1. |
Beats 90% || Linked List || Python | merge-in-between-linked-lists | 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. $$O(n)$$ -->\nO((length(list1))+(length(list2)))\n\n- Space complexity:\n<!-- Add your spa... | 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. |
Beats 90% || Linked List || Python | merge-in-between-linked-lists | 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. $$O(n)$$ -->\nO((length(list1))+(length(list2)))\n\n- Space complexity:\n<!-- Add your spa... | 1 | You are given an integer matrix `isWater` of size `m x n` that represents a map of **land** and **water** cells.
* If `isWater[i][j] == 0`, cell `(i, j)` is a **land** cell.
* If `isWater[i][j] == 1`, cell `(i, j)` is a **water** cell.
You must assign each cell a height in a way that follows these rules:
* The... | Check which edges need to be changed. Let the next node of the (a-1)th node of list1 be the 0-th node in list 2. Let the next node of the last node of list2 be the (b+1)-th node in list 1. |
Python3 || Easy solution. | merge-in-between-linked-lists | 0 | 1 | # It would be really great if you guys upvote, if found the solution helpful.\n# Code\n```\n# Definition for singly-linked list.\n# class ListNode:\n# def __init__(self, val=0, next=None):\n# self.val = val\n# self.next = next\nclass Solution:\n def mergeInBetween(self, list1: ListNode, a: int, b... | 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. |
Python3 || Easy solution. | merge-in-between-linked-lists | 0 | 1 | # It would be really great if you guys upvote, if found the solution helpful.\n# Code\n```\n# Definition for singly-linked list.\n# class ListNode:\n# def __init__(self, val=0, next=None):\n# self.val = val\n# self.next = next\nclass Solution:\n def mergeInBetween(self, list1: ListNode, a: int, b... | 1 | You are given an integer matrix `isWater` of size `m x n` that represents a map of **land** and **water** cells.
* If `isWater[i][j] == 0`, cell `(i, j)` is a **land** cell.
* If `isWater[i][j] == 1`, cell `(i, j)` is a **water** cell.
You must assign each cell a height in a way that follows these rules:
* The... | Check which edges need to be changed. Let the next node of the (a-1)th node of list1 be the 0-th node in list 2. Let the next node of the last node of list2 be the (b+1)-th node in list 1. |
Super Easy and Efficient Python | merge-in-between-linked-lists | 0 | 1 | # Complexity\n- Time complexity: O(n)\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```\n# Definition for singly-linked list.\n# class ListNode:\n# def __init__(self, val=0, next=None):\n# self.val = val\... | 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. |
Super Easy and Efficient Python | merge-in-between-linked-lists | 0 | 1 | # Complexity\n- Time complexity: O(n)\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```\n# Definition for singly-linked list.\n# class ListNode:\n# def __init__(self, val=0, next=None):\n# self.val = val\... | 1 | You are given an integer matrix `isWater` of size `m x n` that represents a map of **land** and **water** cells.
* If `isWater[i][j] == 0`, cell `(i, j)` is a **land** cell.
* If `isWater[i][j] == 1`, cell `(i, j)` is a **water** cell.
You must assign each cell a height in a way that follows these rules:
* The... | Check which edges need to be changed. Let the next node of the (a-1)th node of list1 be the 0-th node in list 2. Let the next node of the last node of list2 be the (b+1)-th node in list 1. |
[Python 3] - 2 pointers approach code with comments - O(m+n) | merge-in-between-linked-lists | 0 | 1 | This is my approach to the problem using 2 pointers :\n```\nclass Solution:\n def mergeInBetween(self, list1: ListNode, a: int, b: int, list2: ListNode) -> ListNode:\n # Define 2 pointers\n\t\tptr1 = list1\n ptr2 = list1\n i, j = 0, 0\n\t\t\n\t\t# Loop for pointer 1 to reach previous node from a... | 16 | 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 3] - 2 pointers approach code with comments - O(m+n) | merge-in-between-linked-lists | 0 | 1 | This is my approach to the problem using 2 pointers :\n```\nclass Solution:\n def mergeInBetween(self, list1: ListNode, a: int, b: int, list2: ListNode) -> ListNode:\n # Define 2 pointers\n\t\tptr1 = list1\n ptr2 = list1\n i, j = 0, 0\n\t\t\n\t\t# Loop for pointer 1 to reach previous node from a... | 16 | You are given an integer matrix `isWater` of size `m x n` that represents a map of **land** and **water** cells.
* If `isWater[i][j] == 0`, cell `(i, j)` is a **land** cell.
* If `isWater[i][j] == 1`, cell `(i, j)` is a **water** cell.
You must assign each cell a height in a way that follows these rules:
* The... | Check which edges need to be changed. Let the next node of the (a-1)th node of list1 be the 0-th node in list 2. Let the next node of the last node of list2 be the (b+1)-th node in list 1. |
Easy Solution | design-front-middle-back-queue | 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 | Design a queue that supports `push` and `pop` operations in the front, middle, and back.
Implement the `FrontMiddleBack` class:
* `FrontMiddleBack()` Initializes the queue.
* `void pushFront(int val)` Adds `val` to the **front** of the queue.
* `void pushMiddle(int val)` Adds `val` to the **middle** of the queu... | null |
Design Front Middle Back Queue | design-front-middle-back-queue | 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 | Design a queue that supports `push` and `pop` operations in the front, middle, and back.
Implement the `FrontMiddleBack` class:
* `FrontMiddleBack()` Initializes the queue.
* `void pushFront(int val)` Adds `val` to the **front** of the queue.
* `void pushMiddle(int val)` Adds `val` to the **middle** of the queu... | null |
O(1) with simple two deques | design-front-middle-back-queue | 0 | 1 | # Intuition\ntwo deque with the number of elements [x,x] or [x,x+1] will be good\n\n# Approach\nmaintain two deque, make sure len(right)-len(left) in [0,1]\n\n# Complexity\n- Time complexity:\nO(1)\n\n- Space complexity:\nO(n)\n\n# Code\n```\nfrom collections import deque\nclass FrontMiddleBackQueue:\n\n def __init_... | 0 | Design a queue that supports `push` and `pop` operations in the front, middle, and back.
Implement the `FrontMiddleBack` class:
* `FrontMiddleBack()` Initializes the queue.
* `void pushFront(int val)` Adds `val` to the **front** of the queue.
* `void pushMiddle(int val)` Adds `val` to the **middle** of the queu... | null |
Minor Modification to List | Explained | design-front-middle-back-queue | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nMost of a pythonic list in fact has this functionality already. \nAs such, we can just extend and logically implement much from there. Only real catch is middle push and pop. \n\n# Approach\n<!-- Describe your approach to solving the prob... | 0 | Design a queue that supports `push` and `pop` operations in the front, middle, and back.
Implement the `FrontMiddleBack` class:
* `FrontMiddleBack()` Initializes the queue.
* `void pushFront(int val)` Adds `val` to the **front** of the queue.
* `void pushMiddle(int val)` Adds `val` to the **middle** of the queu... | null |
Solution | design-front-middle-back-queue | 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 | Design a queue that supports `push` and `pop` operations in the front, middle, and back.
Implement the `FrontMiddleBack` class:
* `FrontMiddleBack()` Initializes the queue.
* `void pushFront(int val)` Adds `val` to the **front** of the queue.
* `void pushMiddle(int val)` Adds `val` to the **middle** of the queu... | null |
✅using list || python | design-front-middle-back-queue | 0 | 1 | \n# Code\n```\nclass FrontMiddleBackQueue:\n\n def __init__(self):\n self.q=[] \n\n def pushFront(self, val: int) -> None:\n self.q=[val]+self.q[:]\n\n def pushMiddle(self, val: int) -> None:\n i=(len(self.q))//2\n self.q=self.q[:i]+[val]+self.q[i:]\n\n def pushBack(self, val: in... | 0 | Design a queue that supports `push` and `pop` operations in the front, middle, and back.
Implement the `FrontMiddleBack` class:
* `FrontMiddleBack()` Initializes the queue.
* `void pushFront(int val)` Adds `val` to the **front** of the queue.
* `void pushMiddle(int val)` Adds `val` to the **middle** of the queu... | null |
[Python] LIS from Left and Right and Combine Result | Easy Understanding | minimum-number-of-removals-to-make-mountain-array | 0 | 1 | \n# Code\n```\nclass Solution:\n def minimumMountainRemovals(self, nums: List[int]) -> int:\n N = len(nums)\n LIS = [1] * N\n LDS = [1] * N\n\n # Longest Increasing Subsequences from Left\n for i in range(N):\n for j in range(i):\n if nums[j] < nums[i]:\n ... | 1 | You may recall that an array `arr` is a **mountain array** if and only if:
* `arr.length >= 3`
* There exists some index `i` (**0-indexed**) with `0 < i < arr.length - 1` such that:
* `arr[0] < arr[1] < ... < arr[i - 1] < arr[i]`
* `arr[i] > arr[i + 1] > ... > arr[arr.length - 1]`
Given an integer arr... | null |
[Python] LIS from Left and Right and Combine Result | Easy Understanding | minimum-number-of-removals-to-make-mountain-array | 0 | 1 | \n# Code\n```\nclass Solution:\n def minimumMountainRemovals(self, nums: List[int]) -> int:\n N = len(nums)\n LIS = [1] * N\n LDS = [1] * N\n\n # Longest Increasing Subsequences from Left\n for i in range(N):\n for j in range(i):\n if nums[j] < nums[i]:\n ... | 1 | There is a tree (i.e., a connected, undirected graph that has no cycles) consisting of `n` nodes numbered from `0` to `n - 1` and exactly `n - 1` edges. Each node has a value associated with it, and the **root** of the tree is node `0`.
To represent this tree, you are given an integer array `nums` and a 2D array `edge... | Think the opposite direction instead of minimum elements to remove the maximum mountain subsequence Think of LIS it's kind of close |
cout << "Solution C++ and Python3" << endl; | richest-customer-wealth | 0 | 1 | # Solution in C++\n```\nclass Solution {\npublic:\n int maximumWealth(vector<vector<int>>& accounts) {\n int max = 0;\n int sum = 0;\n for (int i = 0; i < accounts.size(); i++ ){\n for (int j = 0; j < accounts.at(i).size(); j++){\n sum += accounts.at(i).at(j); \n ... | 2 | You are given an `m x n` integer grid `accounts` where `accounts[i][j]` is the amount of money the `ith` customer has in the `jth` bank. Return _the **wealth** that the richest customer has._
A customer's **wealth** is the amount of money they have in all their bank accounts. The richest ... | Do a binary search over the array, exclude the half of the array that doesn't contain the largest number. Keep shrinking the search space till it reaches the size of 2 where you can easily determine which one has the largest integer. |
cout << "Solution C++ and Python3" << endl; | richest-customer-wealth | 0 | 1 | # Solution in C++\n```\nclass Solution {\npublic:\n int maximumWealth(vector<vector<int>>& accounts) {\n int max = 0;\n int sum = 0;\n for (int i = 0; i < accounts.size(); i++ ){\n for (int j = 0; j < accounts.at(i).size(); j++){\n sum += accounts.at(i).at(j); \n ... | 2 | There is an undirected **star** graph consisting of `n` nodes labeled from `1` to `n`. A star graph is a graph where there is one **center** node and **exactly** `n - 1` edges that connect the center node with every other node.
You are given a 2D integer array `edges` where each `edges[i] = [ui, vi]` indicates that th... | Calculate the wealth of each customer Find the maximum element in array. |
Python | 1-liner simple solution | richest-customer-wealth | 0 | 1 | ```\nclass Solution:\n def maximumWealth(self, accounts: List[List[int]]) -> int:\n return max([sum(acc) for acc in accounts])\n``` | 192 | You are given an `m x n` integer grid `accounts` where `accounts[i][j]` is the amount of money the `ith` customer has in the `jth` bank. Return _the **wealth** that the richest customer has._
A customer's **wealth** is the amount of money they have in all their bank accounts. The richest ... | Do a binary search over the array, exclude the half of the array that doesn't contain the largest number. Keep shrinking the search space till it reaches the size of 2 where you can easily determine which one has the largest integer. |
Python | 1-liner simple solution | richest-customer-wealth | 0 | 1 | ```\nclass Solution:\n def maximumWealth(self, accounts: List[List[int]]) -> int:\n return max([sum(acc) for acc in accounts])\n``` | 192 | There is an undirected **star** graph consisting of `n` nodes labeled from `1` to `n`. A star graph is a graph where there is one **center** node and **exactly** `n - 1` edges that connect the center node with every other node.
You are given a 2D integer array `edges` where each `edges[i] = [ui, vi]` indicates that th... | Calculate the wealth of each customer Find the maximum element in array. |
solution for beginners | richest-customer-wealth | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ --... | 2 | You are given an `m x n` integer grid `accounts` where `accounts[i][j]` is the amount of money the `ith` customer has in the `jth` bank. Return _the **wealth** that the richest customer has._
A customer's **wealth** is the amount of money they have in all their bank accounts. The richest ... | Do a binary search over the array, exclude the half of the array that doesn't contain the largest number. Keep shrinking the search space till it reaches the size of 2 where you can easily determine which one has the largest integer. |
solution for beginners | richest-customer-wealth | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ --... | 2 | There is an undirected **star** graph consisting of `n` nodes labeled from `1` to `n`. A star graph is a graph where there is one **center** node and **exactly** `n - 1` edges that connect the center node with every other node.
You are given a 2D integer array `edges` where each `edges[i] = [ui, vi]` indicates that th... | Calculate the wealth of each customer Find the maximum element in array. |
Simple and elegant result in Python3! | richest-customer-wealth | 0 | 1 | # Intuition\nSimple and elegant one-liner.\nHave a nice day!\n\n# Code\n```\nclass Solution:\n def maximumWealth(self, accounts: List[List[int]]) -> int:\n return max(map(sum, accounts))\n\n``` | 3 | You are given an `m x n` integer grid `accounts` where `accounts[i][j]` is the amount of money the `ith` customer has in the `jth` bank. Return _the **wealth** that the richest customer has._
A customer's **wealth** is the amount of money they have in all their bank accounts. The richest ... | Do a binary search over the array, exclude the half of the array that doesn't contain the largest number. Keep shrinking the search space till it reaches the size of 2 where you can easily determine which one has the largest integer. |
Simple and elegant result in Python3! | richest-customer-wealth | 0 | 1 | # Intuition\nSimple and elegant one-liner.\nHave a nice day!\n\n# Code\n```\nclass Solution:\n def maximumWealth(self, accounts: List[List[int]]) -> int:\n return max(map(sum, accounts))\n\n``` | 3 | There is an undirected **star** graph consisting of `n` nodes labeled from `1` to `n`. A star graph is a graph where there is one **center** node and **exactly** `n - 1` edges that connect the center node with every other node.
You are given a 2D integer array `edges` where each `edges[i] = [ui, vi]` indicates that th... | Calculate the wealth of each customer Find the maximum element in array. |
Python 3 -> O(1) space. Simple solution with explanation | richest-customer-wealth | 0 | 1 | **Suggestions to make it better are always welcomed.**\n\nApproach:\n1. For every person, add his wealth from all the banks ie. for every row, sum all the column values and store in totalWealth.\n2. If totalWealth is more than maxWealth, then update maxWealth.\n\n```\ndef maximumWealth(self, accounts: List[List[int]]) ... | 146 | You are given an `m x n` integer grid `accounts` where `accounts[i][j]` is the amount of money the `ith` customer has in the `jth` bank. Return _the **wealth** that the richest customer has._
A customer's **wealth** is the amount of money they have in all their bank accounts. The richest ... | Do a binary search over the array, exclude the half of the array that doesn't contain the largest number. Keep shrinking the search space till it reaches the size of 2 where you can easily determine which one has the largest integer. |
Python 3 -> O(1) space. Simple solution with explanation | richest-customer-wealth | 0 | 1 | **Suggestions to make it better are always welcomed.**\n\nApproach:\n1. For every person, add his wealth from all the banks ie. for every row, sum all the column values and store in totalWealth.\n2. If totalWealth is more than maxWealth, then update maxWealth.\n\n```\ndef maximumWealth(self, accounts: List[List[int]]) ... | 146 | There is an undirected **star** graph consisting of `n` nodes labeled from `1` to `n`. A star graph is a graph where there is one **center** node and **exactly** `n - 1` edges that connect the center node with every other node.
You are given a 2D integer array `edges` where each `edges[i] = [ui, vi]` indicates that th... | Calculate the wealth of each customer Find the maximum element in array. |
Python | Easy Solution✅ | richest-customer-wealth | 0 | 1 | Solution 1:\n```\ndef maximumWealth(self, accounts: List[List[int]]) -> int:\n return max(map(sum,accounts))\n```\nSolution 2:\n```\ndef maximumWealth(self, accounts: List[List[int]]) -> int:\n large = 0\n for money in accounts:\n large = max(sum(money), large)\n return large\n``` | 7 | You are given an `m x n` integer grid `accounts` where `accounts[i][j]` is the amount of money the `ith` customer has in the `jth` bank. Return _the **wealth** that the richest customer has._
A customer's **wealth** is the amount of money they have in all their bank accounts. The richest ... | Do a binary search over the array, exclude the half of the array that doesn't contain the largest number. Keep shrinking the search space till it reaches the size of 2 where you can easily determine which one has the largest integer. |
Python | Easy Solution✅ | richest-customer-wealth | 0 | 1 | Solution 1:\n```\ndef maximumWealth(self, accounts: List[List[int]]) -> int:\n return max(map(sum,accounts))\n```\nSolution 2:\n```\ndef maximumWealth(self, accounts: List[List[int]]) -> int:\n large = 0\n for money in accounts:\n large = max(sum(money), large)\n return large\n``` | 7 | There is an undirected **star** graph consisting of `n` nodes labeled from `1` to `n`. A star graph is a graph where there is one **center** node and **exactly** `n - 1` edges that connect the center node with every other node.
You are given a 2D integer array `edges` where each `edges[i] = [ui, vi]` indicates that th... | Calculate the wealth of each customer Find the maximum element in array. |
Beats 97.25% #python3 | richest-customer-wealth | 0 | 1 | \n\n# Approach\nAt first, sum the first user\'s wealth and store it to a variable.\nThen, traverse through the users and calculate their wealth and check if the wealth is greater than that variable assign it again or pass.\n\n\n# Code\n```\nclass Solution:\n def maximumWealth(self, accounts: List[List[int]]) -> int:... | 8 | You are given an `m x n` integer grid `accounts` where `accounts[i][j]` is the amount of money the `ith` customer has in the `jth` bank. Return _the **wealth** that the richest customer has._
A customer's **wealth** is the amount of money they have in all their bank accounts. The richest ... | Do a binary search over the array, exclude the half of the array that doesn't contain the largest number. Keep shrinking the search space till it reaches the size of 2 where you can easily determine which one has the largest integer. |
Beats 97.25% #python3 | richest-customer-wealth | 0 | 1 | \n\n# Approach\nAt first, sum the first user\'s wealth and store it to a variable.\nThen, traverse through the users and calculate their wealth and check if the wealth is greater than that variable assign it again or pass.\n\n\n# Code\n```\nclass Solution:\n def maximumWealth(self, accounts: List[List[int]]) -> int:... | 8 | There is an undirected **star** graph consisting of `n` nodes labeled from `1` to `n`. A star graph is a graph where there is one **center** node and **exactly** `n - 1` edges that connect the center node with every other node.
You are given a 2D integer array `edges` where each `edges[i] = [ui, vi]` indicates that th... | Calculate the wealth of each customer Find the maximum element in array. |
✔️ [Python3] ONE-LINER, Explained | richest-customer-wealth | 0 | 1 | **UPVOTE if you like (\uD83C\uDF38\u25E0\u203F\u25E0), If you have any question, feel free to ask.**\n\nTwo simple steps: \n1. Iterate over accounts and for every account calculate the sum.\n2. Iterate over the resulting list and find the maximum.\n\nTime: **O(n)** - for scan\nSpace: **O(1)** - can be implemented witho... | 54 | You are given an `m x n` integer grid `accounts` where `accounts[i][j]` is the amount of money the `ith` customer has in the `jth` bank. Return _the **wealth** that the richest customer has._
A customer's **wealth** is the amount of money they have in all their bank accounts. The richest ... | Do a binary search over the array, exclude the half of the array that doesn't contain the largest number. Keep shrinking the search space till it reaches the size of 2 where you can easily determine which one has the largest integer. |
✔️ [Python3] ONE-LINER, Explained | richest-customer-wealth | 0 | 1 | **UPVOTE if you like (\uD83C\uDF38\u25E0\u203F\u25E0), If you have any question, feel free to ask.**\n\nTwo simple steps: \n1. Iterate over accounts and for every account calculate the sum.\n2. Iterate over the resulting list and find the maximum.\n\nTime: **O(n)** - for scan\nSpace: **O(1)** - can be implemented witho... | 54 | There is an undirected **star** graph consisting of `n` nodes labeled from `1` to `n`. A star graph is a graph where there is one **center** node and **exactly** `n - 1` edges that connect the center node with every other node.
You are given a 2D integer array `edges` where each `edges[i] = [ui, vi]` indicates that th... | Calculate the wealth of each customer Find the maximum element in array. |
Python 3 approach with detailed explanation , algorithm and dry run using Monotonic Stack. | find-the-most-competitive-subsequence | 0 | 1 | # Intuition\nHere, we take into consideration, the concept that, whether we will have enough surplus elements to add in the result(stack), if we pop/remove the current element from the result(stack).\n\n# Approach\nALGORITHM - \n \n 1. Initialize the \'stack\' and a variable \'stack_len\' to keep track \n ... | 12 | Given an integer array `nums` and a positive integer `k`, return _the most **competitive** subsequence of_ `nums` _of size_ `k`.
An array's subsequence is a resulting sequence obtained by erasing some (possibly zero) elements from the array.
We define that a subsequence `a` is more **competitive** than a subsequence ... | null |
Python 3 approach with detailed explanation , algorithm and dry run using Monotonic Stack. | find-the-most-competitive-subsequence | 0 | 1 | # Intuition\nHere, we take into consideration, the concept that, whether we will have enough surplus elements to add in the result(stack), if we pop/remove the current element from the result(stack).\n\n# Approach\nALGORITHM - \n \n 1. Initialize the \'stack\' and a variable \'stack_len\' to keep track \n ... | 12 | There is a school that has classes of students and each class will be having a final exam. You are given a 2D integer array `classes`, where `classes[i] = [passi, totali]`. You know beforehand that in the `ith` class, there are `totali` total students, but only `passi` number of students will pass the exam.
You are al... | In lexicographical order, the elements to the left have higher priority than those that come after. Can you think of a strategy that incrementally builds the answer from left to right? |
Python3, monotonic stack | find-the-most-competitive-subsequence | 0 | 1 | # Intuition\nInitially I thought its going to be a LIS problem but optimized LIS requires binary search and this might be an overkill for the problem\nIn discussion section someone mentioned lexicographically smallest subsequence and I recall this problem [1081](https://leetcode.com/problems/smallest-subsequence-of-dis... | 2 | Given an integer array `nums` and a positive integer `k`, return _the most **competitive** subsequence of_ `nums` _of size_ `k`.
An array's subsequence is a resulting sequence obtained by erasing some (possibly zero) elements from the array.
We define that a subsequence `a` is more **competitive** than a subsequence ... | null |
Python3, monotonic stack | find-the-most-competitive-subsequence | 0 | 1 | # Intuition\nInitially I thought its going to be a LIS problem but optimized LIS requires binary search and this might be an overkill for the problem\nIn discussion section someone mentioned lexicographically smallest subsequence and I recall this problem [1081](https://leetcode.com/problems/smallest-subsequence-of-dis... | 2 | There is a school that has classes of students and each class will be having a final exam. You are given a 2D integer array `classes`, where `classes[i] = [passi, totali]`. You know beforehand that in the `ith` class, there are `totali` total students, but only `passi` number of students will pass the exam.
You are al... | In lexicographical order, the elements to the left have higher priority than those that come after. Can you think of a strategy that incrementally builds the answer from left to right? |
Easy peasy python solution | find-the-most-competitive-subsequence | 0 | 1 | \n```\ndef mostCompetitive(self, nums: List[int], k: int) -> List[int]:\n st = []\n for idx, num in enumerate(nums):\n if not st:\n st.append(num)\n else:\n # pop element from st if last element is greater than the current element\n # and ... | 14 | Given an integer array `nums` and a positive integer `k`, return _the most **competitive** subsequence of_ `nums` _of size_ `k`.
An array's subsequence is a resulting sequence obtained by erasing some (possibly zero) elements from the array.
We define that a subsequence `a` is more **competitive** than a subsequence ... | null |
Easy peasy python solution | find-the-most-competitive-subsequence | 0 | 1 | \n```\ndef mostCompetitive(self, nums: List[int], k: int) -> List[int]:\n st = []\n for idx, num in enumerate(nums):\n if not st:\n st.append(num)\n else:\n # pop element from st if last element is greater than the current element\n # and ... | 14 | There is a school that has classes of students and each class will be having a final exam. You are given a 2D integer array `classes`, where `classes[i] = [passi, totali]`. You know beforehand that in the `ith` class, there are `totali` total students, but only `passi` number of students will pass the exam.
You are al... | In lexicographical order, the elements to the left have higher priority than those that come after. Can you think of a strategy that incrementally builds the answer from left to right? |
[Python3] greedy O(N) | find-the-most-competitive-subsequence | 0 | 1 | **Approach 1** - stack `O(N)`\nHere, we maintain an increasing mono-stack. The trick is that there is a capacity constraint of `k`. You are only allowed to pop out of the stack if the sum of elements on stack `len(stack)` and remaining elements in `nums` (`len(nums) - i`) is more than enough for `k`. \n\nImplementation... | 6 | Given an integer array `nums` and a positive integer `k`, return _the most **competitive** subsequence of_ `nums` _of size_ `k`.
An array's subsequence is a resulting sequence obtained by erasing some (possibly zero) elements from the array.
We define that a subsequence `a` is more **competitive** than a subsequence ... | null |
[Python3] greedy O(N) | find-the-most-competitive-subsequence | 0 | 1 | **Approach 1** - stack `O(N)`\nHere, we maintain an increasing mono-stack. The trick is that there is a capacity constraint of `k`. You are only allowed to pop out of the stack if the sum of elements on stack `len(stack)` and remaining elements in `nums` (`len(nums) - i`) is more than enough for `k`. \n\nImplementation... | 6 | There is a school that has classes of students and each class will be having a final exam. You are given a 2D integer array `classes`, where `classes[i] = [passi, totali]`. You know beforehand that in the `ith` class, there are `totali` total students, but only `passi` number of students will pass the exam.
You are al... | In lexicographical order, the elements to the left have higher priority than those that come after. Can you think of a strategy that incrementally builds the answer from left to right? |
python3: easy to understand | find-the-most-competitive-subsequence | 0 | 1 | Very silimar to [**Monotonous Stack**](https://labuladong.gitbook.io/algo-en/ii.-data-structure/monotonicstack)\n\nWe should maintance a decreasing monotone stack from top to botom, so we should choose the numbers as small as possible into the stack. \nFor example: \nnums = [3,5,2,6,8], k = 2\nstep1: stack = [3] `appe... | 5 | Given an integer array `nums` and a positive integer `k`, return _the most **competitive** subsequence of_ `nums` _of size_ `k`.
An array's subsequence is a resulting sequence obtained by erasing some (possibly zero) elements from the array.
We define that a subsequence `a` is more **competitive** than a subsequence ... | null |
python3: easy to understand | find-the-most-competitive-subsequence | 0 | 1 | Very silimar to [**Monotonous Stack**](https://labuladong.gitbook.io/algo-en/ii.-data-structure/monotonicstack)\n\nWe should maintance a decreasing monotone stack from top to botom, so we should choose the numbers as small as possible into the stack. \nFor example: \nnums = [3,5,2,6,8], k = 2\nstep1: stack = [3] `appe... | 5 | There is a school that has classes of students and each class will be having a final exam. You are given a 2D integer array `classes`, where `classes[i] = [passi, totali]`. You know beforehand that in the `ith` class, there are `totali` total students, but only `passi` number of students will pass the exam.
You are al... | In lexicographical order, the elements to the left have higher priority than those that come after. Can you think of a strategy that incrementally builds the answer from left to right? |
Python. clear & cool solution. O(n). Using stack | find-the-most-competitive-subsequence | 0 | 1 | \tclass Solution:\n\t\tdef mostCompetitive(self, nums: List[int], k: int) -> List[int]:\n\t\t\tend = len(nums) - k\n\t\t\tans = []\n\t\t\tfor num in nums:\n\t\t\t\twhile end and ans and num < ans[-1] :\n\t\t\t\t\tans.pop()\n\t\t\t\t\tend -= 1\n\t\t\t\tans.append(num)\n\t\t\t\n\t\t\treturn ans[:k] | 8 | Given an integer array `nums` and a positive integer `k`, return _the most **competitive** subsequence of_ `nums` _of size_ `k`.
An array's subsequence is a resulting sequence obtained by erasing some (possibly zero) elements from the array.
We define that a subsequence `a` is more **competitive** than a subsequence ... | null |
Python. clear & cool solution. O(n). Using stack | find-the-most-competitive-subsequence | 0 | 1 | \tclass Solution:\n\t\tdef mostCompetitive(self, nums: List[int], k: int) -> List[int]:\n\t\t\tend = len(nums) - k\n\t\t\tans = []\n\t\t\tfor num in nums:\n\t\t\t\twhile end and ans and num < ans[-1] :\n\t\t\t\t\tans.pop()\n\t\t\t\t\tend -= 1\n\t\t\t\tans.append(num)\n\t\t\t\n\t\t\treturn ans[:k] | 8 | There is a school that has classes of students and each class will be having a final exam. You are given a 2D integer array `classes`, where `classes[i] = [passi, totali]`. You know beforehand that in the `ith` class, there are `totali` total students, but only `passi` number of students will pass the exam.
You are al... | In lexicographical order, the elements to the left have higher priority than those that come after. Can you think of a strategy that incrementally builds the answer from left to right? |
Monotonic stack + condition on subsequence length | find-the-most-competitive-subsequence | 0 | 1 | # Intuition\nWe can use the greedy approach and to keep at the begenning of the subsequence the smallest elements, as long as we have enough elements left.\n\n# Approach\nUse monotonic stack and keep track of the length of subsequence. You should break the invariant of monotonic non-decrease when you have just enough e... | 0 | Given an integer array `nums` and a positive integer `k`, return _the most **competitive** subsequence of_ `nums` _of size_ `k`.
An array's subsequence is a resulting sequence obtained by erasing some (possibly zero) elements from the array.
We define that a subsequence `a` is more **competitive** than a subsequence ... | null |
Monotonic stack + condition on subsequence length | find-the-most-competitive-subsequence | 0 | 1 | # Intuition\nWe can use the greedy approach and to keep at the begenning of the subsequence the smallest elements, as long as we have enough elements left.\n\n# Approach\nUse monotonic stack and keep track of the length of subsequence. You should break the invariant of monotonic non-decrease when you have just enough e... | 0 | There is a school that has classes of students and each class will be having a final exam. You are given a 2D integer array `classes`, where `classes[i] = [passi, totali]`. You know beforehand that in the `ith` class, there are `totali` total students, but only `passi` number of students will pass the exam.
You are al... | In lexicographical order, the elements to the left have higher priority than those that come after. Can you think of a strategy that incrementally builds the answer from left to right? |
Very easy Python solution with detailed expanation | minimum-moves-to-make-array-complementary | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nAs other solution describe:\nIf `x + y = T`, the cost is 0.\nIf `min(x, y) + 1 > T`, the cost is 2 -- we need to decrease both x and y.\nIf `max(x, y) + limit < T`, the cost is 2 -- we need to increase both x and y.\nOtherwise the cost is... | 3 | You are given an integer array `nums` of **even** length `n` and an integer `limit`. In one move, you can replace any integer from `nums` with another integer between `1` and `limit`, inclusive.
The array `nums` is **complementary** if for all indices `i` (**0-indexed**), `nums[i] + nums[n - 1 - i]` equals the same nu... | Build the array arr using the given formula, define target = sum(arr) / n What is the number of operations needed to convert arr so that all elements equal target ? |
Very easy Python solution with detailed expanation | minimum-moves-to-make-array-complementary | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nAs other solution describe:\nIf `x + y = T`, the cost is 0.\nIf `min(x, y) + 1 > T`, the cost is 2 -- we need to decrease both x and y.\nIf `max(x, y) + limit < T`, the cost is 2 -- we need to increase both x and y.\nOtherwise the cost is... | 3 | You are given an array of integers `nums` **(0-indexed)** and an integer `k`.
The **score** of a subarray `(i, j)` is defined as `min(nums[i], nums[i+1], ..., nums[j]) * (j - i + 1)`. A **good** subarray is a subarray where `i <= k <= j`.
Return _the maximum possible **score** of a **good** subarray._
**Example 1:**... | Given a target sum x, each pair of nums[i] and nums[n-1-i] would either need 0, 1, or 2 modifications. Can you find the optimal target sum x value such that the sum of modifications is minimized? Create a difference array to efficiently sum all the modifications. |
Sweep Algorithm | Explained [Python] | minimum-moves-to-make-array-complementary | 0 | 1 | Let\'s start with a few observations:\n1. Let\'s say x = target number (sum for each complimentary numbers) => We need to change complimentary values to achieve this target\n2. The min target number achievable is 2 \n\t* 1 <= nums[i] <= limit <= 105\n\t* so either the nums[i] already equal to 1 each or we replace them ... | 6 | You are given an integer array `nums` of **even** length `n` and an integer `limit`. In one move, you can replace any integer from `nums` with another integer between `1` and `limit`, inclusive.
The array `nums` is **complementary** if for all indices `i` (**0-indexed**), `nums[i] + nums[n - 1 - i]` equals the same nu... | Build the array arr using the given formula, define target = sum(arr) / n What is the number of operations needed to convert arr so that all elements equal target ? |
Sweep Algorithm | Explained [Python] | minimum-moves-to-make-array-complementary | 0 | 1 | Let\'s start with a few observations:\n1. Let\'s say x = target number (sum for each complimentary numbers) => We need to change complimentary values to achieve this target\n2. The min target number achievable is 2 \n\t* 1 <= nums[i] <= limit <= 105\n\t* so either the nums[i] already equal to 1 each or we replace them ... | 6 | You are given an array of integers `nums` **(0-indexed)** and an integer `k`.
The **score** of a subarray `(i, j)` is defined as `min(nums[i], nums[i+1], ..., nums[j]) * (j - i + 1)`. A **good** subarray is a subarray where `i <= k <= j`.
Return _the maximum possible **score** of a **good** subarray._
**Example 1:**... | Given a target sum x, each pair of nums[i] and nums[n-1-i] would either need 0, 1, or 2 modifications. Can you find the optimal target sum x value such that the sum of modifications is minimized? Create a difference array to efficiently sum all the modifications. |
Python3: Sweep Line with Comments | minimum-moves-to-make-array-complementary | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ --... | 0 | You are given an integer array `nums` of **even** length `n` and an integer `limit`. In one move, you can replace any integer from `nums` with another integer between `1` and `limit`, inclusive.
The array `nums` is **complementary** if for all indices `i` (**0-indexed**), `nums[i] + nums[n - 1 - i]` equals the same nu... | Build the array arr using the given formula, define target = sum(arr) / n What is the number of operations needed to convert arr so that all elements equal target ? |
Python3: Sweep Line with Comments | minimum-moves-to-make-array-complementary | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ --... | 0 | You are given an array of integers `nums` **(0-indexed)** and an integer `k`.
The **score** of a subarray `(i, j)` is defined as `min(nums[i], nums[i+1], ..., nums[j]) * (j - i + 1)`. A **good** subarray is a subarray where `i <= k <= j`.
Return _the maximum possible **score** of a **good** subarray._
**Example 1:**... | Given a target sum x, each pair of nums[i] and nums[n-1-i] would either need 0, 1, or 2 modifications. Can you find the optimal target sum x value such that the sum of modifications is minimized? Create a difference array to efficiently sum all the modifications. |
Sweep line | minimum-moves-to-make-array-complementary | 0 | 1 | # Code\n```\nclass Solution:\n def minMoves(self, nums: List[int], limit: int) -> int:\n cnt = Counter()\n i, n = 0, len(nums)\n p = [0] * (2 * limit + 2)\n while i < n // 2:\n cnt[nums[i] + nums[n-1-i]] += 1\n p[min(nums[i], nums[n-1-i]) + 1] += 1\n p[max... | 0 | You are given an integer array `nums` of **even** length `n` and an integer `limit`. In one move, you can replace any integer from `nums` with another integer between `1` and `limit`, inclusive.
The array `nums` is **complementary** if for all indices `i` (**0-indexed**), `nums[i] + nums[n - 1 - i]` equals the same nu... | Build the array arr using the given formula, define target = sum(arr) / n What is the number of operations needed to convert arr so that all elements equal target ? |
Sweep line | minimum-moves-to-make-array-complementary | 0 | 1 | # Code\n```\nclass Solution:\n def minMoves(self, nums: List[int], limit: int) -> int:\n cnt = Counter()\n i, n = 0, len(nums)\n p = [0] * (2 * limit + 2)\n while i < n // 2:\n cnt[nums[i] + nums[n-1-i]] += 1\n p[min(nums[i], nums[n-1-i]) + 1] += 1\n p[max... | 0 | You are given an array of integers `nums` **(0-indexed)** and an integer `k`.
The **score** of a subarray `(i, j)` is defined as `min(nums[i], nums[i+1], ..., nums[j]) * (j - i + 1)`. A **good** subarray is a subarray where `i <= k <= j`.
Return _the maximum possible **score** of a **good** subarray._
**Example 1:**... | Given a target sum x, each pair of nums[i] and nums[n-1-i] would either need 0, 1, or 2 modifications. Can you find the optimal target sum x value such that the sum of modifications is minimized? Create a difference array to efficiently sum all the modifications. |
clean code with good explanation | minimum-moves-to-make-array-complementary | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n##### Let `l` be the `nums[i]`, and `r` be the `nums[n-i-1]`\n1. All pair sum must in the range `[2, limit*2]`.\n2. By one step modification, the max sum that we can get is `max(l, r) + limit` and the min sum that we can get is `min(l, r)... | 0 | You are given an integer array `nums` of **even** length `n` and an integer `limit`. In one move, you can replace any integer from `nums` with another integer between `1` and `limit`, inclusive.
The array `nums` is **complementary** if for all indices `i` (**0-indexed**), `nums[i] + nums[n - 1 - i]` equals the same nu... | Build the array arr using the given formula, define target = sum(arr) / n What is the number of operations needed to convert arr so that all elements equal target ? |
clean code with good explanation | minimum-moves-to-make-array-complementary | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n##### Let `l` be the `nums[i]`, and `r` be the `nums[n-i-1]`\n1. All pair sum must in the range `[2, limit*2]`.\n2. By one step modification, the max sum that we can get is `max(l, r) + limit` and the min sum that we can get is `min(l, r)... | 0 | You are given an array of integers `nums` **(0-indexed)** and an integer `k`.
The **score** of a subarray `(i, j)` is defined as `min(nums[i], nums[i+1], ..., nums[j]) * (j - i + 1)`. A **good** subarray is a subarray where `i <= k <= j`.
Return _the maximum possible **score** of a **good** subarray._
**Example 1:**... | Given a target sum x, each pair of nums[i] and nums[n-1-i] would either need 0, 1, or 2 modifications. Can you find the optimal target sum x value such that the sum of modifications is minimized? Create a difference array to efficiently sum all the modifications. |
Solution | minimum-moves-to-make-array-complementary | 1 | 1 | ```C++ []\nstatic const auto fast = []() {ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0); return 0; } ();\nclass Solution {\npublic:\n int minMoves(vector<int>& nums, int limit) {\n int n=nums.size(),c2[2*limit+2],c3[2*limit+1],ans=n,t1,t2;\n memset(c2,0,sizeof(c2));\n memset(c3,0,siz... | 0 | You are given an integer array `nums` of **even** length `n` and an integer `limit`. In one move, you can replace any integer from `nums` with another integer between `1` and `limit`, inclusive.
The array `nums` is **complementary** if for all indices `i` (**0-indexed**), `nums[i] + nums[n - 1 - i]` equals the same nu... | Build the array arr using the given formula, define target = sum(arr) / n What is the number of operations needed to convert arr so that all elements equal target ? |
Solution | minimum-moves-to-make-array-complementary | 1 | 1 | ```C++ []\nstatic const auto fast = []() {ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0); return 0; } ();\nclass Solution {\npublic:\n int minMoves(vector<int>& nums, int limit) {\n int n=nums.size(),c2[2*limit+2],c3[2*limit+1],ans=n,t1,t2;\n memset(c2,0,sizeof(c2));\n memset(c3,0,siz... | 0 | You are given an array of integers `nums` **(0-indexed)** and an integer `k`.
The **score** of a subarray `(i, j)` is defined as `min(nums[i], nums[i+1], ..., nums[j]) * (j - i + 1)`. A **good** subarray is a subarray where `i <= k <= j`.
Return _the maximum possible **score** of a **good** subarray._
**Example 1:**... | Given a target sum x, each pair of nums[i] and nums[n-1-i] would either need 0, 1, or 2 modifications. Can you find the optimal target sum x value such that the sum of modifications is minimized? Create a difference array to efficiently sum all the modifications. |
Day 55 || Priority_Queue || Easiest Beginner Friendly Sol | minimize-deviation-in-array | 1 | 1 | # Intuition of this Problem:\nThe intuition behind this logic is to reduce the maximum difference between any two elements in the array by either decreasing the maximum value or increasing the minimum value.\n\nBy transforming all odd numbers to even numbers, we can always divide even numbers by 2, so the maximum value... | 306 | You are given an array `nums` of `n` positive integers.
You can perform two types of operations on any element of the array any number of times:
* If the element is **even**, **divide** it by `2`.
* For example, if the array is `[1,2,3,4]`, then you can do this operation on the last element, and the array wil... | If you can place balls such that the answer is x then you can do it for y where y < x. Similarly if you cannot place balls such that the answer is x then you can do it for y where y > x. Binary search on the answer and greedily see if it is possible. |
Day 55 || Priority_Queue || Easiest Beginner Friendly Sol | minimize-deviation-in-array | 1 | 1 | # Intuition of this Problem:\nThe intuition behind this logic is to reduce the maximum difference between any two elements in the array by either decreasing the maximum value or increasing the minimum value.\n\nBy transforming all odd numbers to even numbers, we can always divide even numbers by 2, so the maximum value... | 306 | You are given two strings `firstString` and `secondString` that are **0-indexed** and consist only of lowercase English letters. Count the number of index quadruples `(i,j,a,b)` that satisfy the following conditions:
* `0 <= i <= j < firstString.length`
* `0 <= a <= b < secondString.length`
* The substring of `f... | Assume you start with the minimum possible value for each number so you can only multiply a number by 2 till it reaches its maximum possible value. If there is a better solution than the current one, then it must have either its maximum value less than the current maximum value, or the minimum value larger than the cur... |
Python easy to understand | Max Heap | minimize-deviation-in-array | 0 | 1 | # Approach\n<!-- Describe your approach to solving the problem. -->\n###### Create a `Max_Heap` having maximum possible value for each element. Repeatedly pop the `top` (max element) of the heap and divide it by 2 until the `top` of the heap is even and update the minimum deviation for each step.\n\n# Complexity\n- Tim... | 1 | You are given an array `nums` of `n` positive integers.
You can perform two types of operations on any element of the array any number of times:
* If the element is **even**, **divide** it by `2`.
* For example, if the array is `[1,2,3,4]`, then you can do this operation on the last element, and the array wil... | If you can place balls such that the answer is x then you can do it for y where y < x. Similarly if you cannot place balls such that the answer is x then you can do it for y where y > x. Binary search on the answer and greedily see if it is possible. |
Python easy to understand | Max Heap | minimize-deviation-in-array | 0 | 1 | # Approach\n<!-- Describe your approach to solving the problem. -->\n###### Create a `Max_Heap` having maximum possible value for each element. Repeatedly pop the `top` (max element) of the heap and divide it by 2 until the `top` of the heap is even and update the minimum deviation for each step.\n\n# Complexity\n- Tim... | 1 | You are given two strings `firstString` and `secondString` that are **0-indexed** and consist only of lowercase English letters. Count the number of index quadruples `(i,j,a,b)` that satisfy the following conditions:
* `0 <= i <= j < firstString.length`
* `0 <= a <= b < secondString.length`
* The substring of `f... | Assume you start with the minimum possible value for each number so you can only multiply a number by 2 till it reaches its maximum possible value. If there is a better solution than the current one, then it must have either its maximum value less than the current maximum value, or the minimum value larger than the cur... |
Minimize Deviation in Array - Python Heap solution | minimize-deviation-in-array | 0 | 1 | # Approach\nWe will use priority heap.\n\n1. Multiply all odd numbers by 2 => we get maximum possible values of all numbers in nums. Now all numbers are even.\n\n2. We will use priotrity heap. But we need multiply all numbers by (-1), because we will take maximum element of heap, not minimum.\n\n3. Keep track min numbe... | 1 | You are given an array `nums` of `n` positive integers.
You can perform two types of operations on any element of the array any number of times:
* If the element is **even**, **divide** it by `2`.
* For example, if the array is `[1,2,3,4]`, then you can do this operation on the last element, and the array wil... | If you can place balls such that the answer is x then you can do it for y where y < x. Similarly if you cannot place balls such that the answer is x then you can do it for y where y > x. Binary search on the answer and greedily see if it is possible. |
Minimize Deviation in Array - Python Heap solution | minimize-deviation-in-array | 0 | 1 | # Approach\nWe will use priority heap.\n\n1. Multiply all odd numbers by 2 => we get maximum possible values of all numbers in nums. Now all numbers are even.\n\n2. We will use priotrity heap. But we need multiply all numbers by (-1), because we will take maximum element of heap, not minimum.\n\n3. Keep track min numbe... | 1 | You are given two strings `firstString` and `secondString` that are **0-indexed** and consist only of lowercase English letters. Count the number of index quadruples `(i,j,a,b)` that satisfy the following conditions:
* `0 <= i <= j < firstString.length`
* `0 <= a <= b < secondString.length`
* The substring of `f... | Assume you start with the minimum possible value for each number so you can only multiply a number by 2 till it reaches its maximum possible value. If there is a better solution than the current one, then it must have either its maximum value less than the current maximum value, or the minimum value larger than the cur... |
String Replacement Approach in Python | goal-parser-interpretation | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nWe can solve this problem by replacing specific substrings in the input string with their corresponding interpretations.\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\nWe can return a copy of the given address wit... | 2 | You own a **Goal Parser** that can interpret a string `command`. The `command` consists of an alphabet of `"G "`, `"() "` and/or `"(al) "` in some order. The Goal Parser will interpret `"G "` as the string `"G "`, `"() "` as the string `"o "`, and `"(al) "` as the string `"al "`. The interpreted strings are then concat... | There is no way if the sum (number of '1's) is not divisible by the number of splits. So sum%3 should be 0. Preffix s1 , and suffix s3 should have sum/3 characters '1'. Follow up: Can you generalize the problem with numbers between [-10^9, 10^9] such the sum between subarrays s1, s2, s3 are the same? |
String Replacement Approach in Python | goal-parser-interpretation | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nWe can solve this problem by replacing specific substrings in the input string with their corresponding interpretations.\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\nWe can return a copy of the given address wit... | 2 | There is an authentication system that works with authentication tokens. For each session, the user will receive a new authentication token that will expire `timeToLive` seconds after the `currentTime`. If the token is renewed, the expiry time will be **extended** to expire `timeToLive` seconds after the (potentially d... | You need to check at most 2 characters to determine which character comes next. |
my variant | goal-parser-interpretation | 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 own a **Goal Parser** that can interpret a string `command`. The `command` consists of an alphabet of `"G "`, `"() "` and/or `"(al) "` in some order. The Goal Parser will interpret `"G "` as the string `"G "`, `"() "` as the string `"o "`, and `"(al) "` as the string `"al "`. The interpreted strings are then concat... | There is no way if the sum (number of '1's) is not divisible by the number of splits. So sum%3 should be 0. Preffix s1 , and suffix s3 should have sum/3 characters '1'. Follow up: Can you generalize the problem with numbers between [-10^9, 10^9] such the sum between subarrays s1, s2, s3 are the same? |
my variant | goal-parser-interpretation | 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 | There is an authentication system that works with authentication tokens. For each session, the user will receive a new authentication token that will expire `timeToLive` seconds after the `currentTime`. If the token is renewed, the expiry time will be **extended** to expire `timeToLive` seconds after the (potentially d... | You need to check at most 2 characters to determine which character comes next. |
Goal Parser Interpretation (string replacement ) | goal-parser-interpretation | 0 | 1 | # No Loops - using in-built Methods\n\n\n# Complexity\n- Time complexity:\n- O(n)\nn is the length of the input IP address string. In the worst case, we may need to traverse the whole string.\n\n\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n- O(n)\nn is the length of the input IP addr... | 1 | You own a **Goal Parser** that can interpret a string `command`. The `command` consists of an alphabet of `"G "`, `"() "` and/or `"(al) "` in some order. The Goal Parser will interpret `"G "` as the string `"G "`, `"() "` as the string `"o "`, and `"(al) "` as the string `"al "`. The interpreted strings are then concat... | There is no way if the sum (number of '1's) is not divisible by the number of splits. So sum%3 should be 0. Preffix s1 , and suffix s3 should have sum/3 characters '1'. Follow up: Can you generalize the problem with numbers between [-10^9, 10^9] such the sum between subarrays s1, s2, s3 are the same? |
Goal Parser Interpretation (string replacement ) | goal-parser-interpretation | 0 | 1 | # No Loops - using in-built Methods\n\n\n# Complexity\n- Time complexity:\n- O(n)\nn is the length of the input IP address string. In the worst case, we may need to traverse the whole string.\n\n\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n- O(n)\nn is the length of the input IP addr... | 1 | There is an authentication system that works with authentication tokens. For each session, the user will receive a new authentication token that will expire `timeToLive` seconds after the `currentTime`. If the token is renewed, the expiry time will be **extended** to expire `timeToLive` seconds after the (potentially d... | You need to check at most 2 characters to determine which character comes next. |
Python solution using regex | goal-parser-interpretation | 0 | 1 | # Regex\n\n# Code\n```\nclass Solution:\n def interpret(self, command: str) -> str:\n\n def replacing(_str: str) -> str:\n content = _str.group(1)\n \n if content == "()":\n return "o"\n else:\n return "al"\n\n return sub(r"(\\(.... | 2 | You own a **Goal Parser** that can interpret a string `command`. The `command` consists of an alphabet of `"G "`, `"() "` and/or `"(al) "` in some order. The Goal Parser will interpret `"G "` as the string `"G "`, `"() "` as the string `"o "`, and `"(al) "` as the string `"al "`. The interpreted strings are then concat... | There is no way if the sum (number of '1's) is not divisible by the number of splits. So sum%3 should be 0. Preffix s1 , and suffix s3 should have sum/3 characters '1'. Follow up: Can you generalize the problem with numbers between [-10^9, 10^9] such the sum between subarrays s1, s2, s3 are the same? |
Python solution using regex | goal-parser-interpretation | 0 | 1 | # Regex\n\n# Code\n```\nclass Solution:\n def interpret(self, command: str) -> str:\n\n def replacing(_str: str) -> str:\n content = _str.group(1)\n \n if content == "()":\n return "o"\n else:\n return "al"\n\n return sub(r"(\\(.... | 2 | There is an authentication system that works with authentication tokens. For each session, the user will receive a new authentication token that will expire `timeToLive` seconds after the `currentTime`. If the token is renewed, the expiry time will be **extended** to expire `timeToLive` seconds after the (potentially d... | You need to check at most 2 characters to determine which character comes next. |
Python3 easiest and understable solution FIZMAT ON TOP | goal-parser-interpretation | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. --> I started to think that i need to append smth to my string, after i did code with "for", but it didnt work as well, bc there was default count i, like i+=1, so i did with while, that work enough well.\n\n# Approach\n<!-- Describe your appr... | 2 | You own a **Goal Parser** that can interpret a string `command`. The `command` consists of an alphabet of `"G "`, `"() "` and/or `"(al) "` in some order. The Goal Parser will interpret `"G "` as the string `"G "`, `"() "` as the string `"o "`, and `"(al) "` as the string `"al "`. The interpreted strings are then concat... | There is no way if the sum (number of '1's) is not divisible by the number of splits. So sum%3 should be 0. Preffix s1 , and suffix s3 should have sum/3 characters '1'. Follow up: Can you generalize the problem with numbers between [-10^9, 10^9] such the sum between subarrays s1, s2, s3 are the same? |
Python3 easiest and understable solution FIZMAT ON TOP | goal-parser-interpretation | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. --> I started to think that i need to append smth to my string, after i did code with "for", but it didnt work as well, bc there was default count i, like i+=1, so i did with while, that work enough well.\n\n# Approach\n<!-- Describe your appr... | 2 | There is an authentication system that works with authentication tokens. For each session, the user will receive a new authentication token that will expire `timeToLive` seconds after the `currentTime`. If the token is renewed, the expiry time will be **extended** to expire `timeToLive` seconds after the (potentially d... | You need to check at most 2 characters to determine which character comes next. |
Python one-liner | goal-parser-interpretation | 0 | 1 | ```\nclass Solution:\n def interpret(self, command: str) -> str:\n return command.replace(\'()\',\'o\').replace(\'(al)\',\'al\')\n``` | 119 | You own a **Goal Parser** that can interpret a string `command`. The `command` consists of an alphabet of `"G "`, `"() "` and/or `"(al) "` in some order. The Goal Parser will interpret `"G "` as the string `"G "`, `"() "` as the string `"o "`, and `"(al) "` as the string `"al "`. The interpreted strings are then concat... | There is no way if the sum (number of '1's) is not divisible by the number of splits. So sum%3 should be 0. Preffix s1 , and suffix s3 should have sum/3 characters '1'. Follow up: Can you generalize the problem with numbers between [-10^9, 10^9] such the sum between subarrays s1, s2, s3 are the same? |
Python one-liner | goal-parser-interpretation | 0 | 1 | ```\nclass Solution:\n def interpret(self, command: str) -> str:\n return command.replace(\'()\',\'o\').replace(\'(al)\',\'al\')\n``` | 119 | There is an authentication system that works with authentication tokens. For each session, the user will receive a new authentication token that will expire `timeToLive` seconds after the `currentTime`. If the token is renewed, the expiry time will be **extended** to expire `timeToLive` seconds after the (potentially d... | You need to check at most 2 characters to determine which character comes next. |
[Python \ C++ ] Simple solution using dictionary | goal-parser-interpretation | 0 | 1 | Python\n```\n def interpret(self, s: str) -> str:\n d = {"(al)":"al", "()":"o","G":"G"}\n tmp= ""\n res=""\n for i in range(len(s)):\n tmp+=s[i]\n if(tmp in d):\n res+=d[tmp]\n tmp=""\n return res\n```\n\nC++\n\n```\nstring interpret... | 83 | You own a **Goal Parser** that can interpret a string `command`. The `command` consists of an alphabet of `"G "`, `"() "` and/or `"(al) "` in some order. The Goal Parser will interpret `"G "` as the string `"G "`, `"() "` as the string `"o "`, and `"(al) "` as the string `"al "`. The interpreted strings are then concat... | There is no way if the sum (number of '1's) is not divisible by the number of splits. So sum%3 should be 0. Preffix s1 , and suffix s3 should have sum/3 characters '1'. Follow up: Can you generalize the problem with numbers between [-10^9, 10^9] such the sum between subarrays s1, s2, s3 are the same? |
[Python \ C++ ] Simple solution using dictionary | goal-parser-interpretation | 0 | 1 | Python\n```\n def interpret(self, s: str) -> str:\n d = {"(al)":"al", "()":"o","G":"G"}\n tmp= ""\n res=""\n for i in range(len(s)):\n tmp+=s[i]\n if(tmp in d):\n res+=d[tmp]\n tmp=""\n return res\n```\n\nC++\n\n```\nstring interpret... | 83 | There is an authentication system that works with authentication tokens. For each session, the user will receive a new authentication token that will expire `timeToLive` seconds after the `currentTime`. If the token is renewed, the expiry time will be **extended** to expire `timeToLive` seconds after the (potentially d... | You need to check at most 2 characters to determine which character comes next. |
A very simple approach in Python | goal-parser-interpretation | 0 | 1 | \n\n# Code\n```\nclass Solution:\n def interpret(self, command: str) -> str:\n newstr = command.replace(\'()\',\'o\')\n return newstr.replace(\'(al)\',\'al\')\n``` | 4 | You own a **Goal Parser** that can interpret a string `command`. The `command` consists of an alphabet of `"G "`, `"() "` and/or `"(al) "` in some order. The Goal Parser will interpret `"G "` as the string `"G "`, `"() "` as the string `"o "`, and `"(al) "` as the string `"al "`. The interpreted strings are then concat... | There is no way if the sum (number of '1's) is not divisible by the number of splits. So sum%3 should be 0. Preffix s1 , and suffix s3 should have sum/3 characters '1'. Follow up: Can you generalize the problem with numbers between [-10^9, 10^9] such the sum between subarrays s1, s2, s3 are the same? |
A very simple approach in Python | goal-parser-interpretation | 0 | 1 | \n\n# Code\n```\nclass Solution:\n def interpret(self, command: str) -> str:\n newstr = command.replace(\'()\',\'o\')\n return newstr.replace(\'(al)\',\'al\')\n``` | 4 | There is an authentication system that works with authentication tokens. For each session, the user will receive a new authentication token that will expire `timeToLive` seconds after the `currentTime`. If the token is renewed, the expiry time will be **extended** to expire `timeToLive` seconds after the (potentially d... | You need to check at most 2 characters to determine which character comes next. |
Python Solution 3 Ways!! For Loop | Regex | Replace Method | goal-parser-interpretation | 0 | 1 | # Using For Loops - No Modules, No Methods\n```\nclass Solution:\n def interpret(self, command: str) -> str:\n \n message: str = ""\n \n for index, char in enumerate(command):\n if command[index: index + 4] == "(al)":\n message += "al"\n elif char == "(":\... | 2 | You own a **Goal Parser** that can interpret a string `command`. The `command` consists of an alphabet of `"G "`, `"() "` and/or `"(al) "` in some order. The Goal Parser will interpret `"G "` as the string `"G "`, `"() "` as the string `"o "`, and `"(al) "` as the string `"al "`. The interpreted strings are then concat... | There is no way if the sum (number of '1's) is not divisible by the number of splits. So sum%3 should be 0. Preffix s1 , and suffix s3 should have sum/3 characters '1'. Follow up: Can you generalize the problem with numbers between [-10^9, 10^9] such the sum between subarrays s1, s2, s3 are the same? |
Python Solution 3 Ways!! For Loop | Regex | Replace Method | goal-parser-interpretation | 0 | 1 | # Using For Loops - No Modules, No Methods\n```\nclass Solution:\n def interpret(self, command: str) -> str:\n \n message: str = ""\n \n for index, char in enumerate(command):\n if command[index: index + 4] == "(al)":\n message += "al"\n elif char == "(":\... | 2 | There is an authentication system that works with authentication tokens. For each session, the user will receive a new authentication token that will expire `timeToLive` seconds after the `currentTime`. If the token is renewed, the expiry time will be **extended** to expire `timeToLive` seconds after the (potentially d... | You need to check at most 2 characters to determine which character comes next. |
[Python] 4 ways to attack | goal-parser-interpretation | 0 | 1 | Just while loop with carefully incrementing counter is good enough:\n```\n# runtime beats 32% ; memory beats 99.76%\ndef interpret(command):\n \n return_string = ""\n i = 0\n while i < len(command):\n if command[i] == "G":\n return_string += \'G\'\n i += 1\n elif command[i:i + 2] == \'()\':\n r... | 38 | You own a **Goal Parser** that can interpret a string `command`. The `command` consists of an alphabet of `"G "`, `"() "` and/or `"(al) "` in some order. The Goal Parser will interpret `"G "` as the string `"G "`, `"() "` as the string `"o "`, and `"(al) "` as the string `"al "`. The interpreted strings are then concat... | There is no way if the sum (number of '1's) is not divisible by the number of splits. So sum%3 should be 0. Preffix s1 , and suffix s3 should have sum/3 characters '1'. Follow up: Can you generalize the problem with numbers between [-10^9, 10^9] such the sum between subarrays s1, s2, s3 are the same? |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.