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 |
|---|---|---|---|---|---|---|---|
Python - dict, deque, and chain star | diagonal-traverse-ii | 0 | 1 | ```\nclass Solution:\n def findDiagonalOrder(self, nums: List[List[int]]) -> List[int]:\n res = defaultdict(deque)\n for i, row in enumerate(nums):\n for j, x in enumerate(row):\n res[i + j].appendleft(x)\n return chain(*res.values())\n``` | 6 | Given a 2D integer array `nums`, return _all elements of_ `nums` _in diagonal order as shown in the below images_.
**Example 1:**
**Input:** nums = \[\[1,2,3\],\[4,5,6\],\[7,8,9\]\]
**Output:** \[1,4,2,7,5,3,8,6,9\]
**Example 2:**
**Input:** nums = \[\[1,2,3,4,5\],\[6,7\],\[8\],\[9,10,11\],\[12,13,14,15,16\]\]
**Ou... | Use Breadth First Search (BFS) to traverse all possible boxes you can open. Only push to the queue the boxes the you have with their keys. |
Python - dict, deque, and chain star | diagonal-traverse-ii | 0 | 1 | ```\nclass Solution:\n def findDiagonalOrder(self, nums: List[List[int]]) -> List[int]:\n res = defaultdict(deque)\n for i, row in enumerate(nums):\n for j, x in enumerate(row):\n res[i + j].appendleft(x)\n return chain(*res.values())\n``` | 6 | 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. |
(●'◡'●) Dicts are OP💕.py | diagonal-traverse-ii | 0 | 1 | # Intuition\nWe can definitely think on something like (i+j) elements are together\n# Code\n```js\nclass Solution:\n def findDiagonalOrder(self, nums: List[List[int]]) -> List[int]:\n mp,ans={},[]\n for i in range(len(nums)):\n for j in range(len(nums[i])):\n if i+j not in mp:... | 11 | Given a 2D integer array `nums`, return _all elements of_ `nums` _in diagonal order as shown in the below images_.
**Example 1:**
**Input:** nums = \[\[1,2,3\],\[4,5,6\],\[7,8,9\]\]
**Output:** \[1,4,2,7,5,3,8,6,9\]
**Example 2:**
**Input:** nums = \[\[1,2,3,4,5\],\[6,7\],\[8\],\[9,10,11\],\[12,13,14,15,16\]\]
**Ou... | Use Breadth First Search (BFS) to traverse all possible boxes you can open. Only push to the queue the boxes the you have with their keys. |
(●'◡'●) Dicts are OP💕.py | diagonal-traverse-ii | 0 | 1 | # Intuition\nWe can definitely think on something like (i+j) elements are together\n# Code\n```js\nclass Solution:\n def findDiagonalOrder(self, nums: List[List[int]]) -> List[int]:\n mp,ans={},[]\n for i in range(len(nums)):\n for j in range(len(nums[i])):\n if i+j not in mp:... | 11 | 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. |
✅ One Line Solution | diagonal-traverse-ii | 0 | 1 | # Complexity\n- Time complexity: $$O(n*log(n))$$.\n\n- Space complexity: $$O(n)$$.\n\n# Code - Oneliner\n```\nclass Solution:\n def findDiagonalOrder(self, nums: List[List[int]]) -> List[int]:\n return [num for _, _, num in sorted((i+j, -i, nums[i][j]) for i in range(len(nums)) for j in range(len(nums[i])))]\... | 6 | Given a 2D integer array `nums`, return _all elements of_ `nums` _in diagonal order as shown in the below images_.
**Example 1:**
**Input:** nums = \[\[1,2,3\],\[4,5,6\],\[7,8,9\]\]
**Output:** \[1,4,2,7,5,3,8,6,9\]
**Example 2:**
**Input:** nums = \[\[1,2,3,4,5\],\[6,7\],\[8\],\[9,10,11\],\[12,13,14,15,16\]\]
**Ou... | Use Breadth First Search (BFS) to traverse all possible boxes you can open. Only push to the queue the boxes the you have with their keys. |
✅ One Line Solution | diagonal-traverse-ii | 0 | 1 | # Complexity\n- Time complexity: $$O(n*log(n))$$.\n\n- Space complexity: $$O(n)$$.\n\n# Code - Oneliner\n```\nclass Solution:\n def findDiagonalOrder(self, nums: List[List[int]]) -> List[int]:\n return [num for _, _, num in sorted((i+j, -i, nums[i][j]) for i in range(len(nums)) for j in range(len(nums[i])))]\... | 6 | 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 Solution || Beginner Friendly || Easy to Understand ✅ | diagonal-traverse-ii | 1 | 1 | # Approach\n<!-- Describe your approach to solving the problem. -->\n$$Initialization:$$\n- Initialize a variable `count` to keep track of the total number of elements in the result array.\n- Create an ArrayList of ArrayLists called `lists` to store the elements diagonally.\n\n$$Traversing the Matrix Diagonally:$$\n- U... | 87 | Given a 2D integer array `nums`, return _all elements of_ `nums` _in diagonal order as shown in the below images_.
**Example 1:**
**Input:** nums = \[\[1,2,3\],\[4,5,6\],\[7,8,9\]\]
**Output:** \[1,4,2,7,5,3,8,6,9\]
**Example 2:**
**Input:** nums = \[\[1,2,3,4,5\],\[6,7\],\[8\],\[9,10,11\],\[12,13,14,15,16\]\]
**Ou... | Use Breadth First Search (BFS) to traverse all possible boxes you can open. Only push to the queue the boxes the you have with their keys. |
Simple Solution || Beginner Friendly || Easy to Understand ✅ | diagonal-traverse-ii | 1 | 1 | # Approach\n<!-- Describe your approach to solving the problem. -->\n$$Initialization:$$\n- Initialize a variable `count` to keep track of the total number of elements in the result array.\n- Create an ArrayList of ArrayLists called `lists` to store the elements diagonally.\n\n$$Traversing the Matrix Diagonally:$$\n- U... | 87 | 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. |
EEzz simple Python solution using Hashmap/deafulatdict(list)! | diagonal-traverse-ii | 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(i+j) for same\xA0diagonal elements is equal. \nThey will be added to a result(res) list after being stored in a hashmap. (To obtain the upward diagonal traversal, rev... | 3 | Given a 2D integer array `nums`, return _all elements of_ `nums` _in diagonal order as shown in the below images_.
**Example 1:**
**Input:** nums = \[\[1,2,3\],\[4,5,6\],\[7,8,9\]\]
**Output:** \[1,4,2,7,5,3,8,6,9\]
**Example 2:**
**Input:** nums = \[\[1,2,3,4,5\],\[6,7\],\[8\],\[9,10,11\],\[12,13,14,15,16\]\]
**Ou... | Use Breadth First Search (BFS) to traverse all possible boxes you can open. Only push to the queue the boxes the you have with their keys. |
EEzz simple Python solution using Hashmap/deafulatdict(list)! | diagonal-traverse-ii | 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(i+j) for same\xA0diagonal elements is equal. \nThey will be added to a result(res) list after being stored in a hashmap. (To obtain the upward diagonal traversal, rev... | 3 | 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, Rust] Elegant & Short | O(n) | Ordered Map | diagonal-traverse-ii | 0 | 1 | # Complexity\n- Time complexity: $$O(n)$$\n- Space complexity: $$O(n)$$\n\n# Code\n\n```python []\nfrom collections import defaultdict, deque\n\nclass Solution:\n def findDiagonalOrder(self, matrix: list[list[int]]) -> list[int]:\n diagonals = defaultdict(deque)\n\n for i, row in enumerate(matrix):\n ... | 2 | Given a 2D integer array `nums`, return _all elements of_ `nums` _in diagonal order as shown in the below images_.
**Example 1:**
**Input:** nums = \[\[1,2,3\],\[4,5,6\],\[7,8,9\]\]
**Output:** \[1,4,2,7,5,3,8,6,9\]
**Example 2:**
**Input:** nums = \[\[1,2,3,4,5\],\[6,7\],\[8\],\[9,10,11\],\[12,13,14,15,16\]\]
**Ou... | Use Breadth First Search (BFS) to traverse all possible boxes you can open. Only push to the queue the boxes the you have with their keys. |
[Python, Rust] Elegant & Short | O(n) | Ordered Map | diagonal-traverse-ii | 0 | 1 | # Complexity\n- Time complexity: $$O(n)$$\n- Space complexity: $$O(n)$$\n\n# Code\n\n```python []\nfrom collections import defaultdict, deque\n\nclass Solution:\n def findDiagonalOrder(self, matrix: list[list[int]]) -> list[int]:\n diagonals = defaultdict(deque)\n\n for i, row in enumerate(matrix):\n ... | 2 | 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. |
【Video】Give me 9 minutes - How we think about a solution | diagonal-traverse-ii | 1 | 1 | # Intuition\nUsing BFS\n\n---\n\n# Solution Video\n\nhttps://youtu.be/4wkt4kGip64\n\n\u25A0 Timeline of the video\n`0:04` How we think about a solution\n`1:08` How can you put numbers into queue with right order?\n`2:34` Demonstrate how it works\n`6:24` Summary of main idea\n`6:47` Coding\n`8:40` Time Complexity and Sp... | 38 | Given a 2D integer array `nums`, return _all elements of_ `nums` _in diagonal order as shown in the below images_.
**Example 1:**
**Input:** nums = \[\[1,2,3\],\[4,5,6\],\[7,8,9\]\]
**Output:** \[1,4,2,7,5,3,8,6,9\]
**Example 2:**
**Input:** nums = \[\[1,2,3,4,5\],\[6,7\],\[8\],\[9,10,11\],\[12,13,14,15,16\]\]
**Ou... | Use Breadth First Search (BFS) to traverse all possible boxes you can open. Only push to the queue the boxes the you have with their keys. |
【Video】Give me 9 minutes - How we think about a solution | diagonal-traverse-ii | 1 | 1 | # Intuition\nUsing BFS\n\n---\n\n# Solution Video\n\nhttps://youtu.be/4wkt4kGip64\n\n\u25A0 Timeline of the video\n`0:04` How we think about a solution\n`1:08` How can you put numbers into queue with right order?\n`2:34` Demonstrate how it works\n`6:24` Summary of main idea\n`6:47` Coding\n`8:40` Time Complexity and Sp... | 38 | 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. |
All pairs of (i+j) that sums up to same value are diagonal | diagonal-traverse-ii | 0 | 1 | # Code\n```\nclass Solution:\n def findDiagonalOrder(self, nums: List[List[int]]) -> List[int]:\n n,m=len(nums),0\n for i in nums:\n m=max(m,len(i))\n ans=[[] for i in range(n+m-1)]\n for i in range(n):\n for j in range(len(nums[i])):\n ans[i+j].append... | 1 | Given a 2D integer array `nums`, return _all elements of_ `nums` _in diagonal order as shown in the below images_.
**Example 1:**
**Input:** nums = \[\[1,2,3\],\[4,5,6\],\[7,8,9\]\]
**Output:** \[1,4,2,7,5,3,8,6,9\]
**Example 2:**
**Input:** nums = \[\[1,2,3,4,5\],\[6,7\],\[8\],\[9,10,11\],\[12,13,14,15,16\]\]
**Ou... | Use Breadth First Search (BFS) to traverse all possible boxes you can open. Only push to the queue the boxes the you have with their keys. |
All pairs of (i+j) that sums up to same value are diagonal | diagonal-traverse-ii | 0 | 1 | # Code\n```\nclass Solution:\n def findDiagonalOrder(self, nums: List[List[int]]) -> List[int]:\n n,m=len(nums),0\n for i in nums:\n m=max(m,len(i))\n ans=[[] for i in range(n+m-1)]\n for i in range(n):\n for j in range(len(nums[i])):\n ans[i+j].append... | 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 Solution for noobs | diagonal-traverse-ii | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nThe code is about processing a 2D list (nums) diagonally and returning the elements in a specific order. It does this by first mapping elements of the 2D list to diagonals, and then sorting these diagonals based on their indices. Finally,... | 1 | Given a 2D integer array `nums`, return _all elements of_ `nums` _in diagonal order as shown in the below images_.
**Example 1:**
**Input:** nums = \[\[1,2,3\],\[4,5,6\],\[7,8,9\]\]
**Output:** \[1,4,2,7,5,3,8,6,9\]
**Example 2:**
**Input:** nums = \[\[1,2,3,4,5\],\[6,7\],\[8\],\[9,10,11\],\[12,13,14,15,16\]\]
**Ou... | Use Breadth First Search (BFS) to traverse all possible boxes you can open. Only push to the queue the boxes the you have with their keys. |
Simple Solution for noobs | diagonal-traverse-ii | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nThe code is about processing a 2D list (nums) diagonally and returning the elements in a specific order. It does this by first mapping elements of the 2D list to diagonals, and then sorting these diagonals based on their indices. Finally,... | 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. |
Ordinary Heap-Based Approach | constrained-subsequence-sum | 0 | 1 | # Complexity\n- Time complexity: $$O(n*log(n))$$.\n\n- Space complexity: $$O(n)$$.\n\n# Code\n```\nclass Solution:\n def constrainedSubsetSum(self, nums: List[int], k: int) -> int:\n SUM, IND = 0, 1\n\n if all(num < 0 for num in nums):\n return max(nums)\n\n heap = [(0, math.inf)]\n ... | 3 | Given an integer array `nums` and an integer `k`, return the maximum sum of a **non-empty** subsequence of that array such that for every two **consecutive** integers in the subsequence, `nums[i]` and `nums[j]`, where `i < j`, the condition `j - i <= k` is satisfied.
A _subsequence_ of an array is obtained by deleting... | null |
Ordinary Heap-Based Approach | constrained-subsequence-sum | 0 | 1 | # Complexity\n- Time complexity: $$O(n*log(n))$$.\n\n- Space complexity: $$O(n)$$.\n\n# Code\n```\nclass Solution:\n def constrainedSubsetSum(self, nums: List[int], k: int) -> int:\n SUM, IND = 0, 1\n\n if all(num < 0 for num in nums):\n return max(nums)\n\n heap = [(0, math.inf)]\n ... | 3 | Design the `CombinationIterator` class:
* `CombinationIterator(string characters, int combinationLength)` Initializes the object with a string `characters` of **sorted distinct** lowercase English letters and a number `combinationLength` as arguments.
* `next()` Returns the next combination of length `combinationL... | Use dynamic programming. Let dp[i] be the solution for the prefix of the array that ends at index i, if the element at index i is in the subsequence. dp[i] = nums[i] + max(0, dp[i-k], dp[i-k+1], ..., dp[i-1]) Use a heap with the sliding window technique to optimize the dp. |
Python3 Solution | constrained-subsequence-sum | 0 | 1 | \n```\nclass Solution:\n def constrainedSubsetSum(self, nums: List[int], k: int) -> int:\n deque=[]\n for i,num in enumerate(nums):\n while deque and deque[0]<i-k:\n deque.pop(0)\n if deque:\n nums[i]=nums[deque[0]]+num\n while deque and nu... | 2 | Given an integer array `nums` and an integer `k`, return the maximum sum of a **non-empty** subsequence of that array such that for every two **consecutive** integers in the subsequence, `nums[i]` and `nums[j]`, where `i < j`, the condition `j - i <= k` is satisfied.
A _subsequence_ of an array is obtained by deleting... | null |
Python3 Solution | constrained-subsequence-sum | 0 | 1 | \n```\nclass Solution:\n def constrainedSubsetSum(self, nums: List[int], k: int) -> int:\n deque=[]\n for i,num in enumerate(nums):\n while deque and deque[0]<i-k:\n deque.pop(0)\n if deque:\n nums[i]=nums[deque[0]]+num\n while deque and nu... | 2 | Design the `CombinationIterator` class:
* `CombinationIterator(string characters, int combinationLength)` Initializes the object with a string `characters` of **sorted distinct** lowercase English letters and a number `combinationLength` as arguments.
* `next()` Returns the next combination of length `combinationL... | Use dynamic programming. Let dp[i] be the solution for the prefix of the array that ends at index i, if the element at index i is in the subsequence. dp[i] = nums[i] + max(0, dp[i-k], dp[i-k+1], ..., dp[i-1]) Use a heap with the sliding window technique to optimize the dp. |
Video Solution | Explanation With Drawings | In Depth | C++ | Java | Python 3 | constrained-subsequence-sum | 1 | 1 | # Intuition, approach and complexity discussed in detail in video solution\nhttps://youtu.be/KIB_v0d9r_E\n\n# Code\nC++\n```\n//TC : O(n), |nums | = n\n//SC : O(n)\nclass Solution {\npublic:\n int constrainedSubsetSum(vector<int>& nums, int k) {\n //maxTillThis will represent the max sum of sub seq ending at ... | 1 | Given an integer array `nums` and an integer `k`, return the maximum sum of a **non-empty** subsequence of that array such that for every two **consecutive** integers in the subsequence, `nums[i]` and `nums[j]`, where `i < j`, the condition `j - i <= k` is satisfied.
A _subsequence_ of an array is obtained by deleting... | null |
Video Solution | Explanation With Drawings | In Depth | C++ | Java | Python 3 | constrained-subsequence-sum | 1 | 1 | # Intuition, approach and complexity discussed in detail in video solution\nhttps://youtu.be/KIB_v0d9r_E\n\n# Code\nC++\n```\n//TC : O(n), |nums | = n\n//SC : O(n)\nclass Solution {\npublic:\n int constrainedSubsetSum(vector<int>& nums, int k) {\n //maxTillThis will represent the max sum of sub seq ending at ... | 1 | Design the `CombinationIterator` class:
* `CombinationIterator(string characters, int combinationLength)` Initializes the object with a string `characters` of **sorted distinct** lowercase English letters and a number `combinationLength` as arguments.
* `next()` Returns the next combination of length `combinationL... | Use dynamic programming. Let dp[i] be the solution for the prefix of the array that ends at index i, if the element at index i is in the subsequence. dp[i] = nums[i] + max(0, dp[i-k], dp[i-k+1], ..., dp[i-1]) Use a heap with the sliding window technique to optimize the dp. |
【Video】Give me 10 minutes - How we think about a solution - Python, JavaScript, Java, C++ | constrained-subsequence-sum | 1 | 1 | # Intuition\nhaving pairs (index and max value) in deque with decreasing order\n\n---\n\n# Solution Video\n\nhttps://youtu.be/KuMkwvvesgo\n\n\u25A0 Timeline of the video\n`0:04` Explain a few basic idea\n`3:21` What data we should put in deque?\n`5:10` Demonstrate how it works\n`13:27` Coding\n`16:03` Time Complexity a... | 51 | Given an integer array `nums` and an integer `k`, return the maximum sum of a **non-empty** subsequence of that array such that for every two **consecutive** integers in the subsequence, `nums[i]` and `nums[j]`, where `i < j`, the condition `j - i <= k` is satisfied.
A _subsequence_ of an array is obtained by deleting... | null |
【Video】Give me 10 minutes - How we think about a solution - Python, JavaScript, Java, C++ | constrained-subsequence-sum | 1 | 1 | # Intuition\nhaving pairs (index and max value) in deque with decreasing order\n\n---\n\n# Solution Video\n\nhttps://youtu.be/KuMkwvvesgo\n\n\u25A0 Timeline of the video\n`0:04` Explain a few basic idea\n`3:21` What data we should put in deque?\n`5:10` Demonstrate how it works\n`13:27` Coding\n`16:03` Time Complexity a... | 51 | Design the `CombinationIterator` class:
* `CombinationIterator(string characters, int combinationLength)` Initializes the object with a string `characters` of **sorted distinct** lowercase English letters and a number `combinationLength` as arguments.
* `next()` Returns the next combination of length `combinationL... | Use dynamic programming. Let dp[i] be the solution for the prefix of the array that ends at index i, if the element at index i is in the subsequence. dp[i] = nums[i] + max(0, dp[i-k], dp[i-k+1], ..., dp[i-1]) Use a heap with the sliding window technique to optimize the dp. |
✅☑[C++/Java/Python/JavaScript] || 3 Approaches || EXPLAINED🔥 | constrained-subsequence-sum | 1 | 1 | # PLEASE UPVOTE IF IT HELPED\n\n---\n\n\n# Approaches\n**(Also explained in the code)**\n\n#### ***Approach 1 (Heap)***\n1. We use a max heap (priority_queue) to efficiently keep track of the maximum sum of a constrained subset and the index at which it ends.\n\n1. We initialize the answer (`ans`) with the value of the... | 20 | Given an integer array `nums` and an integer `k`, return the maximum sum of a **non-empty** subsequence of that array such that for every two **consecutive** integers in the subsequence, `nums[i]` and `nums[j]`, where `i < j`, the condition `j - i <= k` is satisfied.
A _subsequence_ of an array is obtained by deleting... | null |
✅☑[C++/Java/Python/JavaScript] || 3 Approaches || EXPLAINED🔥 | constrained-subsequence-sum | 1 | 1 | # PLEASE UPVOTE IF IT HELPED\n\n---\n\n\n# Approaches\n**(Also explained in the code)**\n\n#### ***Approach 1 (Heap)***\n1. We use a max heap (priority_queue) to efficiently keep track of the maximum sum of a constrained subset and the index at which it ends.\n\n1. We initialize the answer (`ans`) with the value of the... | 20 | Design the `CombinationIterator` class:
* `CombinationIterator(string characters, int combinationLength)` Initializes the object with a string `characters` of **sorted distinct** lowercase English letters and a number `combinationLength` as arguments.
* `next()` Returns the next combination of length `combinationL... | Use dynamic programming. Let dp[i] be the solution for the prefix of the array that ends at index i, if the element at index i is in the subsequence. dp[i] = nums[i] + max(0, dp[i-k], dp[i-k+1], ..., dp[i-1]) Use a heap with the sliding window technique to optimize the dp. |
✅ Beats 100% 🔥 in Both || Easiest Approach || Detailed Explanation || | constrained-subsequence-sum | 1 | 1 | \n# PHP\n\n\n# Python\n\n\n# Intuition\nIf we want to get... | 4 | Given an integer array `nums` and an integer `k`, return the maximum sum of a **non-empty** subsequence of that array such that for every two **consecutive** integers in the subsequence, `nums[i]` and `nums[j]`, where `i < j`, the condition `j - i <= k` is satisfied.
A _subsequence_ of an array is obtained by deleting... | null |
✅ Beats 100% 🔥 in Both || Easiest Approach || Detailed Explanation || | constrained-subsequence-sum | 1 | 1 | \n# PHP\n\n\n# Python\n\n\n# Intuition\nIf we want to get... | 4 | Design the `CombinationIterator` class:
* `CombinationIterator(string characters, int combinationLength)` Initializes the object with a string `characters` of **sorted distinct** lowercase English letters and a number `combinationLength` as arguments.
* `next()` Returns the next combination of length `combinationL... | Use dynamic programming. Let dp[i] be the solution for the prefix of the array that ends at index i, if the element at index i is in the subsequence. dp[i] = nums[i] + max(0, dp[i-k], dp[i-k+1], ..., dp[i-1]) Use a heap with the sliding window technique to optimize the dp. |
Python DP + Max Window Element ( Monotonic Queue) | constrained-subsequence-sum | 0 | 1 | \n```\nclass Solution:\n def constrainedSubsetSum(self, nums: List[int], k: int) -> int:\n dp = copy.deepcopy(nums)\n q = deque()\n q.append(0)\n for i in range(1,len(nums)):\n while(q[-1] < i-k ):\n q.pop()\n dp[i] = max(dp[i],dp[q[-1]]+nums[i])\n ... | 1 | Given an integer array `nums` and an integer `k`, return the maximum sum of a **non-empty** subsequence of that array such that for every two **consecutive** integers in the subsequence, `nums[i]` and `nums[j]`, where `i < j`, the condition `j - i <= k` is satisfied.
A _subsequence_ of an array is obtained by deleting... | null |
Python DP + Max Window Element ( Monotonic Queue) | constrained-subsequence-sum | 0 | 1 | \n```\nclass Solution:\n def constrainedSubsetSum(self, nums: List[int], k: int) -> int:\n dp = copy.deepcopy(nums)\n q = deque()\n q.append(0)\n for i in range(1,len(nums)):\n while(q[-1] < i-k ):\n q.pop()\n dp[i] = max(dp[i],dp[q[-1]]+nums[i])\n ... | 1 | Design the `CombinationIterator` class:
* `CombinationIterator(string characters, int combinationLength)` Initializes the object with a string `characters` of **sorted distinct** lowercase English letters and a number `combinationLength` as arguments.
* `next()` Returns the next combination of length `combinationL... | Use dynamic programming. Let dp[i] be the solution for the prefix of the array that ends at index i, if the element at index i is in the subsequence. dp[i] = nums[i] + max(0, dp[i-k], dp[i-k+1], ..., dp[i-1]) Use a heap with the sliding window technique to optimize the dp. |
Simple Python and C++ solutions. Vectors in C++ | kids-with-the-greatest-number-of-candies | 0 | 1 | # Solution in C++\n```\nclass Solution {\npublic:\n vector<bool> kidsWithCandies(vector<int>& candies, int extraCandies) {\n vector<bool>answer;\n int max_value = 0;\n for (int i = 0; i < candies.size(); i++){\n if (max_value <= candies.at(i)){max_value = candies.at(i);}\n }\n\... | 1 | There are `n` kids with candies. You are given an integer array `candies`, where each `candies[i]` represents the number of candies the `ith` kid has, and an integer `extraCandies`, denoting the number of extra candies that you have.
Return _a boolean array_ `result` _of length_ `n`_, where_ `result[i]` _is_ `true` _i... | Consider how reversing each edge of the graph can help us. How can performing BFS/DFS on the reversed graph help us find the ancestors of every node? |
Simple Python and C++ solutions. Vectors in C++ | kids-with-the-greatest-number-of-candies | 0 | 1 | # Solution in C++\n```\nclass Solution {\npublic:\n vector<bool> kidsWithCandies(vector<int>& candies, int extraCandies) {\n vector<bool>answer;\n int max_value = 0;\n for (int i = 0; i < candies.size(); i++){\n if (max_value <= candies.at(i)){max_value = candies.at(i);}\n }\n\... | 1 | You are given a string `s` and an integer array `indices` of the **same length**. The string `s` will be shuffled such that the character at the `ith` position moves to `indices[i]` in the shuffled string.
Return _the shuffled string_.
**Example 1:**
**Input:** s = "codeleet ", `indices` = \[4,5,6,7,0,2,1,3\]
**Out... | Use greedy approach. For each kid check if candies[i] + extraCandies ≥ maximum in Candies[i]. |
Shortest 1 line solution | kids-with-the-greatest-number-of-candies | 0 | 1 | # Code\n```\nclass Solution:\n def kidsWithCandies(self, c: List[int], eC: int) -> List[bool]:\n return [x+eC >= max(c)for x in c]\n```\nplease upvote :D | 9 | There are `n` kids with candies. You are given an integer array `candies`, where each `candies[i]` represents the number of candies the `ith` kid has, and an integer `extraCandies`, denoting the number of extra candies that you have.
Return _a boolean array_ `result` _of length_ `n`_, where_ `result[i]` _is_ `true` _i... | Consider how reversing each edge of the graph can help us. How can performing BFS/DFS on the reversed graph help us find the ancestors of every node? |
Shortest 1 line solution | kids-with-the-greatest-number-of-candies | 0 | 1 | # Code\n```\nclass Solution:\n def kidsWithCandies(self, c: List[int], eC: int) -> List[bool]:\n return [x+eC >= max(c)for x in c]\n```\nplease upvote :D | 9 | You are given a string `s` and an integer array `indices` of the **same length**. The string `s` will be shuffled such that the character at the `ith` position moves to `indices[i]` in the shuffled string.
Return _the shuffled string_.
**Example 1:**
**Input:** s = "codeleet ", `indices` = \[4,5,6,7,0,2,1,3\]
**Out... | Use greedy approach. For each kid check if candies[i] + extraCandies ≥ maximum in Candies[i]. |
[Java/Python 3] O(digits) w/ brief explanation and analysis. | max-difference-you-can-get-from-changing-an-integer | 1 | 1 | Convert to char array/list then process.\n\nThe following explains the algorithm of the code: -- credit to **@leetcodeCR97 and @martin20**.\n\nIf your first character `x` is \n1. neither `1` nor `9` then you will be replacing `a` with `9 `and `b` with `1 `if it matched the first character of the given number. \n2. `1` ... | 25 | You are given an integer `num`. You will apply the following steps exactly **two** times:
* Pick a digit `x (0 <= x <= 9)`.
* Pick another digit `y (0 <= y <= 9)`. The digit `y` can be equal to `x`.
* Replace all the occurrences of `x` in the decimal representation of `num` by `y`.
* The new integer **cannot**... | Depth-first search (DFS) with the parameters: current node in the binary tree and current position in the array of integers. When reaching at final position check if it is a leaf node. |
[Java/Python 3] O(digits) w/ brief explanation and analysis. | max-difference-you-can-get-from-changing-an-integer | 1 | 1 | Convert to char array/list then process.\n\nThe following explains the algorithm of the code: -- credit to **@leetcodeCR97 and @martin20**.\n\nIf your first character `x` is \n1. neither `1` nor `9` then you will be replacing `a` with `9 `and `b` with `1 `if it matched the first character of the given number. \n2. `1` ... | 25 | You are given a **0-indexed** binary string `target` of length `n`. You have another binary string `s` of length `n` that is initially set to all zeros. You want to make `s` equal to `target`.
In one operation, you can pick an index `i` where `0 <= i < n` and flip all bits in the **inclusive** range `[i, n - 1]`. Flip... | We need to get the max and min value after changing num and the answer is max - min. Use brute force, try all possible changes and keep the minimum and maximum values. |
Pure Heuristics Python O(n) | max-difference-you-can-get-from-changing-an-integer | 0 | 1 | # Approach\nBrute Force\n\n# Code\n```\nclass Solution:\n def maxDiff(self, num: int) -> int:\n l = [i for i in str(num) if i!=\'9\' ]\n if len(l)!=0: \n f = l[0]\n max_ = int(\'\'.join([n if n!=f else \'9\' for n in str(num)]))\n else: max_ = num\n\n if str(num)[0] ... | 0 | You are given an integer `num`. You will apply the following steps exactly **two** times:
* Pick a digit `x (0 <= x <= 9)`.
* Pick another digit `y (0 <= y <= 9)`. The digit `y` can be equal to `x`.
* Replace all the occurrences of `x` in the decimal representation of `num` by `y`.
* The new integer **cannot**... | Depth-first search (DFS) with the parameters: current node in the binary tree and current position in the array of integers. When reaching at final position check if it is a leaf node. |
Pure Heuristics Python O(n) | max-difference-you-can-get-from-changing-an-integer | 0 | 1 | # Approach\nBrute Force\n\n# Code\n```\nclass Solution:\n def maxDiff(self, num: int) -> int:\n l = [i for i in str(num) if i!=\'9\' ]\n if len(l)!=0: \n f = l[0]\n max_ = int(\'\'.join([n if n!=f else \'9\' for n in str(num)]))\n else: max_ = num\n\n if str(num)[0] ... | 0 | You are given a **0-indexed** binary string `target` of length `n`. You have another binary string `s` of length `n` that is initially set to all zeros. You want to make `s` equal to `target`.
In one operation, you can pick an index `i` where `0 <= i < n` and flip all bits in the **inclusive** range `[i, n - 1]`. Flip... | We need to get the max and min value after changing num and the answer is max - min. Use brute force, try all possible changes and keep the minimum and maximum values. |
Easy To Understand Solution !! Beats 98% and solved in O(n) with extra space complexity of O(n). | max-difference-you-can-get-from-changing-an-integer | 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 `num`. You will apply the following steps exactly **two** times:
* Pick a digit `x (0 <= x <= 9)`.
* Pick another digit `y (0 <= y <= 9)`. The digit `y` can be equal to `x`.
* Replace all the occurrences of `x` in the decimal representation of `num` by `y`.
* The new integer **cannot**... | Depth-first search (DFS) with the parameters: current node in the binary tree and current position in the array of integers. When reaching at final position check if it is a leaf node. |
Easy To Understand Solution !! Beats 98% and solved in O(n) with extra space complexity of O(n). | max-difference-you-can-get-from-changing-an-integer | 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 **0-indexed** binary string `target` of length `n`. You have another binary string `s` of length `n` that is initially set to all zeros. You want to make `s` equal to `target`.
In one operation, you can pick an index `i` where `0 <= i < n` and flip all bits in the **inclusive** range `[i, n - 1]`. Flip... | We need to get the max and min value after changing num and the answer is max - min. Use brute force, try all possible changes and keep the minimum and maximum values. |
simplest sol in python | max-difference-you-can-get-from-changing-an-integer | 0 | 1 | # Intuition\nMax length of string representation of num can be 9, so can go with for loop\n\n# Code\n```\nclass Solution:\n def maxDiff(self, num: int) -> int:\n a,b = str(num),str(num)\n for i in range(len(a)):\n if a[i] != a[0] and a[i] != "0": \n a=a.replace(a[i],"0",-1);br... | 0 | You are given an integer `num`. You will apply the following steps exactly **two** times:
* Pick a digit `x (0 <= x <= 9)`.
* Pick another digit `y (0 <= y <= 9)`. The digit `y` can be equal to `x`.
* Replace all the occurrences of `x` in the decimal representation of `num` by `y`.
* The new integer **cannot**... | Depth-first search (DFS) with the parameters: current node in the binary tree and current position in the array of integers. When reaching at final position check if it is a leaf node. |
simplest sol in python | max-difference-you-can-get-from-changing-an-integer | 0 | 1 | # Intuition\nMax length of string representation of num can be 9, so can go with for loop\n\n# Code\n```\nclass Solution:\n def maxDiff(self, num: int) -> int:\n a,b = str(num),str(num)\n for i in range(len(a)):\n if a[i] != a[0] and a[i] != "0": \n a=a.replace(a[i],"0",-1);br... | 0 | You are given a **0-indexed** binary string `target` of length `n`. You have another binary string `s` of length `n` that is initially set to all zeros. You want to make `s` equal to `target`.
In one operation, you can pick an index `i` where `0 <= i < n` and flip all bits in the **inclusive** range `[i, n - 1]`. Flip... | We need to get the max and min value after changing num and the answer is max - min. Use brute force, try all possible changes and keep the minimum and maximum values. |
Python3 Naive Solution | max-difference-you-can-get-from-changing-an-integer | 0 | 1 | \n\n# Code\n```\nclass Solution:\n def maxDiff(self, num: int) -> int:\n \n s1=str(num)\n s2=str(num)\n \n if s1[0]!="1":\n s1=s1.replace(s1[0],"1")\n else:\n i=0\n while i<len(s1):\n if s1[i]!="1"and s2[i]!="0":\n ... | 0 | You are given an integer `num`. You will apply the following steps exactly **two** times:
* Pick a digit `x (0 <= x <= 9)`.
* Pick another digit `y (0 <= y <= 9)`. The digit `y` can be equal to `x`.
* Replace all the occurrences of `x` in the decimal representation of `num` by `y`.
* The new integer **cannot**... | Depth-first search (DFS) with the parameters: current node in the binary tree and current position in the array of integers. When reaching at final position check if it is a leaf node. |
Python3 Naive Solution | max-difference-you-can-get-from-changing-an-integer | 0 | 1 | \n\n# Code\n```\nclass Solution:\n def maxDiff(self, num: int) -> int:\n \n s1=str(num)\n s2=str(num)\n \n if s1[0]!="1":\n s1=s1.replace(s1[0],"1")\n else:\n i=0\n while i<len(s1):\n if s1[i]!="1"and s2[i]!="0":\n ... | 0 | You are given a **0-indexed** binary string `target` of length `n`. You have another binary string `s` of length `n` that is initially set to all zeros. You want to make `s` equal to `target`.
In one operation, you can pick an index `i` where `0 <= i < n` and flip all bits in the **inclusive** range `[i, n - 1]`. Flip... | We need to get the max and min value after changing num and the answer is max - min. Use brute force, try all possible changes and keep the minimum and maximum values. |
Easy Python solution | max-difference-you-can-get-from-changing-an-integer | 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 `num`. You will apply the following steps exactly **two** times:
* Pick a digit `x (0 <= x <= 9)`.
* Pick another digit `y (0 <= y <= 9)`. The digit `y` can be equal to `x`.
* Replace all the occurrences of `x` in the decimal representation of `num` by `y`.
* The new integer **cannot**... | Depth-first search (DFS) with the parameters: current node in the binary tree and current position in the array of integers. When reaching at final position check if it is a leaf node. |
Easy Python solution | max-difference-you-can-get-from-changing-an-integer | 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 **0-indexed** binary string `target` of length `n`. You have another binary string `s` of length `n` that is initially set to all zeros. You want to make `s` equal to `target`.
In one operation, you can pick an index `i` where `0 <= i < n` and flip all bits in the **inclusive** range `[i, n - 1]`. Flip... | We need to get the max and min value after changing num and the answer is max - min. Use brute force, try all possible changes and keep the minimum and maximum values. |
Solution in Python | max-difference-you-can-get-from-changing-an-integer | 0 | 1 | # Code\n```\nclass Solution:\n def maxDiff(self, num: int) -> int:\n x = int(num)\n mostSignificantDigit = -1\n while x>=1:\n if x%10 != 9 or mostSignificantDigit==-1 or x%10==mostSignificantDigit: \n mostSignificantDigit = int(x%10)\n x=int(x/10)\n x=... | 0 | You are given an integer `num`. You will apply the following steps exactly **two** times:
* Pick a digit `x (0 <= x <= 9)`.
* Pick another digit `y (0 <= y <= 9)`. The digit `y` can be equal to `x`.
* Replace all the occurrences of `x` in the decimal representation of `num` by `y`.
* The new integer **cannot**... | Depth-first search (DFS) with the parameters: current node in the binary tree and current position in the array of integers. When reaching at final position check if it is a leaf node. |
Solution in Python | max-difference-you-can-get-from-changing-an-integer | 0 | 1 | # Code\n```\nclass Solution:\n def maxDiff(self, num: int) -> int:\n x = int(num)\n mostSignificantDigit = -1\n while x>=1:\n if x%10 != 9 or mostSignificantDigit==-1 or x%10==mostSignificantDigit: \n mostSignificantDigit = int(x%10)\n x=int(x/10)\n x=... | 0 | You are given a **0-indexed** binary string `target` of length `n`. You have another binary string `s` of length `n` that is initially set to all zeros. You want to make `s` equal to `target`.
In one operation, you can pick an index `i` where `0 <= i < n` and flip all bits in the **inclusive** range `[i, n - 1]`. Flip... | We need to get the max and min value after changing num and the answer is max - min. Use brute force, try all possible changes and keep the minimum and maximum values. |
(num as string)||Simplest Python solution beats 92% | max-difference-you-can-get-from-changing-an-integer | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n**Max value of N is 10^8**\n**if We think here of N as Strig then the Max len is just 9 chars**\n# Code\n```\nclass Solution:\n def maxDiff(self, num: int) -> int:\n a,b = str(num),str(num)\n for i in range(len(a)):\n ... | 0 | You are given an integer `num`. You will apply the following steps exactly **two** times:
* Pick a digit `x (0 <= x <= 9)`.
* Pick another digit `y (0 <= y <= 9)`. The digit `y` can be equal to `x`.
* Replace all the occurrences of `x` in the decimal representation of `num` by `y`.
* The new integer **cannot**... | Depth-first search (DFS) with the parameters: current node in the binary tree and current position in the array of integers. When reaching at final position check if it is a leaf node. |
(num as string)||Simplest Python solution beats 92% | max-difference-you-can-get-from-changing-an-integer | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n**Max value of N is 10^8**\n**if We think here of N as Strig then the Max len is just 9 chars**\n# Code\n```\nclass Solution:\n def maxDiff(self, num: int) -> int:\n a,b = str(num),str(num)\n for i in range(len(a)):\n ... | 0 | You are given a **0-indexed** binary string `target` of length `n`. You have another binary string `s` of length `n` that is initially set to all zeros. You want to make `s` equal to `target`.
In one operation, you can pick an index `i` where `0 <= i < n` and flip all bits in the **inclusive** range `[i, n - 1]`. Flip... | We need to get the max and min value after changing num and the answer is max - min. Use brute force, try all possible changes and keep the minimum and maximum values. |
Python Easy Solution , O(n) | max-difference-you-can-get-from-changing-an-integer | 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 `num`. You will apply the following steps exactly **two** times:
* Pick a digit `x (0 <= x <= 9)`.
* Pick another digit `y (0 <= y <= 9)`. The digit `y` can be equal to `x`.
* Replace all the occurrences of `x` in the decimal representation of `num` by `y`.
* The new integer **cannot**... | Depth-first search (DFS) with the parameters: current node in the binary tree and current position in the array of integers. When reaching at final position check if it is a leaf node. |
Python Easy Solution , O(n) | max-difference-you-can-get-from-changing-an-integer | 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 **0-indexed** binary string `target` of length `n`. You have another binary string `s` of length `n` that is initially set to all zeros. You want to make `s` equal to `target`.
In one operation, you can pick an index `i` where `0 <= i < n` and flip all bits in the **inclusive** range `[i, n - 1]`. Flip... | We need to get the max and min value after changing num and the answer is max - min. Use brute force, try all possible changes and keep the minimum and maximum values. |
Python solution | check-if-a-string-can-break-another-string | 0 | 1 | \n\n# Approach\nSort the strings in descending order and check the characters\n\n# Complexity\n- Time complexity:\nO(nLogn)\n\n- Space complexity:\nO(n)\n\n# Code\n```\nclass Solution:\n def checkIfCanBreak(self, s1: str, s2: str) -> bool:\n l1 = [i for i in s1]\n l2 = [i for i in s2]\n l1.sort(... | 0 | Given two strings: `s1` and `s2` with the same size, check if some permutation of string `s1` can break some permutation of string `s2` or vice-versa. In other words `s2` can break `s1` or vice-versa.
A string `x` can break string `y` (both of size `n`) if `x[i] >= y[i]` (in alphabetical order) for all `i` between `0`... | For encryption, use hashmap to map each char of word1 to its value. For decryption, use trie to prune when necessary. |
Python solution | check-if-a-string-can-break-another-string | 0 | 1 | \n\n# Approach\nSort the strings in descending order and check the characters\n\n# Complexity\n- Time complexity:\nO(nLogn)\n\n- Space complexity:\nO(n)\n\n# Code\n```\nclass Solution:\n def checkIfCanBreak(self, s1: str, s2: str) -> bool:\n l1 = [i for i in s1]\n l2 = [i for i in s2]\n l1.sort(... | 0 | You are given the `root` of a binary tree and an integer `distance`. A pair of two different **leaf** nodes of a binary tree is said to be good if the length of **the shortest path** between them is less than or equal to `distance`.
Return _the number of good leaf node pairs_ in the tree.
**Example 1:**
**Input:** r... | Sort both strings and then check if one of them can break the other. |
One Liner Solution | check-if-a-string-can-break-another-string | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ --... | 0 | Given two strings: `s1` and `s2` with the same size, check if some permutation of string `s1` can break some permutation of string `s2` or vice-versa. In other words `s2` can break `s1` or vice-versa.
A string `x` can break string `y` (both of size `n`) if `x[i] >= y[i]` (in alphabetical order) for all `i` between `0`... | For encryption, use hashmap to map each char of word1 to its value. For decryption, use trie to prune when necessary. |
One Liner Solution | check-if-a-string-can-break-another-string | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ --... | 0 | You are given the `root` of a binary tree and an integer `distance`. A pair of two different **leaf** nodes of a binary tree is said to be good if the length of **the shortest path** between them is less than or equal to `distance`.
Return _the number of good leaf node pairs_ in the tree.
**Example 1:**
**Input:** r... | Sort both strings and then check if one of them can break the other. |
Python3 Solution | check-if-a-string-can-break-another-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. -->\nsort the two strings and check if one breaks the other linearly\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\nO(nlogn + mlo... | 0 | Given two strings: `s1` and `s2` with the same size, check if some permutation of string `s1` can break some permutation of string `s2` or vice-versa. In other words `s2` can break `s1` or vice-versa.
A string `x` can break string `y` (both of size `n`) if `x[i] >= y[i]` (in alphabetical order) for all `i` between `0`... | For encryption, use hashmap to map each char of word1 to its value. For decryption, use trie to prune when necessary. |
Python3 Solution | check-if-a-string-can-break-another-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. -->\nsort the two strings and check if one breaks the other linearly\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\nO(nlogn + mlo... | 0 | You are given the `root` of a binary tree and an integer `distance`. A pair of two different **leaf** nodes of a binary tree is said to be good if the length of **the shortest path** between them is less than or equal to `distance`.
Return _the number of good leaf node pairs_ in the tree.
**Example 1:**
**Input:** r... | Sort both strings and then check if one of them can break the other. |
Python3 Clean and Short Solution | check-if-a-string-can-break-another-string | 0 | 1 | \n# Code\n```\nclass Solution:\n def checkIfCanBreak(self, s1: str, s2: str) -> bool:\n \n \n sort1=sorted(s1)\n sort2=sorted(s2)\n \n return all(a<=b for a,b in zip(sort1,sort2)) or all(a>=b for a,b in zip(sort1,sort2))\n \n \n``` | 0 | Given two strings: `s1` and `s2` with the same size, check if some permutation of string `s1` can break some permutation of string `s2` or vice-versa. In other words `s2` can break `s1` or vice-versa.
A string `x` can break string `y` (both of size `n`) if `x[i] >= y[i]` (in alphabetical order) for all `i` between `0`... | For encryption, use hashmap to map each char of word1 to its value. For decryption, use trie to prune when necessary. |
Python3 Clean and Short Solution | check-if-a-string-can-break-another-string | 0 | 1 | \n# Code\n```\nclass Solution:\n def checkIfCanBreak(self, s1: str, s2: str) -> bool:\n \n \n sort1=sorted(s1)\n sort2=sorted(s2)\n \n return all(a<=b for a,b in zip(sort1,sort2)) or all(a>=b for a,b in zip(sort1,sort2))\n \n \n``` | 0 | You are given the `root` of a binary tree and an integer `distance`. A pair of two different **leaf** nodes of a binary tree is said to be good if the length of **the shortest path** between them is less than or equal to `distance`.
Return _the number of good leaf node pairs_ in the tree.
**Example 1:**
**Input:** r... | Sort both strings and then check if one of them can break the other. |
Easy Python Solution Using Loops | check-if-a-string-can-break-another-string | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ --... | 0 | Given two strings: `s1` and `s2` with the same size, check if some permutation of string `s1` can break some permutation of string `s2` or vice-versa. In other words `s2` can break `s1` or vice-versa.
A string `x` can break string `y` (both of size `n`) if `x[i] >= y[i]` (in alphabetical order) for all `i` between `0`... | For encryption, use hashmap to map each char of word1 to its value. For decryption, use trie to prune when necessary. |
Easy Python Solution Using Loops | check-if-a-string-can-break-another-string | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ --... | 0 | You are given the `root` of a binary tree and an integer `distance`. A pair of two different **leaf** nodes of a binary tree is said to be good if the length of **the shortest path** between them is less than or equal to `distance`.
Return _the number of good leaf node pairs_ in the tree.
**Example 1:**
**Input:** r... | Sort both strings and then check if one of them can break the other. |
O(n) two pointer :D | check-if-a-string-can-break-another-string | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ --... | 0 | Given two strings: `s1` and `s2` with the same size, check if some permutation of string `s1` can break some permutation of string `s2` or vice-versa. In other words `s2` can break `s1` or vice-versa.
A string `x` can break string `y` (both of size `n`) if `x[i] >= y[i]` (in alphabetical order) for all `i` between `0`... | For encryption, use hashmap to map each char of word1 to its value. For decryption, use trie to prune when necessary. |
O(n) two pointer :D | check-if-a-string-can-break-another-string | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ --... | 0 | You are given the `root` of a binary tree and an integer `distance`. A pair of two different **leaf** nodes of a binary tree is said to be good if the length of **the shortest path** between them is less than or equal to `distance`.
Return _the number of good leaf node pairs_ in the tree.
**Example 1:**
**Input:** r... | Sort both strings and then check if one of them can break the other. |
Python beginner friendly solution O(2*N) with sort nlogn | check-if-a-string-can-break-another-string | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ --... | 0 | Given two strings: `s1` and `s2` with the same size, check if some permutation of string `s1` can break some permutation of string `s2` or vice-versa. In other words `s2` can break `s1` or vice-versa.
A string `x` can break string `y` (both of size `n`) if `x[i] >= y[i]` (in alphabetical order) for all `i` between `0`... | For encryption, use hashmap to map each char of word1 to its value. For decryption, use trie to prune when necessary. |
Python beginner friendly solution O(2*N) with sort nlogn | check-if-a-string-can-break-another-string | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ --... | 0 | You are given the `root` of a binary tree and an integer `distance`. A pair of two different **leaf** nodes of a binary tree is said to be good if the length of **the shortest path** between them is less than or equal to `distance`.
Return _the number of good leaf node pairs_ in the tree.
**Example 1:**
**Input:** r... | Sort both strings and then check if one of them can break the other. |
[Python3] dp & bitmasking | number-of-ways-to-wear-different-hats-to-each-other | 0 | 1 | Algorithm:\nDefine function `fn(h, mask)` to indicate the number of ways to wear `h` (starting from 0) to last hat among people whose availability is indicated by set bit in mask. \n\nImplementation (time complexity `O(40 * 2^10)` | space complexity `O(40 * 2^10)` \n```\nfrom functools import lru_cache\n\nclass Solutio... | 11 | There are `n` people and `40` types of hats labeled from `1` to `40`.
Given a 2D integer array `hats`, where `hats[i]` is a list of all hats preferred by the `ith` person.
Return _the number of ways that the `n` people wear different hats to each other_.
Since the answer may be too large, return it modulo `109 + 7`.... | Scan from right to left, in each step of the scanning check whether there is a trailing "#" 2 indexes away. |
[Python3] dp & bitmasking | number-of-ways-to-wear-different-hats-to-each-other | 0 | 1 | Algorithm:\nDefine function `fn(h, mask)` to indicate the number of ways to wear `h` (starting from 0) to last hat among people whose availability is indicated by set bit in mask. \n\nImplementation (time complexity `O(40 * 2^10)` | space complexity `O(40 * 2^10)` \n```\nfrom functools import lru_cache\n\nclass Solutio... | 11 | [Run-length encoding](http://en.wikipedia.org/wiki/Run-length_encoding) is a string compression method that works by replacing consecutive identical characters (repeated 2 or more times) with the concatenation of the character and the number marking the count of the characters (length of the run). For example, to compr... | Dynamic programming + bitmask. dp(peopleMask, idHat) number of ways to wear different hats given a bitmask (people visited) and used hats from 1 to idHat-1. |
Updated Process | O(p_i) Time and Space | Commented / Explained | number-of-ways-to-wear-different-hats-to-each-other | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nPriorly I had worked this with a limit based on needing to consider all hats even if not present. New update allows for consideration only on number of hats as well as early stopping conditional. This gives great boost in average speed an... | 0 | There are `n` people and `40` types of hats labeled from `1` to `40`.
Given a 2D integer array `hats`, where `hats[i]` is a list of all hats preferred by the `ith` person.
Return _the number of ways that the `n` people wear different hats to each other_.
Since the answer may be too large, return it modulo `109 + 7`.... | Scan from right to left, in each step of the scanning check whether there is a trailing "#" 2 indexes away. |
Updated Process | O(p_i) Time and Space | Commented / Explained | number-of-ways-to-wear-different-hats-to-each-other | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nPriorly I had worked this with a limit based on needing to consider all hats even if not present. New update allows for consideration only on number of hats as well as early stopping conditional. This gives great boost in average speed an... | 0 | [Run-length encoding](http://en.wikipedia.org/wiki/Run-length_encoding) is a string compression method that works by replacing consecutive identical characters (repeated 2 or more times) with the concatenation of the character and the number marking the count of the characters (length of the run). For example, to compr... | Dynamic programming + bitmask. dp(peopleMask, idHat) number of ways to wear different hats given a bitmask (people visited) and used hats from 1 to idHat-1. |
Python (Simple DP + Bitmask) | number-of-ways-to-wear-different-hats-to-each-other | 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 are `n` people and `40` types of hats labeled from `1` to `40`.
Given a 2D integer array `hats`, where `hats[i]` is a list of all hats preferred by the `ith` person.
Return _the number of ways that the `n` people wear different hats to each other_.
Since the answer may be too large, return it modulo `109 + 7`.... | Scan from right to left, in each step of the scanning check whether there is a trailing "#" 2 indexes away. |
Python (Simple DP + Bitmask) | number-of-ways-to-wear-different-hats-to-each-other | 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 | [Run-length encoding](http://en.wikipedia.org/wiki/Run-length_encoding) is a string compression method that works by replacing consecutive identical characters (repeated 2 or more times) with the concatenation of the character and the number marking the count of the characters (length of the run). For example, to compr... | Dynamic programming + bitmask. dp(peopleMask, idHat) number of ways to wear different hats given a bitmask (people visited) and used hats from 1 to idHat-1. |
💯Faster✅💯 Lesser✅3 Methods🔥Using Set Operations🔥Using HashSet🔥Using Lists🔥Python🐍Java☕C++✅C📈 | destination-city | 1 | 1 | # \uD83D\uDE80 Hi, I\'m [Mohammed Raziullah Ansari](https://leetcode.com/Mohammed_Raziullah_Ansari/), and I\'m excited to share 3 ways to solve this question with detailed explanation of each approach:\n\n# Problem Explaination: \nThe `Destination City` problem typically involves a scenario where you are given a list o... | 25 | You are given the array `paths`, where `paths[i] = [cityAi, cityBi]` means there exists a direct path going from `cityAi` to `cityBi`. _Return the destination city, that is, the city without any path outgoing to another city._
It is guaranteed that the graph of paths forms a line without any loop, therefore, there wil... | Do BFS to find the kth level friends. Then collect movies saw by kth level friends and sort them accordingly. |
💯Faster✅💯 Lesser✅3 Methods🔥Using Set Operations🔥Using HashSet🔥Using Lists🔥Python🐍Java☕C++✅C📈 | destination-city | 1 | 1 | # \uD83D\uDE80 Hi, I\'m [Mohammed Raziullah Ansari](https://leetcode.com/Mohammed_Raziullah_Ansari/), and I\'m excited to share 3 ways to solve this question with detailed explanation of each approach:\n\n# Problem Explaination: \nThe `Destination City` problem typically involves a scenario where you are given a list o... | 25 | 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. |
✅ Navigating Urban Labyrinths: Elegant Solution to Find the Destination City 🔥 | destination-city | 1 | 1 | # Description\n\n## Intuition\nThe problem requires finding the destination city, which is the city without any outgoing path to another city. The graph is guaranteed to form a line without any loops.\n\n## Approach\nThe solution uses a dictionary (`check`) to keep track of whether a city is an "out" city (has an outgo... | 9 | You are given the array `paths`, where `paths[i] = [cityAi, cityBi]` means there exists a direct path going from `cityAi` to `cityBi`. _Return the destination city, that is, the city without any path outgoing to another city._
It is guaranteed that the graph of paths forms a line without any loop, therefore, there wil... | Do BFS to find the kth level friends. Then collect movies saw by kth level friends and sort them accordingly. |
✅ Navigating Urban Labyrinths: Elegant Solution to Find the Destination City 🔥 | destination-city | 1 | 1 | # Description\n\n## Intuition\nThe problem requires finding the destination city, which is the city without any outgoing path to another city. The graph is guaranteed to form a line without any loops.\n\n## Approach\nThe solution uses a dictionary (`check`) to keep track of whether a city is an "out" city (has an outgo... | 9 | 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. |
✅☑[C++/C/Java/Python/JavaScript] || 2 Approaches || EXPLAINED🔥 | destination-city | 1 | 1 | # PLEASE UPVOTE IF IT HELPED\n\n---\n\n\n# Approaches\n**(Also explained in the code)**\n\n#### ***Approach 1(Brute Force)***\n1. **Iterate Through Paths:**\n\n - The code iterates through each path in the `paths` vector.\n - It selects a candidate destination city (`candidate = paths[i][1]`) from the current pat... | 15 | You are given the array `paths`, where `paths[i] = [cityAi, cityBi]` means there exists a direct path going from `cityAi` to `cityBi`. _Return the destination city, that is, the city without any path outgoing to another city._
It is guaranteed that the graph of paths forms a line without any loop, therefore, there wil... | Do BFS to find the kth level friends. Then collect movies saw by kth level friends and sort them accordingly. |
✅☑[C++/C/Java/Python/JavaScript] || 2 Approaches || EXPLAINED🔥 | destination-city | 1 | 1 | # PLEASE UPVOTE IF IT HELPED\n\n---\n\n\n# Approaches\n**(Also explained in the code)**\n\n#### ***Approach 1(Brute Force)***\n1. **Iterate Through Paths:**\n\n - The code iterates through each path in the `paths` vector.\n - It selects a candidate destination city (`candidate = paths[i][1]`) from the current pat... | 15 | 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. |
【Video】Give me 3 minutes - How we think about a solution | destination-city | 0 | 1 | # Intuition\nWe use subtraction.\n\n---\n\n# Solution Video\n\nhttps://youtu.be/tjILACCjQI0\n\n\u25A0 Timeline of the video\n\n`0:04` Explain how we can solve this question\n`1:29` Coding\n`2:40` Time Complexity and Space Complexity\n`3:13` Step by step algorithm of my solution code\n\n### \u2B50\uFE0F\u2B50\uFE0F Don\... | 48 | You are given the array `paths`, where `paths[i] = [cityAi, cityBi]` means there exists a direct path going from `cityAi` to `cityBi`. _Return the destination city, that is, the city without any path outgoing to another city._
It is guaranteed that the graph of paths forms a line without any loop, therefore, there wil... | Do BFS to find the kth level friends. Then collect movies saw by kth level friends and sort them accordingly. |
【Video】Give me 3 minutes - How we think about a solution | destination-city | 0 | 1 | # Intuition\nWe use subtraction.\n\n---\n\n# Solution Video\n\nhttps://youtu.be/tjILACCjQI0\n\n\u25A0 Timeline of the video\n\n`0:04` Explain how we can solve this question\n`1:29` Coding\n`2:40` Time Complexity and Space Complexity\n`3:13` Step by step algorithm of my solution code\n\n### \u2B50\uFE0F\u2B50\uFE0F Don\... | 48 | 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. |
1436. Destination City => Stepwise Explained 😎| Easy To Understand 🚀 | destination-city | 1 | 1 | # Intuition\nThe code aims to find the destination city given a list of paths. It uses an unordered map to represent the paths between cities and a set to store all cities. The destination city is the one that does not have an outgoing path.\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Appr... | 4 | You are given the array `paths`, where `paths[i] = [cityAi, cityBi]` means there exists a direct path going from `cityAi` to `cityBi`. _Return the destination city, that is, the city without any path outgoing to another city._
It is guaranteed that the graph of paths forms a line without any loop, therefore, there wil... | Do BFS to find the kth level friends. Then collect movies saw by kth level friends and sort them accordingly. |
1436. Destination City => Stepwise Explained 😎| Easy To Understand 🚀 | destination-city | 1 | 1 | # Intuition\nThe code aims to find the destination city given a list of paths. It uses an unordered map to represent the paths between cities and a set to store all cities. The destination city is the one that does not have an outgoing path.\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Appr... | 4 | 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. |
"🧠 Intuitive Solution in C++ | ⏳ 0ms Runtime (Top 5%) | ✅ Beginner-Friendly Explanation" | destination-city | 0 | 1 | # Intuition\n<h2>\nI thought this as topological sort problem but since there is only one node with no outgoing edge therefore the solution becomes much much easier to think.<br><br>\nBut first Step is to think this as topological sort problem.\n<br><br>\n\nEdit : Topological Sort got nothing to do with the provided so... | 3 | You are given the array `paths`, where `paths[i] = [cityAi, cityBi]` means there exists a direct path going from `cityAi` to `cityBi`. _Return the destination city, that is, the city without any path outgoing to another city._
It is guaranteed that the graph of paths forms a line without any loop, therefore, there wil... | Do BFS to find the kth level friends. Then collect movies saw by kth level friends and sort them accordingly. |
"🧠 Intuitive Solution in C++ | ⏳ 0ms Runtime (Top 5%) | ✅ Beginner-Friendly Explanation" | destination-city | 0 | 1 | # Intuition\n<h2>\nI thought this as topological sort problem but since there is only one node with no outgoing edge therefore the solution becomes much much easier to think.<br><br>\nBut first Step is to think this as topological sort problem.\n<br><br>\n\nEdit : Topological Sort got nothing to do with the provided so... | 3 | 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. |
pYthon3 Solution | destination-city | 0 | 1 | \n```\nclass Solution:\n def destCity(self, paths: List[List[str]]) -> str:\n frm,to=zip(*paths)\n return (set(to)-set(frm)).pop()\n``` | 3 | You are given the array `paths`, where `paths[i] = [cityAi, cityBi]` means there exists a direct path going from `cityAi` to `cityBi`. _Return the destination city, that is, the city without any path outgoing to another city._
It is guaranteed that the graph of paths forms a line without any loop, therefore, there wil... | Do BFS to find the kth level friends. Then collect movies saw by kth level friends and sort them accordingly. |
pYthon3 Solution | destination-city | 0 | 1 | \n```\nclass Solution:\n def destCity(self, paths: List[List[str]]) -> str:\n frm,to=zip(*paths)\n return (set(to)-set(frm)).pop()\n``` | 3 | 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. |
🤯 CRAZY ONE-LINE SOLUTION WITH EXPLANATION | destination-city | 0 | 1 | # Approach\n\n1. `paths`\n\n ```\n [\n ["London", "New York"],\n ["New York", "Lima"],\n ["Lima", "Sao Paulo"]\n ] \n ```\n\n2. `zip(*paths)`\n\n ```\n [\n ["London", "New York", "Lima"], \n ["New York", "Lima", "Sao Paulo"]\n ]\n ```\n\n3. `map(set, zip(*pat... | 2 | You are given the array `paths`, where `paths[i] = [cityAi, cityBi]` means there exists a direct path going from `cityAi` to `cityBi`. _Return the destination city, that is, the city without any path outgoing to another city._
It is guaranteed that the graph of paths forms a line without any loop, therefore, there wil... | Do BFS to find the kth level friends. Then collect movies saw by kth level friends and sort them accordingly. |
🤯 CRAZY ONE-LINE SOLUTION WITH EXPLANATION | destination-city | 0 | 1 | # Approach\n\n1. `paths`\n\n ```\n [\n ["London", "New York"],\n ["New York", "Lima"],\n ["Lima", "Sao Paulo"]\n ] \n ```\n\n2. `zip(*paths)`\n\n ```\n [\n ["London", "New York", "Lima"], \n ["New York", "Lima", "Sao Paulo"]\n ]\n ```\n\n3. `map(set, zip(*pat... | 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. |
Beats 99.44% runtime | Python3 | Sets | destination-city | 0 | 1 | # Approach\nWe need to find the "end destination", ie the place from where you can go any further. In the test cases, you can notice that the "end destination" is never a start (by "start" I mean the first elem of a sublist. By "destination", I mean the second elem of a sublist)\n\nSo I just find the place which is nev... | 2 | You are given the array `paths`, where `paths[i] = [cityAi, cityBi]` means there exists a direct path going from `cityAi` to `cityBi`. _Return the destination city, that is, the city without any path outgoing to another city._
It is guaranteed that the graph of paths forms a line without any loop, therefore, there wil... | Do BFS to find the kth level friends. Then collect movies saw by kth level friends and sort them accordingly. |
Beats 99.44% runtime | Python3 | Sets | destination-city | 0 | 1 | # Approach\nWe need to find the "end destination", ie the place from where you can go any further. In the test cases, you can notice that the "end destination" is never a start (by "start" I mean the first elem of a sublist. By "destination", I mean the second elem of a sublist)\n\nSo I just find the place which is nev... | 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. |
✅ One Line Solution | destination-city | 0 | 1 | # Code #1\nTime complexity: $$O(n)$$. Space complexity: $$O(n)$$.\n```\nclass Solution:\n def destCity(self, paths: List[List[str]]) -> str:\n return max({B for _,B in paths} - {A for A,_ in paths})\n```\n\n# Code #2\nTime complexity: $$O(n)$$. Space complexity: $$O(n)$$.\n```\nclass Solution:\n def destCi... | 2 | You are given the array `paths`, where `paths[i] = [cityAi, cityBi]` means there exists a direct path going from `cityAi` to `cityBi`. _Return the destination city, that is, the city without any path outgoing to another city._
It is guaranteed that the graph of paths forms a line without any loop, therefore, there wil... | Do BFS to find the kth level friends. Then collect movies saw by kth level friends and sort them accordingly. |
✅ One Line Solution | destination-city | 0 | 1 | # Code #1\nTime complexity: $$O(n)$$. Space complexity: $$O(n)$$.\n```\nclass Solution:\n def destCity(self, paths: List[List[str]]) -> str:\n return max({B for _,B in paths} - {A for A,_ in paths})\n```\n\n# Code #2\nTime complexity: $$O(n)$$. Space complexity: $$O(n)$$.\n```\nclass Solution:\n def destCi... | 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. |
Eezz Solulu without zip function | destination-city | 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* Initialize two sets dc and ac to store destination cities and arrival cities, respectively.\n* Iterate through the paths list and add each city to the corresponding ... | 2 | You are given the array `paths`, where `paths[i] = [cityAi, cityBi]` means there exists a direct path going from `cityAi` to `cityBi`. _Return the destination city, that is, the city without any path outgoing to another city._
It is guaranteed that the graph of paths forms a line without any loop, therefore, there wil... | Do BFS to find the kth level friends. Then collect movies saw by kth level friends and sort them accordingly. |
Eezz Solulu without zip function | destination-city | 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* Initialize two sets dc and ac to store destination cities and arrival cities, respectively.\n* Iterate through the paths list and add each city to the corresponding ... | 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. |
⭐One-Liner Hack in 🐍|| Two Approaches⭐ | destination-city | 0 | 1 | # Code - 1\n```py\nclass Solution:\n def destCity(self, paths: List[List[str]]) -> str:\n return (set(city[1] for city in paths) - set(city[0] for city in paths)).pop()\n```\n---\n# Code - 2\n```py\nclass Solution:\n def destCity(self, paths: List[List[str]]) -> str:\n return next(city[1] for city i... | 2 | You are given the array `paths`, where `paths[i] = [cityAi, cityBi]` means there exists a direct path going from `cityAi` to `cityBi`. _Return the destination city, that is, the city without any path outgoing to another city._
It is guaranteed that the graph of paths forms a line without any loop, therefore, there wil... | Do BFS to find the kth level friends. Then collect movies saw by kth level friends and sort them accordingly. |
⭐One-Liner Hack in 🐍|| Two Approaches⭐ | destination-city | 0 | 1 | # Code - 1\n```py\nclass Solution:\n def destCity(self, paths: List[List[str]]) -> str:\n return (set(city[1] for city in paths) - set(city[0] for city in paths)).pop()\n```\n---\n# Code - 2\n```py\nclass Solution:\n def destCity(self, paths: List[List[str]]) -> str:\n return next(city[1] for city i... | 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. |
✅ code in 2mins || simple || Python | destination-city | 0 | 1 | # Intuition\nThe problem asks to find the destination city, which is the city that is not the starting city in any route. One intuitive way to solve this is to create sets for the starting cities and destination cities, and then find the city that is in the destination set but not in the starting set.\n\n\n# Approach\n... | 6 | You are given the array `paths`, where `paths[i] = [cityAi, cityBi]` means there exists a direct path going from `cityAi` to `cityBi`. _Return the destination city, that is, the city without any path outgoing to another city._
It is guaranteed that the graph of paths forms a line without any loop, therefore, there wil... | Do BFS to find the kth level friends. Then collect movies saw by kth level friends and sort them accordingly. |
✅ code in 2mins || simple || Python | destination-city | 0 | 1 | # Intuition\nThe problem asks to find the destination city, which is the city that is not the starting city in any route. One intuitive way to solve this is to create sets for the starting cities and destination cities, and then find the city that is in the destination set but not in the starting set.\n\n\n# Approach\n... | 6 | 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.