title stringlengths 1 100 | titleSlug stringlengths 3 77 | Java int64 0 1 | Python3 int64 1 1 | content stringlengths 28 44.4k | voteCount int64 0 3.67k | question_content stringlengths 65 5k | question_hints stringclasses 970
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DFS/Greedy unique solution | number-of-ways-to-reconstruct-a-tree | 0 | 1 | # Code\n```\nfrom collections import defaultdict\n\nclass Solution:\n def checkWays(self, pairs):\n nodes, graph, degree = set(), defaultdict(set), defaultdict(int)\n\n for i, j in pairs:\n nodes |= {i, j}\n graph[i].add(j)\n graph[j].add(i)\n degree[i] += 1\... | 0 | You are given an array `pairs`, where `pairs[i] = [xi, yi]`, and:
* There are no duplicates.
* `xi < yi`
Let `ways` be the number of rooted trees that satisfy the following conditions:
* The tree consists of nodes whose values appeared in `pairs`.
* A pair `[xi, yi]` exists in `pairs` **if and only if** `xi`... | Try to put at least one box in the house pushing it from either side. Once you put one box to the house, you can solve the problem with the same logic used to solve version I. You have a warehouse open from the left only and a warehouse open from the right only. |
DFS/Greedy unique solution | number-of-ways-to-reconstruct-a-tree | 0 | 1 | # Code\n```\nfrom collections import defaultdict\n\nclass Solution:\n def checkWays(self, pairs):\n nodes, graph, degree = set(), defaultdict(set), defaultdict(int)\n\n for i, j in pairs:\n nodes |= {i, j}\n graph[i].add(j)\n graph[j].add(i)\n degree[i] += 1\... | 0 | There are `m` boys and `n` girls in a class attending an upcoming party.
You are given an `m x n` integer matrix `grid`, where `grid[i][j]` equals `0` or `1`. If `grid[i][j] == 1`, then that means the `ith` boy can invite the `jth` girl to the party. A boy can invite at most **one girl**, and a girl can accept at most... | Think inductively. The first step is to get the root. Obviously, the root should be in pairs with all the nodes. If there isn't exactly one such node, then there are 0 ways. The number of pairs involving a node must be less than or equal to that number of its parent. Actually, if it's equal, then there is not exactly 1... |
Python (Simple Maths) | number-of-ways-to-reconstruct-a-tree | 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 `pairs`, where `pairs[i] = [xi, yi]`, and:
* There are no duplicates.
* `xi < yi`
Let `ways` be the number of rooted trees that satisfy the following conditions:
* The tree consists of nodes whose values appeared in `pairs`.
* A pair `[xi, yi]` exists in `pairs` **if and only if** `xi`... | Try to put at least one box in the house pushing it from either side. Once you put one box to the house, you can solve the problem with the same logic used to solve version I. You have a warehouse open from the left only and a warehouse open from the right only. |
Python (Simple Maths) | number-of-ways-to-reconstruct-a-tree | 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 `m` boys and `n` girls in a class attending an upcoming party.
You are given an `m x n` integer matrix `grid`, where `grid[i][j]` equals `0` or `1`. If `grid[i][j] == 1`, then that means the `ith` boy can invite the `jth` girl to the party. A boy can invite at most **one girl**, and a girl can accept at most... | Think inductively. The first step is to get the root. Obviously, the root should be in pairs with all the nodes. If there isn't exactly one such node, then there are 0 ways. The number of pairs involving a node must be less than or equal to that number of its parent. Actually, if it's equal, then there is not exactly 1... |
[Python/Java/CPP/C++] Easy Solution with Explanation [ACCEPTED] | decode-xored-array | 1 | 1 | \n\n**Explanation**\n**a XOR b = c**, we know the values of **a** and **c**. we use the formula to find **b** -> **a XOR c = b** \n**Complexity**\n\nTime ```O(N)```\nSpace ```O(10)```\n\n**Python:**\n```\ndef decode(self, encoded: List[int], first: int) -> List[int]:\n r = [first]\n for i in encoded:\n... | 41 | There is a **hidden** integer array `arr` that consists of `n` non-negative integers.
It was encoded into another integer array `encoded` of length `n - 1`, such that `encoded[i] = arr[i] XOR arr[i + 1]`. For example, if `arr = [1,0,2,1]`, then `encoded = [1,2,3]`.
You are given the `encoded` array. You are also give... | Simulate the process but don’t move the pointer beyond the main folder. Simulate the process but don’t move the pointer beyond the main folder. |
[Python/Java/CPP/C++] Easy Solution with Explanation [ACCEPTED] | decode-xored-array | 1 | 1 | \n\n**Explanation**\n**a XOR b = c**, we know the values of **a** and **c**. we use the formula to find **b** -> **a XOR c = b** \n**Complexity**\n\nTime ```O(N)```\nSpace ```O(10)```\n\n**Python:**\n```\ndef decode(self, encoded: List[int], first: int) -> List[int]:\n r = [first]\n for i in encoded:\n... | 41 | A string is considered **beautiful** if it satisfies the following conditions:
* Each of the 5 English vowels (`'a'`, `'e'`, `'i'`, `'o'`, `'u'`) must appear **at least once** in it.
* The letters must be sorted in **alphabetical order** (i.e. all `'a'`s before `'e'`s, all `'e'`s before `'i'`s, etc.).
For example... | Since that encoded[i] = arr[i] XOR arr[i+1], then arr[i+1] = encoded[i] XOR arr[i]. Iterate on i from beginning to end, and set arr[i+1] = encoded[i] XOR arr[i]. |
Easy Solution: Explained! | decode-xored-array | 0 | 1 | \n\n# Code\n```\nclass Solution:\n def decode(self, encoded: List[int], first: int) -> List[int]:\n result = []\n for i in range(0, len(encoded)+1):\n if(i == 0):\n result.append(0^first)\n else:\n result.append(encoded[i-1]^result[i-1])\n retu... | 3 | There is a **hidden** integer array `arr` that consists of `n` non-negative integers.
It was encoded into another integer array `encoded` of length `n - 1`, such that `encoded[i] = arr[i] XOR arr[i + 1]`. For example, if `arr = [1,0,2,1]`, then `encoded = [1,2,3]`.
You are given the `encoded` array. You are also give... | Simulate the process but don’t move the pointer beyond the main folder. Simulate the process but don’t move the pointer beyond the main folder. |
Easy Solution: Explained! | decode-xored-array | 0 | 1 | \n\n# Code\n```\nclass Solution:\n def decode(self, encoded: List[int], first: int) -> List[int]:\n result = []\n for i in range(0, len(encoded)+1):\n if(i == 0):\n result.append(0^first)\n else:\n result.append(encoded[i-1]^result[i-1])\n retu... | 3 | A string is considered **beautiful** if it satisfies the following conditions:
* Each of the 5 English vowels (`'a'`, `'e'`, `'i'`, `'o'`, `'u'`) must appear **at least once** in it.
* The letters must be sorted in **alphabetical order** (i.e. all `'a'`s before `'e'`s, all `'e'`s before `'i'`s, etc.).
For example... | Since that encoded[i] = arr[i] XOR arr[i+1], then arr[i+1] = encoded[i] XOR arr[i]. Iterate on i from beginning to end, and set arr[i+1] = encoded[i] XOR arr[i]. |
python solution O(N) | decode-xored-array | 0 | 1 | Time and Space Complexcity O(N)\n```\nclass Solution:\n def decode(self, A: List[int], first: int) -> List[int]:\n ans=[first]\n n=len(A)\n for i in range(n):\n a=ans[-1]^A[i]\n ans.append(a)\n return ans\n \n``` | 1 | There is a **hidden** integer array `arr` that consists of `n` non-negative integers.
It was encoded into another integer array `encoded` of length `n - 1`, such that `encoded[i] = arr[i] XOR arr[i + 1]`. For example, if `arr = [1,0,2,1]`, then `encoded = [1,2,3]`.
You are given the `encoded` array. You are also give... | Simulate the process but don’t move the pointer beyond the main folder. Simulate the process but don’t move the pointer beyond the main folder. |
python solution O(N) | decode-xored-array | 0 | 1 | Time and Space Complexcity O(N)\n```\nclass Solution:\n def decode(self, A: List[int], first: int) -> List[int]:\n ans=[first]\n n=len(A)\n for i in range(n):\n a=ans[-1]^A[i]\n ans.append(a)\n return ans\n \n``` | 1 | A string is considered **beautiful** if it satisfies the following conditions:
* Each of the 5 English vowels (`'a'`, `'e'`, `'i'`, `'o'`, `'u'`) must appear **at least once** in it.
* The letters must be sorted in **alphabetical order** (i.e. all `'a'`s before `'e'`s, all `'e'`s before `'i'`s, etc.).
For example... | Since that encoded[i] = arr[i] XOR arr[i+1], then arr[i+1] = encoded[i] XOR arr[i]. Iterate on i from beginning to end, and set arr[i+1] = encoded[i] XOR arr[i]. |
Python - 1 Liner (List Comprehension with Assignment Expresion) | decode-xored-array | 0 | 1 | ```\nclass Solution:\n def decode(self, encoded: List[int], first: int) -> List[int]:\n return [first] + [first:= first ^ x for x in encoded]\n``` | 18 | There is a **hidden** integer array `arr` that consists of `n` non-negative integers.
It was encoded into another integer array `encoded` of length `n - 1`, such that `encoded[i] = arr[i] XOR arr[i + 1]`. For example, if `arr = [1,0,2,1]`, then `encoded = [1,2,3]`.
You are given the `encoded` array. You are also give... | Simulate the process but don’t move the pointer beyond the main folder. Simulate the process but don’t move the pointer beyond the main folder. |
Python - 1 Liner (List Comprehension with Assignment Expresion) | decode-xored-array | 0 | 1 | ```\nclass Solution:\n def decode(self, encoded: List[int], first: int) -> List[int]:\n return [first] + [first:= first ^ x for x in encoded]\n``` | 18 | A string is considered **beautiful** if it satisfies the following conditions:
* Each of the 5 English vowels (`'a'`, `'e'`, `'i'`, `'o'`, `'u'`) must appear **at least once** in it.
* The letters must be sorted in **alphabetical order** (i.e. all `'a'`s before `'e'`s, all `'e'`s before `'i'`s, etc.).
For example... | Since that encoded[i] = arr[i] XOR arr[i+1], then arr[i+1] = encoded[i] XOR arr[i]. Iterate on i from beginning to end, and set arr[i+1] = encoded[i] XOR arr[i]. |
PYTHON|C Linear Time Complexity and O(1) space complexity | swapping-nodes-in-a-linked-list | 0 | 1 | # Intuition\nIterate list two times first to get the first index to change second to get the seocnd index while searching for first index we can traverse to get the length also then in second traverse using length find the index number for second element and get to that element. \nJust swap these two values.\n\n# Appro... | 1 | You are given the `head` of a linked list, and an integer `k`.
Return _the head of the linked list after **swapping** the values of the_ `kth` _node from the beginning and the_ `kth` _node from the end (the list is **1-indexed**)._
**Example 1:**
**Input:** head = \[1,2,3,4,5\], k = 2
**Output:** \[1,4,3,2,5\]
**Ex... | Think simulation Note that the number of turns will never be more than 50 / 4 * n |
PYTHON|C Linear Time Complexity and O(1) space complexity | swapping-nodes-in-a-linked-list | 0 | 1 | # Intuition\nIterate list two times first to get the first index to change second to get the seocnd index while searching for first index we can traverse to get the length also then in second traverse using length find the index number for second element and get to that element. \nJust swap these two values.\n\n# Appro... | 1 | You are given a **0-indexed** array of positive integers `w` where `w[i]` describes the **weight** of the `ith` index.
You need to implement the function `pickIndex()`, which **randomly** picks an index in the range `[0, w.length - 1]` (**inclusive**) and returns it. The **probability** of picking an index `i` is `w[i... | We can transform the linked list to an array this should ease things up After transforming the linked list to an array it becomes as easy as swapping two integers in an array then rebuilding the linked list |
Solution | swapping-nodes-in-a-linked-list | 1 | 1 | ```C++ []\nclass Solution {\npublic:\n ListNode* swapNodes(ListNode* head, int k) {\n ListNode* temp=head;\n ListNode* first=head;\n ListNode* second=head;\n int count=0,i=k-1;\n while (temp!=NULL)\n {\n if(i==0)\n {\n first=temp;\n ... | 1 | You are given the `head` of a linked list, and an integer `k`.
Return _the head of the linked list after **swapping** the values of the_ `kth` _node from the beginning and the_ `kth` _node from the end (the list is **1-indexed**)._
**Example 1:**
**Input:** head = \[1,2,3,4,5\], k = 2
**Output:** \[1,4,3,2,5\]
**Ex... | Think simulation Note that the number of turns will never be more than 50 / 4 * n |
Solution | swapping-nodes-in-a-linked-list | 1 | 1 | ```C++ []\nclass Solution {\npublic:\n ListNode* swapNodes(ListNode* head, int k) {\n ListNode* temp=head;\n ListNode* first=head;\n ListNode* second=head;\n int count=0,i=k-1;\n while (temp!=NULL)\n {\n if(i==0)\n {\n first=temp;\n ... | 1 | You are given a **0-indexed** array of positive integers `w` where `w[i]` describes the **weight** of the `ith` index.
You need to implement the function `pickIndex()`, which **randomly** picks an index in the range `[0, w.length - 1]` (**inclusive**) and returns it. The **probability** of picking an index `i` is `w[i... | We can transform the linked list to an array this should ease things up After transforming the linked list to an array it becomes as easy as swapping two integers in an array then rebuilding the linked list |
Python short and clean. Functional programming. | swapping-nodes-in-a-linked-list | 0 | 1 | # Complexity\n- Time complexity: $$O(n)$$\n\n- Space complexity: $$O(n)$$\n\nwhere, `n is the number of nodes in the linkedlist`\n\n# Code\nImperative:\n```python\nclass Solution:\n def swapNodes(self, head: ListNode | None, k: int) -> ListNode | None:\n sentinal_head = i = j = ListNode(next=head)\n \n... | 2 | You are given the `head` of a linked list, and an integer `k`.
Return _the head of the linked list after **swapping** the values of the_ `kth` _node from the beginning and the_ `kth` _node from the end (the list is **1-indexed**)._
**Example 1:**
**Input:** head = \[1,2,3,4,5\], k = 2
**Output:** \[1,4,3,2,5\]
**Ex... | Think simulation Note that the number of turns will never be more than 50 / 4 * n |
Python short and clean. Functional programming. | swapping-nodes-in-a-linked-list | 0 | 1 | # Complexity\n- Time complexity: $$O(n)$$\n\n- Space complexity: $$O(n)$$\n\nwhere, `n is the number of nodes in the linkedlist`\n\n# Code\nImperative:\n```python\nclass Solution:\n def swapNodes(self, head: ListNode | None, k: int) -> ListNode | None:\n sentinal_head = i = j = ListNode(next=head)\n \n... | 2 | You are given a **0-indexed** array of positive integers `w` where `w[i]` describes the **weight** of the `ith` index.
You need to implement the function `pickIndex()`, which **randomly** picks an index in the range `[0, w.length - 1]` (**inclusive**) and returns it. The **probability** of picking an index `i` is `w[i... | We can transform the linked list to an array this should ease things up After transforming the linked list to an array it becomes as easy as swapping two integers in an array then rebuilding the linked list |
Python simple solution beats 99.55% with explanation & key takeaway explained! O(n) time, O(1) space | swapping-nodes-in-a-linked-list | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nLet n be the length of the array. Such that m = n - k\nk + m = n: We need to find linklist index at -k. Therefore, it\'s translate to n - k index. Therefore, m is the position. \n\n***\nKey takeaway: m = n - k\n***\n\n._
**Example 1:**
**Input:** head = \[1,2,3,4,5\], k = 2
**Output:** \[1,4,3,2,5\]
**Ex... | Think simulation Note that the number of turns will never be more than 50 / 4 * n |
Python simple solution beats 99.55% with explanation & key takeaway explained! O(n) time, O(1) space | swapping-nodes-in-a-linked-list | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nLet n be the length of the array. Such that m = n - k\nk + m = n: We need to find linklist index at -k. Therefore, it\'s translate to n - k index. Therefore, m is the position. \n\n***\nKey takeaway: m = n - k\n***\n\n`, which **randomly** picks an index in the range `[0, w.length - 1]` (**inclusive**) and returns it. The **probability** of picking an index `i` is `w[i... | We can transform the linked list to an array this should ease things up After transforming the linked list to an array it becomes as easy as swapping two integers in an array then rebuilding the linked list |
Python 3 | DFS, Greedy O(N) | Explanations | minimize-hamming-distance-after-swap-operations | 0 | 1 | ### Explanation\n- First, build a graph\n- Then, traverse each index `i`\n\t- If `i` is not visited, then DFS on it and its neighbors, ultimately\n\t\t- `d[i]` will be a **counter** in type of `defaultdict()`, which represents the neighbors\' values & their frequencies\n\t\t- for any direct/indirect neighbor `nei` of `... | 3 | You are given two integer arrays, `source` and `target`, both of length `n`. You are also given an array `allowedSwaps` where each `allowedSwaps[i] = [ai, bi]` indicates that you are allowed to swap the elements at index `ai` and index `bi` **(0-indexed)** of array `source`. Note that you can swap elements at a specifi... | Create a tree structure of the family. Without deaths, the order of inheritance is simply a pre-order traversal of the tree. Mark the dead family members tree nodes and don't include them in the final order. |
Python 3 | DFS, Greedy O(N) | Explanations | minimize-hamming-distance-after-swap-operations | 0 | 1 | ### Explanation\n- First, build a graph\n- Then, traverse each index `i`\n\t- If `i` is not visited, then DFS on it and its neighbors, ultimately\n\t\t- `d[i]` will be a **counter** in type of `defaultdict()`, which represents the neighbors\' values & their frequencies\n\t\t- for any direct/indirect neighbor `nei` of `... | 3 | You want to build `n` new buildings in a city. The new buildings will be built in a line and are labeled from `1` to `n`.
However, there are city restrictions on the heights of the new buildings:
* The height of each building must be a non-negative integer.
* The height of the first building **must** be `0`.
* ... | The source array can be imagined as a graph where each index is a node and each allowedSwaps[i] is an edge. Nodes within the same component can be freely swapped with each other. For each component, find the number of common elements. The elements that are not in common will contribute to the total Hamming distance. |
Fancy Union-Find Approach | minimize-hamming-distance-after-swap-operations | 0 | 1 | # Complexity\n- Time complexity: $$O(n*log(n))$$.\n\n- Space complexity: $$O(n)$$.\n\n# Code\n```\nclass UnionFind:\n def __init__(self):\n self.par = {}\n \n def union(self, el1, el2):\n self.par[self.find(el2)] = self.find(el1)\n \n\n def find(self, el):\n p = self.par.get(el, el)\... | 0 | You are given two integer arrays, `source` and `target`, both of length `n`. You are also given an array `allowedSwaps` where each `allowedSwaps[i] = [ai, bi]` indicates that you are allowed to swap the elements at index `ai` and index `bi` **(0-indexed)** of array `source`. Note that you can swap elements at a specifi... | Create a tree structure of the family. Without deaths, the order of inheritance is simply a pre-order traversal of the tree. Mark the dead family members tree nodes and don't include them in the final order. |
Fancy Union-Find Approach | minimize-hamming-distance-after-swap-operations | 0 | 1 | # Complexity\n- Time complexity: $$O(n*log(n))$$.\n\n- Space complexity: $$O(n)$$.\n\n# Code\n```\nclass UnionFind:\n def __init__(self):\n self.par = {}\n \n def union(self, el1, el2):\n self.par[self.find(el2)] = self.find(el1)\n \n\n def find(self, el):\n p = self.par.get(el, el)\... | 0 | You want to build `n` new buildings in a city. The new buildings will be built in a line and are labeled from `1` to `n`.
However, there are city restrictions on the heights of the new buildings:
* The height of each building must be a non-negative integer.
* The height of the first building **must** be `0`.
* ... | The source array can be imagined as a graph where each index is a node and each allowedSwaps[i] is an edge. Nodes within the same component can be freely swapped with each other. For each component, find the number of common elements. The elements that are not in common will contribute to the total Hamming distance. |
Simple and clear python3 solutions | 1040 ms - faster than 90.82% solutions | minimize-hamming-distance-after-swap-operations | 0 | 1 | # Complexity\n- Time complexity: $$O(n + m \\cdot \\alpha(2 \\cdot m))$$\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity: $$O(m)$$\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\nwhere `n = source.length`, `m = allowedSwaps.length`, $\\alpha(n)$ is inverse Ackermann function ([wi... | 0 | You are given two integer arrays, `source` and `target`, both of length `n`. You are also given an array `allowedSwaps` where each `allowedSwaps[i] = [ai, bi]` indicates that you are allowed to swap the elements at index `ai` and index `bi` **(0-indexed)** of array `source`. Note that you can swap elements at a specifi... | Create a tree structure of the family. Without deaths, the order of inheritance is simply a pre-order traversal of the tree. Mark the dead family members tree nodes and don't include them in the final order. |
Simple and clear python3 solutions | 1040 ms - faster than 90.82% solutions | minimize-hamming-distance-after-swap-operations | 0 | 1 | # Complexity\n- Time complexity: $$O(n + m \\cdot \\alpha(2 \\cdot m))$$\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity: $$O(m)$$\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\nwhere `n = source.length`, `m = allowedSwaps.length`, $\\alpha(n)$ is inverse Ackermann function ([wi... | 0 | You want to build `n` new buildings in a city. The new buildings will be built in a line and are labeled from `1` to `n`.
However, there are city restrictions on the heights of the new buildings:
* The height of each building must be a non-negative integer.
* The height of the first building **must** be `0`.
* ... | The source array can be imagined as a graph where each index is a node and each allowedSwaps[i] is an edge. Nodes within the same component can be freely swapped with each other. For each component, find the number of common elements. The elements that are not in common will contribute to the total Hamming distance. |
Python UnionFind | minimize-hamming-distance-after-swap-operations | 0 | 1 | # Intuition\nThe idea is that if (0, 1) is exchangeable and (0, 2) is exchangeable,\nthen any apair in (0, 1, 2) can be exchangeble.\n\nThis means, we could build connected components where\neach component is a list of indices that can be exchangeable with any of them.\n\nIn Union find terms, we simply iterate through ... | 0 | You are given two integer arrays, `source` and `target`, both of length `n`. You are also given an array `allowedSwaps` where each `allowedSwaps[i] = [ai, bi]` indicates that you are allowed to swap the elements at index `ai` and index `bi` **(0-indexed)** of array `source`. Note that you can swap elements at a specifi... | Create a tree structure of the family. Without deaths, the order of inheritance is simply a pre-order traversal of the tree. Mark the dead family members tree nodes and don't include them in the final order. |
Python UnionFind | minimize-hamming-distance-after-swap-operations | 0 | 1 | # Intuition\nThe idea is that if (0, 1) is exchangeable and (0, 2) is exchangeable,\nthen any apair in (0, 1, 2) can be exchangeble.\n\nThis means, we could build connected components where\neach component is a list of indices that can be exchangeable with any of them.\n\nIn Union find terms, we simply iterate through ... | 0 | You want to build `n` new buildings in a city. The new buildings will be built in a line and are labeled from `1` to `n`.
However, there are city restrictions on the heights of the new buildings:
* The height of each building must be a non-negative integer.
* The height of the first building **must** be `0`.
* ... | The source array can be imagined as a graph where each index is a node and each allowedSwaps[i] is an edge. Nodes within the same component can be freely swapped with each other. For each component, find the number of common elements. The elements that are not in common will contribute to the total Hamming distance. |
Python3 Connected Subgraph - Runtime 100% | minimize-hamming-distance-after-swap-operations | 0 | 1 | # Approach\nSince we can swaps as many times as possible, swaps can be seen a connected subgraphs. For example if we can swap [0,1] and [1,2], we can also swap [0, 2]. This means {0, 1, 2} is a connected subgraph where any of the values can be swapped in any of the indexes in that subgraph.\n\nSo the idea is that we fi... | 0 | You are given two integer arrays, `source` and `target`, both of length `n`. You are also given an array `allowedSwaps` where each `allowedSwaps[i] = [ai, bi]` indicates that you are allowed to swap the elements at index `ai` and index `bi` **(0-indexed)** of array `source`. Note that you can swap elements at a specifi... | Create a tree structure of the family. Without deaths, the order of inheritance is simply a pre-order traversal of the tree. Mark the dead family members tree nodes and don't include them in the final order. |
Python3 Connected Subgraph - Runtime 100% | minimize-hamming-distance-after-swap-operations | 0 | 1 | # Approach\nSince we can swaps as many times as possible, swaps can be seen a connected subgraphs. For example if we can swap [0,1] and [1,2], we can also swap [0, 2]. This means {0, 1, 2} is a connected subgraph where any of the values can be swapped in any of the indexes in that subgraph.\n\nSo the idea is that we fi... | 0 | You want to build `n` new buildings in a city. The new buildings will be built in a line and are labeled from `1` to `n`.
However, there are city restrictions on the heights of the new buildings:
* The height of each building must be a non-negative integer.
* The height of the first building **must** be `0`.
* ... | The source array can be imagined as a graph where each index is a node and each allowedSwaps[i] is an edge. Nodes within the same component can be freely swapped with each other. For each component, find the number of common elements. The elements that are not in common will contribute to the total Hamming distance. |
Python3 || DSU Algorithm | minimize-hamming-distance-after-swap-operations | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nFor all the swaps operations run union find search. Which will components on the the indexes which can be swapped within themselves.\n\nNow, for each index find its parent (find operation of parent) and create a hashmap for all the values... | 0 | You are given two integer arrays, `source` and `target`, both of length `n`. You are also given an array `allowedSwaps` where each `allowedSwaps[i] = [ai, bi]` indicates that you are allowed to swap the elements at index `ai` and index `bi` **(0-indexed)** of array `source`. Note that you can swap elements at a specifi... | Create a tree structure of the family. Without deaths, the order of inheritance is simply a pre-order traversal of the tree. Mark the dead family members tree nodes and don't include them in the final order. |
Python3 || DSU Algorithm | minimize-hamming-distance-after-swap-operations | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nFor all the swaps operations run union find search. Which will components on the the indexes which can be swapped within themselves.\n\nNow, for each index find its parent (find operation of parent) and create a hashmap for all the values... | 0 | You want to build `n` new buildings in a city. The new buildings will be built in a line and are labeled from `1` to `n`.
However, there are city restrictions on the heights of the new buildings:
* The height of each building must be a non-negative integer.
* The height of the first building **must** be `0`.
* ... | The source array can be imagined as a graph where each index is a node and each allowedSwaps[i] is an edge. Nodes within the same component can be freely swapped with each other. For each component, find the number of common elements. The elements that are not in common will contribute to the total Hamming distance. |
Python Union Find Solution 90 % Faster | minimize-hamming-distance-after-swap-operations | 0 | 1 | My code initializes the parent dictionary, which maps each index to itself, and the rank list, which stores the ranks of the parent elements in the union-find algorithm.\n\nThe find function implements path compression in the union-find algorithm. It iterativaley finds the root parent of an element and updates the pare... | 0 | You are given two integer arrays, `source` and `target`, both of length `n`. You are also given an array `allowedSwaps` where each `allowedSwaps[i] = [ai, bi]` indicates that you are allowed to swap the elements at index `ai` and index `bi` **(0-indexed)** of array `source`. Note that you can swap elements at a specifi... | Create a tree structure of the family. Without deaths, the order of inheritance is simply a pre-order traversal of the tree. Mark the dead family members tree nodes and don't include them in the final order. |
Python Union Find Solution 90 % Faster | minimize-hamming-distance-after-swap-operations | 0 | 1 | My code initializes the parent dictionary, which maps each index to itself, and the rank list, which stores the ranks of the parent elements in the union-find algorithm.\n\nThe find function implements path compression in the union-find algorithm. It iterativaley finds the root parent of an element and updates the pare... | 0 | You want to build `n` new buildings in a city. The new buildings will be built in a line and are labeled from `1` to `n`.
However, there are city restrictions on the heights of the new buildings:
* The height of each building must be a non-negative integer.
* The height of the first building **must** be `0`.
* ... | The source array can be imagined as a graph where each index is a node and each allowedSwaps[i] is an edge. Nodes within the same component can be freely swapped with each other. For each component, find the number of common elements. The elements that are not in common will contribute to the total Hamming distance. |
[Python3] backtracking | find-minimum-time-to-finish-all-jobs | 0 | 1 | **Algo**\nHere, we simply try out all possibilities but eliminate those that are apparently a waste of time. \n\n**Implementation**\n```\nclass Solution:\n def minimumTimeRequired(self, jobs: List[int], k: int) -> int:\n jobs.sort(reverse=True)\n \n def fn(i):\n """Assign jobs to work... | 6 | You are given an integer array `jobs`, where `jobs[i]` is the amount of time it takes to complete the `ith` job.
There are `k` workers that you can assign jobs to. Each job should be assigned to **exactly** one worker. The **working time** of a worker is the sum of the time it takes to complete all jobs assigned to th... | Think brute force When is a subset of requests okay? |
[Python3] backtracking | find-minimum-time-to-finish-all-jobs | 0 | 1 | **Algo**\nHere, we simply try out all possibilities but eliminate those that are apparently a waste of time. \n\n**Implementation**\n```\nclass Solution:\n def minimumTimeRequired(self, jobs: List[int], k: int) -> int:\n jobs.sort(reverse=True)\n \n def fn(i):\n """Assign jobs to work... | 6 | You are given two integers, `m` and `k`, and a stream of integers. You are tasked to implement a data structure that calculates the **MKAverage** for the stream.
The **MKAverage** can be calculated using these steps:
1. If the number of the elements in the stream is less than `m` you should consider the **MKAverage*... | We can select a subset of tasks and assign it to a worker then solve the subproblem on the remaining tasks |
Python implementation of the theoretically optimal O(3^N) solution | find-minimum-time-to-finish-all-jobs | 0 | 1 | Here\'s an implementation in python of the theoretically optimal solution in O(3^N). All the other posted python solutions at this time are not theoretically optimal and just implement backtracking with different pruning techniques to fit the time limit.\n\nI struggled a bit to get the solution to fit into the time lim... | 2 | You are given an integer array `jobs`, where `jobs[i]` is the amount of time it takes to complete the `ith` job.
There are `k` workers that you can assign jobs to. Each job should be assigned to **exactly** one worker. The **working time** of a worker is the sum of the time it takes to complete all jobs assigned to th... | Think brute force When is a subset of requests okay? |
Python implementation of the theoretically optimal O(3^N) solution | find-minimum-time-to-finish-all-jobs | 0 | 1 | Here\'s an implementation in python of the theoretically optimal solution in O(3^N). All the other posted python solutions at this time are not theoretically optimal and just implement backtracking with different pruning techniques to fit the time limit.\n\nI struggled a bit to get the solution to fit into the time lim... | 2 | You are given two integers, `m` and `k`, and a stream of integers. You are tasked to implement a data structure that calculates the **MKAverage** for the stream.
The **MKAverage** can be calculated using these steps:
1. If the number of the elements in the stream is less than `m` you should consider the **MKAverage*... | We can select a subset of tasks and assign it to a worker then solve the subproblem on the remaining tasks |
Branch & Bound + Greedy beats %98 | find-minimum-time-to-finish-all-jobs | 0 | 1 | # Intuition\nThis is a NP-complete problem. We can use Branch & Bound algorithm to solve it. However, a naive implementation will have TLE error for some test cases.\n\nInitially, I use the sum of all jobs\' time as upper bound. Due to the TLE, I sorted the jobs in decending order. It helped a little, but still cause T... | 0 | You are given an integer array `jobs`, where `jobs[i]` is the amount of time it takes to complete the `ith` job.
There are `k` workers that you can assign jobs to. Each job should be assigned to **exactly** one worker. The **working time** of a worker is the sum of the time it takes to complete all jobs assigned to th... | Think brute force When is a subset of requests okay? |
Branch & Bound + Greedy beats %98 | find-minimum-time-to-finish-all-jobs | 0 | 1 | # Intuition\nThis is a NP-complete problem. We can use Branch & Bound algorithm to solve it. However, a naive implementation will have TLE error for some test cases.\n\nInitially, I use the sum of all jobs\' time as upper bound. Due to the TLE, I sorted the jobs in decending order. It helped a little, but still cause T... | 0 | You are given two integers, `m` and `k`, and a stream of integers. You are tasked to implement a data structure that calculates the **MKAverage** for the stream.
The **MKAverage** can be calculated using these steps:
1. If the number of the elements in the stream is less than `m` you should consider the **MKAverage*... | We can select a subset of tasks and assign it to a worker then solve the subproblem on the remaining tasks |
4 optimizations that can be made. Beats 99.53% in runtime | find-minimum-time-to-finish-all-jobs | 0 | 1 | \n\n# Approach\nThis can be done by as a brute-force search with early termination. We find that the naive solution will TLE and certain optimisations are necessary.\n\nThis solution acts as a summary of the... | 0 | You are given an integer array `jobs`, where `jobs[i]` is the amount of time it takes to complete the `ith` job.
There are `k` workers that you can assign jobs to. Each job should be assigned to **exactly** one worker. The **working time** of a worker is the sum of the time it takes to complete all jobs assigned to th... | Think brute force When is a subset of requests okay? |
4 optimizations that can be made. Beats 99.53% in runtime | find-minimum-time-to-finish-all-jobs | 0 | 1 | \n\n# Approach\nThis can be done by as a brute-force search with early termination. We find that the naive solution will TLE and certain optimisations are necessary.\n\nThis solution acts as a summary of the... | 0 | You are given two integers, `m` and `k`, and a stream of integers. You are tasked to implement a data structure that calculates the **MKAverage** for the stream.
The **MKAverage** can be calculated using these steps:
1. If the number of the elements in the stream is less than `m` you should consider the **MKAverage*... | We can select a subset of tasks and assign it to a worker then solve the subproblem on the remaining tasks |
Simple solution with HashMap in Python3 | number-of-rectangles-that-can-form-the-largest-square | 0 | 1 | # Intuition\nHere we have:\n- list of `rectangles` with height and width\n- our goal is to find **maximized count** of rects **we can cut**, to have a **perfect square** \n\nAt each step we should somehow **cut** all of the rects to find a `maxLen`.\n\nFor this problem we\'re going to use **HashMap** to track **frequen... | 1 | You are given an array `rectangles` where `rectangles[i] = [li, wi]` represents the `ith` rectangle of length `li` and width `wi`.
You can cut the `ith` rectangle to form a square with a side length of `k` if both `k <= li` and `k <= wi`. For example, if you have a rectangle `[4,6]`, you can cut it to get a square wit... | Try to use dynamic programming where the current index and remaining number of line segments to form can describe any intermediate state. To make the computation of each state in constant time, we could add another flag to the state that indicates whether or not we are in the middle of placing a line (placed start poin... |
With basic of python code | number-of-rectangles-that-can-form-the-largest-square | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ --... | 1 | You are given an array `rectangles` where `rectangles[i] = [li, wi]` represents the `ith` rectangle of length `li` and width `wi`.
You can cut the `ith` rectangle to form a square with a side length of `k` if both `k <= li` and `k <= wi`. For example, if you have a rectangle `[4,6]`, you can cut it to get a square wit... | Try to use dynamic programming where the current index and remaining number of line segments to form can describe any intermediate state. To make the computation of each state in constant time, we could add another flag to the state that indicates whether or not we are in the middle of placing a line (placed start poin... |
Python - Runtime O(N) and O(1) space [ACCEPTED] | number-of-rectangles-that-can-form-the-largest-square | 0 | 1 | This can be done just by tracking the max length and its count.\n\n```\ndef countGoodRectangles(rectangles):\n\tmax_len = float(\'-inf\')\n\tcount = 0\n\tfor item in rectangles:\n\t\tmin_len = min(item)\n\t\tif min_len == max_len:\n\t\t\tcount += 1\n\t\telif min_len > max_len:\n\t\t\tmax_len = min_len\n\t\t\tcount = 1\... | 18 | You are given an array `rectangles` where `rectangles[i] = [li, wi]` represents the `ith` rectangle of length `li` and width `wi`.
You can cut the `ith` rectangle to form a square with a side length of `k` if both `k <= li` and `k <= wi`. For example, if you have a rectangle `[4,6]`, you can cut it to get a square wit... | Try to use dynamic programming where the current index and remaining number of line segments to form can describe any intermediate state. To make the computation of each state in constant time, we could add another flag to the state that indicates whether or not we are in the middle of placing a line (placed start poin... |
[python|| O(N)] | number-of-rectangles-that-can-form-the-largest-square | 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- O(N)\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n- O(1)\n<!-- Add your space complexity here, ... | 2 | You are given an array `rectangles` where `rectangles[i] = [li, wi]` represents the `ith` rectangle of length `li` and width `wi`.
You can cut the `ith` rectangle to form a square with a side length of `k` if both `k <= li` and `k <= wi`. For example, if you have a rectangle `[4,6]`, you can cut it to get a square wit... | Try to use dynamic programming where the current index and remaining number of line segments to form can describe any intermediate state. To make the computation of each state in constant time, we could add another flag to the state that indicates whether or not we are in the middle of placing a line (placed start poin... |
✅✔ SIMPLE PYTHON3 SOLUTION ✅✔easy to understand | number-of-rectangles-that-can-form-the-largest-square | 0 | 1 | ***UPVOTE*** if it is helpful\n``` \nclass Solution:\n def countGoodRectangles(self, rectangles: List[List[int]]) -> int:\n c = []\n count = 0\n for i in rectangles:\n c.append(min(i))\n ans = max(c)\n for j in range(len(c)):\n if c[j] == ans:\n ... | 0 | You are given an array `rectangles` where `rectangles[i] = [li, wi]` represents the `ith` rectangle of length `li` and width `wi`.
You can cut the `ith` rectangle to form a square with a side length of `k` if both `k <= li` and `k <= wi`. For example, if you have a rectangle `[4,6]`, you can cut it to get a square wit... | Try to use dynamic programming where the current index and remaining number of line segments to form can describe any intermediate state. To make the computation of each state in constant time, we could add another flag to the state that indicates whether or not we are in the middle of placing a line (placed start poin... |
Python easy solution | number-of-rectangles-that-can-form-the-largest-square | 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. -->\nGei min in each rectangles pair as max side, then sort and get the count of max rectangle\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g.... | 0 | You are given an array `rectangles` where `rectangles[i] = [li, wi]` represents the `ith` rectangle of length `li` and width `wi`.
You can cut the `ith` rectangle to form a square with a side length of `k` if both `k <= li` and `k <= wi`. For example, if you have a rectangle `[4,6]`, you can cut it to get a square wit... | Try to use dynamic programming where the current index and remaining number of line segments to form can describe any intermediate state. To make the computation of each state in constant time, we could add another flag to the state that indicates whether or not we are in the middle of placing a line (placed start poin... |
[Python3] freq table | tuple-with-same-product | 0 | 1 | **Algo**\nUse a nested loop to scan possible `x*y` and keep a frequency table. When a product has been seen before, update `ans` with counting before the current occurrence. Return `8*ans`. \n\n**Implementation**\n```\nclass Solution:\n def tupleSameProduct(self, nums: List[int]) -> int:\n ans = 0\n fr... | 34 | Given an array `nums` of **distinct** positive integers, return _the number of tuples_ `(a, b, c, d)` _such that_ `a * b = c * d` _where_ `a`_,_ `b`_,_ `c`_, and_ `d` _are elements of_ `nums`_, and_ `a != b != c != d`_._
**Example 1:**
**Input:** nums = \[2,3,4,6\]
**Output:** 8
**Explanation:** There are 8 valid tup... | The constraints are small enough to consider every possible coordinate and calculate its quality. |
[PYTHON3] [0 LINER] [FASTER THAN 100.00%] [NO ITERTOOLS] [ENTIRE FILE IS 2 LINES] Fast 0-liner | tuple-with-same-product | 0 | 1 | ```\nclass Solution:\n tupleSameProduct = lambda _, nums: sum(count*(count-1) for _, count in Counter([nums[i] * nums[j] for j in range(1, len(nums)) for i in range(len(nums) - 1) if j > i]).items()) * 4\n``` | 1 | Given an array `nums` of **distinct** positive integers, return _the number of tuples_ `(a, b, c, d)` _such that_ `a * b = c * d` _where_ `a`_,_ `b`_,_ `c`_, and_ `d` _are elements of_ `nums`_, and_ `a != b != c != d`_._
**Example 1:**
**Input:** nums = \[2,3,4,6\]
**Output:** 8
**Explanation:** There are 8 valid tup... | The constraints are small enough to consider every possible coordinate and calculate its quality. |
Simple Python3 | 6 Lines | Detailed Explanation | tuple-with-same-product | 0 | 1 | Logic:\na * b = c * d; there are 8 permutations of tuples for every (a,b,c,d) that satisfy this i.e (a,b,c,d), (a,b,d,c), (b,a,c,d), (b,a,d,c), (c,d,a,b), (c,d,b,a), (d,c,a,b), (d,c,b,a). So for every two pairs (a,b) and (c,d) that have the same product, we have 8 tuples and hence whenever we find a tuple (a,b) we chec... | 14 | Given an array `nums` of **distinct** positive integers, return _the number of tuples_ `(a, b, c, d)` _such that_ `a * b = c * d` _where_ `a`_,_ `b`_,_ `c`_, and_ `d` _are elements of_ `nums`_, and_ `a != b != c != d`_._
**Example 1:**
**Input:** nums = \[2,3,4,6\]
**Output:** 8
**Explanation:** There are 8 valid tup... | The constraints are small enough to consider every possible coordinate and calculate its quality. |
[Python, Java] Elegant & Short | Counter | tuple-with-same-product | 0 | 1 | # Complexity\n- Time complexity: $$O(n^{2})$$\n- Space complexity: $$O(n^{2})$$\n\n# Code\n\n\n```python []\nclass Solution:\n def tupleSameProduct(self, nums: List[int]) -> int:\n n = len(nums)\n\n products = Counter(\n nums[i] * nums[j]\n for i in range(n)\n for j in ... | 2 | Given an array `nums` of **distinct** positive integers, return _the number of tuples_ `(a, b, c, d)` _such that_ `a * b = c * d` _where_ `a`_,_ `b`_,_ `c`_, and_ `d` _are elements of_ `nums`_, and_ `a != b != c != d`_._
**Example 1:**
**Input:** nums = \[2,3,4,6\]
**Output:** 8
**Explanation:** There are 8 valid tup... | The constraints are small enough to consider every possible coordinate and calculate its quality. |
[Python] Hashmap with explanation | tuple-with-same-product | 0 | 1 | \nFor example: [1,2,4,5,10]\n\nwe traverse the array:\n2 = nums[0] * nums[1]\n4 = nums[0] * nums[2]\n5 = nums[0] * nums[3]\n10 = nums[0] * nums[4] # (1,10)\n8 = nums[1] * nums[2]\n10 = nums[1] * nums[3] # (2,5)\n20 = nums[1] * nums[4] # (2,10)\n20 = nums[2] * nums[3] # (4,5)\n40 = nums[2] * nums[4]\n50 = nums[3] * nums... | 10 | Given an array `nums` of **distinct** positive integers, return _the number of tuples_ `(a, b, c, d)` _such that_ `a * b = c * d` _where_ `a`_,_ `b`_,_ `c`_, and_ `d` _are elements of_ `nums`_, and_ `a != b != c != d`_._
**Example 1:**
**Input:** nums = \[2,3,4,6\]
**Output:** 8
**Explanation:** There are 8 valid tup... | The constraints are small enough to consider every possible coordinate and calculate its quality. |
Python O(N^2) [ACCEPTED] 100% better | tuple-with-same-product | 0 | 1 | The idea is to count the no. of time the product occurred, its already mentioned in the question no. are disctinct, we just need to the all the combinations.\n\nFor a,b,c,d there can be 8 combinations, lets take (a,b) as 1 and (c,d) as 2 so there it can be written as "12" or "21" so 2 ways and then the 1- which is ab c... | 5 | Given an array `nums` of **distinct** positive integers, return _the number of tuples_ `(a, b, c, d)` _such that_ `a * b = c * d` _where_ `a`_,_ `b`_,_ `c`_, and_ `d` _are elements of_ `nums`_, and_ `a != b != c != d`_._
**Example 1:**
**Input:** nums = \[2,3,4,6\]
**Output:** 8
**Explanation:** There are 8 valid tup... | The constraints are small enough to consider every possible coordinate and calculate its quality. |
simple fast python | tuple-with-same-product | 0 | 1 | each pair creates 8 results. \nwe index each possible pair permutation with their product \n# Code\n```\nfrom collections import defaultdict\nclass Solution:\n def tupleSameProduct(self, nums: List[int]) -> int:\n prods = defaultdict(int)\n l = len(nums)\n count = 0 \n for i in range(l): ... | 0 | Given an array `nums` of **distinct** positive integers, return _the number of tuples_ `(a, b, c, d)` _such that_ `a * b = c * d` _where_ `a`_,_ `b`_,_ `c`_, and_ `d` _are elements of_ `nums`_, and_ `a != b != c != d`_._
**Example 1:**
**Input:** nums = \[2,3,4,6\]
**Output:** 8
**Explanation:** There are 8 valid tup... | The constraints are small enough to consider every possible coordinate and calculate its quality. |
python3 math + hashmap | tuple-with-same-product | 0 | 1 | \n\n# Code\n```\nfrom collections import defaultdict\nclass Solution:\n def tupleSameProduct(self, nums: List[int]) -> int:\n d = defaultdict(int)\n for i in range(len(nums)):\n for j in range(i+1,len(nums)):\n d[nums[i]*nums[j]]+=1\n \n ans = 0\n for k in... | 0 | Given an array `nums` of **distinct** positive integers, return _the number of tuples_ `(a, b, c, d)` _such that_ `a * b = c * d` _where_ `a`_,_ `b`_,_ `c`_, and_ `d` _are elements of_ `nums`_, and_ `a != b != c != d`_._
**Example 1:**
**Input:** nums = \[2,3,4,6\]
**Output:** 8
**Explanation:** There are 8 valid tup... | The constraints are small enough to consider every possible coordinate and calculate its quality. |
Python - one line | tuple-with-same-product | 0 | 1 | # Intuition\n"Distinct numbers" is the main insight here. This allows us to skip checking for same indices in matching pairs.\nSo just count same value occurence for different pairs.\nmultiply combination every pair with every other pair (n*(n-1), *triangular number* can be used here).\nThat number has to be multiplied... | 0 | Given an array `nums` of **distinct** positive integers, return _the number of tuples_ `(a, b, c, d)` _such that_ `a * b = c * d` _where_ `a`_,_ `b`_,_ `c`_, and_ `d` _are elements of_ `nums`_, and_ `a != b != c != d`_._
**Example 1:**
**Input:** nums = \[2,3,4,6\]
**Output:** 8
**Explanation:** There are 8 valid tup... | The constraints are small enough to consider every possible coordinate and calculate its quality. |
python code | tuple-with-same-product | 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 an array `nums` of **distinct** positive integers, return _the number of tuples_ `(a, b, c, d)` _such that_ `a * b = c * d` _where_ `a`_,_ `b`_,_ `c`_, and_ `d` _are elements of_ `nums`_, and_ `a != b != c != d`_._
**Example 1:**
**Input:** nums = \[2,3,4,6\]
**Output:** 8
**Explanation:** There are 8 valid tup... | The constraints are small enough to consider every possible coordinate and calculate its quality. |
Easy to understand solution. | tuple-with-same-product | 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: Enlighten me!\n<!-- Add your time complexity here, e.g. $$O(n)$$ --> \n\n- Space complexity: Nevermind!\n<!-- Add your space complex... | 0 | Given an array `nums` of **distinct** positive integers, return _the number of tuples_ `(a, b, c, d)` _such that_ `a * b = c * d` _where_ `a`_,_ `b`_,_ `c`_, and_ `d` _are elements of_ `nums`_, and_ `a != b != c != d`_._
**Example 1:**
**Input:** nums = \[2,3,4,6\]
**Output:** 8
**Explanation:** There are 8 valid tup... | The constraints are small enough to consider every possible coordinate and calculate its quality. |
[Python 3] Two-line Pythonic solution, beats 100%, O(n^2) | tuple-with-same-product | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n- Instead of finding each combination of $(a, b, c, d)$, look for the number of possible pairs of products $p = q$ where $p = a \\times b$ and $q = c \\times d$, and multiply by the number of permutations of $a$ and $b$ and $c$ and $d$.\n... | 0 | Given an array `nums` of **distinct** positive integers, return _the number of tuples_ `(a, b, c, d)` _such that_ `a * b = c * d` _where_ `a`_,_ `b`_,_ `c`_, and_ `d` _are elements of_ `nums`_, and_ `a != b != c != d`_._
**Example 1:**
**Input:** nums = \[2,3,4,6\]
**Output:** 8
**Explanation:** There are 8 valid tup... | The constraints are small enough to consider every possible coordinate and calculate its quality. |
【Video】Give me 10 minutes - how we think about a solution | largest-submatrix-with-rearrangements | 1 | 1 | # Intuition\nUsing sort for width.\n\n---\n\n# Solution Video\n\nhttps://youtu.be/odAv92zWKqs\n\nThese days, many people watch my video more than 70% of whole time, which is good to my channel reputation. Thank you!\n\nI\'m creating this video or post with the goal of ensuring that viewers, whether they read the post o... | 77 | You are given a binary matrix `matrix` of size `m x n`, and you are allowed to rearrange the **columns** of the `matrix` in any order.
Return _the area of the largest submatrix within_ `matrix` _where **every** element of the submatrix is_ `1` _after reordering the columns optimally._
**Example 1:**
**Input:** matri... | Try working backward: consider all trivial states you know to be winning or losing, and work backward to determine which other states can be labeled as winning or losing. |
【Video】Give me 10 minutes - how we think about a solution | largest-submatrix-with-rearrangements | 1 | 1 | # Intuition\nUsing sort for width.\n\n---\n\n# Solution Video\n\nhttps://youtu.be/odAv92zWKqs\n\nThese days, many people watch my video more than 70% of whole time, which is good to my channel reputation. Thank you!\n\nI\'m creating this video or post with the goal of ensuring that viewers, whether they read the post o... | 77 | Design a system that manages the reservation state of `n` seats that are numbered from `1` to `n`.
Implement the `SeatManager` class:
* `SeatManager(int n)` Initializes a `SeatManager` object that will manage `n` seats numbered from `1` to `n`. All seats are initially available.
* `int reserve()` Fetches the **sm... | For each column, find the number of consecutive ones ending at each position. For each row, sort the cumulative ones in non-increasing order and "fit" the largest submatrix. |
16 line Python solution (998ms) | largest-submatrix-with-rearrangements | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nIterate over rows while keeping track of the height of columns of 1\'s above the row. Sort the heights after each iteration and check the area of all possible rectangles that would fit into the graph of heights.\n# Approach\n<!-- Describe... | 2 | You are given a binary matrix `matrix` of size `m x n`, and you are allowed to rearrange the **columns** of the `matrix` in any order.
Return _the area of the largest submatrix within_ `matrix` _where **every** element of the submatrix is_ `1` _after reordering the columns optimally._
**Example 1:**
**Input:** matri... | Try working backward: consider all trivial states you know to be winning or losing, and work backward to determine which other states can be labeled as winning or losing. |
16 line Python solution (998ms) | largest-submatrix-with-rearrangements | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nIterate over rows while keeping track of the height of columns of 1\'s above the row. Sort the heights after each iteration and check the area of all possible rectangles that would fit into the graph of heights.\n# Approach\n<!-- Describe... | 2 | Design a system that manages the reservation state of `n` seats that are numbered from `1` to `n`.
Implement the `SeatManager` class:
* `SeatManager(int n)` Initializes a `SeatManager` object that will manage `n` seats numbered from `1` to `n`. All seats are initially available.
* `int reserve()` Fetches the **sm... | For each column, find the number of consecutive ones ending at each position. For each row, sort the cumulative ones in non-increasing order and "fit" the largest submatrix. |
✅☑[C++/Java/Python/JavaScript] || 3 Approaches || EXPLAINED🔥 | largest-submatrix-with-rearrangements | 1 | 1 | # PLEASE UPVOTE IF IT HELPED\n\n---\n\n\n# Approaches\n**(Also explained in the code)**\n\n#### ***Approach 1(Sort By Height On Each Baseline Row)***\n1. **Matrix traversal:**\n - The code iterates through each row of the matrix.\n1. **Accumulating consecutive ones:**\n - For each non-zero element in the matrix (... | 2 | You are given a binary matrix `matrix` of size `m x n`, and you are allowed to rearrange the **columns** of the `matrix` in any order.
Return _the area of the largest submatrix within_ `matrix` _where **every** element of the submatrix is_ `1` _after reordering the columns optimally._
**Example 1:**
**Input:** matri... | Try working backward: consider all trivial states you know to be winning or losing, and work backward to determine which other states can be labeled as winning or losing. |
✅☑[C++/Java/Python/JavaScript] || 3 Approaches || EXPLAINED🔥 | largest-submatrix-with-rearrangements | 1 | 1 | # PLEASE UPVOTE IF IT HELPED\n\n---\n\n\n# Approaches\n**(Also explained in the code)**\n\n#### ***Approach 1(Sort By Height On Each Baseline Row)***\n1. **Matrix traversal:**\n - The code iterates through each row of the matrix.\n1. **Accumulating consecutive ones:**\n - For each non-zero element in the matrix (... | 2 | Design a system that manages the reservation state of `n` seats that are numbered from `1` to `n`.
Implement the `SeatManager` class:
* `SeatManager(int n)` Initializes a `SeatManager` object that will manage `n` seats numbered from `1` to `n`. All seats are initially available.
* `int reserve()` Fetches the **sm... | For each column, find the number of consecutive ones ending at each position. For each row, sort the cumulative ones in non-increasing order and "fit" the largest submatrix. |
Python3 Solution | largest-submatrix-with-rearrangements | 0 | 1 | \n```\nclass Solution:\n def largestSubmatrix(self, matrix: List[List[int]]) -> int:\n row=len(matrix)\n col=len(matrix[0])\n ans=0\n for r in range(1,row):\n for c in range(col):\n matrix[r][c]+=matrix[r-1][c] if matrix[r][c] else 0\n\n for r in range(row... | 2 | You are given a binary matrix `matrix` of size `m x n`, and you are allowed to rearrange the **columns** of the `matrix` in any order.
Return _the area of the largest submatrix within_ `matrix` _where **every** element of the submatrix is_ `1` _after reordering the columns optimally._
**Example 1:**
**Input:** matri... | Try working backward: consider all trivial states you know to be winning or losing, and work backward to determine which other states can be labeled as winning or losing. |
Python3 Solution | largest-submatrix-with-rearrangements | 0 | 1 | \n```\nclass Solution:\n def largestSubmatrix(self, matrix: List[List[int]]) -> int:\n row=len(matrix)\n col=len(matrix[0])\n ans=0\n for r in range(1,row):\n for c in range(col):\n matrix[r][c]+=matrix[r-1][c] if matrix[r][c] else 0\n\n for r in range(row... | 2 | Design a system that manages the reservation state of `n` seats that are numbered from `1` to `n`.
Implement the `SeatManager` class:
* `SeatManager(int n)` Initializes a `SeatManager` object that will manage `n` seats numbered from `1` to `n`. All seats are initially available.
* `int reserve()` Fetches the **sm... | For each column, find the number of consecutive ones ending at each position. For each row, sort the cumulative ones in non-increasing order and "fit" the largest submatrix. |
One line solution. Runtime 100%!!! | largest-submatrix-with-rearrangements | 0 | 1 | \n```\nclass Solution:\n def largestSubmatrix(self, matrix: List[List[int]]) -> int:\n return max(max... | 1 | You are given a binary matrix `matrix` of size `m x n`, and you are allowed to rearrange the **columns** of the `matrix` in any order.
Return _the area of the largest submatrix within_ `matrix` _where **every** element of the submatrix is_ `1` _after reordering the columns optimally._
**Example 1:**
**Input:** matri... | Try working backward: consider all trivial states you know to be winning or losing, and work backward to determine which other states can be labeled as winning or losing. |
One line solution. Runtime 100%!!! | largest-submatrix-with-rearrangements | 0 | 1 | \n```\nclass Solution:\n def largestSubmatrix(self, matrix: List[List[int]]) -> int:\n return max(max... | 1 | Design a system that manages the reservation state of `n` seats that are numbered from `1` to `n`.
Implement the `SeatManager` class:
* `SeatManager(int n)` Initializes a `SeatManager` object that will manage `n` seats numbered from `1` to `n`. All seats are initially available.
* `int reserve()` Fetches the **sm... | For each column, find the number of consecutive ones ending at each position. For each row, sort the cumulative ones in non-increasing order and "fit" the largest submatrix. |
[We💕Simple] Counting Sort, Dynamic Programming w/ Explanations | largest-submatrix-with-rearrangements | 0 | 1 | I always like to find solutions that are very simple and easy, yet do not sacrifice time complexity. I keep creating more solutions\n\n# Code\n``` Kotlin []\nclass Solution {\n fun largestSubmatrix(matrix: Array<IntArray>): Int {\n val heights = IntArray(matrix[0].size)\n var maxArea = 0\n for (... | 1 | You are given a binary matrix `matrix` of size `m x n`, and you are allowed to rearrange the **columns** of the `matrix` in any order.
Return _the area of the largest submatrix within_ `matrix` _where **every** element of the submatrix is_ `1` _after reordering the columns optimally._
**Example 1:**
**Input:** matri... | Try working backward: consider all trivial states you know to be winning or losing, and work backward to determine which other states can be labeled as winning or losing. |
[We💕Simple] Counting Sort, Dynamic Programming w/ Explanations | largest-submatrix-with-rearrangements | 0 | 1 | I always like to find solutions that are very simple and easy, yet do not sacrifice time complexity. I keep creating more solutions\n\n# Code\n``` Kotlin []\nclass Solution {\n fun largestSubmatrix(matrix: Array<IntArray>): Int {\n val heights = IntArray(matrix[0].size)\n var maxArea = 0\n for (... | 1 | Design a system that manages the reservation state of `n` seats that are numbered from `1` to `n`.
Implement the `SeatManager` class:
* `SeatManager(int n)` Initializes a `SeatManager` object that will manage `n` seats numbered from `1` to `n`. All seats are initially available.
* `int reserve()` Fetches the **sm... | For each column, find the number of consecutive ones ending at each position. For each row, sort the cumulative ones in non-increasing order and "fit" the largest submatrix. |
Python3 Clean & Commented Top-down DP with the early stopping trick | cat-and-mouse-ii | 0 | 1 | The question is composed of sub-problems, which can be formatted as the dp problem.\n\nThe tricky part is that we don\'t need to do 1000 turns to determine the result.\nLet\'s say the cat and the mouse decide to take only 1 step each time, how many turns it takes to search the whole grid?\nYes, m * n * 2 (mouse and cat... | 53 | A game is played by a cat and a mouse named Cat and Mouse.
The environment is represented by a `grid` of size `rows x cols`, where each element is a wall, floor, player (Cat, Mouse), or food.
* Players are represented by the characters `'C'`(Cat)`,'M'`(Mouse).
* Floors are represented by the character `'.'` and c... | Use two arrays to save the cumulative multipliers at each time point and cumulative sums adjusted by the current multiplier. The function getIndex(idx) ask to the current value modulo 10^9+7. Use modular inverse and both arrays to calculate this value. |
Python3 Clean & Commented Top-down DP with the early stopping trick | cat-and-mouse-ii | 0 | 1 | The question is composed of sub-problems, which can be formatted as the dp problem.\n\nThe tricky part is that we don\'t need to do 1000 turns to determine the result.\nLet\'s say the cat and the mouse decide to take only 1 step each time, how many turns it takes to search the whole grid?\nYes, m * n * 2 (mouse and cat... | 53 | You are given a binary matrix `matrix` of size `m x n`, and you are allowed to rearrange the **columns** of the `matrix` in any order.
Return _the area of the largest submatrix within_ `matrix` _where **every** element of the submatrix is_ `1` _after reordering the columns optimally._
**Example 1:**
**Input:** matri... | Try working backward: consider all trivial states you know to be winning or losing, and work backward to determine which other states can be labeled as winning or losing. |
1000 turns is too big a threshold | cat-and-mouse-ii | 0 | 1 | # Code\n```\nclass Solution:\n def canMouseWin(self, grid: List[str], catJump: int, mouseJump: int) -> bool:\n m = len(grid)\n n = len(grid[0])\n for i in range(m):\n for j in range(n):\n if grid[i][j] == \'F\':\n food = (i,j)\n elif gr... | 0 | A game is played by a cat and a mouse named Cat and Mouse.
The environment is represented by a `grid` of size `rows x cols`, where each element is a wall, floor, player (Cat, Mouse), or food.
* Players are represented by the characters `'C'`(Cat)`,'M'`(Mouse).
* Floors are represented by the character `'.'` and c... | Use two arrays to save the cumulative multipliers at each time point and cumulative sums adjusted by the current multiplier. The function getIndex(idx) ask to the current value modulo 10^9+7. Use modular inverse and both arrays to calculate this value. |
1000 turns is too big a threshold | cat-and-mouse-ii | 0 | 1 | # Code\n```\nclass Solution:\n def canMouseWin(self, grid: List[str], catJump: int, mouseJump: int) -> bool:\n m = len(grid)\n n = len(grid[0])\n for i in range(m):\n for j in range(n):\n if grid[i][j] == \'F\':\n food = (i,j)\n elif gr... | 0 | You are given a binary matrix `matrix` of size `m x n`, and you are allowed to rearrange the **columns** of the `matrix` in any order.
Return _the area of the largest submatrix within_ `matrix` _where **every** element of the submatrix is_ `1` _after reordering the columns optimally._
**Example 1:**
**Input:** matri... | Try working backward: consider all trivial states you know to be winning or losing, and work backward to determine which other states can be labeled as winning or losing. |
Clean DP + Shortest path heuristic + Early stopping | cat-and-mouse-ii | 0 | 1 | # Approach\nUses DP to cache the expansion of the game tree. It also pre-computes the shortest distance from any reachable point to the food and uses it to explore those branches first. Finally, it also uses a couple of early stopping conditions including if there exists a path from the mouse or cat to the food, and th... | 0 | A game is played by a cat and a mouse named Cat and Mouse.
The environment is represented by a `grid` of size `rows x cols`, where each element is a wall, floor, player (Cat, Mouse), or food.
* Players are represented by the characters `'C'`(Cat)`,'M'`(Mouse).
* Floors are represented by the character `'.'` and c... | Use two arrays to save the cumulative multipliers at each time point and cumulative sums adjusted by the current multiplier. The function getIndex(idx) ask to the current value modulo 10^9+7. Use modular inverse and both arrays to calculate this value. |
Clean DP + Shortest path heuristic + Early stopping | cat-and-mouse-ii | 0 | 1 | # Approach\nUses DP to cache the expansion of the game tree. It also pre-computes the shortest distance from any reachable point to the food and uses it to explore those branches first. Finally, it also uses a couple of early stopping conditions including if there exists a path from the mouse or cat to the food, and th... | 0 | You are given a binary matrix `matrix` of size `m x n`, and you are allowed to rearrange the **columns** of the `matrix` in any order.
Return _the area of the largest submatrix within_ `matrix` _where **every** element of the submatrix is_ `1` _after reordering the columns optimally._
**Example 1:**
**Input:** matri... | Try working backward: consider all trivial states you know to be winning or losing, and work backward to determine which other states can be labeled as winning or losing. |
Python solution | cat-and-mouse-ii | 0 | 1 | # Intuition\n\nThe code is relatively long. It does the following. First you need to get a graph, one for the cat one for the mouse. They differ as the max jump size may be different. Then we have a two player game, on the product of these two graphs.\n\nThe function `alice_wins` compute if the first player (the mouse)... | 0 | A game is played by a cat and a mouse named Cat and Mouse.
The environment is represented by a `grid` of size `rows x cols`, where each element is a wall, floor, player (Cat, Mouse), or food.
* Players are represented by the characters `'C'`(Cat)`,'M'`(Mouse).
* Floors are represented by the character `'.'` and c... | Use two arrays to save the cumulative multipliers at each time point and cumulative sums adjusted by the current multiplier. The function getIndex(idx) ask to the current value modulo 10^9+7. Use modular inverse and both arrays to calculate this value. |
Python solution | cat-and-mouse-ii | 0 | 1 | # Intuition\n\nThe code is relatively long. It does the following. First you need to get a graph, one for the cat one for the mouse. They differ as the max jump size may be different. Then we have a two player game, on the product of these two graphs.\n\nThe function `alice_wins` compute if the first player (the mouse)... | 0 | You are given a binary matrix `matrix` of size `m x n`, and you are allowed to rearrange the **columns** of the `matrix` in any order.
Return _the area of the largest submatrix within_ `matrix` _where **every** element of the submatrix is_ `1` _after reordering the columns optimally._
**Example 1:**
**Input:** matri... | Try working backward: consider all trivial states you know to be winning or losing, and work backward to determine which other states can be labeled as winning or losing. |
[Python3] dp | cat-and-mouse-ii | 0 | 1 | **Algo**\nI wrote this implementation based on this [post](https://leetcode.com/problems/cat-and-mouse-ii/discuss/1020616/Python3-Clean-and-Commented-Top-down-DP-with-the-early-stopping-trick). Although the 1000 turns limit looks fishy as the grid is at most 8x8, I couldn\'t come up with a bound as tight as `m*n*2`. In... | 6 | A game is played by a cat and a mouse named Cat and Mouse.
The environment is represented by a `grid` of size `rows x cols`, where each element is a wall, floor, player (Cat, Mouse), or food.
* Players are represented by the characters `'C'`(Cat)`,'M'`(Mouse).
* Floors are represented by the character `'.'` and c... | Use two arrays to save the cumulative multipliers at each time point and cumulative sums adjusted by the current multiplier. The function getIndex(idx) ask to the current value modulo 10^9+7. Use modular inverse and both arrays to calculate this value. |
[Python3] dp | cat-and-mouse-ii | 0 | 1 | **Algo**\nI wrote this implementation based on this [post](https://leetcode.com/problems/cat-and-mouse-ii/discuss/1020616/Python3-Clean-and-Commented-Top-down-DP-with-the-early-stopping-trick). Although the 1000 turns limit looks fishy as the grid is at most 8x8, I couldn\'t come up with a bound as tight as `m*n*2`. In... | 6 | You are given a binary matrix `matrix` of size `m x n`, and you are allowed to rearrange the **columns** of the `matrix` in any order.
Return _the area of the largest submatrix within_ `matrix` _where **every** element of the submatrix is_ `1` _after reordering the columns optimally._
**Example 1:**
**Input:** matri... | Try working backward: consider all trivial states you know to be winning or losing, and work backward to determine which other states can be labeled as winning or losing. |
[Python] Top Down DP | cat-and-mouse-ii | 0 | 1 | **Approach:**\n\n1. For every space that is not a wall, find which spaces the cat and mouse can jump to from this space. (```cat_moves``` and ```mouse_moves```)\n\n2. Find the starting locations of the cat, the mouse, and the food. (```cat```, ```mouse```, and ```food```)\n\n3. Take an Alice and Bob style game approach... | 3 | A game is played by a cat and a mouse named Cat and Mouse.
The environment is represented by a `grid` of size `rows x cols`, where each element is a wall, floor, player (Cat, Mouse), or food.
* Players are represented by the characters `'C'`(Cat)`,'M'`(Mouse).
* Floors are represented by the character `'.'` and c... | Use two arrays to save the cumulative multipliers at each time point and cumulative sums adjusted by the current multiplier. The function getIndex(idx) ask to the current value modulo 10^9+7. Use modular inverse and both arrays to calculate this value. |
[Python] Top Down DP | cat-and-mouse-ii | 0 | 1 | **Approach:**\n\n1. For every space that is not a wall, find which spaces the cat and mouse can jump to from this space. (```cat_moves``` and ```mouse_moves```)\n\n2. Find the starting locations of the cat, the mouse, and the food. (```cat```, ```mouse```, and ```food```)\n\n3. Take an Alice and Bob style game approach... | 3 | You are given a binary matrix `matrix` of size `m x n`, and you are allowed to rearrange the **columns** of the `matrix` in any order.
Return _the area of the largest submatrix within_ `matrix` _where **every** element of the submatrix is_ `1` _after reordering the columns optimally._
**Example 1:**
**Input:** matri... | Try working backward: consider all trivial states you know to be winning or losing, and work backward to determine which other states can be labeled as winning or losing. |
Modern Pandas [Method Chaining] ✅ | the-number-of-employees-which-report-to-each-employee | 0 | 1 | # Code\n\n```python\nimport pandas as pd\n\ndef count_employees(employees: pd.DataFrame) -> pd.DataFrame:\n return (\n employees\n .groupby("reports_to", as_index=False)\n .aggregate(\n reports_count=pd.NamedAgg("employee_id", "nunique"),\n average_age = pd.NamedAgg("age", ... | 0 | You are given two **0-indexed** integer arrays `servers` and `tasks` of lengths `n` and `m` respectively. `servers[i]` is the **weight** of the `ith` server, and `tasks[j]` is the **time needed** to process the `jth` task **in seconds**.
Tasks are assigned to the servers using a **task ... | null |
Easy to understand python solution | find-the-highest-altitude | 0 | 1 | # Approach\nTo solve this problem, we need to store the original element and move to the next element by adding the difference. Thus, cur_num stores the value from the list with heights. At each iteration we look at the maximum value.\n\n# Complexity\n- Time complexity:\nThe complexity of this algorithm is O(n) and it ... | 3 | There is a biker going on a road trip. The road trip consists of `n + 1` points at different altitudes. The biker starts his trip on point `0` with altitude equal `0`.
You are given an integer array `gain` of length `n` where `gain[i]` is the **net gain in altitude** between points `i` and `i + 1` for all (`0 <=... | The fastest way to convert n to zero is to remove all set bits starting from the leftmost one. Try some simple examples to learn the rule of how many steps are needed to remove one set bit. consider n=2^k case first, then solve for all n. |
Easy to understand python solution | find-the-highest-altitude | 0 | 1 | # Approach\nTo solve this problem, we need to store the original element and move to the next element by adding the difference. Thus, cur_num stores the value from the list with heights. At each iteration we look at the maximum value.\n\n# Complexity\n- Time complexity:\nThe complexity of this algorithm is O(n) and it ... | 3 | It is a sweltering summer day, and a boy wants to buy some ice cream bars.
At the store, there are `n` ice cream bars. You are given an array `costs` of length `n`, where `costs[i]` is the price of the `ith` ice cream bar in coins. The boy initially has `coins` coins to spend, and he wants to buy as many ice cream bar... | Let's note that the altitude of an element is the sum of gains of all the elements behind it Getting the altitudes can be done by getting the prefix sum array of the given array |
Detailed understandable beginner explanation | find-the-highest-altitude | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\nThe given code is implementing a method largestAltitude in the Solution class. This method takes a list gain as input, which represents the altitude gain or loss at each point of a journey. The goal is to calculate the highest altitude ... | 1 | There is a biker going on a road trip. The road trip consists of `n + 1` points at different altitudes. The biker starts his trip on point `0` with altitude equal `0`.
You are given an integer array `gain` of length `n` where `gain[i]` is the **net gain in altitude** between points `i` and `i + 1` for all (`0 <=... | The fastest way to convert n to zero is to remove all set bits starting from the leftmost one. Try some simple examples to learn the rule of how many steps are needed to remove one set bit. consider n=2^k case first, then solve for all n. |
Detailed understandable beginner explanation | find-the-highest-altitude | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\nThe given code is implementing a method largestAltitude in the Solution class. This method takes a list gain as input, which represents the altitude gain or loss at each point of a journey. The goal is to calculate the highest altitude ... | 1 | It is a sweltering summer day, and a boy wants to buy some ice cream bars.
At the store, there are `n` ice cream bars. You are given an array `costs` of length `n`, where `costs[i]` is the price of the `ith` ice cream bar in coins. The boy initially has `coins` coins to spend, and he wants to buy as many ice cream bar... | Let's note that the altitude of an element is the sum of gains of all the elements behind it Getting the altitudes can be done by getting the prefix sum array of the given array |
Simple Solution in both Python and C++ languages😊 | find-the-highest-altitude | 0 | 1 | # Solution \n```Python3 []\nclass Solution:\n def largestAltitude(self, gain: List[int]) -> int:\n highest_point = 0\n prev_altitude = 0\n for i in gain:\n prev_altitude += i\n highest_point = max(highest_point, prev_altitude)\n\n return highest_point\n```\n```C++ []... | 1 | There is a biker going on a road trip. The road trip consists of `n + 1` points at different altitudes. The biker starts his trip on point `0` with altitude equal `0`.
You are given an integer array `gain` of length `n` where `gain[i]` is the **net gain in altitude** between points `i` and `i + 1` for all (`0 <=... | The fastest way to convert n to zero is to remove all set bits starting from the leftmost one. Try some simple examples to learn the rule of how many steps are needed to remove one set bit. consider n=2^k case first, then solve for all n. |
Simple Solution in both Python and C++ languages😊 | find-the-highest-altitude | 0 | 1 | # Solution \n```Python3 []\nclass Solution:\n def largestAltitude(self, gain: List[int]) -> int:\n highest_point = 0\n prev_altitude = 0\n for i in gain:\n prev_altitude += i\n highest_point = max(highest_point, prev_altitude)\n\n return highest_point\n```\n```C++ []... | 1 | It is a sweltering summer day, and a boy wants to buy some ice cream bars.
At the store, there are `n` ice cream bars. You are given an array `costs` of length `n`, where `costs[i]` is the price of the `ith` ice cream bar in coins. The boy initially has `coins` coins to spend, and he wants to buy as many ice cream bar... | Let's note that the altitude of an element is the sum of gains of all the elements behind it Getting the altitudes can be done by getting the prefix sum array of the given array |
Python Very Easy Solution || O(n) || Java beats 100% || C# | find-the-highest-altitude | 1 | 1 | # Complexity\n- Time complexity:\nO(n)\n\n- Space complexity:\nO(1)\n\n# Code\n\n1) Python\n\n```\nclass Solution:\n def largestAltitude(self, gain: List[int]) -> int:\n maxVal=0\n alt=0\n for i in gain:\n alt+=i\n maxVal=max(alt,maxVal)\n return maxVal\n```\n\n2) Ja... | 1 | There is a biker going on a road trip. The road trip consists of `n + 1` points at different altitudes. The biker starts his trip on point `0` with altitude equal `0`.
You are given an integer array `gain` of length `n` where `gain[i]` is the **net gain in altitude** between points `i` and `i + 1` for all (`0 <=... | The fastest way to convert n to zero is to remove all set bits starting from the leftmost one. Try some simple examples to learn the rule of how many steps are needed to remove one set bit. consider n=2^k case first, then solve for all n. |
Python Very Easy Solution || O(n) || Java beats 100% || C# | find-the-highest-altitude | 1 | 1 | # Complexity\n- Time complexity:\nO(n)\n\n- Space complexity:\nO(1)\n\n# Code\n\n1) Python\n\n```\nclass Solution:\n def largestAltitude(self, gain: List[int]) -> int:\n maxVal=0\n alt=0\n for i in gain:\n alt+=i\n maxVal=max(alt,maxVal)\n return maxVal\n```\n\n2) Ja... | 1 | It is a sweltering summer day, and a boy wants to buy some ice cream bars.
At the store, there are `n` ice cream bars. You are given an array `costs` of length `n`, where `costs[i]` is the price of the `ith` ice cream bar in coins. The boy initially has `coins` coins to spend, and he wants to buy as many ice cream bar... | Let's note that the altitude of an element is the sum of gains of all the elements behind it Getting the altitudes can be done by getting the prefix sum array of the given array |
Python 3 || 5 lines, w/ explanation and example || T/M: 75% / 67% | minimum-number-of-people-to-teach | 0 | 1 | Here\'s the plan:\n- Determine the group of users who do not have a common language with at least one friend.\n- Determine the most common language in that group.\n- Return the number in the group who do not speak that language.\n```\nclass Solution:\n def minimumTeachings(self, n, languages, friendships):\n ... | 3 | On a social network consisting of `m` users and some friendships between users, two users can communicate with each other if they know a common language.
You are given an integer `n`, an array `languages`, and an array `friendships` where:
* There are `n` languages numbered `1` through `n`,
* `languages[i]` is th... | Sort the points by polar angle with the original position. Now only a consecutive collection of points would be visible from any coordinate. We can use two pointers to keep track of visible points for each start point For handling the cyclic condition, it’d be helpful to append the point list to itself after sorting. |
Python 3 || 5 lines, w/ explanation and example || T/M: 75% / 67% | minimum-number-of-people-to-teach | 0 | 1 | Here\'s the plan:\n- Determine the group of users who do not have a common language with at least one friend.\n- Determine the most common language in that group.\n- Return the number in the group who do not speak that language.\n```\nclass Solution:\n def minimumTeachings(self, n, languages, friendships):\n ... | 3 | You are given `n` tasks labeled from `0` to `n - 1` represented by a 2D integer array `tasks`, where `tasks[i] = [enqueueTimei, processingTimei]` means that the `ith` task will be available to process at `enqueueTimei` and will take `processingTimei` to finish processing.
You have a single-threaded CPU... | You can just use brute force and find out for each language the number of users you need to teach Note that a user can appear in multiple friendships but you need to teach that user only once |
Python, 3 steps | minimum-number-of-people-to-teach | 0 | 1 | First, find those who can\'t communicate with each other.\nThen, find the most popular language among them.\nThen teach that language to the minority who doesn\'t speak it: \n```\ndef minimumTeachings(self, n: int, languages: List[List[int]], friendships: List[List[int]]) -> int:\n\tlanguages = [set(l) for l in languag... | 93 | On a social network consisting of `m` users and some friendships between users, two users can communicate with each other if they know a common language.
You are given an integer `n`, an array `languages`, and an array `friendships` where:
* There are `n` languages numbered `1` through `n`,
* `languages[i]` is th... | Sort the points by polar angle with the original position. Now only a consecutive collection of points would be visible from any coordinate. We can use two pointers to keep track of visible points for each start point For handling the cyclic condition, it’d be helpful to append the point list to itself after sorting. |
Python, 3 steps | minimum-number-of-people-to-teach | 0 | 1 | First, find those who can\'t communicate with each other.\nThen, find the most popular language among them.\nThen teach that language to the minority who doesn\'t speak it: \n```\ndef minimumTeachings(self, n: int, languages: List[List[int]], friendships: List[List[int]]) -> int:\n\tlanguages = [set(l) for l in languag... | 93 | You are given `n` tasks labeled from `0` to `n - 1` represented by a 2D integer array `tasks`, where `tasks[i] = [enqueueTimei, processingTimei]` means that the `ith` task will be available to process at `enqueueTimei` and will take `processingTimei` to finish processing.
You have a single-threaded CPU... | You can just use brute force and find out for each language the number of users you need to teach Note that a user can appear in multiple friendships but you need to teach that user only once |
Python3 solution: logic + brute force | minimum-number-of-people-to-teach | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nFirst intuition was graph, because the problem was framed like a graph problem\n# Approach\n<!-- Describe your approach to solving the problem. -->\nSince constraints are rather small, brute force trial and error would suffice\n# Complexi... | 0 | On a social network consisting of `m` users and some friendships between users, two users can communicate with each other if they know a common language.
You are given an integer `n`, an array `languages`, and an array `friendships` where:
* There are `n` languages numbered `1` through `n`,
* `languages[i]` is th... | Sort the points by polar angle with the original position. Now only a consecutive collection of points would be visible from any coordinate. We can use two pointers to keep track of visible points for each start point For handling the cyclic condition, it’d be helpful to append the point list to itself after sorting. |
Python3 solution: logic + brute force | minimum-number-of-people-to-teach | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nFirst intuition was graph, because the problem was framed like a graph problem\n# Approach\n<!-- Describe your approach to solving the problem. -->\nSince constraints are rather small, brute force trial and error would suffice\n# Complexi... | 0 | You are given `n` tasks labeled from `0` to `n - 1` represented by a 2D integer array `tasks`, where `tasks[i] = [enqueueTimei, processingTimei]` means that the `ith` task will be available to process at `enqueueTimei` and will take `processingTimei` to finish processing.
You have a single-threaded CPU... | You can just use brute force and find out for each language the number of users you need to teach Note that a user can appear in multiple friendships but you need to teach that user only once |
clean commented code | minimum-number-of-people-to-teach | 0 | 1 | the question is tricky to understand but easy to implement when you overcome that intial hurdle. \n\n```\nclass Solution:\n def minimumTeachings(self, n: int, languages: List[List[int]], friendships: List[List[int]]) -> int:\n # modify languages to be a set for quick intersection \n languages = {person... | 0 | On a social network consisting of `m` users and some friendships between users, two users can communicate with each other if they know a common language.
You are given an integer `n`, an array `languages`, and an array `friendships` where:
* There are `n` languages numbered `1` through `n`,
* `languages[i]` is th... | Sort the points by polar angle with the original position. Now only a consecutive collection of points would be visible from any coordinate. We can use two pointers to keep track of visible points for each start point For handling the cyclic condition, it’d be helpful to append the point list to itself after sorting. |
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