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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Clean Code || O(N) | count-good-meals | 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: O(22N)\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity: O(N)\n<!-- Add your space complexity here, e.g.... | 2 | You are given a string `s` (**0-indexed**). You are asked to perform the following operation on `s` until you get a sorted string:
1. Find **the largest index** `i` such that `1 <= i < s.length` and `s[i] < s[i - 1]`.
2. Find **the largest index** `j` such that `i <= j < s.length` and `s[k] < s[i - 1]` f... | Note that the number of powers of 2 is at most 21 so this turns the problem to a classic find the number of pairs that sum to a certain value but for 21 values You need to use something fasters than the NlogN approach since there is already the log of iterating over the powers so one idea is two pointers |
Count good meals - python | count-good-meals | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nexpending the 2sum question and noting there is only a small number of 2 powers (21) so it is not efficient to caculate it each time.\n# Approach\n<!-- Describe your approach to solving the problem. -->\nfor items such i that i+i = 2^n we... | 0 | A **good meal** is a meal that contains **exactly two different food items** with a sum of deliciousness equal to a power of two.
You can pick **any** two different foods to make a good meal.
Given an array of integers `deliciousness` where `deliciousness[i]` is the deliciousness of the `ith` item of fo... | Find the smallest rowSum or colSum, and let it be x. Place that number in the grid, and subtract x from rowSum and colSum. Continue until all the sums are satisfied. |
Count good meals - python | count-good-meals | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nexpending the 2sum question and noting there is only a small number of 2 powers (21) so it is not efficient to caculate it each time.\n# Approach\n<!-- Describe your approach to solving the problem. -->\nfor items such i that i+i = 2^n we... | 0 | You are given a string `s` (**0-indexed**). You are asked to perform the following operation on `s` until you get a sorted string:
1. Find **the largest index** `i` such that `1 <= i < s.length` and `s[i] < s[i - 1]`.
2. Find **the largest index** `j` such that `i <= j < s.length` and `s[k] < s[i - 1]` f... | Note that the number of powers of 2 is at most 21 so this turns the problem to a classic find the number of pairs that sum to a certain value but for 21 values You need to use something fasters than the NlogN approach since there is already the log of iterating over the powers so one idea is two pointers |
Python3 solution | count-good-meals | 0 | 1 | \n# Code\n```\nclass Solution:\n def countPairs(self, deliciousness: List[int]) -> int:\n MOD = 10**9 + 7\n maxSum = max(deliciousness) * 2\n deliciousCount = {}\n deliciousMeals = 0\n\n for d in deliciousness:\n powerOfTwo = 1\n while powerOfTwo <= maxSum:\n ... | 0 | A **good meal** is a meal that contains **exactly two different food items** with a sum of deliciousness equal to a power of two.
You can pick **any** two different foods to make a good meal.
Given an array of integers `deliciousness` where `deliciousness[i]` is the deliciousness of the `ith` item of fo... | Find the smallest rowSum or colSum, and let it be x. Place that number in the grid, and subtract x from rowSum and colSum. Continue until all the sums are satisfied. |
Python3 solution | count-good-meals | 0 | 1 | \n# Code\n```\nclass Solution:\n def countPairs(self, deliciousness: List[int]) -> int:\n MOD = 10**9 + 7\n maxSum = max(deliciousness) * 2\n deliciousCount = {}\n deliciousMeals = 0\n\n for d in deliciousness:\n powerOfTwo = 1\n while powerOfTwo <= maxSum:\n ... | 0 | You are given a string `s` (**0-indexed**). You are asked to perform the following operation on `s` until you get a sorted string:
1. Find **the largest index** `i` such that `1 <= i < s.length` and `s[i] < s[i - 1]`.
2. Find **the largest index** `j` such that `i <= j < s.length` and `s[k] < s[i - 1]` f... | Note that the number of powers of 2 is at most 21 so this turns the problem to a classic find the number of pairs that sum to a certain value but for 21 values You need to use something fasters than the NlogN approach since there is already the log of iterating over the powers so one idea is two pointers |
Simple python code beated 98% | count-good-meals | 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 | A **good meal** is a meal that contains **exactly two different food items** with a sum of deliciousness equal to a power of two.
You can pick **any** two different foods to make a good meal.
Given an array of integers `deliciousness` where `deliciousness[i]` is the deliciousness of the `ith` item of fo... | Find the smallest rowSum or colSum, and let it be x. Place that number in the grid, and subtract x from rowSum and colSum. Continue until all the sums are satisfied. |
Simple python code beated 98% | count-good-meals | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ --... | 0 | You are given a string `s` (**0-indexed**). You are asked to perform the following operation on `s` until you get a sorted string:
1. Find **the largest index** `i` such that `1 <= i < s.length` and `s[i] < s[i - 1]`.
2. Find **the largest index** `j` such that `i <= j < s.length` and `s[k] < s[i - 1]` f... | Note that the number of powers of 2 is at most 21 so this turns the problem to a classic find the number of pairs that sum to a certain value but for 21 values You need to use something fasters than the NlogN approach since there is already the log of iterating over the powers so one idea is two pointers |
Python - 5 lines, variation of 2 sum | count-good-meals | 0 | 1 | # Intuition\nSame idea as in hash table based 2 sum. Difference is, here one needs to find previously seen value that sum up to some power of 2, To speed up power of 2 match, precompute powers of 2.\n\n# Complexity\n- Time complexity:\n$$O(n)$$\n\n- Space complexity:\n$$O(n)$$\n\n# Code\n```\nclass Solution:\n def c... | 0 | A **good meal** is a meal that contains **exactly two different food items** with a sum of deliciousness equal to a power of two.
You can pick **any** two different foods to make a good meal.
Given an array of integers `deliciousness` where `deliciousness[i]` is the deliciousness of the `ith` item of fo... | Find the smallest rowSum or colSum, and let it be x. Place that number in the grid, and subtract x from rowSum and colSum. Continue until all the sums are satisfied. |
Python - 5 lines, variation of 2 sum | count-good-meals | 0 | 1 | # Intuition\nSame idea as in hash table based 2 sum. Difference is, here one needs to find previously seen value that sum up to some power of 2, To speed up power of 2 match, precompute powers of 2.\n\n# Complexity\n- Time complexity:\n$$O(n)$$\n\n- Space complexity:\n$$O(n)$$\n\n# Code\n```\nclass Solution:\n def c... | 0 | You are given a string `s` (**0-indexed**). You are asked to perform the following operation on `s` until you get a sorted string:
1. Find **the largest index** `i` such that `1 <= i < s.length` and `s[i] < s[i - 1]`.
2. Find **the largest index** `j` such that `i <= j < s.length` and `s[k] < s[i - 1]` f... | Note that the number of powers of 2 is at most 21 so this turns the problem to a classic find the number of pairs that sum to a certain value but for 21 values You need to use something fasters than the NlogN approach since there is already the log of iterating over the powers so one idea is two pointers |
Python Very Simple | count-good-meals | 0 | 1 | # Intuition / Approach\nLike two sum, iterate though the ints and keep a hash map of old ints to check if current int will sum with previous int to equal a target.Iterate through the targets which are possible powers of 2 to sum up matches for each of the ints.\n\n# Complexity\n$$O(n)$$\n\n# Code\n```\nclass Solution:\... | 0 | A **good meal** is a meal that contains **exactly two different food items** with a sum of deliciousness equal to a power of two.
You can pick **any** two different foods to make a good meal.
Given an array of integers `deliciousness` where `deliciousness[i]` is the deliciousness of the `ith` item of fo... | Find the smallest rowSum or colSum, and let it be x. Place that number in the grid, and subtract x from rowSum and colSum. Continue until all the sums are satisfied. |
Python Very Simple | count-good-meals | 0 | 1 | # Intuition / Approach\nLike two sum, iterate though the ints and keep a hash map of old ints to check if current int will sum with previous int to equal a target.Iterate through the targets which are possible powers of 2 to sum up matches for each of the ints.\n\n# Complexity\n$$O(n)$$\n\n# Code\n```\nclass Solution:\... | 0 | You are given a string `s` (**0-indexed**). You are asked to perform the following operation on `s` until you get a sorted string:
1. Find **the largest index** `i` such that `1 <= i < s.length` and `s[i] < s[i - 1]`.
2. Find **the largest index** `j` such that `i <= j < s.length` and `s[k] < s[i - 1]` f... | Note that the number of powers of 2 is at most 21 so this turns the problem to a classic find the number of pairs that sum to a certain value but for 21 values You need to use something fasters than the NlogN approach since there is already the log of iterating over the powers so one idea is two pointers |
How much is this different from 2sum? Hashtable runnig up to 22 times | count-good-meals | 0 | 1 | # Code\n```\nclass Solution:\n def countPairs(self, deliciousness: List[int]) -> int:\n deliciousness.sort()\n res = 0\n for i in range(22):\n target = 2**i\n if max(deliciousness)*2 < target: break \n freq = {}\n for a in deliciousness:\n ... | 0 | A **good meal** is a meal that contains **exactly two different food items** with a sum of deliciousness equal to a power of two.
You can pick **any** two different foods to make a good meal.
Given an array of integers `deliciousness` where `deliciousness[i]` is the deliciousness of the `ith` item of fo... | Find the smallest rowSum or colSum, and let it be x. Place that number in the grid, and subtract x from rowSum and colSum. Continue until all the sums are satisfied. |
How much is this different from 2sum? Hashtable runnig up to 22 times | count-good-meals | 0 | 1 | # Code\n```\nclass Solution:\n def countPairs(self, deliciousness: List[int]) -> int:\n deliciousness.sort()\n res = 0\n for i in range(22):\n target = 2**i\n if max(deliciousness)*2 < target: break \n freq = {}\n for a in deliciousness:\n ... | 0 | You are given a string `s` (**0-indexed**). You are asked to perform the following operation on `s` until you get a sorted string:
1. Find **the largest index** `i` such that `1 <= i < s.length` and `s[i] < s[i - 1]`.
2. Find **the largest index** `j` such that `i <= j < s.length` and `s[k] < s[i - 1]` f... | Note that the number of powers of 2 is at most 21 so this turns the problem to a classic find the number of pairs that sum to a certain value but for 21 values You need to use something fasters than the NlogN approach since there is already the log of iterating over the powers so one idea is two pointers |
very simple solution like Leetcode 1 || Python | count-good-meals | 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:\nO(n)\n\n- Space complexity:\nO(1)\n\n# Code\n```\nclass Solution:\n def countPairs(self, deliciousness: List[int]) -> int:\n ... | 0 | A **good meal** is a meal that contains **exactly two different food items** with a sum of deliciousness equal to a power of two.
You can pick **any** two different foods to make a good meal.
Given an array of integers `deliciousness` where `deliciousness[i]` is the deliciousness of the `ith` item of fo... | Find the smallest rowSum or colSum, and let it be x. Place that number in the grid, and subtract x from rowSum and colSum. Continue until all the sums are satisfied. |
very simple solution like Leetcode 1 || Python | count-good-meals | 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:\nO(n)\n\n- Space complexity:\nO(1)\n\n# Code\n```\nclass Solution:\n def countPairs(self, deliciousness: List[int]) -> int:\n ... | 0 | You are given a string `s` (**0-indexed**). You are asked to perform the following operation on `s` until you get a sorted string:
1. Find **the largest index** `i` such that `1 <= i < s.length` and `s[i] < s[i - 1]`.
2. Find **the largest index** `j` such that `i <= j < s.length` and `s[k] < s[i - 1]` f... | Note that the number of powers of 2 is at most 21 so this turns the problem to a classic find the number of pairs that sum to a certain value but for 21 values You need to use something fasters than the NlogN approach since there is already the log of iterating over the powers so one idea is two pointers |
[Python3] binary search & 2-pointer | ways-to-split-array-into-three-subarrays | 0 | 1 | **Algo**\nCompute prefix array of `nums`. Any at index `i`, you want to find index `j` such at \n`prefix[i] <= prefix[j] - prefix[i] <= prefix[-1] - prefix[j]`\nThe rest comes out naturally. \n\n**Implementation** \n```\nclass Solution:\n def waysToSplit(self, nums: List[int]) -> int:\n prefix = [0]\n ... | 76 | A split of an integer array is **good** if:
* The array is split into three **non-empty** contiguous subarrays - named `left`, `mid`, `right` respectively from left to right.
* The sum of the elements in `left` is less than or equal to the sum of the elements in `mid`, and the sum of the elements in `mid` is less ... | null |
Python 3 | Binary Search | Explanation | ways-to-split-array-into-three-subarrays | 0 | 1 | ### Explanation\n- Calculate prefix sum and store at `pre_sum`\n- Iterate each index `i` and consider it as the ending index of the 1st segment\n- Do _**binary search**_ on the 2nd segment ending index. This is doable because `pre_sum` is a non-decreasing array\n\t- `[i+1, j]` is the shortest possible segment of the 2n... | 23 | A split of an integer array is **good** if:
* The array is split into three **non-empty** contiguous subarrays - named `left`, `mid`, `right` respectively from left to right.
* The sum of the elements in `left` is less than or equal to the sum of the elements in `mid`, and the sum of the elements in `mid` is less ... | null |
📌📌 Well-Coded || Both Method || Two-Pointer || Binary Search 🐍 | ways-to-split-array-into-three-subarrays | 0 | 1 | ## IDEA:\n\n**Two Pointers :**\n\'\'\'\n\n\tclass Solution:\n def waysToSplit(self, nums: List[int]) -> int:\n \n MOD = 10**9 + 7\n n = len(nums)\n for i in range(1,n):\n nums[i]+=nums[i-1]\n \n res,j,k = 0,0,0\n for i in range(n-2):\n if j<=i:\n... | 3 | A split of an integer array is **good** if:
* The array is split into three **non-empty** contiguous subarrays - named `left`, `mid`, `right` respectively from left to right.
* The sum of the elements in `left` is less than or equal to the sum of the elements in `mid`, and the sum of the elements in `mid` is less ... | null |
intuitive (at least relatively intuitive) binary search | ways-to-split-array-into-three-subarrays | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nThis is actually a really hard problem with many edge cases (in my opinion). \n\n# Approach\n<!-- Describe your approach to solving the problem. -->\nFirst thing we can note is that the elements of the array are going to be non-negative. ... | 0 | A split of an integer array is **good** if:
* The array is split into three **non-empty** contiguous subarrays - named `left`, `mid`, `right` respectively from left to right.
* The sum of the elements in `left` is less than or equal to the sum of the elements in `mid`, and the sum of the elements in `mid` is less ... | null |
Python 3 solution | ways-to-split-array-into-three-subarrays | 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 | A split of an integer array is **good** if:
* The array is split into three **non-empty** contiguous subarrays - named `left`, `mid`, `right` respectively from left to right.
* The sum of the elements in `left` is less than or equal to the sum of the elements in `mid`, and the sum of the elements in `mid` is less ... | null |
✔ Python3 Solution | LIS | minimum-operations-to-make-a-subsequence | 0 | 1 | ```\nclass Solution:\n def minOperations(self, target, arr):\n n, nums = len(target), []\n D = {target[i]: i for i in range(n)}\n res = [D[i] for i in arr if i in D.keys()]\n for i in res:\n j = bisect.bisect_left(nums, i)\n if j == len(nums): nums.append(i)\n ... | 1 | You are given an array `target` that consists of **distinct** integers and another integer array `arr` that **can** have duplicates.
In one operation, you can insert any integer at any position in `arr`. For example, if `arr = [1,4,1,2]`, you can add `3` in the middle and make it `[1,4,3,1,2]`. Note that you can inser... | Because the vector is sparse, use a data structure that stores the index and value where the element is nonzero. |
✔ Python3 Solution | LIS | minimum-operations-to-make-a-subsequence | 0 | 1 | ```\nclass Solution:\n def minOperations(self, target, arr):\n n, nums = len(target), []\n D = {target[i]: i for i in range(n)}\n res = [D[i] for i in arr if i in D.keys()]\n for i in res:\n j = bisect.bisect_left(nums, i)\n if j == len(nums): nums.append(i)\n ... | 1 | A **pangram** is a sentence where every letter of the English alphabet appears at least once.
Given a string `sentence` containing only lowercase English letters, return `true` _if_ `sentence` _is a **pangram**, or_ `false` _otherwise._
**Example 1:**
**Input:** sentence = "thequickbrownfoxjumpsoverthelazydog "
**O... | The problem can be reduced to computing Longest Common Subsequence between both arrays. Since one of the arrays has distinct elements, we can consider that these elements describe an arrangement of numbers, and we can replace each element in the other array with the index it appeared at in the first array. Then the pro... |
Python 3 || 6 lines, dct and bisect, w/ example || T/M: 96%/96% | minimum-operations-to-make-a-subsequence | 0 | 1 | \n```\nclass Solution:\n def minOperations(self, target: List[int], arr: List[int]) -> int:\n # Example: target = [6,4,8,1,3,2]\n dct, lst = {t: i for i,t in enumerate(target)}, [-1] # arr = [4,7,6,2,3,8,6,1]\n ... | 5 | You are given an array `target` that consists of **distinct** integers and another integer array `arr` that **can** have duplicates.
In one operation, you can insert any integer at any position in `arr`. For example, if `arr = [1,4,1,2]`, you can add `3` in the middle and make it `[1,4,3,1,2]`. Note that you can inser... | Because the vector is sparse, use a data structure that stores the index and value where the element is nonzero. |
Python 3 || 6 lines, dct and bisect, w/ example || T/M: 96%/96% | minimum-operations-to-make-a-subsequence | 0 | 1 | \n```\nclass Solution:\n def minOperations(self, target: List[int], arr: List[int]) -> int:\n # Example: target = [6,4,8,1,3,2]\n dct, lst = {t: i for i,t in enumerate(target)}, [-1] # arr = [4,7,6,2,3,8,6,1]\n ... | 5 | A **pangram** is a sentence where every letter of the English alphabet appears at least once.
Given a string `sentence` containing only lowercase English letters, return `true` _if_ `sentence` _is a **pangram**, or_ `false` _otherwise._
**Example 1:**
**Input:** sentence = "thequickbrownfoxjumpsoverthelazydog "
**O... | The problem can be reduced to computing Longest Common Subsequence between both arrays. Since one of the arrays has distinct elements, we can consider that these elements describe an arrangement of numbers, and we can replace each element in the other array with the index it appeared at in the first array. Then the pro... |
[Python3] binary search | minimum-operations-to-make-a-subsequence | 0 | 1 | **Algo**\nMaintain an increasing array of indices of values in `target`. \n\n**Implementation** \n```\nclass Solution:\n def minOperations(self, target: List[int], arr: List[int]) -> int:\n loc = {x: i for i, x in enumerate(target)}\n stack = []\n for x in arr: \n if x in loc: \n ... | 5 | You are given an array `target` that consists of **distinct** integers and another integer array `arr` that **can** have duplicates.
In one operation, you can insert any integer at any position in `arr`. For example, if `arr = [1,4,1,2]`, you can add `3` in the middle and make it `[1,4,3,1,2]`. Note that you can inser... | Because the vector is sparse, use a data structure that stores the index and value where the element is nonzero. |
[Python3] binary search | minimum-operations-to-make-a-subsequence | 0 | 1 | **Algo**\nMaintain an increasing array of indices of values in `target`. \n\n**Implementation** \n```\nclass Solution:\n def minOperations(self, target: List[int], arr: List[int]) -> int:\n loc = {x: i for i, x in enumerate(target)}\n stack = []\n for x in arr: \n if x in loc: \n ... | 5 | A **pangram** is a sentence where every letter of the English alphabet appears at least once.
Given a string `sentence` containing only lowercase English letters, return `true` _if_ `sentence` _is a **pangram**, or_ `false` _otherwise._
**Example 1:**
**Input:** sentence = "thequickbrownfoxjumpsoverthelazydog "
**O... | The problem can be reduced to computing Longest Common Subsequence between both arrays. Since one of the arrays has distinct elements, we can consider that these elements describe an arrangement of numbers, and we can replace each element in the other array with the index it appeared at in the first array. Then the pro... |
LCS - Russian Doll Approach with dict 100% | minimum-operations-to-make-a-subsequence | 0 | 1 | # Intuition \nBasically a variation of LCS(longest common subsequence), even simpler as target items are uniqle.\nSo dict can be used to get index of items in target, and making searching in arr even faster.\n\nGet all the [item]:index in target, search if each item in arr is also in target and make a list(loc) of inde... | 0 | You are given an array `target` that consists of **distinct** integers and another integer array `arr` that **can** have duplicates.
In one operation, you can insert any integer at any position in `arr`. For example, if `arr = [1,4,1,2]`, you can add `3` in the middle and make it `[1,4,3,1,2]`. Note that you can inser... | Because the vector is sparse, use a data structure that stores the index and value where the element is nonzero. |
LCS - Russian Doll Approach with dict 100% | minimum-operations-to-make-a-subsequence | 0 | 1 | # Intuition \nBasically a variation of LCS(longest common subsequence), even simpler as target items are uniqle.\nSo dict can be used to get index of items in target, and making searching in arr even faster.\n\nGet all the [item]:index in target, search if each item in arr is also in target and make a list(loc) of inde... | 0 | A **pangram** is a sentence where every letter of the English alphabet appears at least once.
Given a string `sentence` containing only lowercase English letters, return `true` _if_ `sentence` _is a **pangram**, or_ `false` _otherwise._
**Example 1:**
**Input:** sentence = "thequickbrownfoxjumpsoverthelazydog "
**O... | The problem can be reduced to computing Longest Common Subsequence between both arrays. Since one of the arrays has distinct elements, we can consider that these elements describe an arrangement of numbers, and we can replace each element in the other array with the index it appeared at in the first array. Then the pro... |
Python |DP vs Bisect | Short clean code | minimum-operations-to-make-a-subsequence | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nTwo solutions are presented: DP $O(N^2)$ (TLE) and Bisect $O(NlogN)$\nThe DP solution works for any kind of target array; it doesn\'t use the condition: the numbers in the target array are distinct.\n\nThe Bisect solution, by using the di... | 0 | You are given an array `target` that consists of **distinct** integers and another integer array `arr` that **can** have duplicates.
In one operation, you can insert any integer at any position in `arr`. For example, if `arr = [1,4,1,2]`, you can add `3` in the middle and make it `[1,4,3,1,2]`. Note that you can inser... | Because the vector is sparse, use a data structure that stores the index and value where the element is nonzero. |
Python |DP vs Bisect | Short clean code | minimum-operations-to-make-a-subsequence | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nTwo solutions are presented: DP $O(N^2)$ (TLE) and Bisect $O(NlogN)$\nThe DP solution works for any kind of target array; it doesn\'t use the condition: the numbers in the target array are distinct.\n\nThe Bisect solution, by using the di... | 0 | A **pangram** is a sentence where every letter of the English alphabet appears at least once.
Given a string `sentence` containing only lowercase English letters, return `true` _if_ `sentence` _is a **pangram**, or_ `false` _otherwise._
**Example 1:**
**Input:** sentence = "thequickbrownfoxjumpsoverthelazydog "
**O... | The problem can be reduced to computing Longest Common Subsequence between both arrays. Since one of the arrays has distinct elements, we can consider that these elements describe an arrangement of numbers, and we can replace each element in the other array with the index it appeared at in the first array. Then the pro... |
Python (Simple Binary Search) | minimum-operations-to-make-a-subsequence | 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 `target` that consists of **distinct** integers and another integer array `arr` that **can** have duplicates.
In one operation, you can insert any integer at any position in `arr`. For example, if `arr = [1,4,1,2]`, you can add `3` in the middle and make it `[1,4,3,1,2]`. Note that you can inser... | Because the vector is sparse, use a data structure that stores the index and value where the element is nonzero. |
Python (Simple Binary Search) | minimum-operations-to-make-a-subsequence | 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 | A **pangram** is a sentence where every letter of the English alphabet appears at least once.
Given a string `sentence` containing only lowercase English letters, return `true` _if_ `sentence` _is a **pangram**, or_ `false` _otherwise._
**Example 1:**
**Input:** sentence = "thequickbrownfoxjumpsoverthelazydog "
**O... | The problem can be reduced to computing Longest Common Subsequence between both arrays. Since one of the arrays has distinct elements, we can consider that these elements describe an arrangement of numbers, and we can replace each element in the other array with the index it appeared at in the first array. Then the pro... |
Python easy to read and understand | lcs-lis | minimum-operations-to-make-a-subsequence | 0 | 1 | **LCS(TLE):**\n```\nclass Solution:\n def lcs(self, nums1, nums2, m, n):\n if m == 0 or n == 0:\n return 0\n if (m, n) in self.d:\n return self.d[(m, n)]\n if nums1[m-1] == nums2[n-1]:\n self.d[(m, n)] = self.lcs(nums1, nums2, m-1, n-1) + 1\n return se... | 0 | You are given an array `target` that consists of **distinct** integers and another integer array `arr` that **can** have duplicates.
In one operation, you can insert any integer at any position in `arr`. For example, if `arr = [1,4,1,2]`, you can add `3` in the middle and make it `[1,4,3,1,2]`. Note that you can inser... | Because the vector is sparse, use a data structure that stores the index and value where the element is nonzero. |
Python easy to read and understand | lcs-lis | minimum-operations-to-make-a-subsequence | 0 | 1 | **LCS(TLE):**\n```\nclass Solution:\n def lcs(self, nums1, nums2, m, n):\n if m == 0 or n == 0:\n return 0\n if (m, n) in self.d:\n return self.d[(m, n)]\n if nums1[m-1] == nums2[n-1]:\n self.d[(m, n)] = self.lcs(nums1, nums2, m-1, n-1) + 1\n return se... | 0 | A **pangram** is a sentence where every letter of the English alphabet appears at least once.
Given a string `sentence` containing only lowercase English letters, return `true` _if_ `sentence` _is a **pangram**, or_ `false` _otherwise._
**Example 1:**
**Input:** sentence = "thequickbrownfoxjumpsoverthelazydog "
**O... | The problem can be reduced to computing Longest Common Subsequence between both arrays. Since one of the arrays has distinct elements, we can consider that these elements describe an arrangement of numbers, and we can replace each element in the other array with the index it appeared at in the first array. Then the pro... |
Python LIS with SortedList | minimum-operations-to-make-a-subsequence | 0 | 1 | # Intuition\nBasically we are looking for the longest subsequence of `target` that can be found within `arr`. Since we\'re told that `target` has unique values, we can consider the index within `target` of each value in `arr`, and then solve the longest increasing subsequence problem.\n\nOnce we know the length of the... | 0 | You are given an array `target` that consists of **distinct** integers and another integer array `arr` that **can** have duplicates.
In one operation, you can insert any integer at any position in `arr`. For example, if `arr = [1,4,1,2]`, you can add `3` in the middle and make it `[1,4,3,1,2]`. Note that you can inser... | Because the vector is sparse, use a data structure that stores the index and value where the element is nonzero. |
Python LIS with SortedList | minimum-operations-to-make-a-subsequence | 0 | 1 | # Intuition\nBasically we are looking for the longest subsequence of `target` that can be found within `arr`. Since we\'re told that `target` has unique values, we can consider the index within `target` of each value in `arr`, and then solve the longest increasing subsequence problem.\n\nOnce we know the length of the... | 0 | A **pangram** is a sentence where every letter of the English alphabet appears at least once.
Given a string `sentence` containing only lowercase English letters, return `true` _if_ `sentence` _is a **pangram**, or_ `false` _otherwise._
**Example 1:**
**Input:** sentence = "thequickbrownfoxjumpsoverthelazydog "
**O... | The problem can be reduced to computing Longest Common Subsequence between both arrays. Since one of the arrays has distinct elements, we can consider that these elements describe an arrangement of numbers, and we can replace each element in the other array with the index it appeared at in the first array. Then the pro... |
🚀✅ Beats 100% , Simple Maths | calculate-money-in-leetcode-bank | 1 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nCalculate the Saving from each Week and observe a pattern between them\n```\nWeek 1 -> {1,2,3,4,5,6,7} = 28\nWeek 2 -> {2,3,4,5,6,7,8} = 28 + 7\nand so on\n```\nEach week the Saving increases by an amount of 7\n\n---\n\n\n# Approach\n<!--... | 11 | Hercy wants to save money for his first car. He puts money in the Leetcode bank **every day**.
He starts by putting in `$1` on Monday, the first day. Every day from Tuesday to Sunday, he will put in `$1` more than the day before. On every subsequent Monday, he will put in `$1` more than the **previous Monday**.
Given... | Use Dynamic programming. Keep the highest value and lowest value you can achieve up to a point. |
🚀✅ Beats 100% , Simple Maths | calculate-money-in-leetcode-bank | 1 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nCalculate the Saving from each Week and observe a pattern between them\n```\nWeek 1 -> {1,2,3,4,5,6,7} = 28\nWeek 2 -> {2,3,4,5,6,7,8} = 28 + 7\nand so on\n```\nEach week the Saving increases by an amount of 7\n\n---\n\n\n# Approach\n<!--... | 11 | You are given the logs for users' actions on LeetCode, and an integer `k`. The logs are represented by a 2D integer array `logs` where each `logs[i] = [IDi, timei]` indicates that the user with `IDi` performed an action at the minute `timei`.
**Multiple users** can perform actions simultaneously, and a single user can... | Simulate the process by keeping track of how much money John is putting in and which day of the week it is, and use this information to deduce how much money John will put in the next day. |
🥇 O (1) math explained ; ] ✅ | calculate-money-in-leetcode-bank | 1 | 1 | \n**UPVOTE IF HELPFuuL**\n\n# Problem Statement\n- Start by `$1` on first day `[MONDAY]`.\n- For every day in the week put `$1` more than yesterday.\n- On next Monday, start again by increasing `$1` from previous Monday.\n\n# Approach\n```\nWeek 1 money : [1, 2, 3, 4, 5, 6, 7] -> 28\nWeek 2 money : [2, 3, 4, 5, 6, 7, 8... | 72 | Hercy wants to save money for his first car. He puts money in the Leetcode bank **every day**.
He starts by putting in `$1` on Monday, the first day. Every day from Tuesday to Sunday, he will put in `$1` more than the day before. On every subsequent Monday, he will put in `$1` more than the **previous Monday**.
Given... | Use Dynamic programming. Keep the highest value and lowest value you can achieve up to a point. |
🥇 O (1) math explained ; ] ✅ | calculate-money-in-leetcode-bank | 1 | 1 | \n**UPVOTE IF HELPFuuL**\n\n# Problem Statement\n- Start by `$1` on first day `[MONDAY]`.\n- For every day in the week put `$1` more than yesterday.\n- On next Monday, start again by increasing `$1` from previous Monday.\n\n# Approach\n```\nWeek 1 money : [1, 2, 3, 4, 5, 6, 7] -> 28\nWeek 2 money : [2, 3, 4, 5, 6, 7, 8... | 72 | You are given the logs for users' actions on LeetCode, and an integer `k`. The logs are represented by a 2D integer array `logs` where each `logs[i] = [IDi, timei]` indicates that the user with `IDi` performed an action at the minute `timei`.
**Multiple users** can perform actions simultaneously, and a single user can... | Simulate the process by keeping track of how much money John is putting in and which day of the week it is, and use this information to deduce how much money John will put in the next day. |
Simple Iteration | calculate-money-in-leetcode-bank | 1 | 1 | > If the remainder of the division of the ordinal number = 1, then it is Monday. \n\n> The amount of money on Monday is equal to the number of the week, on other days is equal to the previous one + 1\n\n### C++\n```cpp\npublic:\n int totalMoney(int n) {\n int totalMoney = 0;\n\n for (int day = 1, dailySum = 0,... | 9 | Hercy wants to save money for his first car. He puts money in the Leetcode bank **every day**.
He starts by putting in `$1` on Monday, the first day. Every day from Tuesday to Sunday, he will put in `$1` more than the day before. On every subsequent Monday, he will put in `$1` more than the **previous Monday**.
Given... | Use Dynamic programming. Keep the highest value and lowest value you can achieve up to a point. |
Simple Iteration | calculate-money-in-leetcode-bank | 1 | 1 | > If the remainder of the division of the ordinal number = 1, then it is Monday. \n\n> The amount of money on Monday is equal to the number of the week, on other days is equal to the previous one + 1\n\n### C++\n```cpp\npublic:\n int totalMoney(int n) {\n int totalMoney = 0;\n\n for (int day = 1, dailySum = 0,... | 9 | You are given the logs for users' actions on LeetCode, and an integer `k`. The logs are represented by a 2D integer array `logs` where each `logs[i] = [IDi, timei]` indicates that the user with `IDi` performed an action at the minute `timei`.
**Multiple users** can perform actions simultaneously, and a single user can... | Simulate the process by keeping track of how much money John is putting in and which day of the week it is, and use this information to deduce how much money John will put in the next day. |
✅☑[C++/Java/Python/JavaScript] || 3 Approaches || Beats 100% || EXPLAINED🔥 | calculate-money-in-leetcode-bank | 1 | 1 | # PLEASE UPVOTE IF IT HELPED\n\n---\n\n\n# Approaches\n**(Also explained in the code)**\n\n#### ***Approach 1(Simulate)***\n1. **Variables Initialization:** `ans` is initialized to zero to store the total money earned. `monday` represents the amount earned on Mondays, starting from 1.\n\n1. **Main Loop:** The `while` l... | 17 | Hercy wants to save money for his first car. He puts money in the Leetcode bank **every day**.
He starts by putting in `$1` on Monday, the first day. Every day from Tuesday to Sunday, he will put in `$1` more than the day before. On every subsequent Monday, he will put in `$1` more than the **previous Monday**.
Given... | Use Dynamic programming. Keep the highest value and lowest value you can achieve up to a point. |
✅☑[C++/Java/Python/JavaScript] || 3 Approaches || Beats 100% || EXPLAINED🔥 | calculate-money-in-leetcode-bank | 1 | 1 | # PLEASE UPVOTE IF IT HELPED\n\n---\n\n\n# Approaches\n**(Also explained in the code)**\n\n#### ***Approach 1(Simulate)***\n1. **Variables Initialization:** `ans` is initialized to zero to store the total money earned. `monday` represents the amount earned on Mondays, starting from 1.\n\n1. **Main Loop:** The `while` l... | 17 | You are given the logs for users' actions on LeetCode, and an integer `k`. The logs are represented by a 2D integer array `logs` where each `logs[i] = [IDi, timei]` indicates that the user with `IDi` performed an action at the minute `timei`.
**Multiple users** can perform actions simultaneously, and a single user can... | Simulate the process by keeping track of how much money John is putting in and which day of the week it is, and use this information to deduce how much money John will put in the next day. |
Arithmetic Progression O(1)Handwriting||0ms Beats 100% | calculate-money-in-leetcode-bank | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nArithmetic Progression\nConsider the arithmetic progression \n$$\na, a+d, a+2d,\\cdots, e=a+(r-1)d\n$$\nThe sum is $\\frac{(a+e)r}{2}$. \n# Approach\n[Please turn on English subtitles if necessary]\n[https://youtu.be/TqKZGAHLKdw?si=FzWxIk... | 7 | Hercy wants to save money for his first car. He puts money in the Leetcode bank **every day**.
He starts by putting in `$1` on Monday, the first day. Every day from Tuesday to Sunday, he will put in `$1` more than the day before. On every subsequent Monday, he will put in `$1` more than the **previous Monday**.
Given... | Use Dynamic programming. Keep the highest value and lowest value you can achieve up to a point. |
Arithmetic Progression O(1)Handwriting||0ms Beats 100% | calculate-money-in-leetcode-bank | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nArithmetic Progression\nConsider the arithmetic progression \n$$\na, a+d, a+2d,\\cdots, e=a+(r-1)d\n$$\nThe sum is $\\frac{(a+e)r}{2}$. \n# Approach\n[Please turn on English subtitles if necessary]\n[https://youtu.be/TqKZGAHLKdw?si=FzWxIk... | 7 | You are given the logs for users' actions on LeetCode, and an integer `k`. The logs are represented by a 2D integer array `logs` where each `logs[i] = [IDi, timei]` indicates that the user with `IDi` performed an action at the minute `timei`.
**Multiple users** can perform actions simultaneously, and a single user can... | Simulate the process by keeping track of how much money John is putting in and which day of the week it is, and use this information to deduce how much money John will put in the next day. |
💯Faster✅💯 Easy solution to understand [python, c#, Java, Rust] 🔥🔥 | calculate-money-in-leetcode-bank | 1 | 1 | # Description\n\nHercy wants to save money for his first car. He puts money in the Leetcode bank every day.\n\nHe starts by putting in 1 dollar on Monday, the first day. Every day from Tuesday to Sunday, he will put in 1 dollar more than the day before. On every subsequent Monday, he will put in $1 more than the previo... | 7 | Hercy wants to save money for his first car. He puts money in the Leetcode bank **every day**.
He starts by putting in `$1` on Monday, the first day. Every day from Tuesday to Sunday, he will put in `$1` more than the day before. On every subsequent Monday, he will put in `$1` more than the **previous Monday**.
Given... | Use Dynamic programming. Keep the highest value and lowest value you can achieve up to a point. |
💯Faster✅💯 Easy solution to understand [python, c#, Java, Rust] 🔥🔥 | calculate-money-in-leetcode-bank | 1 | 1 | # Description\n\nHercy wants to save money for his first car. He puts money in the Leetcode bank every day.\n\nHe starts by putting in 1 dollar on Monday, the first day. Every day from Tuesday to Sunday, he will put in 1 dollar more than the day before. On every subsequent Monday, he will put in $1 more than the previo... | 7 | You are given the logs for users' actions on LeetCode, and an integer `k`. The logs are represented by a 2D integer array `logs` where each `logs[i] = [IDi, timei]` indicates that the user with `IDi` performed an action at the minute `timei`.
**Multiple users** can perform actions simultaneously, and a single user can... | Simulate the process by keeping track of how much money John is putting in and which day of the week it is, and use this information to deduce how much money John will put in the next day. |
✅ C/C++/Java/Rust/Python/C# - O(1) solution with two approaches | Beats 100 % | 0 ms time | calculate-money-in-leetcode-bank | 1 | 1 | ### I solved this problem with two different approaches. In the first, I did what came to my mind first, and in the second one, I tried to optimize it.\n\n# 1st approach\n- Initialize sum and currentValue to zero and one, respectively.\n- Iterate from day 1 to day n.\n- On each day, add the currentValue to the sum.\n- ... | 5 | Hercy wants to save money for his first car. He puts money in the Leetcode bank **every day**.
He starts by putting in `$1` on Monday, the first day. Every day from Tuesday to Sunday, he will put in `$1` more than the day before. On every subsequent Monday, he will put in `$1` more than the **previous Monday**.
Given... | Use Dynamic programming. Keep the highest value and lowest value you can achieve up to a point. |
✅ C/C++/Java/Rust/Python/C# - O(1) solution with two approaches | Beats 100 % | 0 ms time | calculate-money-in-leetcode-bank | 1 | 1 | ### I solved this problem with two different approaches. In the first, I did what came to my mind first, and in the second one, I tried to optimize it.\n\n# 1st approach\n- Initialize sum and currentValue to zero and one, respectively.\n- Iterate from day 1 to day n.\n- On each day, add the currentValue to the sum.\n- ... | 5 | You are given the logs for users' actions on LeetCode, and an integer `k`. The logs are represented by a 2D integer array `logs` where each `logs[i] = [IDi, timei]` indicates that the user with `IDi` performed an action at the minute `timei`.
**Multiple users** can perform actions simultaneously, and a single user can... | Simulate the process by keeping track of how much money John is putting in and which day of the week it is, and use this information to deduce how much money John will put in the next day. |
Python3 Solution | calculate-money-in-leetcode-bank | 0 | 1 | \n```\nclass Solution:\n def totalMoney(self, n: int) -> int:\n ans=0\n offset=0\n for i in range(1,n+1):\n res=i%7\n if res%7==0:\n res+=7\n ans+=offset+res\n if i%7==0:\n offset+=1\n return ans \n``` | 4 | Hercy wants to save money for his first car. He puts money in the Leetcode bank **every day**.
He starts by putting in `$1` on Monday, the first day. Every day from Tuesday to Sunday, he will put in `$1` more than the day before. On every subsequent Monday, he will put in `$1` more than the **previous Monday**.
Given... | Use Dynamic programming. Keep the highest value and lowest value you can achieve up to a point. |
Python3 Solution | calculate-money-in-leetcode-bank | 0 | 1 | \n```\nclass Solution:\n def totalMoney(self, n: int) -> int:\n ans=0\n offset=0\n for i in range(1,n+1):\n res=i%7\n if res%7==0:\n res+=7\n ans+=offset+res\n if i%7==0:\n offset+=1\n return ans \n``` | 4 | You are given the logs for users' actions on LeetCode, and an integer `k`. The logs are represented by a 2D integer array `logs` where each `logs[i] = [IDi, timei]` indicates that the user with `IDi` performed an action at the minute `timei`.
**Multiple users** can perform actions simultaneously, and a single user can... | Simulate the process by keeping track of how much money John is putting in and which day of the week it is, and use this information to deduce how much money John will put in the next day. |
Python Solution for beginners | calculate-money-in-leetcode-bank | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nEasy approach\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n- Initialize variables: n is set to 20 (the number of integers to sum), res is the result initialized to 0, i is a counter starting at 1, j is a loop co... | 2 | Hercy wants to save money for his first car. He puts money in the Leetcode bank **every day**.
He starts by putting in `$1` on Monday, the first day. Every day from Tuesday to Sunday, he will put in `$1` more than the day before. On every subsequent Monday, he will put in `$1` more than the **previous Monday**.
Given... | Use Dynamic programming. Keep the highest value and lowest value you can achieve up to a point. |
Python Solution for beginners | calculate-money-in-leetcode-bank | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nEasy approach\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n- Initialize variables: n is set to 20 (the number of integers to sum), res is the result initialized to 0, i is a counter starting at 1, j is a loop co... | 2 | You are given the logs for users' actions on LeetCode, and an integer `k`. The logs are represented by a 2D integer array `logs` where each `logs[i] = [IDi, timei]` indicates that the user with `IDi` performed an action at the minute `timei`.
**Multiple users** can perform actions simultaneously, and a single user can... | Simulate the process by keeping track of how much money John is putting in and which day of the week it is, and use this information to deduce how much money John will put in the next day. |
O(1) Formula solution | calculate-money-in-leetcode-bank | 0 | 1 | # Approach\n\nFirst week we put `28` dollars.\nNext week we put `28 + 7` dollars.\nThen we put `28 + 7 + 7` dollars and so on...\n\nThe formula for $m^{th}$ week is $28 + 7*(m-1)$.\n\nUsing arithmetic progression sum formula $\\frac{2a_1+t(m-1)}{2}*m$, we can find the sum of `n` weeks.\n\n`a_1` is `28`\n`t` is `7`\n`m`... | 2 | Hercy wants to save money for his first car. He puts money in the Leetcode bank **every day**.
He starts by putting in `$1` on Monday, the first day. Every day from Tuesday to Sunday, he will put in `$1` more than the day before. On every subsequent Monday, he will put in `$1` more than the **previous Monday**.
Given... | Use Dynamic programming. Keep the highest value and lowest value you can achieve up to a point. |
O(1) Formula solution | calculate-money-in-leetcode-bank | 0 | 1 | # Approach\n\nFirst week we put `28` dollars.\nNext week we put `28 + 7` dollars.\nThen we put `28 + 7 + 7` dollars and so on...\n\nThe formula for $m^{th}$ week is $28 + 7*(m-1)$.\n\nUsing arithmetic progression sum formula $\\frac{2a_1+t(m-1)}{2}*m$, we can find the sum of `n` weeks.\n\n`a_1` is `28`\n`t` is `7`\n`m`... | 2 | You are given the logs for users' actions on LeetCode, and an integer `k`. The logs are represented by a 2D integer array `logs` where each `logs[i] = [IDi, timei]` indicates that the user with `IDi` performed an action at the minute `timei`.
**Multiple users** can perform actions simultaneously, and a single user can... | Simulate the process by keeping track of how much money John is putting in and which day of the week it is, and use this information to deduce how much money John will put in the next day. |
Rust/Python constant time solution with explanation | calculate-money-in-leetcode-bank | 0 | 1 | # Intuition\n\nOf course you can just simulate it, but it will take linear time. You can do better by using math.\n\nFirst lets look at whole weeks 7, 14, 21 and so on days. First day week consumes `1 + 2 + ... + 7 = 28`. Next day will be `2 + 3 + ... + 8 = (1 + 2 + ... + 7) + 7`. For i-th week (starts with 0th week) t... | 2 | Hercy wants to save money for his first car. He puts money in the Leetcode bank **every day**.
He starts by putting in `$1` on Monday, the first day. Every day from Tuesday to Sunday, he will put in `$1` more than the day before. On every subsequent Monday, he will put in `$1` more than the **previous Monday**.
Given... | Use Dynamic programming. Keep the highest value and lowest value you can achieve up to a point. |
Rust/Python constant time solution with explanation | calculate-money-in-leetcode-bank | 0 | 1 | # Intuition\n\nOf course you can just simulate it, but it will take linear time. You can do better by using math.\n\nFirst lets look at whole weeks 7, 14, 21 and so on days. First day week consumes `1 + 2 + ... + 7 = 28`. Next day will be `2 + 3 + ... + 8 = (1 + 2 + ... + 7) + 7`. For i-th week (starts with 0th week) t... | 2 | You are given the logs for users' actions on LeetCode, and an integer `k`. The logs are represented by a 2D integer array `logs` where each `logs[i] = [IDi, timei]` indicates that the user with `IDi` performed an action at the minute `timei`.
**Multiple users** can perform actions simultaneously, and a single user can... | Simulate the process by keeping track of how much money John is putting in and which day of the week it is, and use this information to deduce how much money John will put in the next day. |
🔥🔥 Simplest Code | 2ms | For Beginners | Fully Explained 🚀🔥 | calculate-money-in-leetcode-bank | 1 | 1 | # Intuition\n\n\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n - Initialize `ans` to 0 for aggregating savings and `mon` to 1 for the current Monday\'s deposit. \uD83C\uD... | 11 | Hercy wants to save money for his first car. He puts money in the Leetcode bank **every day**.
He starts by putting in `$1` on Monday, the first day. Every day from Tuesday to Sunday, he will put in `$1` more than the day before. On every subsequent Monday, he will put in `$1` more than the **previous Monday**.
Given... | Use Dynamic programming. Keep the highest value and lowest value you can achieve up to a point. |
🔥🔥 Simplest Code | 2ms | For Beginners | Fully Explained 🚀🔥 | calculate-money-in-leetcode-bank | 1 | 1 | # Intuition\n\n\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n - Initialize `ans` to 0 for aggregating savings and `mon` to 1 for the current Monday\'s deposit. \uD83C\uD... | 11 | You are given the logs for users' actions on LeetCode, and an integer `k`. The logs are represented by a 2D integer array `logs` where each `logs[i] = [IDi, timei]` indicates that the user with `IDi` performed an action at the minute `timei`.
**Multiple users** can perform actions simultaneously, and a single user can... | Simulate the process by keeping track of how much money John is putting in and which day of the week it is, and use this information to deduce how much money John will put in the next day. |
Python3 / Rust integer division and remainder approach | calculate-money-in-leetcode-bank | 0 | 1 | # Intuition\nThe problem describes a scenario where Hercy saves money in the Leetcode bank. He starts with \\$1 on the first day (Monday) and adds an additional dollar each day. The pattern resets every Monday, with each Monday having $1 more than the previous Monday. The goal is to calculate the total amount of money ... | 1 | Hercy wants to save money for his first car. He puts money in the Leetcode bank **every day**.
He starts by putting in `$1` on Monday, the first day. Every day from Tuesday to Sunday, he will put in `$1` more than the day before. On every subsequent Monday, he will put in `$1` more than the **previous Monday**.
Given... | Use Dynamic programming. Keep the highest value and lowest value you can achieve up to a point. |
Python3 / Rust integer division and remainder approach | calculate-money-in-leetcode-bank | 0 | 1 | # Intuition\nThe problem describes a scenario where Hercy saves money in the Leetcode bank. He starts with \\$1 on the first day (Monday) and adds an additional dollar each day. The pattern resets every Monday, with each Monday having $1 more than the previous Monday. The goal is to calculate the total amount of money ... | 1 | You are given the logs for users' actions on LeetCode, and an integer `k`. The logs are represented by a 2D integer array `logs` where each `logs[i] = [IDi, timei]` indicates that the user with `IDi` performed an action at the minute `timei`.
**Multiple users** can perform actions simultaneously, and a single user can... | Simulate the process by keeping track of how much money John is putting in and which day of the week it is, and use this information to deduce how much money John will put in the next day. |
✅ One Line Solution | calculate-money-in-leetcode-bank | 0 | 1 | # Code #1.1 - Oneliner\nTime complexity: $$O(1)$$. Space complexity: $$O(1)$$.\n```\nclass Solution:\n def totalMoney(self, n: int) -> int:\n return (s:=lambda a,d,k:(2*a+d*(k-1))*k//2) and (s(1,1,7)*(n//7) + s(0,7,n//7) + s(n//7+1,1,n%7))\n```\n\n# Code #1.2 - Unwrapped\n```\nclass Solution:\n def totalMo... | 4 | Hercy wants to save money for his first car. He puts money in the Leetcode bank **every day**.
He starts by putting in `$1` on Monday, the first day. Every day from Tuesday to Sunday, he will put in `$1` more than the day before. On every subsequent Monday, he will put in `$1` more than the **previous Monday**.
Given... | Use Dynamic programming. Keep the highest value and lowest value you can achieve up to a point. |
✅ One Line Solution | calculate-money-in-leetcode-bank | 0 | 1 | # Code #1.1 - Oneliner\nTime complexity: $$O(1)$$. Space complexity: $$O(1)$$.\n```\nclass Solution:\n def totalMoney(self, n: int) -> int:\n return (s:=lambda a,d,k:(2*a+d*(k-1))*k//2) and (s(1,1,7)*(n//7) + s(0,7,n//7) + s(n//7+1,1,n%7))\n```\n\n# Code #1.2 - Unwrapped\n```\nclass Solution:\n def totalMo... | 4 | You are given the logs for users' actions on LeetCode, and an integer `k`. The logs are represented by a 2D integer array `logs` where each `logs[i] = [IDi, timei]` indicates that the user with `IDi` performed an action at the minute `timei`.
**Multiple users** can perform actions simultaneously, and a single user can... | Simulate the process by keeping track of how much money John is putting in and which day of the week it is, and use this information to deduce how much money John will put in the next day. |
Money in Leetcode bank. | calculate-money-in-leetcode-bank | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nKeen observation of the pattern of the numbers.\n\n\n# Complexity\n- Time complexity: O(1)\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n\n- Space complexity: O(1)\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# C... | 1 | Hercy wants to save money for his first car. He puts money in the Leetcode bank **every day**.
He starts by putting in `$1` on Monday, the first day. Every day from Tuesday to Sunday, he will put in `$1` more than the day before. On every subsequent Monday, he will put in `$1` more than the **previous Monday**.
Given... | Use Dynamic programming. Keep the highest value and lowest value you can achieve up to a point. |
Money in Leetcode bank. | calculate-money-in-leetcode-bank | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nKeen observation of the pattern of the numbers.\n\n\n# Complexity\n- Time complexity: O(1)\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n\n- Space complexity: O(1)\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# C... | 1 | You are given the logs for users' actions on LeetCode, and an integer `k`. The logs are represented by a 2D integer array `logs` where each `logs[i] = [IDi, timei]` indicates that the user with `IDi` performed an action at the minute `timei`.
**Multiple users** can perform actions simultaneously, and a single user can... | Simulate the process by keeping track of how much money John is putting in and which day of the week it is, and use this information to deduce how much money John will put in the next day. |
calculateMoneyInLeetcodeBank | calculate-money-in-leetcode-bank | 0 | 1 | \n# Code\n```\n# time complexity: O(n)\nclass Solution:\n def totalMoney(self, n: int) -> int:\n count = 0 # value that gets added to the list\n end = 0 # value to keep track when the value of n becomes multiple of 7 (for every monday)\n start = 1 # to keep track of the value to be increased ... | 1 | Hercy wants to save money for his first car. He puts money in the Leetcode bank **every day**.
He starts by putting in `$1` on Monday, the first day. Every day from Tuesday to Sunday, he will put in `$1` more than the day before. On every subsequent Monday, he will put in `$1` more than the **previous Monday**.
Given... | Use Dynamic programming. Keep the highest value and lowest value you can achieve up to a point. |
calculateMoneyInLeetcodeBank | calculate-money-in-leetcode-bank | 0 | 1 | \n# Code\n```\n# time complexity: O(n)\nclass Solution:\n def totalMoney(self, n: int) -> int:\n count = 0 # value that gets added to the list\n end = 0 # value to keep track when the value of n becomes multiple of 7 (for every monday)\n start = 1 # to keep track of the value to be increased ... | 1 | You are given the logs for users' actions on LeetCode, and an integer `k`. The logs are represented by a 2D integer array `logs` where each `logs[i] = [IDi, timei]` indicates that the user with `IDi` performed an action at the minute `timei`.
**Multiple users** can perform actions simultaneously, and a single user can... | Simulate the process by keeping track of how much money John is putting in and which day of the week it is, and use this information to deduce how much money John will put in the next day. |
Mathematical Approach | T.C. - O(1) | Beginner Friendly | Explanation - [Video] | Easy Solution | calculate-money-in-leetcode-bank | 0 | 1 | For better understanding of the question you can also check out this video. Hope it helps.\n\nhttps://youtu.be/3Pjay2_v-os\n\n# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n1. **Contribution from complete weeks:** The code calculates the total contribution from complete weeks using ... | 1 | Hercy wants to save money for his first car. He puts money in the Leetcode bank **every day**.
He starts by putting in `$1` on Monday, the first day. Every day from Tuesday to Sunday, he will put in `$1` more than the day before. On every subsequent Monday, he will put in `$1` more than the **previous Monday**.
Given... | Use Dynamic programming. Keep the highest value and lowest value you can achieve up to a point. |
Mathematical Approach | T.C. - O(1) | Beginner Friendly | Explanation - [Video] | Easy Solution | calculate-money-in-leetcode-bank | 0 | 1 | For better understanding of the question you can also check out this video. Hope it helps.\n\nhttps://youtu.be/3Pjay2_v-os\n\n# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n1. **Contribution from complete weeks:** The code calculates the total contribution from complete weeks using ... | 1 | You are given the logs for users' actions on LeetCode, and an integer `k`. The logs are represented by a 2D integer array `logs` where each `logs[i] = [IDi, timei]` indicates that the user with `IDi` performed an action at the minute `timei`.
**Multiple users** can perform actions simultaneously, and a single user can... | Simulate the process by keeping track of how much money John is putting in and which day of the week it is, and use this information to deduce how much money John will put in the next day. |
✅Unlocking Wealth Accumulation Patterns: Navigating LeetCode's Total Money Earned Challenge | calculate-money-in-leetcode-bank | 1 | 1 | # Exploring Wealth Accumulation Patterns: LeetCode\'s Total Money Earned Challenge\n\n## \uD83D\uDCB0 Problem Overview\nLeetCode introduces us to a fascinating financial challenge with the "Total Money Earned" problem. In this blog post, we\'ll delve into the problem\'s intricacies and provide a comprehensive guide to ... | 1 | Hercy wants to save money for his first car. He puts money in the Leetcode bank **every day**.
He starts by putting in `$1` on Monday, the first day. Every day from Tuesday to Sunday, he will put in `$1` more than the day before. On every subsequent Monday, he will put in `$1` more than the **previous Monday**.
Given... | Use Dynamic programming. Keep the highest value and lowest value you can achieve up to a point. |
✅Unlocking Wealth Accumulation Patterns: Navigating LeetCode's Total Money Earned Challenge | calculate-money-in-leetcode-bank | 1 | 1 | # Exploring Wealth Accumulation Patterns: LeetCode\'s Total Money Earned Challenge\n\n## \uD83D\uDCB0 Problem Overview\nLeetCode introduces us to a fascinating financial challenge with the "Total Money Earned" problem. In this blog post, we\'ll delve into the problem\'s intricacies and provide a comprehensive guide to ... | 1 | You are given the logs for users' actions on LeetCode, and an integer `k`. The logs are represented by a 2D integer array `logs` where each `logs[i] = [IDi, timei]` indicates that the user with `IDi` performed an action at the minute `timei`.
**Multiple users** can perform actions simultaneously, and a single user can... | Simulate the process by keeping track of how much money John is putting in and which day of the week it is, and use this information to deduce how much money John will put in the next day. |
Python solution with explanation💃🏻 | maximum-score-from-removing-substrings | 0 | 1 | ```\nclass Solution:\n def maximumGain(self, s: str, x: int, y: int) -> int:\n\t\t# to calculate first, high value of x or y\n a, b = \'ab\', \'ba\'\n if y > x:\n b, a, y, x = a, b, x, y\n\n answer = 0\n \n for word in [a, b]:\n stack = []\n\n i = 0... | 9 | You are given a string `s` and two integers `x` and `y`. You can perform two types of operations any number of times.
* Remove substring `"ab "` and gain `x` points.
* For example, when removing `"ab "` from `"cabxbae "` it becomes `"cxbae "`.
* Remove substring `"ba "` and gain `y` points.
* For examp... | Each point on the left would either be connected to exactly point already connected to some left node, or a subset of the nodes on the right which are not connected to any node Use dynamic programming with bitmasking, where the state will be (number of points assigned in first group, bitmask of points assigned in secon... |
Python solution with explanation💃🏻 | maximum-score-from-removing-substrings | 0 | 1 | ```\nclass Solution:\n def maximumGain(self, s: str, x: int, y: int) -> int:\n\t\t# to calculate first, high value of x or y\n a, b = \'ab\', \'ba\'\n if y > x:\n b, a, y, x = a, b, x, y\n\n answer = 0\n \n for word in [a, b]:\n stack = []\n\n i = 0... | 9 | You are given two positive integer arrays `nums1` and `nums2`, both of length `n`.
The **absolute sum difference** of arrays `nums1` and `nums2` is defined as the **sum** of `|nums1[i] - nums2[i]|` for each `0 <= i < n` (**0-indexed**).
You can replace **at most one** element of `nums1` with **any** other element in ... | Note that it is always more optimal to take one type of substring before another You can use a stack to handle erasures |
Solution using stack and map - Python! | maximum-score-from-removing-substrings | 0 | 1 | # Code\n```\nclass Solution:\n def maximumGain(self, s: str, x: int, y: int) -> int:\n m = {\'ab\':x, \'ba\':y}\n ab = m.keys()\n stack = sorted(ab, key=lambda x:m[x], reverse=True)\n score = 0\n \n bin = []\n \n for st in stack:\n for char in s:\n if bin and b... | 0 | You are given a string `s` and two integers `x` and `y`. You can perform two types of operations any number of times.
* Remove substring `"ab "` and gain `x` points.
* For example, when removing `"ab "` from `"cabxbae "` it becomes `"cxbae "`.
* Remove substring `"ba "` and gain `y` points.
* For examp... | Each point on the left would either be connected to exactly point already connected to some left node, or a subset of the nodes on the right which are not connected to any node Use dynamic programming with bitmasking, where the state will be (number of points assigned in first group, bitmask of points assigned in secon... |
Solution using stack and map - Python! | maximum-score-from-removing-substrings | 0 | 1 | # Code\n```\nclass Solution:\n def maximumGain(self, s: str, x: int, y: int) -> int:\n m = {\'ab\':x, \'ba\':y}\n ab = m.keys()\n stack = sorted(ab, key=lambda x:m[x], reverse=True)\n score = 0\n \n bin = []\n \n for st in stack:\n for char in s:\n if bin and b... | 0 | You are given two positive integer arrays `nums1` and `nums2`, both of length `n`.
The **absolute sum difference** of arrays `nums1` and `nums2` is defined as the **sum** of `|nums1[i] - nums2[i]|` for each `0 <= i < n` (**0-indexed**).
You can replace **at most one** element of `nums1` with **any** other element in ... | Note that it is always more optimal to take one type of substring before another You can use a stack to handle erasures |
Python O(n) solution (greedy approach) | maximum-score-from-removing-substrings | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nif x>y try to remove all \'ab\' first then all \'ba\'. Else vice-verca. \n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\nO(... | 0 | You are given a string `s` and two integers `x` and `y`. You can perform two types of operations any number of times.
* Remove substring `"ab "` and gain `x` points.
* For example, when removing `"ab "` from `"cabxbae "` it becomes `"cxbae "`.
* Remove substring `"ba "` and gain `y` points.
* For examp... | Each point on the left would either be connected to exactly point already connected to some left node, or a subset of the nodes on the right which are not connected to any node Use dynamic programming with bitmasking, where the state will be (number of points assigned in first group, bitmask of points assigned in secon... |
Python O(n) solution (greedy approach) | maximum-score-from-removing-substrings | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nif x>y try to remove all \'ab\' first then all \'ba\'. Else vice-verca. \n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\nO(... | 0 | You are given two positive integer arrays `nums1` and `nums2`, both of length `n`.
The **absolute sum difference** of arrays `nums1` and `nums2` is defined as the **sum** of `|nums1[i] - nums2[i]|` for each `0 <= i < n` (**0-indexed**).
You can replace **at most one** element of `nums1` with **any** other element in ... | Note that it is always more optimal to take one type of substring before another You can use a stack to handle erasures |
Python 3 || 15 lines, backtrack || T/M: 77% / 78% | construct-the-lexicographically-largest-valid-sequence | 0 | 1 | ```\nclass Solution:\n def constructDistancedSequence(self, n: int) -> List[int]:\n\n def backtrack(idx1 = 0): \n\n if not unseen: return True\n\n if ans[idx1]: return backtrack(idx1+1)\n\n for num in reversed(range(1,n+1)):\n\n idx2 = idx1 + num if num != 1 els... | 3 | Given an integer `n`, find a sequence that satisfies all of the following:
* The integer `1` occurs once in the sequence.
* Each integer between `2` and `n` occurs twice in the sequence.
* For every integer `i` between `2` and `n`, the **distance** between the two occurrences of `i` is exactly `i`.
The **distan... | null |
Python 3 || 15 lines, backtrack || T/M: 77% / 78% | construct-the-lexicographically-largest-valid-sequence | 0 | 1 | ```\nclass Solution:\n def constructDistancedSequence(self, n: int) -> List[int]:\n\n def backtrack(idx1 = 0): \n\n if not unseen: return True\n\n if ans[idx1]: return backtrack(idx1+1)\n\n for num in reversed(range(1,n+1)):\n\n idx2 = idx1 + num if num != 1 els... | 3 | You are given an array `nums` that consists of positive integers.
The **GCD** of a sequence of numbers is defined as the greatest integer that divides **all** the numbers in the sequence evenly.
* For example, the GCD of the sequence `[4,6,16]` is `2`.
A **subsequence** of an array is a sequence that can be formed... | Heuristic algorithm may work. |
Python Greedy+Backtracking | Well Explained | Comments | construct-the-lexicographically-largest-valid-sequence | 0 | 1 | The array contains `2*n-1` elements since `1` occurs once and others twice.\n\nThe basic idea is, we try to put numbers into the array in a greedy way, since it requires **the only "Lexicographically Largest"** one.\nWe try numbers from `n` to `1` from beginning of the array.\nEach time we put `x` into the array at ind... | 25 | Given an integer `n`, find a sequence that satisfies all of the following:
* The integer `1` occurs once in the sequence.
* Each integer between `2` and `n` occurs twice in the sequence.
* For every integer `i` between `2` and `n`, the **distance** between the two occurrences of `i` is exactly `i`.
The **distan... | null |
Python Greedy+Backtracking | Well Explained | Comments | construct-the-lexicographically-largest-valid-sequence | 0 | 1 | The array contains `2*n-1` elements since `1` occurs once and others twice.\n\nThe basic idea is, we try to put numbers into the array in a greedy way, since it requires **the only "Lexicographically Largest"** one.\nWe try numbers from `n` to `1` from beginning of the array.\nEach time we put `x` into the array at ind... | 25 | You are given an array `nums` that consists of positive integers.
The **GCD** of a sequence of numbers is defined as the greatest integer that divides **all** the numbers in the sequence evenly.
* For example, the GCD of the sequence `[4,6,16]` is `2`.
A **subsequence** of an array is a sequence that can be formed... | Heuristic algorithm may work. |
Python3 solution: ternary operators | construct-the-lexicographically-largest-valid-sequence | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nPut the integers in from left to right, and from large to small. Whenver we can fill the array, we have the answer\n# Approach\n<!-- Describe your approach to solving the problem. -->\nStandard backtracking. The answer is found whenever w... | 0 | Given an integer `n`, find a sequence that satisfies all of the following:
* The integer `1` occurs once in the sequence.
* Each integer between `2` and `n` occurs twice in the sequence.
* For every integer `i` between `2` and `n`, the **distance** between the two occurrences of `i` is exactly `i`.
The **distan... | null |
Python3 solution: ternary operators | construct-the-lexicographically-largest-valid-sequence | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nPut the integers in from left to right, and from large to small. Whenver we can fill the array, we have the answer\n# Approach\n<!-- Describe your approach to solving the problem. -->\nStandard backtracking. The answer is found whenever w... | 0 | You are given an array `nums` that consists of positive integers.
The **GCD** of a sequence of numbers is defined as the greatest integer that divides **all** the numbers in the sequence evenly.
* For example, the GCD of the sequence `[4,6,16]` is `2`.
A **subsequence** of an array is a sequence that can be formed... | Heuristic algorithm may work. |
Easy Python Solution Using Recursion | construct-the-lexicographically-largest-valid-sequence | 0 | 1 | # Code\n```\nclass Solution:\n def constructDistancedSequence(self, n: int) -> List[int]:\n arr=[-1 for _ in range(2*n-1)]\n status=self.get(arr,0,0,n)\n return arr\n def get(self,arr,i,bit,n):\n if(i>=len(arr)):\n return True\n elif(arr[i]!=-1):\n return s... | 0 | Given an integer `n`, find a sequence that satisfies all of the following:
* The integer `1` occurs once in the sequence.
* Each integer between `2` and `n` occurs twice in the sequence.
* For every integer `i` between `2` and `n`, the **distance** between the two occurrences of `i` is exactly `i`.
The **distan... | null |
Easy Python Solution Using Recursion | construct-the-lexicographically-largest-valid-sequence | 0 | 1 | # Code\n```\nclass Solution:\n def constructDistancedSequence(self, n: int) -> List[int]:\n arr=[-1 for _ in range(2*n-1)]\n status=self.get(arr,0,0,n)\n return arr\n def get(self,arr,i,bit,n):\n if(i>=len(arr)):\n return True\n elif(arr[i]!=-1):\n return s... | 0 | You are given an array `nums` that consists of positive integers.
The **GCD** of a sequence of numbers is defined as the greatest integer that divides **all** the numbers in the sequence evenly.
* For example, the GCD of the sequence `[4,6,16]` is `2`.
A **subsequence** of an array is a sequence that can be formed... | Heuristic algorithm may work. |
Python | Backtracking | Clean Code | construct-the-lexicographically-largest-valid-sequence | 0 | 1 | # Code\n```\nclass Solution:\n def constructDistancedSequence(self, n: int) -> List[int]:\n m = 2*n - 1\n A = [0]*m\n path = set()\n def dfs(i):\n if i == m:\n return all(A)\n if A[i]:\n return dfs(i+1)\n for v in range(n, 0, ... | 0 | Given an integer `n`, find a sequence that satisfies all of the following:
* The integer `1` occurs once in the sequence.
* Each integer between `2` and `n` occurs twice in the sequence.
* For every integer `i` between `2` and `n`, the **distance** between the two occurrences of `i` is exactly `i`.
The **distan... | null |
Python | Backtracking | Clean Code | construct-the-lexicographically-largest-valid-sequence | 0 | 1 | # Code\n```\nclass Solution:\n def constructDistancedSequence(self, n: int) -> List[int]:\n m = 2*n - 1\n A = [0]*m\n path = set()\n def dfs(i):\n if i == m:\n return all(A)\n if A[i]:\n return dfs(i+1)\n for v in range(n, 0, ... | 0 | You are given an array `nums` that consists of positive integers.
The **GCD** of a sequence of numbers is defined as the greatest integer that divides **all** the numbers in the sequence evenly.
* For example, the GCD of the sequence `[4,6,16]` is `2`.
A **subsequence** of an array is a sequence that can be formed... | Heuristic algorithm may work. |
Python3 DFS w/ backtracking | construct-the-lexicographically-largest-valid-sequence | 0 | 1 | # Intuition\nWe start with an array of all 0s of length $$2n - 1$$. Using depth first search we place the higher number possible in the first cell in the array that is a 0. We repeat this process until we\'ve either found a solution or there are no more possible placements, in which case we backtrack to the next possib... | 0 | Given an integer `n`, find a sequence that satisfies all of the following:
* The integer `1` occurs once in the sequence.
* Each integer between `2` and `n` occurs twice in the sequence.
* For every integer `i` between `2` and `n`, the **distance** between the two occurrences of `i` is exactly `i`.
The **distan... | null |
Python3 DFS w/ backtracking | construct-the-lexicographically-largest-valid-sequence | 0 | 1 | # Intuition\nWe start with an array of all 0s of length $$2n - 1$$. Using depth first search we place the higher number possible in the first cell in the array that is a 0. We repeat this process until we\'ve either found a solution or there are no more possible placements, in which case we backtrack to the next possib... | 0 | You are given an array `nums` that consists of positive integers.
The **GCD** of a sequence of numbers is defined as the greatest integer that divides **all** the numbers in the sequence evenly.
* For example, the GCD of the sequence `[4,6,16]` is `2`.
A **subsequence** of an array is a sequence that can be formed... | Heuristic algorithm may work. |
Python Backtracking Solution, Faster than 91% | construct-the-lexicographically-largest-valid-sequence | 0 | 1 | ```\nclass Solution:\n def constructDistancedSequence(self, n: int) -> List[int]:\n ans, visited = [0] * (2 * n - 1), [False] * (n + 1)\n \n def backtrack(idx: int) -> bool:\n nonlocal ans, visited, n\n if idx == len(ans):\n return True\n if ans[id... | 0 | Given an integer `n`, find a sequence that satisfies all of the following:
* The integer `1` occurs once in the sequence.
* Each integer between `2` and `n` occurs twice in the sequence.
* For every integer `i` between `2` and `n`, the **distance** between the two occurrences of `i` is exactly `i`.
The **distan... | null |
Python Backtracking Solution, Faster than 91% | construct-the-lexicographically-largest-valid-sequence | 0 | 1 | ```\nclass Solution:\n def constructDistancedSequence(self, n: int) -> List[int]:\n ans, visited = [0] * (2 * n - 1), [False] * (n + 1)\n \n def backtrack(idx: int) -> bool:\n nonlocal ans, visited, n\n if idx == len(ans):\n return True\n if ans[id... | 0 | You are given an array `nums` that consists of positive integers.
The **GCD** of a sequence of numbers is defined as the greatest integer that divides **all** the numbers in the sequence evenly.
* For example, the GCD of the sequence `[4,6,16]` is `2`.
A **subsequence** of an array is a sequence that can be formed... | Heuristic algorithm may work. |
[Python3] backtracking | construct-the-lexicographically-largest-valid-sequence | 0 | 1 | **Algo**\nThis a typical backtracking problem. Essentially, we have an array of `2*n-1` positions for which we progressively fill in numbers of `1, ..., n` of which `1` has multiplicity 1 and the others have 2. \n\nHere the strategy is that we aggresively fill in large numbers in high position if possible. \n\n**Implem... | 2 | Given an integer `n`, find a sequence that satisfies all of the following:
* The integer `1` occurs once in the sequence.
* Each integer between `2` and `n` occurs twice in the sequence.
* For every integer `i` between `2` and `n`, the **distance** between the two occurrences of `i` is exactly `i`.
The **distan... | null |
[Python3] backtracking | construct-the-lexicographically-largest-valid-sequence | 0 | 1 | **Algo**\nThis a typical backtracking problem. Essentially, we have an array of `2*n-1` positions for which we progressively fill in numbers of `1, ..., n` of which `1` has multiplicity 1 and the others have 2. \n\nHere the strategy is that we aggresively fill in large numbers in high position if possible. \n\n**Implem... | 2 | You are given an array `nums` that consists of positive integers.
The **GCD** of a sequence of numbers is defined as the greatest integer that divides **all** the numbers in the sequence evenly.
* For example, the GCD of the sequence `[4,6,16]` is `2`.
A **subsequence** of an array is a sequence that can be formed... | Heuristic algorithm may work. |
Python | DFS | Backtracking | construct-the-lexicographically-largest-valid-sequence | 0 | 1 | ```\nclass Solution:\n def constructDistancedSequence(self, n: int) -> List[int]:\n arr=[0]*((n*2)-1)\n self.ans=None\n def dfs(arr,i):\n for num in range(n,0,-1):#Traverse Reverse as we need highest lexicographic number\n if num not in arr:\n if num=... | 1 | Given an integer `n`, find a sequence that satisfies all of the following:
* The integer `1` occurs once in the sequence.
* Each integer between `2` and `n` occurs twice in the sequence.
* For every integer `i` between `2` and `n`, the **distance** between the two occurrences of `i` is exactly `i`.
The **distan... | null |
Python | DFS | Backtracking | construct-the-lexicographically-largest-valid-sequence | 0 | 1 | ```\nclass Solution:\n def constructDistancedSequence(self, n: int) -> List[int]:\n arr=[0]*((n*2)-1)\n self.ans=None\n def dfs(arr,i):\n for num in range(n,0,-1):#Traverse Reverse as we need highest lexicographic number\n if num not in arr:\n if num=... | 1 | You are given an array `nums` that consists of positive integers.
The **GCD** of a sequence of numbers is defined as the greatest integer that divides **all** the numbers in the sequence evenly.
* For example, the GCD of the sequence `[4,6,16]` is `2`.
A **subsequence** of an array is a sequence that can be formed... | Heuristic algorithm may work. |
[Python3] greedy | number-of-ways-to-reconstruct-a-tree | 0 | 1 | \n```\nclass Solution:\n def checkWays(self, pairs: List[List[int]]) -> int:\n graph = {}\n for x, y in pairs: \n graph.setdefault(x, set()).add(y)\n graph.setdefault(y, set()).add(x)\n \n ans = 1 \n ancestors = set()\n for n in sorted(graph, key=lambda... | 5 | 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. |
[Python3] greedy | number-of-ways-to-reconstruct-a-tree | 0 | 1 | \n```\nclass Solution:\n def checkWays(self, pairs: List[List[int]]) -> int:\n graph = {}\n for x, y in pairs: \n graph.setdefault(x, set()).add(y)\n graph.setdefault(y, set()).add(x)\n \n ans = 1 \n ancestors = set()\n for n in sorted(graph, key=lambda... | 5 | 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... |
SOLID Principles | Abstracting the DSU possibilities | Commented and Explained | 100% run time | number-of-ways-to-reconstruct-a-tree | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nThis question is basically asking for a subset of the disjointed set union of the nodes in the graph to be a tree. This can be abstracted out by simple comparisons of the number of neighbors, and whether or not a specific node has neighbo... | 1 | 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. |
SOLID Principles | Abstracting the DSU possibilities | Commented and Explained | 100% run time | number-of-ways-to-reconstruct-a-tree | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nThis question is basically asking for a subset of the disjointed set union of the nodes in the graph to be a tree. This can be abstracted out by simple comparisons of the number of neighbors, and whether or not a specific node has neighbo... | 1 | 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... |
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