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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[Python] 4 ways to attack | goal-parser-interpretation | 0 | 1 | Just while loop with carefully incrementing counter is good enough:\n```\n# runtime beats 32% ; memory beats 99.76%\ndef interpret(command):\n \n return_string = ""\n i = 0\n while i < len(command):\n if command[i] == "G":\n return_string += \'G\'\n i += 1\n elif command[i:i + 2] == \'()\':\n r... | 38 | There is an authentication system that works with authentication tokens. For each session, the user will receive a new authentication token that will expire `timeToLive` seconds after the `currentTime`. If the token is renewed, the expiry time will be **extended** to expire `timeToLive` seconds after the (potentially d... | You need to check at most 2 characters to determine which character comes next. |
Python Two pointer well explain solution quick and still mantain O(1) space complexity | max-number-of-k-sum-pairs | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nThe intuition behind this solution is to use a `two-pointer` approach on a sorted array. Sorting the array allows us to efficiently find pairs of numbers that sum up to the target value `k`. The two pointers (`left` and `right`) are initi... | 10 | You are given an integer array `nums` and an integer `k`.
In one operation, you can pick two numbers from the array whose sum equals `k` and remove them from the array.
Return _the maximum number of operations you can perform on the array_.
**Example 1:**
**Input:** nums = \[1,2,3,4\], k = 5
**Output:** 2
**Explana... | The key is to find the longest non-decreasing subarray starting with the first element or ending with the last element, respectively. After removing some subarray, the result is the concatenation of a sorted prefix and a sorted suffix, where the last element of the prefix is smaller than the first element of the suffix... |
Python Two pointer well explain solution quick and still mantain O(1) space complexity | max-number-of-k-sum-pairs | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nThe intuition behind this solution is to use a `two-pointer` approach on a sorted array. Sorting the array allows us to efficiently find pairs of numbers that sum up to the target value `k`. The two pointers (`left` and `right`) are initi... | 10 | You are given an integer array `coins` of length `n` which represents the `n` coins that you own. The value of the `ith` coin is `coins[i]`. You can **make** some value `x` if you can choose some of your `n` coins such that their values sum up to `x`.
Return the _maximum number of consecutive integer values that you *... | The abstract problem asks to count the number of disjoint pairs with a given sum k. For each possible value x, it can be paired up with k - x. The number of such pairs equals to min(count(x), count(k-x)), unless that x = k / 2, where the number of such pairs will be floor(count(x) / 2). |
[Python 3] Two solution with Counter and two points || beats 99% | max-number-of-k-sum-pairs | 0 | 1 | ```python3 []\nclass Solution:\n def maxOperations(self, nums: List[int], k: int) -> int:\n res, d = 0, Counter(nums)\n for val1, cnt in d.items():\n val2 = k - val1\n # it\'s trick to get rid of duplicates pairs, consider only pairs where val1 >= val2\n if val2 < val1 ... | 16 | You are given an integer array `nums` and an integer `k`.
In one operation, you can pick two numbers from the array whose sum equals `k` and remove them from the array.
Return _the maximum number of operations you can perform on the array_.
**Example 1:**
**Input:** nums = \[1,2,3,4\], k = 5
**Output:** 2
**Explana... | The key is to find the longest non-decreasing subarray starting with the first element or ending with the last element, respectively. After removing some subarray, the result is the concatenation of a sorted prefix and a sorted suffix, where the last element of the prefix is smaller than the first element of the suffix... |
[Python 3] Two solution with Counter and two points || beats 99% | max-number-of-k-sum-pairs | 0 | 1 | ```python3 []\nclass Solution:\n def maxOperations(self, nums: List[int], k: int) -> int:\n res, d = 0, Counter(nums)\n for val1, cnt in d.items():\n val2 = k - val1\n # it\'s trick to get rid of duplicates pairs, consider only pairs where val1 >= val2\n if val2 < val1 ... | 16 | You are given an integer array `coins` of length `n` which represents the `n` coins that you own. The value of the `ith` coin is `coins[i]`. You can **make** some value `x` if you can choose some of your `n` coins such that their values sum up to `x`.
Return the _maximum number of consecutive integer values that you *... | The abstract problem asks to count the number of disjoint pairs with a given sum k. For each possible value x, it can be paired up with k - x. The number of such pairs equals to min(count(x), count(k-x)), unless that x = k / 2, where the number of such pairs will be floor(count(x) / 2). |
99% CPU, 92% RAM:🐱🚀 Suboptimal but Purr-fectly Playful: Sort + Binary Search 🚀 | max-number-of-k-sum-pairs | 0 | 1 | # \uD83D\uDC31 Cat-tastic Intuition \uD83D\uDC31\nWhy chase the optimal mouse when we can have some fun with a playful, yet speedy, kitty solution? Let\'s use some purr-fect library functions to give us a little boost!\n\n# \uD83D\uDC3E Feline Approach \uD83D\uDC3E\nSure, sorting is like the classic cat nap \u2013 ever... | 7 | You are given an integer array `nums` and an integer `k`.
In one operation, you can pick two numbers from the array whose sum equals `k` and remove them from the array.
Return _the maximum number of operations you can perform on the array_.
**Example 1:**
**Input:** nums = \[1,2,3,4\], k = 5
**Output:** 2
**Explana... | The key is to find the longest non-decreasing subarray starting with the first element or ending with the last element, respectively. After removing some subarray, the result is the concatenation of a sorted prefix and a sorted suffix, where the last element of the prefix is smaller than the first element of the suffix... |
99% CPU, 92% RAM:🐱🚀 Suboptimal but Purr-fectly Playful: Sort + Binary Search 🚀 | max-number-of-k-sum-pairs | 0 | 1 | # \uD83D\uDC31 Cat-tastic Intuition \uD83D\uDC31\nWhy chase the optimal mouse when we can have some fun with a playful, yet speedy, kitty solution? Let\'s use some purr-fect library functions to give us a little boost!\n\n# \uD83D\uDC3E Feline Approach \uD83D\uDC3E\nSure, sorting is like the classic cat nap \u2013 ever... | 7 | You are given an integer array `coins` of length `n` which represents the `n` coins that you own. The value of the `ith` coin is `coins[i]`. You can **make** some value `x` if you can choose some of your `n` coins such that their values sum up to `x`.
Return the _maximum number of consecutive integer values that you *... | The abstract problem asks to count the number of disjoint pairs with a given sum k. For each possible value x, it can be paired up with k - x. The number of such pairs equals to min(count(x), count(k-x)), unless that x = k / 2, where the number of such pairs will be floor(count(x) / 2). |
Easy HashMap Solution : Java 🍀 | max-number-of-k-sum-pairs | 1 | 1 | # Easy HashMap Solution : \n# Java Code : \n```\nclass Solution {\n public int maxOperations(int[] nums, int target) {\n HashMap<Integer,Integer> hmap = new HashMap<>();\n int count = 0;\n for(int i = 0;i<nums.length;i++)\n {\n if(hmap.containsKey(target-nums[i]) == true)\n ... | 10 | You are given an integer array `nums` and an integer `k`.
In one operation, you can pick two numbers from the array whose sum equals `k` and remove them from the array.
Return _the maximum number of operations you can perform on the array_.
**Example 1:**
**Input:** nums = \[1,2,3,4\], k = 5
**Output:** 2
**Explana... | The key is to find the longest non-decreasing subarray starting with the first element or ending with the last element, respectively. After removing some subarray, the result is the concatenation of a sorted prefix and a sorted suffix, where the last element of the prefix is smaller than the first element of the suffix... |
Easy HashMap Solution : Java 🍀 | max-number-of-k-sum-pairs | 1 | 1 | # Easy HashMap Solution : \n# Java Code : \n```\nclass Solution {\n public int maxOperations(int[] nums, int target) {\n HashMap<Integer,Integer> hmap = new HashMap<>();\n int count = 0;\n for(int i = 0;i<nums.length;i++)\n {\n if(hmap.containsKey(target-nums[i]) == true)\n ... | 10 | You are given an integer array `coins` of length `n` which represents the `n` coins that you own. The value of the `ith` coin is `coins[i]`. You can **make** some value `x` if you can choose some of your `n` coins such that their values sum up to `x`.
Return the _maximum number of consecutive integer values that you *... | The abstract problem asks to count the number of disjoint pairs with a given sum k. For each possible value x, it can be paired up with k - x. The number of such pairs equals to min(count(x), count(k-x)), unless that x = k / 2, where the number of such pairs will be floor(count(x) / 2). |
Using Python Dictionary | max-number-of-k-sum-pairs | 0 | 1 | # Code\n```\nclass Solution:\n def maxOperations(self, nums: List[int], k: int) -> int:\n\n d = {}\n count = 0\n\n for x in nums:\n if d.get(k - x, 0) > 0:\n count+=1\n d[k-x]-=1\n else:\n d[x] = d.get(x, 0) + 1\n\n return... | 7 | You are given an integer array `nums` and an integer `k`.
In one operation, you can pick two numbers from the array whose sum equals `k` and remove them from the array.
Return _the maximum number of operations you can perform on the array_.
**Example 1:**
**Input:** nums = \[1,2,3,4\], k = 5
**Output:** 2
**Explana... | The key is to find the longest non-decreasing subarray starting with the first element or ending with the last element, respectively. After removing some subarray, the result is the concatenation of a sorted prefix and a sorted suffix, where the last element of the prefix is smaller than the first element of the suffix... |
Using Python Dictionary | max-number-of-k-sum-pairs | 0 | 1 | # Code\n```\nclass Solution:\n def maxOperations(self, nums: List[int], k: int) -> int:\n\n d = {}\n count = 0\n\n for x in nums:\n if d.get(k - x, 0) > 0:\n count+=1\n d[k-x]-=1\n else:\n d[x] = d.get(x, 0) + 1\n\n return... | 7 | You are given an integer array `coins` of length `n` which represents the `n` coins that you own. The value of the `ith` coin is `coins[i]`. You can **make** some value `x` if you can choose some of your `n` coins such that their values sum up to `x`.
Return the _maximum number of consecutive integer values that you *... | The abstract problem asks to count the number of disjoint pairs with a given sum k. For each possible value x, it can be paired up with k - x. The number of such pairs equals to min(count(x), count(k-x)), unless that x = k / 2, where the number of such pairs will be floor(count(x) / 2). |
Simplest Python solution | concatenation-of-consecutive-binary-numbers | 0 | 1 | \n\n# Code\n```\nclass Solution:\n def concatenatedBinary(self, n: int) -> int:\n res = ""\n for x in range(1, n+1):\n res += bin(x)[2:]\n return int(res, 2) % (10**9+7)\n``` | 1 | Given an integer `n`, return _the **decimal value** of the binary string formed by concatenating the binary representations of_ `1` _to_ `n` _in order, **modulo**_ `109 + 7`.
**Example 1:**
**Input:** n = 1
**Output:** 1
**Explanation: ** "1 " in binary corresponds to the decimal value 1.
**Example 2:**
**Input:**... | Use dynamic programming to solve this problem with each state defined by the city index and fuel left. Since the array contains distinct integers fuel will always be spent in each move and so there can be no cycles. |
Simplest Python solution | concatenation-of-consecutive-binary-numbers | 0 | 1 | \n\n# Code\n```\nclass Solution:\n def concatenatedBinary(self, n: int) -> int:\n res = ""\n for x in range(1, n+1):\n res += bin(x)[2:]\n return int(res, 2) % (10**9+7)\n``` | 1 | Given an array of positive integers `nums`, return the _maximum possible sum of an **ascending** subarray in_ `nums`.
A subarray is defined as a contiguous sequence of numbers in an array.
A subarray `[numsl, numsl+1, ..., numsr-1, numsr]` is **ascending** if for all `i` where `l <= i < r`, `numsi < numsi+1`. Note th... | Express the nth number value in a recursion formula and think about how we can do a fast evaluation. |
🔥 [LeetCode The Hard Way] 🔥 Explained Line By Line | concatenation-of-consecutive-binary-numbers | 1 | 1 | Please check out [LeetCode The Hard Way](https://wingkwong.github.io/leetcode-the-hard-way/) for more solution explanations and tutorials. \nI\'ll explain my solution line by line daily and you can find the full list in my [Discord](https://discord.gg/Nqm4jJcyBf).\nIf you like it, please give a star, watch my [Github R... | 71 | Given an integer `n`, return _the **decimal value** of the binary string formed by concatenating the binary representations of_ `1` _to_ `n` _in order, **modulo**_ `109 + 7`.
**Example 1:**
**Input:** n = 1
**Output:** 1
**Explanation: ** "1 " in binary corresponds to the decimal value 1.
**Example 2:**
**Input:**... | Use dynamic programming to solve this problem with each state defined by the city index and fuel left. Since the array contains distinct integers fuel will always be spent in each move and so there can be no cycles. |
🔥 [LeetCode The Hard Way] 🔥 Explained Line By Line | concatenation-of-consecutive-binary-numbers | 1 | 1 | Please check out [LeetCode The Hard Way](https://wingkwong.github.io/leetcode-the-hard-way/) for more solution explanations and tutorials. \nI\'ll explain my solution line by line daily and you can find the full list in my [Discord](https://discord.gg/Nqm4jJcyBf).\nIf you like it, please give a star, watch my [Github R... | 71 | Given an array of positive integers `nums`, return the _maximum possible sum of an **ascending** subarray in_ `nums`.
A subarray is defined as a contiguous sequence of numbers in an array.
A subarray `[numsl, numsl+1, ..., numsr-1, numsr]` is **ascending** if for all `i` where `l <= i < r`, `numsi < numsi+1`. Note th... | Express the nth number value in a recursion formula and think about how we can do a fast evaluation. |
[ Python ] ✅✅ Simple Python Solution Using Bin Function 🥳✌👍 | concatenation-of-consecutive-binary-numbers | 0 | 1 | # If You like the Solution, Don\'t Forget To UpVote Me, Please UpVote! \uD83D\uDD3C\uD83D\uDE4F\n# Runtime: 3110 ms, faster than 38.58% of Python3 online submissions for Concatenation of Consecutive Binary Numbers.\n# Memory Usage: 15.9 MB, less than 22.05% of Python3 online submissions for Concatenation of Consecutive... | 3 | Given an integer `n`, return _the **decimal value** of the binary string formed by concatenating the binary representations of_ `1` _to_ `n` _in order, **modulo**_ `109 + 7`.
**Example 1:**
**Input:** n = 1
**Output:** 1
**Explanation: ** "1 " in binary corresponds to the decimal value 1.
**Example 2:**
**Input:**... | Use dynamic programming to solve this problem with each state defined by the city index and fuel left. Since the array contains distinct integers fuel will always be spent in each move and so there can be no cycles. |
[ Python ] ✅✅ Simple Python Solution Using Bin Function 🥳✌👍 | concatenation-of-consecutive-binary-numbers | 0 | 1 | # If You like the Solution, Don\'t Forget To UpVote Me, Please UpVote! \uD83D\uDD3C\uD83D\uDE4F\n# Runtime: 3110 ms, faster than 38.58% of Python3 online submissions for Concatenation of Consecutive Binary Numbers.\n# Memory Usage: 15.9 MB, less than 22.05% of Python3 online submissions for Concatenation of Consecutive... | 3 | Given an array of positive integers `nums`, return the _maximum possible sum of an **ascending** subarray in_ `nums`.
A subarray is defined as a contiguous sequence of numbers in an array.
A subarray `[numsl, numsl+1, ..., numsr-1, numsr]` is **ascending** if for all `i` where `l <= i < r`, `numsi < numsi+1`. Note th... | Express the nth number value in a recursion formula and think about how we can do a fast evaluation. |
✅✔ SIMPLE PYTHON3 SOLUTION ✅✔ O)n log n ) time | concatenation-of-consecutive-binary-numbers | 0 | 1 | ***UPVOTE*** if it is helpfull\n```\n\nclass Solution:\n def countBits(self,x):\n ab = 0\n while x:\n x >>= 1\n ab += 1\n return ab\n def concatenatedBinary(self, n: int) -> int:\n res = 0\n mod = 10**9 + 7\n for i in range(1, n+1):\n res ... | 2 | Given an integer `n`, return _the **decimal value** of the binary string formed by concatenating the binary representations of_ `1` _to_ `n` _in order, **modulo**_ `109 + 7`.
**Example 1:**
**Input:** n = 1
**Output:** 1
**Explanation: ** "1 " in binary corresponds to the decimal value 1.
**Example 2:**
**Input:**... | Use dynamic programming to solve this problem with each state defined by the city index and fuel left. Since the array contains distinct integers fuel will always be spent in each move and so there can be no cycles. |
✅✔ SIMPLE PYTHON3 SOLUTION ✅✔ O)n log n ) time | concatenation-of-consecutive-binary-numbers | 0 | 1 | ***UPVOTE*** if it is helpfull\n```\n\nclass Solution:\n def countBits(self,x):\n ab = 0\n while x:\n x >>= 1\n ab += 1\n return ab\n def concatenatedBinary(self, n: int) -> int:\n res = 0\n mod = 10**9 + 7\n for i in range(1, n+1):\n res ... | 2 | Given an array of positive integers `nums`, return the _maximum possible sum of an **ascending** subarray in_ `nums`.
A subarray is defined as a contiguous sequence of numbers in an array.
A subarray `[numsl, numsl+1, ..., numsr-1, numsr]` is **ascending** if for all `i` where `l <= i < r`, `numsi < numsi+1`. Note th... | Express the nth number value in a recursion formula and think about how we can do a fast evaluation. |
Python Elegant & Short | One line | Reducing | concatenation-of-consecutive-binary-numbers | 0 | 1 | ```\nclass Solution:\n """\n Time: O(n*log(n))\n Memory: O(1)\n """\n\n MOD = 10 ** 9 + 7\n\n def concatenatedBinary(self, n: int) -> int:\n return reduce(lambda x, y: ((x << y.bit_length()) | y) % self.MOD, range(1, n + 1))\n```\n\nIf you like this solution remember to **upvote it** to let m... | 2 | Given an integer `n`, return _the **decimal value** of the binary string formed by concatenating the binary representations of_ `1` _to_ `n` _in order, **modulo**_ `109 + 7`.
**Example 1:**
**Input:** n = 1
**Output:** 1
**Explanation: ** "1 " in binary corresponds to the decimal value 1.
**Example 2:**
**Input:**... | Use dynamic programming to solve this problem with each state defined by the city index and fuel left. Since the array contains distinct integers fuel will always be spent in each move and so there can be no cycles. |
Python Elegant & Short | One line | Reducing | concatenation-of-consecutive-binary-numbers | 0 | 1 | ```\nclass Solution:\n """\n Time: O(n*log(n))\n Memory: O(1)\n """\n\n MOD = 10 ** 9 + 7\n\n def concatenatedBinary(self, n: int) -> int:\n return reduce(lambda x, y: ((x << y.bit_length()) | y) % self.MOD, range(1, n + 1))\n```\n\nIf you like this solution remember to **upvote it** to let m... | 2 | Given an array of positive integers `nums`, return the _maximum possible sum of an **ascending** subarray in_ `nums`.
A subarray is defined as a contiguous sequence of numbers in an array.
A subarray `[numsl, numsl+1, ..., numsr-1, numsr]` is **ascending** if for all `i` where `l <= i < r`, `numsi < numsi+1`. Note th... | Express the nth number value in a recursion formula and think about how we can do a fast evaluation. |
💚One-Liner explanation and code || Python3 Solution and Explanation | concatenation-of-consecutive-binary-numbers | 0 | 1 | ##### Just in one-line, the answer is completed\uD83D\uDE01\n#### **Explanation:**\nAll we have to do is converting every number in the range of 1 to n using bin() function and make them as a list\n```[bin(i) for i in range(1,n+1)]```\n\nIn that, we have to remove the first 2 charfacters because every bin conversion wi... | 1 | Given an integer `n`, return _the **decimal value** of the binary string formed by concatenating the binary representations of_ `1` _to_ `n` _in order, **modulo**_ `109 + 7`.
**Example 1:**
**Input:** n = 1
**Output:** 1
**Explanation: ** "1 " in binary corresponds to the decimal value 1.
**Example 2:**
**Input:**... | Use dynamic programming to solve this problem with each state defined by the city index and fuel left. Since the array contains distinct integers fuel will always be spent in each move and so there can be no cycles. |
💚One-Liner explanation and code || Python3 Solution and Explanation | concatenation-of-consecutive-binary-numbers | 0 | 1 | ##### Just in one-line, the answer is completed\uD83D\uDE01\n#### **Explanation:**\nAll we have to do is converting every number in the range of 1 to n using bin() function and make them as a list\n```[bin(i) for i in range(1,n+1)]```\n\nIn that, we have to remove the first 2 charfacters because every bin conversion wi... | 1 | Given an array of positive integers `nums`, return the _maximum possible sum of an **ascending** subarray in_ `nums`.
A subarray is defined as a contiguous sequence of numbers in an array.
A subarray `[numsl, numsl+1, ..., numsr-1, numsr]` is **ascending** if for all `i` where `l <= i < r`, `numsi < numsi+1`. Note th... | Express the nth number value in a recursion formula and think about how we can do a fast evaluation. |
Python (Simple DP) | minimum-incompatibility | 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 `nums`, an array of positive integers of size `2 * n`. You must perform `n` operations on this array.
In the `ith` operation **(1-indexed)**, you will:
* Choose two elements, `x` and `y`.
* Receive a score of `i * gcd(x, y)`.
* Remove `x` and `y` from `nums`.
Return _the maximum score you can rec... | The constraints are small enough for a backtrack solution but not any backtrack solution If we use a naive n^k don't you think it can be optimized |
[Python] simple backtracking beats 100% | minimum-incompatibility | 0 | 1 | \n```python\nclass Solution(object):\n def minimumIncompatibility(self, nums, k):\n nums.sort(reverse = True)\n upperbound = len(nums) // k\n arr = [[] for _ in range(k)]\n self.r... | 8 | You are given `nums`, an array of positive integers of size `2 * n`. You must perform `n` operations on this array.
In the `ith` operation **(1-indexed)**, you will:
* Choose two elements, `x` and `y`.
* Receive a score of `i * gcd(x, y)`.
* Remove `x` and `y` from `nums`.
Return _the maximum score you can rec... | The constraints are small enough for a backtrack solution but not any backtrack solution If we use a naive n^k don't you think it can be optimized |
python3 dfs backtrack with pruning | minimum-incompatibility | 0 | 1 | Easy to think of backtrack, but pruning is crucial to avoid TLE. Two pruning steps are adopted: first cut all branches with incompatibility larger than already known minimum; second we consider only empty buckets two remove the symmetry between buckets.\n\n# Code\n```\nclass Solution:\n def minimumIncompatibility(se... | 0 | You are given `nums`, an array of positive integers of size `2 * n`. You must perform `n` operations on this array.
In the `ith` operation **(1-indexed)**, you will:
* Choose two elements, `x` and `y`.
* Receive a score of `i * gcd(x, y)`.
* Remove `x` and `y` from `nums`.
Return _the maximum score you can rec... | The constraints are small enough for a backtrack solution but not any backtrack solution If we use a naive n^k don't you think it can be optimized |
Bitmask DP - [33%, 6%] | minimum-incompatibility | 0 | 1 | # Complexity\nTime complexity: $O(2^n + 2^{n/k})$\nSpace complexity: $O(2^n + 2^{n/k})$\n\n# Code\n```python\nclass Solution:\n def minimumIncompatibility(self, A: List[int], K: int) -> int:\n S = len(A) // K\n # pre-calc the bitmasks\n M = reduce(lambda M, i: {\n n | (1 << i) for... | 0 | You are given `nums`, an array of positive integers of size `2 * n`. You must perform `n` operations on this array.
In the `ith` operation **(1-indexed)**, you will:
* Choose two elements, `x` and `y`.
* Receive a score of `i * gcd(x, y)`.
* Remove `x` and `y` from `nums`.
Return _the maximum score you can rec... | The constraints are small enough for a backtrack solution but not any backtrack solution If we use a naive n^k don't you think it can be optimized |
[Python3] backtracking | minimum-incompatibility | 0 | 1 | **Algo**\nThis is an organized brute force approach. We simply put the elements following the rule into stacks and compute incompatibility. Here, the two tricks are \n1) `if cand + len(nums) - i > ans: return` meaning that if there is no chance for the current run to get smaller incompability we simply return;\n2) `len... | 4 | You are given `nums`, an array of positive integers of size `2 * n`. You must perform `n` operations on this array.
In the `ith` operation **(1-indexed)**, you will:
* Choose two elements, `x` and `y`.
* Receive a score of `i * gcd(x, y)`.
* Remove `x` and `y` from `nums`.
Return _the maximum score you can rec... | The constraints are small enough for a backtrack solution but not any backtrack solution If we use a naive n^k don't you think it can be optimized |
[Python] Easy-Understanding DFS with Memo | minimum-incompatibility | 0 | 1 | **Idea:**\nLet\'s assume the problem is for dictionary `d,` we need to divide them into `k` groups with each group size as `count`, we can pick a group of numbers first and transform the problem to a sub-problem of `(new_d, k - 1, count)`\n**Implementation**\nUse `frozenset` to hash\n```\nimport collections\nclass Solu... | 2 | You are given `nums`, an array of positive integers of size `2 * n`. You must perform `n` operations on this array.
In the `ith` operation **(1-indexed)**, you will:
* Choose two elements, `x` and `y`.
* Receive a score of `i * gcd(x, y)`.
* Remove `x` and `y` from `nums`.
Return _the maximum score you can rec... | The constraints are small enough for a backtrack solution but not any backtrack solution If we use a naive n^k don't you think it can be optimized |
Backtracking + smart optimizations = crazy performance (31ms for Python) | minimum-incompatibility | 0 | 1 | # Intuition\n\nThe base idea is backtracking. All possible combinations of subsets are checked using a recursion.\n\nIdeas used to implement major backtracking optimizations:\n\n1. Backtrack greedily: try combinations with close (to each other) numbers first.\n2. Exit the current backtracking branch as soon as it\'s cl... | 0 | You are given `nums`, an array of positive integers of size `2 * n`. You must perform `n` operations on this array.
In the `ith` operation **(1-indexed)**, you will:
* Choose two elements, `x` and `y`.
* Receive a score of `i * gcd(x, y)`.
* Remove `x` and `y` from `nums`.
Return _the maximum score you can rec... | The constraints are small enough for a backtrack solution but not any backtrack solution If we use a naive n^k don't you think it can be optimized |
Pandas vs SQL | Elegant & Short | All 30 Days of Pandas solutions ✅ | invalid-tweets | 0 | 1 | # Complexity\n- Time complexity: $$O(n)$$\n- Space complexity: $$O(n)$$\n\n# Code\n```Python []\ndef invalid_tweets(tweets: pd.DataFrame) -> pd.DataFrame:\n return tweets[\n tweets[\'content\'].str.len() > 15\n ][[\'tweet_id\']]\n```\n```SQL []\nSELECT tweet_id\n FROM Tweets\n WHERE length(content) >= 16;... | 16 | You are given an integer array `nums` (**0-indexed**). In one operation, you can choose an element of the array and increment it by `1`.
* For example, if `nums = [1,2,3]`, you can choose to increment `nums[1]` to make `nums = [1,**3**,3]`.
Return _the **minimum** number of operations needed to make_ `nums` _**stri... | null |
🔥Simple using Set, Subset🔥 | count-the-number-of-consistent-strings | 0 | 1 | # Approach\n\n1. Given a list of words and a string allowed containing characters that are considered allowed.\n2. Utilize a list comprehension to iterate through each word x in the words list.\n3. For each word, create a set of its characters (`set(x)`) and check if this set is a ***subset*** of the set of allowed cha... | 2 | You are given a string `allowed` consisting of **distinct** characters and an array of strings `words`. A string is **consistent** if all characters in the string appear in the string `allowed`.
Return _the number of **consistent** strings in the array_ `words`.
**Example 1:**
**Input:** allowed = "ab ", words = \[... | Since the problem asks for the latest step, can you start the searching from the end of arr? Use a map to store the current “1” groups. At each step (going backwards) you need to split one group and update the map. |
🔥Simple using Set, Subset🔥 | count-the-number-of-consistent-strings | 0 | 1 | # Approach\n\n1. Given a list of words and a string allowed containing characters that are considered allowed.\n2. Utilize a list comprehension to iterate through each word x in the words list.\n3. For each word, create a set of its characters (`set(x)`) and check if this set is a ***subset*** of the set of allowed cha... | 2 | There is an undirected weighted connected graph. You are given a positive integer `n` which denotes that the graph has `n` nodes labeled from `1` to `n`, and an array `edges` where each `edges[i] = [ui, vi, weighti]` denotes that there is an edge between nodes `ui` and `vi` with weight equal to `weighti`.
A path from ... | A string is incorrect if it contains a character that is not allowed Constraints are small enough for brute force |
Python 3 solution 100 faster than any other codes | count-the-number-of-consistent-strings | 0 | 1 | - We know that sets in python are actually implemented by hash tables, which means you can search a member in set by `O(1)`. Therfore, first we create a set from `allowed` string and check that every letter of each word to be in the allowed string. If its not, we increase the `count`. By the end of the program we know ... | 82 | You are given a string `allowed` consisting of **distinct** characters and an array of strings `words`. A string is **consistent** if all characters in the string appear in the string `allowed`.
Return _the number of **consistent** strings in the array_ `words`.
**Example 1:**
**Input:** allowed = "ab ", words = \[... | Since the problem asks for the latest step, can you start the searching from the end of arr? Use a map to store the current “1” groups. At each step (going backwards) you need to split one group and update the map. |
Python 3 solution 100 faster than any other codes | count-the-number-of-consistent-strings | 0 | 1 | - We know that sets in python are actually implemented by hash tables, which means you can search a member in set by `O(1)`. Therfore, first we create a set from `allowed` string and check that every letter of each word to be in the allowed string. If its not, we increase the `count`. By the end of the program we know ... | 82 | There is an undirected weighted connected graph. You are given a positive integer `n` which denotes that the graph has `n` nodes labeled from `1` to `n`, and an array `edges` where each `edges[i] = [ui, vi, weighti]` denotes that there is an edge between nodes `ui` and `vi` with weight equal to `weighti`.
A path from ... | A string is incorrect if it contains a character that is not allowed Constraints are small enough for brute force |
🤯Optimized-Without-Extra-Space.py😎 | sum-of-absolute-differences-in-a-sorted-array | 0 | 1 | \n# Complexity\n- Time complexity:\n $$O(n)$$ \n- Space complexity:\n $$O(n)$$ --> $$O(1)$$\n`Just use one variable for Prefix sum instead of an arr`\n`Suffix sum = Sum - PrefixSum`\n# Code\n```js\nclass Solution:\n def getSumAbsoluteDifferences(self, nums: List[int]) -> List[int]:\n pr=[None]*len(nums)\n ... | 8 | You are given an integer array `nums` sorted in **non-decreasing** order.
Build and return _an integer array_ `result` _with the same length as_ `nums` _such that_ `result[i]` _is equal to the **summation of absolute differences** between_ `nums[i]` _and all the other elements in the array._
In other words, `result[i... | We need to try all possible divisions for the current row to get the max score. As calculating all possible divisions will lead us to calculate some sub-problems more than once, we need to think of dynamic programming. |
🤯Optimized-Without-Extra-Space.py😎 | sum-of-absolute-differences-in-a-sorted-array | 0 | 1 | \n# Complexity\n- Time complexity:\n $$O(n)$$ \n- Space complexity:\n $$O(n)$$ --> $$O(1)$$\n`Just use one variable for Prefix sum instead of an arr`\n`Suffix sum = Sum - PrefixSum`\n# Code\n```js\nclass Solution:\n def getSumAbsoluteDifferences(self, nums: List[int]) -> List[int]:\n pr=[None]*len(nums)\n ... | 8 | You are given an array `nums` and an integer `k`. The XOR of a segment `[left, right]` where `left <= right` is the `XOR` of all the elements with indices between `left` and `right`, inclusive: `nums[left] XOR nums[left+1] XOR ... XOR nums[right]`.
Return _the minimum number of elements to change in the array_... | Absolute difference is the same as max(a, b) - min(a, b). How can you use this fact with the fact that the array is sorted? For nums[i], the answer is (nums[i] - nums[0]) + (nums[i] - nums[1]) + ... + (nums[i] - nums[i-1]) + (nums[i+1] - nums[i]) + (nums[i+2] - nums[i]) + ... + (nums[n-1] - nums[i]). It can be simplifi... |
🚀 PrefixSum & SuffixSum || Explained Intuition 🚀 | sum-of-absolute-differences-in-a-sorted-array | 1 | 1 | # Problem Description\n\nGiven a **sorted** integer array `nums`, create and return an array result of the same length, where each element `result[i]` represents the sum of absolute differences between `nums[i]` and all other elements in the array, **excluding** itself.\n\n- Constraints:\n - `result[i] = |nums[0]-nu... | 93 | You are given an integer array `nums` sorted in **non-decreasing** order.
Build and return _an integer array_ `result` _with the same length as_ `nums` _such that_ `result[i]` _is equal to the **summation of absolute differences** between_ `nums[i]` _and all the other elements in the array._
In other words, `result[i... | We need to try all possible divisions for the current row to get the max score. As calculating all possible divisions will lead us to calculate some sub-problems more than once, we need to think of dynamic programming. |
🚀 PrefixSum & SuffixSum || Explained Intuition 🚀 | sum-of-absolute-differences-in-a-sorted-array | 1 | 1 | # Problem Description\n\nGiven a **sorted** integer array `nums`, create and return an array result of the same length, where each element `result[i]` represents the sum of absolute differences between `nums[i]` and all other elements in the array, **excluding** itself.\n\n- Constraints:\n - `result[i] = |nums[0]-nu... | 93 | You are given an array `nums` and an integer `k`. The XOR of a segment `[left, right]` where `left <= right` is the `XOR` of all the elements with indices between `left` and `right`, inclusive: `nums[left] XOR nums[left+1] XOR ... XOR nums[right]`.
Return _the minimum number of elements to change in the array_... | Absolute difference is the same as max(a, b) - min(a, b). How can you use this fact with the fact that the array is sorted? For nums[i], the answer is (nums[i] - nums[0]) + (nums[i] - nums[1]) + ... + (nums[i] - nums[i-1]) + (nums[i+1] - nums[i]) + (nums[i+2] - nums[i]) + ... + (nums[n-1] - nums[i]). It can be simplifi... |
【Video】Give me 10 minutes - How we think about a solution. | sum-of-absolute-differences-in-a-sorted-array | 1 | 1 | # Intuition\nCalcuate left side and right side.\n\n---\n\n# Solution Video\n\nhttps://youtu.be/bwQh44GWME8\n\n\u25A0 Timeline of the video\n\n`0:05` How we can solve this question\n`1:56` Calculate left total\n`6:32` Calculate right total\n`11:12` Create answers\n`11:51` Coding\n`14:14` Time Complexity and Space Comple... | 12 | You are given an integer array `nums` sorted in **non-decreasing** order.
Build and return _an integer array_ `result` _with the same length as_ `nums` _such that_ `result[i]` _is equal to the **summation of absolute differences** between_ `nums[i]` _and all the other elements in the array._
In other words, `result[i... | We need to try all possible divisions for the current row to get the max score. As calculating all possible divisions will lead us to calculate some sub-problems more than once, we need to think of dynamic programming. |
【Video】Give me 10 minutes - How we think about a solution. | sum-of-absolute-differences-in-a-sorted-array | 1 | 1 | # Intuition\nCalcuate left side and right side.\n\n---\n\n# Solution Video\n\nhttps://youtu.be/bwQh44GWME8\n\n\u25A0 Timeline of the video\n\n`0:05` How we can solve this question\n`1:56` Calculate left total\n`6:32` Calculate right total\n`11:12` Create answers\n`11:51` Coding\n`14:14` Time Complexity and Space Comple... | 12 | You are given an array `nums` and an integer `k`. The XOR of a segment `[left, right]` where `left <= right` is the `XOR` of all the elements with indices between `left` and `right`, inclusive: `nums[left] XOR nums[left+1] XOR ... XOR nums[right]`.
Return _the minimum number of elements to change in the array_... | Absolute difference is the same as max(a, b) - min(a, b). How can you use this fact with the fact that the array is sorted? For nums[i], the answer is (nums[i] - nums[0]) + (nums[i] - nums[1]) + ... + (nums[i] - nums[i-1]) + (nums[i+1] - nums[i]) + (nums[i+2] - nums[i]) + ... + (nums[n-1] - nums[i]). It can be simplifi... |
Python3 Solution | sum-of-absolute-differences-in-a-sorted-array | 0 | 1 | \n```\nclass Solution:\n def getSumAbsoluteDifferences(self, nums: List[int]) -> List[int]:\n n=len(nums)\n ans=[]\n total=0\n for num in nums:\n total+=abs(num-nums[0])\n ans.append(total)\n for idx in range(1,n):\n res=nums[idx]-nums[idx-1]\n ... | 5 | You are given an integer array `nums` sorted in **non-decreasing** order.
Build and return _an integer array_ `result` _with the same length as_ `nums` _such that_ `result[i]` _is equal to the **summation of absolute differences** between_ `nums[i]` _and all the other elements in the array._
In other words, `result[i... | We need to try all possible divisions for the current row to get the max score. As calculating all possible divisions will lead us to calculate some sub-problems more than once, we need to think of dynamic programming. |
Python3 Solution | sum-of-absolute-differences-in-a-sorted-array | 0 | 1 | \n```\nclass Solution:\n def getSumAbsoluteDifferences(self, nums: List[int]) -> List[int]:\n n=len(nums)\n ans=[]\n total=0\n for num in nums:\n total+=abs(num-nums[0])\n ans.append(total)\n for idx in range(1,n):\n res=nums[idx]-nums[idx-1]\n ... | 5 | You are given an array `nums` and an integer `k`. The XOR of a segment `[left, right]` where `left <= right` is the `XOR` of all the elements with indices between `left` and `right`, inclusive: `nums[left] XOR nums[left+1] XOR ... XOR nums[right]`.
Return _the minimum number of elements to change in the array_... | Absolute difference is the same as max(a, b) - min(a, b). How can you use this fact with the fact that the array is sorted? For nums[i], the answer is (nums[i] - nums[0]) + (nums[i] - nums[1]) + ... + (nums[i] - nums[i-1]) + (nums[i+1] - nums[i]) + (nums[i+2] - nums[i]) + ... + (nums[n-1] - nums[i]). It can be simplifi... |
✅☑[C++/Java/Python/JavaScript] || Beats 99% || 2 Approaches || EXPLAINED🔥 | sum-of-absolute-differences-in-a-sorted-array | 1 | 1 | # PLEASE UPVOTE IF IT HELPED\n\n---\n\n\n# Approaches\n**(Also explained in the code)**\n\n#### ***Approach 1***\n1. **vector<int> getSumAbsoluteDifferences(vector<int>& nums):** A function that takes a vector of integers as input and returns a vector that contains the sum of absolute differences for each element of th... | 9 | You are given an integer array `nums` sorted in **non-decreasing** order.
Build and return _an integer array_ `result` _with the same length as_ `nums` _such that_ `result[i]` _is equal to the **summation of absolute differences** between_ `nums[i]` _and all the other elements in the array._
In other words, `result[i... | We need to try all possible divisions for the current row to get the max score. As calculating all possible divisions will lead us to calculate some sub-problems more than once, we need to think of dynamic programming. |
✅☑[C++/Java/Python/JavaScript] || Beats 99% || 2 Approaches || EXPLAINED🔥 | sum-of-absolute-differences-in-a-sorted-array | 1 | 1 | # PLEASE UPVOTE IF IT HELPED\n\n---\n\n\n# Approaches\n**(Also explained in the code)**\n\n#### ***Approach 1***\n1. **vector<int> getSumAbsoluteDifferences(vector<int>& nums):** A function that takes a vector of integers as input and returns a vector that contains the sum of absolute differences for each element of th... | 9 | You are given an array `nums` and an integer `k`. The XOR of a segment `[left, right]` where `left <= right` is the `XOR` of all the elements with indices between `left` and `right`, inclusive: `nums[left] XOR nums[left+1] XOR ... XOR nums[right]`.
Return _the minimum number of elements to change in the array_... | Absolute difference is the same as max(a, b) - min(a, b). How can you use this fact with the fact that the array is sorted? For nums[i], the answer is (nums[i] - nums[0]) + (nums[i] - nums[1]) + ... + (nums[i] - nums[i-1]) + (nums[i+1] - nums[i]) + (nums[i+2] - nums[i]) + ... + (nums[n-1] - nums[i]). It can be simplifi... |
O(n) Linear time complexity using Python | sum-of-absolute-differences-in-a-sorted-array | 0 | 1 | # **Solution by** [**Rockposedon**](https://leetcode.com/Rockposedon/)\n#\n## **Approach: PrefixSum & SuffixSum**\n#\n### Intuition\nThis solution utilizes the PrefixSum and SuffixSum techniques to efficiently calculate the sum of absolute differences for each element in the array. The key idea is to precompute the pre... | 4 | You are given an integer array `nums` sorted in **non-decreasing** order.
Build and return _an integer array_ `result` _with the same length as_ `nums` _such that_ `result[i]` _is equal to the **summation of absolute differences** between_ `nums[i]` _and all the other elements in the array._
In other words, `result[i... | We need to try all possible divisions for the current row to get the max score. As calculating all possible divisions will lead us to calculate some sub-problems more than once, we need to think of dynamic programming. |
O(n) Linear time complexity using Python | sum-of-absolute-differences-in-a-sorted-array | 0 | 1 | # **Solution by** [**Rockposedon**](https://leetcode.com/Rockposedon/)\n#\n## **Approach: PrefixSum & SuffixSum**\n#\n### Intuition\nThis solution utilizes the PrefixSum and SuffixSum techniques to efficiently calculate the sum of absolute differences for each element in the array. The key idea is to precompute the pre... | 4 | You are given an array `nums` and an integer `k`. The XOR of a segment `[left, right]` where `left <= right` is the `XOR` of all the elements with indices between `left` and `right`, inclusive: `nums[left] XOR nums[left+1] XOR ... XOR nums[right]`.
Return _the minimum number of elements to change in the array_... | Absolute difference is the same as max(a, b) - min(a, b). How can you use this fact with the fact that the array is sorted? For nums[i], the answer is (nums[i] - nums[0]) + (nums[i] - nums[1]) + ... + (nums[i] - nums[i-1]) + (nums[i+1] - nums[i]) + (nums[i+2] - nums[i]) + ... + (nums[n-1] - nums[i]). It can be simplifi... |
C++/Python Math sum||50ms Beats 100% | sum-of-absolute-differences-in-a-sorted-array | 1 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nLet the non-decreasing sequence is given\n$$\na_0\\leq a_1\\leq\\cdots \\leq a_i\\leq \\cdots a_{n-1}\n$$\nConsider the computation\n$$\n\\sum_j|a_j-a_i|=\\sum_{i<j}a_j-a_i+\\sum_{j<i}a_i-a_j \\\\\n=\\sum_{i<j}a_j-(n-i-1)a_i+ia_i-\\sum_{j... | 17 | You are given an integer array `nums` sorted in **non-decreasing** order.
Build and return _an integer array_ `result` _with the same length as_ `nums` _such that_ `result[i]` _is equal to the **summation of absolute differences** between_ `nums[i]` _and all the other elements in the array._
In other words, `result[i... | We need to try all possible divisions for the current row to get the max score. As calculating all possible divisions will lead us to calculate some sub-problems more than once, we need to think of dynamic programming. |
C++/Python Math sum||50ms Beats 100% | sum-of-absolute-differences-in-a-sorted-array | 1 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nLet the non-decreasing sequence is given\n$$\na_0\\leq a_1\\leq\\cdots \\leq a_i\\leq \\cdots a_{n-1}\n$$\nConsider the computation\n$$\n\\sum_j|a_j-a_i|=\\sum_{i<j}a_j-a_i+\\sum_{j<i}a_i-a_j \\\\\n=\\sum_{i<j}a_j-(n-i-1)a_i+ia_i-\\sum_{j... | 17 | You are given an array `nums` and an integer `k`. The XOR of a segment `[left, right]` where `left <= right` is the `XOR` of all the elements with indices between `left` and `right`, inclusive: `nums[left] XOR nums[left+1] XOR ... XOR nums[right]`.
Return _the minimum number of elements to change in the array_... | Absolute difference is the same as max(a, b) - min(a, b). How can you use this fact with the fact that the array is sorted? For nums[i], the answer is (nums[i] - nums[0]) + (nums[i] - nums[1]) + ... + (nums[i] - nums[i-1]) + (nums[i+1] - nums[i]) + (nums[i+2] - nums[i]) + ... + (nums[n-1] - nums[i]). It can be simplifi... |
Simple solution with Prefix Sum in Python3 | sum-of-absolute-differences-in-a-sorted-array | 0 | 1 | # Intuition\nHere we have:\n- `nums` as list of integers\n- our goal is to find sum of absolute differences in sorted array\n\nWe\'re going to use **Prefix Sum** to calculate `left` and `right` parts (the first is for **positive** diff, and the last if for **negative**).\n\n# Approach\n1. declare and fill `prefix`\n2. ... | 1 | You are given an integer array `nums` sorted in **non-decreasing** order.
Build and return _an integer array_ `result` _with the same length as_ `nums` _such that_ `result[i]` _is equal to the **summation of absolute differences** between_ `nums[i]` _and all the other elements in the array._
In other words, `result[i... | We need to try all possible divisions for the current row to get the max score. As calculating all possible divisions will lead us to calculate some sub-problems more than once, we need to think of dynamic programming. |
Simple solution with Prefix Sum in Python3 | sum-of-absolute-differences-in-a-sorted-array | 0 | 1 | # Intuition\nHere we have:\n- `nums` as list of integers\n- our goal is to find sum of absolute differences in sorted array\n\nWe\'re going to use **Prefix Sum** to calculate `left` and `right` parts (the first is for **positive** diff, and the last if for **negative**).\n\n# Approach\n1. declare and fill `prefix`\n2. ... | 1 | You are given an array `nums` and an integer `k`. The XOR of a segment `[left, right]` where `left <= right` is the `XOR` of all the elements with indices between `left` and `right`, inclusive: `nums[left] XOR nums[left+1] XOR ... XOR nums[right]`.
Return _the minimum number of elements to change in the array_... | Absolute difference is the same as max(a, b) - min(a, b). How can you use this fact with the fact that the array is sorted? For nums[i], the answer is (nums[i] - nums[0]) + (nums[i] - nums[1]) + ... + (nums[i] - nums[i-1]) + (nums[i+1] - nums[i]) + (nums[i+2] - nums[i]) + ... + (nums[n-1] - nums[i]). It can be simplifi... |
Math explained with inductive logic and code with descriptive variable names 🐍 3️⃣.🔟 and 💎 | sum-of-absolute-differences-in-a-sorted-array | 0 | 1 | # Intuition\nThe way the problem is explained it is automatically $$O(n^2)$$ this should send off signal flairs that there will be a trick. If we re-examine the conditions provided, nums is a strictly increasing array of positive numbers. Is there some [sum?] way to use this information to reduce the time complexity?\n... | 1 | You are given an integer array `nums` sorted in **non-decreasing** order.
Build and return _an integer array_ `result` _with the same length as_ `nums` _such that_ `result[i]` _is equal to the **summation of absolute differences** between_ `nums[i]` _and all the other elements in the array._
In other words, `result[i... | We need to try all possible divisions for the current row to get the max score. As calculating all possible divisions will lead us to calculate some sub-problems more than once, we need to think of dynamic programming. |
Math explained with inductive logic and code with descriptive variable names 🐍 3️⃣.🔟 and 💎 | sum-of-absolute-differences-in-a-sorted-array | 0 | 1 | # Intuition\nThe way the problem is explained it is automatically $$O(n^2)$$ this should send off signal flairs that there will be a trick. If we re-examine the conditions provided, nums is a strictly increasing array of positive numbers. Is there some [sum?] way to use this information to reduce the time complexity?\n... | 1 | You are given an array `nums` and an integer `k`. The XOR of a segment `[left, right]` where `left <= right` is the `XOR` of all the elements with indices between `left` and `right`, inclusive: `nums[left] XOR nums[left+1] XOR ... XOR nums[right]`.
Return _the minimum number of elements to change in the array_... | Absolute difference is the same as max(a, b) - min(a, b). How can you use this fact with the fact that the array is sorted? For nums[i], the answer is (nums[i] - nums[0]) + (nums[i] - nums[1]) + ... + (nums[i] - nums[i-1]) + (nums[i+1] - nums[i]) + (nums[i+2] - nums[i]) + ... + (nums[n-1] - nums[i]). It can be simplifi... |
Prefix Sum Approach | sum-of-absolute-differences-in-a-sorted-array | 0 | 1 | # Code\n```\nclass Solution:\n def getSumAbsoluteDifferences(self, nums: List[int]) -> List[int]:\n s,n=0,len(nums)\n ps,ans=[0]*n,[0]*n\n for i in range(n):\n s+=nums[i]\n ps[i]=s\n for i in range(n):\n ans[i] = s - 2*ps[i] - nums[i]*(n-1-i) + nums[i]*(i+... | 1 | You are given an integer array `nums` sorted in **non-decreasing** order.
Build and return _an integer array_ `result` _with the same length as_ `nums` _such that_ `result[i]` _is equal to the **summation of absolute differences** between_ `nums[i]` _and all the other elements in the array._
In other words, `result[i... | We need to try all possible divisions for the current row to get the max score. As calculating all possible divisions will lead us to calculate some sub-problems more than once, we need to think of dynamic programming. |
Prefix Sum Approach | sum-of-absolute-differences-in-a-sorted-array | 0 | 1 | # Code\n```\nclass Solution:\n def getSumAbsoluteDifferences(self, nums: List[int]) -> List[int]:\n s,n=0,len(nums)\n ps,ans=[0]*n,[0]*n\n for i in range(n):\n s+=nums[i]\n ps[i]=s\n for i in range(n):\n ans[i] = s - 2*ps[i] - nums[i]*(n-1-i) + nums[i]*(i+... | 1 | You are given an array `nums` and an integer `k`. The XOR of a segment `[left, right]` where `left <= right` is the `XOR` of all the elements with indices between `left` and `right`, inclusive: `nums[left] XOR nums[left+1] XOR ... XOR nums[right]`.
Return _the minimum number of elements to change in the array_... | Absolute difference is the same as max(a, b) - min(a, b). How can you use this fact with the fact that the array is sorted? For nums[i], the answer is (nums[i] - nums[0]) + (nums[i] - nums[1]) + ... + (nums[i] - nums[i-1]) + (nums[i+1] - nums[i]) + (nums[i+2] - nums[i]) + ... + (nums[n-1] - nums[i]). It can be simplifi... |
Rust/Python/Go linear time/constant space with explanation | sum-of-absolute-differences-in-a-sorted-array | 0 | 1 | # Intuition\nLets look what be the result for the position `i`. We need to look at values before and after this one:\n\n1. Every value at position after `i` is $\\ge$ than `nums[i]` the result will be `sum(nums[i + 1:])` minus `nums[i]` multiplied by the number of items (which will be `len(nums) - i`).\n2. Every value ... | 1 | You are given an integer array `nums` sorted in **non-decreasing** order.
Build and return _an integer array_ `result` _with the same length as_ `nums` _such that_ `result[i]` _is equal to the **summation of absolute differences** between_ `nums[i]` _and all the other elements in the array._
In other words, `result[i... | We need to try all possible divisions for the current row to get the max score. As calculating all possible divisions will lead us to calculate some sub-problems more than once, we need to think of dynamic programming. |
Rust/Python/Go linear time/constant space with explanation | sum-of-absolute-differences-in-a-sorted-array | 0 | 1 | # Intuition\nLets look what be the result for the position `i`. We need to look at values before and after this one:\n\n1. Every value at position after `i` is $\\ge$ than `nums[i]` the result will be `sum(nums[i + 1:])` minus `nums[i]` multiplied by the number of items (which will be `len(nums) - i`).\n2. Every value ... | 1 | You are given an array `nums` and an integer `k`. The XOR of a segment `[left, right]` where `left <= right` is the `XOR` of all the elements with indices between `left` and `right`, inclusive: `nums[left] XOR nums[left+1] XOR ... XOR nums[right]`.
Return _the minimum number of elements to change in the array_... | Absolute difference is the same as max(a, b) - min(a, b). How can you use this fact with the fact that the array is sorted? For nums[i], the answer is (nums[i] - nums[0]) + (nums[i] - nums[1]) + ... + (nums[i] - nums[i-1]) + (nums[i+1] - nums[i]) + (nums[i+2] - nums[i]) + ... + (nums[n-1] - nums[i]). It can be simplifi... |
Beats 99.38 % || PYTHON 3 || Explained | stone-game-vi | 0 | 1 | # Intuition\nIn this problem instead of picking/not Picking largest making sum we need to pick those stones by alice which can hurt bob more.\nSo we combine them and sort them in reverse order\n`why we can do sorting? Because in question its stated we can pick any stone (and not only from left most or right most)`\nNow... | 3 | Alice and Bob take turns playing a game, with Alice starting first.
There are `n` stones in a pile. On each player's turn, they can **remove** a stone from the pile and receive points based on the stone's value. Alice and Bob may **value the stones differently**.
You are given two integer arrays of length `n`, `alice... | null |
Beats 99.38 % || PYTHON 3 || Explained | stone-game-vi | 0 | 1 | # Intuition\nIn this problem instead of picking/not Picking largest making sum we need to pick those stones by alice which can hurt bob more.\nSo we combine them and sort them in reverse order\n`why we can do sorting? Because in question its stated we can pick any stone (and not only from left most or right most)`\nNow... | 3 | There is a garden of `n` flowers, and each flower has an integer beauty value. The flowers are arranged in a line. You are given an integer array `flowers` of size `n` and each `flowers[i]` represents the beauty of the `ith` flower.
A garden is **valid** if it meets these conditions:
* The garden has at least two f... | When one takes the stone, they not only get the points, but they take them away from the other player too. Greedily choose the stone with the maximum aliceValues[i] + bobValues[i]. |
Python Easy understanding | stone-game-vi | 0 | 1 | Simple logic, greedy approach, sort based on maximum sum of pairs.\n* The best approach is not only to pick the stone with maximum value, but the one where you could hurt your opponent\'s chances by picking an optimal value.\n* So by picking a stone, you get x points and you affect bob\'s chances by -y points, so the i... | 15 | Alice and Bob take turns playing a game, with Alice starting first.
There are `n` stones in a pile. On each player's turn, they can **remove** a stone from the pile and receive points based on the stone's value. Alice and Bob may **value the stones differently**.
You are given two integer arrays of length `n`, `alice... | null |
Python Easy understanding | stone-game-vi | 0 | 1 | Simple logic, greedy approach, sort based on maximum sum of pairs.\n* The best approach is not only to pick the stone with maximum value, but the one where you could hurt your opponent\'s chances by picking an optimal value.\n* So by picking a stone, you get x points and you affect bob\'s chances by -y points, so the i... | 15 | There is a garden of `n` flowers, and each flower has an integer beauty value. The flowers are arranged in a line. You are given an integer array `flowers` of size `n` and each `flowers[i]` represents the beauty of the `ith` flower.
A garden is **valid** if it meets these conditions:
* The garden has at least two f... | When one takes the stone, they not only get the points, but they take them away from the other player too. Greedily choose the stone with the maximum aliceValues[i] + bobValues[i]. |
Easy Python | 99% Speed | stone-game-vi | 0 | 1 | **Easy Python | 99% Speed**\n\nEasy Python algorithm, which can be derived by rephrasing the problem statement:\n\nInstead of taking turns starting from a situation where both players have zero score, we first consider that Bob has a maximum score of ```sum(B)```. As a result, Alice starts with an inmense loss ```-sum(... | 5 | Alice and Bob take turns playing a game, with Alice starting first.
There are `n` stones in a pile. On each player's turn, they can **remove** a stone from the pile and receive points based on the stone's value. Alice and Bob may **value the stones differently**.
You are given two integer arrays of length `n`, `alice... | null |
Easy Python | 99% Speed | stone-game-vi | 0 | 1 | **Easy Python | 99% Speed**\n\nEasy Python algorithm, which can be derived by rephrasing the problem statement:\n\nInstead of taking turns starting from a situation where both players have zero score, we first consider that Bob has a maximum score of ```sum(B)```. As a result, Alice starts with an inmense loss ```-sum(... | 5 | There is a garden of `n` flowers, and each flower has an integer beauty value. The flowers are arranged in a line. You are given an integer array `flowers` of size `n` and each `flowers[i]` represents the beauty of the `ith` flower.
A garden is **valid** if it meets these conditions:
* The garden has at least two f... | When one takes the stone, they not only get the points, but they take them away from the other player too. Greedily choose the stone with the maximum aliceValues[i] + bobValues[i]. |
Greedy + Heep | stone-game-vi | 0 | 1 | # Complexity\n- Time complexity: $$O(n*log(n))$$.\n\n- Space complexity: $$O(n)$$.\n\n# Code\n```\nclass Solution:\n def stoneGameVI(self, aliceValues: List[int], bobValues: List[int]) -> int:\n values = [(-a - b, a, b) for a, b in zip(aliceValues, bobValues)]\n heapq.heapify(values)\n aliceScor... | 1 | Alice and Bob take turns playing a game, with Alice starting first.
There are `n` stones in a pile. On each player's turn, they can **remove** a stone from the pile and receive points based on the stone's value. Alice and Bob may **value the stones differently**.
You are given two integer arrays of length `n`, `alice... | null |
Greedy + Heep | stone-game-vi | 0 | 1 | # Complexity\n- Time complexity: $$O(n*log(n))$$.\n\n- Space complexity: $$O(n)$$.\n\n# Code\n```\nclass Solution:\n def stoneGameVI(self, aliceValues: List[int], bobValues: List[int]) -> int:\n values = [(-a - b, a, b) for a, b in zip(aliceValues, bobValues)]\n heapq.heapify(values)\n aliceScor... | 1 | There is a garden of `n` flowers, and each flower has an integer beauty value. The flowers are arranged in a line. You are given an integer array `flowers` of size `n` and each `flowers[i]` represents the beauty of the `ith` flower.
A garden is **valid** if it meets these conditions:
* The garden has at least two f... | When one takes the stone, they not only get the points, but they take them away from the other player too. Greedily choose the stone with the maximum aliceValues[i] + bobValues[i]. |
python3 greedy | stone-game-vi | 0 | 1 | # Code\n```\nclass Solution:\n def stoneGameVI(self, aliceValues: List[int], bobValues: List[int]) -> int:\n \n # intuition: stones are valued by individual points + the number of points the other person loses out on, so we can simply add the two scores in a new array to assign values\n values =... | 0 | Alice and Bob take turns playing a game, with Alice starting first.
There are `n` stones in a pile. On each player's turn, they can **remove** a stone from the pile and receive points based on the stone's value. Alice and Bob may **value the stones differently**.
You are given two integer arrays of length `n`, `alice... | null |
python3 greedy | stone-game-vi | 0 | 1 | # Code\n```\nclass Solution:\n def stoneGameVI(self, aliceValues: List[int], bobValues: List[int]) -> int:\n \n # intuition: stones are valued by individual points + the number of points the other person loses out on, so we can simply add the two scores in a new array to assign values\n values =... | 0 | There is a garden of `n` flowers, and each flower has an integer beauty value. The flowers are arranged in a line. You are given an integer array `flowers` of size `n` and each `flowers[i]` represents the beauty of the `ith` flower.
A garden is **valid** if it meets these conditions:
* The garden has at least two f... | When one takes the stone, they not only get the points, but they take them away from the other player too. Greedily choose the stone with the maximum aliceValues[i] + bobValues[i]. |
Stone Game VI: Python Greedy | stone-game-vi | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nLets say it is alice\'s turn and there are only two options available. Option 1 is to select the stone at index $$i$$ and option 2 is to select the stone at index $$j$$.\nAlice wants to win and knows the stone values for bob as well.\nAli... | 0 | Alice and Bob take turns playing a game, with Alice starting first.
There are `n` stones in a pile. On each player's turn, they can **remove** a stone from the pile and receive points based on the stone's value. Alice and Bob may **value the stones differently**.
You are given two integer arrays of length `n`, `alice... | null |
Stone Game VI: Python Greedy | stone-game-vi | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nLets say it is alice\'s turn and there are only two options available. Option 1 is to select the stone at index $$i$$ and option 2 is to select the stone at index $$j$$.\nAlice wants to win and knows the stone values for bob as well.\nAli... | 0 | There is a garden of `n` flowers, and each flower has an integer beauty value. The flowers are arranged in a line. You are given an integer array `flowers` of size `n` and each `flowers[i]` represents the beauty of the `ith` flower.
A garden is **valid** if it meets these conditions:
* The garden has at least two f... | When one takes the stone, they not only get the points, but they take them away from the other player too. Greedily choose the stone with the maximum aliceValues[i] + bobValues[i]. |
Python3: FT 100%: TC O(N+R), SC O(R): Counting Sort and Iteration | stone-game-vi | 0 | 1 | # Intuition\n\nThe usual game theory approach, of top-down DFS, will NOT work here: there are 1e5 plays, so you\'ll get TLE for certain.\n\nSo I thought about what it really means for Alice to win: she needs to have a sum of plays that is larger than Bob\'s sum: `aliceSum > bobSum`. Still hard, but now we\'ve reduced t... | 0 | Alice and Bob take turns playing a game, with Alice starting first.
There are `n` stones in a pile. On each player's turn, they can **remove** a stone from the pile and receive points based on the stone's value. Alice and Bob may **value the stones differently**.
You are given two integer arrays of length `n`, `alice... | null |
Python3: FT 100%: TC O(N+R), SC O(R): Counting Sort and Iteration | stone-game-vi | 0 | 1 | # Intuition\n\nThe usual game theory approach, of top-down DFS, will NOT work here: there are 1e5 plays, so you\'ll get TLE for certain.\n\nSo I thought about what it really means for Alice to win: she needs to have a sum of plays that is larger than Bob\'s sum: `aliceSum > bobSum`. Still hard, but now we\'ve reduced t... | 0 | There is a garden of `n` flowers, and each flower has an integer beauty value. The flowers are arranged in a line. You are given an integer array `flowers` of size `n` and each `flowers[i]` represents the beauty of the `ith` flower.
A garden is **valid** if it meets these conditions:
* The garden has at least two f... | When one takes the stone, they not only get the points, but they take them away from the other player too. Greedily choose the stone with the maximum aliceValues[i] + bobValues[i]. |
Greedy Python Solution | stone-game-vi | 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 | Alice and Bob take turns playing a game, with Alice starting first.
There are `n` stones in a pile. On each player's turn, they can **remove** a stone from the pile and receive points based on the stone's value. Alice and Bob may **value the stones differently**.
You are given two integer arrays of length `n`, `alice... | null |
Greedy Python Solution | stone-game-vi | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ --... | 0 | There is a garden of `n` flowers, and each flower has an integer beauty value. The flowers are arranged in a line. You are given an integer array `flowers` of size `n` and each `flowers[i]` represents the beauty of the `ith` flower.
A garden is **valid** if it meets these conditions:
* The garden has at least two f... | When one takes the stone, they not only get the points, but they take them away from the other player too. Greedily choose the stone with the maximum aliceValues[i] + bobValues[i]. |
5 line python solution (beat 88% runtime) | stone-game-vi | 0 | 1 | # Code\n```\nclass Solution:\n def stoneGameVI(self, aliceValues: List[int], bobValues: List[int]) -> int:\n combinedArr = sorted(zip(aliceValues, bobValues), key = lambda x: -(x[0]+x[1]))\n alice = sum(aliceValues for aliceValues, bobValues in combinedArr[::2])\n bob = sum(bobValues for aliceVa... | 0 | Alice and Bob take turns playing a game, with Alice starting first.
There are `n` stones in a pile. On each player's turn, they can **remove** a stone from the pile and receive points based on the stone's value. Alice and Bob may **value the stones differently**.
You are given two integer arrays of length `n`, `alice... | null |
5 line python solution (beat 88% runtime) | stone-game-vi | 0 | 1 | # Code\n```\nclass Solution:\n def stoneGameVI(self, aliceValues: List[int], bobValues: List[int]) -> int:\n combinedArr = sorted(zip(aliceValues, bobValues), key = lambda x: -(x[0]+x[1]))\n alice = sum(aliceValues for aliceValues, bobValues in combinedArr[::2])\n bob = sum(bobValues for aliceVa... | 0 | There is a garden of `n` flowers, and each flower has an integer beauty value. The flowers are arranged in a line. You are given an integer array `flowers` of size `n` and each `flowers[i]` represents the beauty of the `ith` flower.
A garden is **valid** if it meets these conditions:
* The garden has at least two f... | When one takes the stone, they not only get the points, but they take them away from the other player too. Greedily choose the stone with the maximum aliceValues[i] + bobValues[i]. |
An almost one line solution | stone-game-vi | 0 | 1 | # Intuition\n\nSuppose Alice selects a subset of stones $S$ and Bob the complement $S^c$. Then the difference of the value obtained by Alice and Bob is:\n$$\nv = \\sum_{i\\in S} a_i-\\sum_{j\\in S^c} b_j\n=\\sum_{i\\in S} (a_i+b_i) - B, \\quad \\text{where $B=\\sum_j b_j$ the total value of stones for Bob}.\n$$\nHere $... | 0 | Alice and Bob take turns playing a game, with Alice starting first.
There are `n` stones in a pile. On each player's turn, they can **remove** a stone from the pile and receive points based on the stone's value. Alice and Bob may **value the stones differently**.
You are given two integer arrays of length `n`, `alice... | null |
An almost one line solution | stone-game-vi | 0 | 1 | # Intuition\n\nSuppose Alice selects a subset of stones $S$ and Bob the complement $S^c$. Then the difference of the value obtained by Alice and Bob is:\n$$\nv = \\sum_{i\\in S} a_i-\\sum_{j\\in S^c} b_j\n=\\sum_{i\\in S} (a_i+b_i) - B, \\quad \\text{where $B=\\sum_j b_j$ the total value of stones for Bob}.\n$$\nHere $... | 0 | There is a garden of `n` flowers, and each flower has an integer beauty value. The flowers are arranged in a line. You are given an integer array `flowers` of size `n` and each `flowers[i]` represents the beauty of the `ith` flower.
A garden is **valid** if it meets these conditions:
* The garden has at least two f... | When one takes the stone, they not only get the points, but they take them away from the other player too. Greedily choose the stone with the maximum aliceValues[i] + bobValues[i]. |
96/94 in speed and memory #2023 1st solution | delivering-boxes-from-storage-to-ports | 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 have the task of delivering some boxes from storage to their ports using only one ship. However, this ship has a **limit** on the **number of boxes** and the **total weight** that it can carry.
You are given an array `boxes`, where `boxes[i] = [portsi, weighti]`, and three integers `portsCount`, `maxBoxes`, and... | Create an array dp where dp[i][j] is the min edit distance for the path starting at node i and compared to index j of the targetPath. Traverse the dp array to obtain a valid answer. |
python solution || dynamic programming + sliding window || faster than 55% | delivering-boxes-from-storage-to-ports | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach: dynamic programming + sliding window\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity: $$O(n)$$\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity: $$O(n)$... | 0 | You have the task of delivering some boxes from storage to their ports using only one ship. However, this ship has a **limit** on the **number of boxes** and the **total weight** that it can carry.
You are given an array `boxes`, where `boxes[i] = [portsi, weighti]`, and three integers `portsCount`, `maxBoxes`, and... | Create an array dp where dp[i][j] is the min edit distance for the path starting at node i and compared to index j of the targetPath. Traverse the dp array to obtain a valid answer. |
[Python3] dp + greedy | delivering-boxes-from-storage-to-ports | 0 | 1 | \n```\nclass Solution:\n def boxDelivering(self, boxes: List[List[int]], portsCount: int, maxBoxes: int, maxWeight: int) -> int:\n dp = [0] + [inf]*len(boxes)\n trips = 2\n ii = 0\n for i in range(len(boxes)):\n maxWeight -= boxes[i][1]\n if i and boxes[i-1][0] != bo... | 2 | You have the task of delivering some boxes from storage to their ports using only one ship. However, this ship has a **limit** on the **number of boxes** and the **total weight** that it can carry.
You are given an array `boxes`, where `boxes[i] = [portsi, weighti]`, and three integers `portsCount`, `maxBoxes`, and... | Create an array dp where dp[i][j] is the min edit distance for the path starting at node i and compared to index j of the targetPath. Traverse the dp array to obtain a valid answer. |
Additional test case for easy debugging (read if test 39 fails) | delivering-boxes-from-storage-to-ports | 0 | 1 | This problem can be solved optimally via a sliding window approach, as explained by others. Something I got stuck on is the optimal unloading of boxes, to me this seems the crux of the problem. Unfortunately for me, my code was wrong only on test case 39 (80000 boxes). I have made a simpler example producing the same e... | 1 | You have the task of delivering some boxes from storage to their ports using only one ship. However, this ship has a **limit** on the **number of boxes** and the **total weight** that it can carry.
You are given an array `boxes`, where `boxes[i] = [portsi, weighti]`, and three integers `portsCount`, `maxBoxes`, and... | Create an array dp where dp[i][j] is the min edit distance for the path starting at node i and compared to index j of the targetPath. Traverse the dp array to obtain a valid answer. |
Except for the winner, each team loses exactly 1 game!||Math explains | count-of-matches-in-tournament | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nExcept for the winner, each team loses exactly 1 game!\n# Approach\n<!-- Describe your approach to solving the problem. -->\n**Theorem.** The are n teams. It needs $n-1$ games to determine the winner.\n*Proof.*\nThe are n teams. 1 team is... | 9 | You are given an integer `n`, the number of teams in a tournament that has strange rules:
* If the current number of teams is **even**, each team gets paired with another team. A total of `n / 2` matches are played, and `n / 2` teams advance to the next round.
* If the current number of teams is **odd**, one team ... | null |
Except for the winner, each team loses exactly 1 game!||Math explains | count-of-matches-in-tournament | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nExcept for the winner, each team loses exactly 1 game!\n# Approach\n<!-- Describe your approach to solving the problem. -->\n**Theorem.** The are n teams. It needs $n-1$ games to determine the winner.\n*Proof.*\nThe are n teams. 1 team is... | 9 | You are given an **even** integer `n`. You initially have a permutation `perm` of size `n` where `perm[i] == i` **(0-indexed)**.
In one operation, you will create a new array `arr`, and for each `i`:
* If `i % 2 == 0`, then `arr[i] = perm[i / 2]`.
* If `i % 2 == 1`, then `arr[i] = perm[n / 2 + (i - 1... | Simulate the tournament as given in the statement. Be careful when handling odd integers. |
🥇 return N-1 ; ] | count-of-matches-in-tournament | 1 | 1 | \n**Upvote If You Like the IDEA**\n\nThere are `n` teams participating in the tournament.\nAfter each match `1` team is eliminated.\n`n-1` teams needs to be eliminated: hence `n-1` matches.\n\n\n# Complexity\n- Time complexity: O(1)\n\n- Space complexity: O(1)\n\n``` Python []\nclass Solution:\n def numberOfMatches(... | 72 | You are given an integer `n`, the number of teams in a tournament that has strange rules:
* If the current number of teams is **even**, each team gets paired with another team. A total of `n / 2` matches are played, and `n / 2` teams advance to the next round.
* If the current number of teams is **odd**, one team ... | null |
🥇 return N-1 ; ] | count-of-matches-in-tournament | 1 | 1 | \n**Upvote If You Like the IDEA**\n\nThere are `n` teams participating in the tournament.\nAfter each match `1` team is eliminated.\n`n-1` teams needs to be eliminated: hence `n-1` matches.\n\n\n# Complexity\n- Time complexity: O(1)\n\n- Space complexity: O(1)\n\n``` Python []\nclass Solution:\n def numberOfMatches(... | 72 | You are given an **even** integer `n`. You initially have a permutation `perm` of size `n` where `perm[i] == i` **(0-indexed)**.
In one operation, you will create a new array `arr`, and for each `i`:
* If `i % 2 == 0`, then `arr[i] = perm[i / 2]`.
* If `i % 2 == 1`, then `arr[i] = perm[n / 2 + (i - 1... | Simulate the tournament as given in the statement. Be careful when handling odd integers. |
🔥🔥 SIMPLE MATH | MOST EASY SOLUTION | STEPWISE EXPLAINATION 🚀🔥 | count-of-matches-in-tournament | 1 | 1 | # Intuition\n\n\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\uD83D\uDD0D **Understanding the Problem:**\n - G... | 4 | You are given an integer `n`, the number of teams in a tournament that has strange rules:
* If the current number of teams is **even**, each team gets paired with another team. A total of `n / 2` matches are played, and `n / 2` teams advance to the next round.
* If the current number of teams is **odd**, one team ... | null |
🔥🔥 SIMPLE MATH | MOST EASY SOLUTION | STEPWISE EXPLAINATION 🚀🔥 | count-of-matches-in-tournament | 1 | 1 | # Intuition\n\n\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\uD83D\uDD0D **Understanding the Problem:**\n - G... | 4 | You are given an **even** integer `n`. You initially have a permutation `perm` of size `n` where `perm[i] == i` **(0-indexed)**.
In one operation, you will create a new array `arr`, and for each `i`:
* If `i % 2 == 0`, then `arr[i] = perm[i / 2]`.
* If `i % 2 == 1`, then `arr[i] = perm[n / 2 + (i - 1... | Simulate the tournament as given in the statement. Be careful when handling odd integers. |
💯Faster✅💯 Lesser✅3 Methods🔥Simple Math🔥Iterative Solution🔥Recursive Solution🔥Visualized Too🔥 | count-of-matches-in-tournament | 1 | 1 | # \uD83D\uDE80 Hi, I\'m [Mohammed Raziullah Ansari](https://leetcode.com/Mohammed_Raziullah_Ansari/), and I\'m excited to share 3 ways to solve this question with detailed explanation of each approach:\n\n# Problem Explaination: \nThe problem involves simulating a tournament with a specific set of rules and determining... | 25 | You are given an integer `n`, the number of teams in a tournament that has strange rules:
* If the current number of teams is **even**, each team gets paired with another team. A total of `n / 2` matches are played, and `n / 2` teams advance to the next round.
* If the current number of teams is **odd**, one team ... | null |
💯Faster✅💯 Lesser✅3 Methods🔥Simple Math🔥Iterative Solution🔥Recursive Solution🔥Visualized Too🔥 | count-of-matches-in-tournament | 1 | 1 | # \uD83D\uDE80 Hi, I\'m [Mohammed Raziullah Ansari](https://leetcode.com/Mohammed_Raziullah_Ansari/), and I\'m excited to share 3 ways to solve this question with detailed explanation of each approach:\n\n# Problem Explaination: \nThe problem involves simulating a tournament with a specific set of rules and determining... | 25 | You are given an **even** integer `n`. You initially have a permutation `perm` of size `n` where `perm[i] == i` **(0-indexed)**.
In one operation, you will create a new array `arr`, and for each `i`:
* If `i % 2 == 0`, then `arr[i] = perm[i / 2]`.
* If `i % 2 == 1`, then `arr[i] = perm[n / 2 + (i - 1... | Simulate the tournament as given in the statement. Be careful when handling odd integers. |
✅☑[C++/Java/Python/JavaScript] || 2 Approaches || EXPLAINED🔥 | count-of-matches-in-tournament | 1 | 1 | # PLEASE UPVOTE IF IT HELPED\n\n---\n\n\n# Approaches\n**(Also explained in the code)**\n\n#### ***Approach 1()***\n1. **Variable Initialization:** Initialize the variable `ans` as 0, which will store the total number of matches played.\n\n1. **While Loop:** Execute a loop until the value of `n` becomes 1 (signifying t... | 2 | You are given an integer `n`, the number of teams in a tournament that has strange rules:
* If the current number of teams is **even**, each team gets paired with another team. A total of `n / 2` matches are played, and `n / 2` teams advance to the next round.
* If the current number of teams is **odd**, one team ... | null |
✅☑[C++/Java/Python/JavaScript] || 2 Approaches || EXPLAINED🔥 | count-of-matches-in-tournament | 1 | 1 | # PLEASE UPVOTE IF IT HELPED\n\n---\n\n\n# Approaches\n**(Also explained in the code)**\n\n#### ***Approach 1()***\n1. **Variable Initialization:** Initialize the variable `ans` as 0, which will store the total number of matches played.\n\n1. **While Loop:** Execute a loop until the value of `n` becomes 1 (signifying t... | 2 | You are given an **even** integer `n`. You initially have a permutation `perm` of size `n` where `perm[i] == i` **(0-indexed)**.
In one operation, you will create a new array `arr`, and for each `i`:
* If `i % 2 == 0`, then `arr[i] = perm[i / 2]`.
* If `i % 2 == 1`, then `arr[i] = perm[n / 2 + (i - 1... | Simulate the tournament as given in the statement. Be careful when handling odd integers. |
(^_^)Python3 and C++ Solutions(100% Beats C++ users)📈 | count-of-matches-in-tournament | 0 | 1 | # Solution in Python3\n```\nclass Solution:\n def numberOfMatches(self, n: int) -> int:\n counter = 0\n while n != 1:\n if n % 2 == 0:\n counter += n/2\n n = n/2\n else:\n counter += (n - 1) / 2\n n = (n - 1) / 2 + 1\n ... | 2 | You are given an integer `n`, the number of teams in a tournament that has strange rules:
* If the current number of teams is **even**, each team gets paired with another team. A total of `n / 2` matches are played, and `n / 2` teams advance to the next round.
* If the current number of teams is **odd**, one team ... | null |
(^_^)Python3 and C++ Solutions(100% Beats C++ users)📈 | count-of-matches-in-tournament | 0 | 1 | # Solution in Python3\n```\nclass Solution:\n def numberOfMatches(self, n: int) -> int:\n counter = 0\n while n != 1:\n if n % 2 == 0:\n counter += n/2\n n = n/2\n else:\n counter += (n - 1) / 2\n n = (n - 1) / 2 + 1\n ... | 2 | You are given an **even** integer `n`. You initially have a permutation `perm` of size `n` where `perm[i] == i` **(0-indexed)**.
In one operation, you will create a new array `arr`, and for each `i`:
* If `i % 2 == 0`, then `arr[i] = perm[i / 2]`.
* If `i % 2 == 1`, then `arr[i] = perm[n / 2 + (i - 1... | Simulate the tournament as given in the statement. Be careful when handling odd integers. |
✅Beginner Friendly✅Mastering Tournament Dynamics: A Comprehensive Guide to Match Counting Algorithms | count-of-matches-in-tournament | 1 | 1 | > \u2753**The way to solve this is DRY RUN and Observe that number of matches is one less than number of teams given**\n## \uD83D\uDCA1Approach 1: Recursive Approach - Brute Force\n\n### \u2728Explanation\nThis approach uses recursion to count the number of matches played. The base cases are when there is only one tea... | 1 | You are given an integer `n`, the number of teams in a tournament that has strange rules:
* If the current number of teams is **even**, each team gets paired with another team. A total of `n / 2` matches are played, and `n / 2` teams advance to the next round.
* If the current number of teams is **odd**, one team ... | null |
✅Beginner Friendly✅Mastering Tournament Dynamics: A Comprehensive Guide to Match Counting Algorithms | count-of-matches-in-tournament | 1 | 1 | > \u2753**The way to solve this is DRY RUN and Observe that number of matches is one less than number of teams given**\n## \uD83D\uDCA1Approach 1: Recursive Approach - Brute Force\n\n### \u2728Explanation\nThis approach uses recursion to count the number of matches played. The base cases are when there is only one tea... | 1 | You are given an **even** integer `n`. You initially have a permutation `perm` of size `n` where `perm[i] == i` **(0-indexed)**.
In one operation, you will create a new array `arr`, and for each `i`:
* If `i % 2 == 0`, then `arr[i] = perm[i / 2]`.
* If `i % 2 == 1`, then `arr[i] = perm[n / 2 + (i - 1... | Simulate the tournament as given in the statement. Be careful when handling odd integers. |
🚀✅🚀 100% beats || One line || Easy peasy || Java, Python, C#, Rust, Ruby, Go, C, C++, Php|| 🔥🔥 | count-of-matches-in-tournament | 1 | 1 | \n\n\n# Intuition\nThe solution is based on the observation that, in each round of the tournament, one team is eliminated, except in the final round where the winner is decided. Theref... | 7 | You are given an integer `n`, the number of teams in a tournament that has strange rules:
* If the current number of teams is **even**, each team gets paired with another team. A total of `n / 2` matches are played, and `n / 2` teams advance to the next round.
* If the current number of teams is **odd**, one team ... | null |
🚀✅🚀 100% beats || One line || Easy peasy || Java, Python, C#, Rust, Ruby, Go, C, C++, Php|| 🔥🔥 | count-of-matches-in-tournament | 1 | 1 | \n\n\n# Intuition\nThe solution is based on the observation that, in each round of the tournament, one team is eliminated, except in the final round where the winner is decided. Theref... | 7 | You are given an **even** integer `n`. You initially have a permutation `perm` of size `n` where `perm[i] == i` **(0-indexed)**.
In one operation, you will create a new array `arr`, and for each `i`:
* If `i % 2 == 0`, then `arr[i] = perm[i / 2]`.
* If `i % 2 == 1`, then `arr[i] = perm[n / 2 + (i - 1... | Simulate the tournament as given in the statement. Be careful when handling odd integers. |
Very easy soln Just find max of given str | partitioning-into-minimum-number-of-deci-binary-numbers | 0 | 1 | # ONe liner Code very easy\n\n# Code\n```\nclass Solution:\n def minPartitions(self, n: str) -> int:\n return max(map(int,n))\n``` | 1 | A decimal number is called **deci-binary** if each of its digits is either `0` or `1` without any leading zeros. For example, `101` and `1100` are **deci-binary**, while `112` and `3001` are not.
Given a string `n` that represents a positive decimal integer, return _the **minimum** number of positive **deci-binary** n... | Use a three-layer loop to check all possible patterns by iterating through all possible starting positions, all indexes less than m, and if the character at the index is repeated k times. |
Very easy soln Just find max of given str | partitioning-into-minimum-number-of-deci-binary-numbers | 0 | 1 | # ONe liner Code very easy\n\n# Code\n```\nclass Solution:\n def minPartitions(self, n: str) -> int:\n return max(map(int,n))\n``` | 1 | You are given a string `s` that contains some bracket pairs, with each pair containing a **non-empty** key.
* For example, in the string `"(name)is(age)yearsold "`, there are **two** bracket pairs that contain the keys `"name "` and `"age "`.
You know the values of a wide range of keys. This is represented by a 2D ... | Think about if the input was only one digit. Then you need to add up as many ones as the value of this digit. If the input has multiple digits, then you can solve for each digit independently, and merge the answers to form numbers that add up to that input. Thus the answer is equal to the max digit. |
[C++/Java/Python]- Easy One liner with explanation | partitioning-into-minimum-number-of-deci-binary-numbers | 1 | 1 | For this one, we would just rely on simple maths.\n> Example: **n = "153"**\n> \nNow let\'s break down each digit with required number of ones as we can use only `0` or `1` in deci-binaries.\n>**1** - 1 0 0 0 0 \n>**5** - 1 1 1 1 1\n>**3** - 1 1 1 0 0 \n> Added zero padding to the tail to align with max number of ... | 70 | A decimal number is called **deci-binary** if each of its digits is either `0` or `1` without any leading zeros. For example, `101` and `1100` are **deci-binary**, while `112` and `3001` are not.
Given a string `n` that represents a positive decimal integer, return _the **minimum** number of positive **deci-binary** n... | Use a three-layer loop to check all possible patterns by iterating through all possible starting positions, all indexes less than m, and if the character at the index is repeated k times. |
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