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minimum-operations-to-reduce-x-to-zero | Change Your Perspective | JAVA Explanation | change-your-perspective-java-explanation-5phd | Logic\nInitially, this question might look like a DP problem: pick either left or right recursively and collect the number of operations along the way. A brute | ciote | NORMAL | 2022-06-11T01:26:17.092317+00:00 | 2022-06-11T08:18:42.701216+00:00 | 18,240 | false | ### Logic\nInitially, this question might look like a DP problem: pick either left or right recursively and collect the number of operations along the way. A brute force approach would result in an `O(2^n)` complexity which isn\'t great. Memoization would improve this but we\'ll exceed the memory limit as I\'ve tested.... | 687 | 2 | ['Two Pointers', 'Java'] | 53 |
minimum-operations-to-reduce-x-to-zero | [Java] Detailed Explanation - O(N) Prefix Sum/Map - Longest Target Sub-Array | java-detailed-explanation-on-prefix-summ-hdew | Key Notes:\n- We could use dfs+memo or BFS, but they are too slow and will TLE (?)\n- If it exists an answer, then it means we have a subarray in the middle of | frankkkkk | NORMAL | 2020-11-15T04:01:47.274207+00:00 | 2020-11-15T04:57:20.483673+00:00 | 24,339 | false | **Key Notes:**\n- We could use dfs+memo or BFS, but they are too slow and will TLE (?)\n- If it exists an answer, then it means we have **a subarray in the middle of original array whose sum is == totalSum - x**\n- If we want to minimize our operations, then we should **maximize the length of the middle subarray.**\n\t... | 492 | 5 | [] | 32 |
minimum-operations-to-reduce-x-to-zero | Are you stuck ? once read this . logic explained clearly | easy-to-understand | are-you-stuck-once-read-this-logic-expla-ugx7 | so , the problem is saying that we can remove either leftmost or rightmost element and we have to remove them in such a way that \n1.remove minimum number of el | kevaljoshi2312 | NORMAL | 2021-02-24T14:04:21.056506+00:00 | 2021-02-24T14:04:21.056548+00:00 | 13,070 | false | so , the problem is saying that we can remove either leftmost or rightmost element and we have to remove them in such a way that \n1.remove minimum number of elements and\n2. sum of all removed element is X \n\nNow main point is that if successfully remove elements such that sum of removed elements is X , then \n1. sum... | 326 | 1 | ['C', 'C++'] | 34 |
minimum-operations-to-reduce-x-to-zero | ✅ [C++/Python] Simple Solution w/ Explanation | Sliding Window | cpython-simple-solution-w-explanation-sl-q7ra | Sometimes, converting a problem into some other familiar one helps a lot. This question is one of them.\nLet me state a different problem, and your task is to r | r0gue_shinobi | NORMAL | 2022-06-11T01:11:30.684302+00:00 | 2022-06-11T02:56:41.180591+00:00 | 15,310 | false | Sometimes, converting a problem into some other familiar one helps a lot. This question is one of them.\nLet me state a different problem, and your task is to relate how solving this problem will help in solving the actual one.\n> Given an array containing integers, your task is to find the length of the longest subarr... | 207 | 1 | ['Greedy', 'C', 'Sliding Window', 'Python', 'Python3'] | 15 |
minimum-operations-to-reduce-x-to-zero | ✅ 96.51% Sliding Window | 9651-sliding-window-by-vanamsen-3lh5 | Comprehensive Guide to Solving "Minimum Operations to Reduce X to Zero"\n\n## Introduction & Problem Statement\n\nGiven an integer array nums and an integer x, | vanAmsen | NORMAL | 2023-09-20T00:12:02.046844+00:00 | 2023-09-20T02:02:46.884474+00:00 | 24,536 | false | # Comprehensive Guide to Solving "Minimum Operations to Reduce X to Zero"\n\n## Introduction & Problem Statement\n\nGiven an integer array `nums` and an integer `x`, the task is to find the minimum number of operations to reduce `x` to exactly 0 by removing either the leftmost or rightmost element from the array `nums`... | 159 | 12 | ['Sliding Window', 'PHP', 'Prefix Sum', 'C++', 'Java', 'Go', 'Python3', 'Rust', 'JavaScript', 'C#'] | 21 |
minimum-operations-to-reduce-x-to-zero | C# Sliding window O(n) Time O(1) Space | c-sliding-window-on-time-o1-space-by-leo-y5zf | This problem is equivalent to finding the longest subarray whose sum is == totalSum - x\n\n\npublic class Solution\n{\n public int MinOperations(int[] nums, | leoooooo | NORMAL | 2020-11-15T04:07:14.553530+00:00 | 2020-11-15T04:36:47.527931+00:00 | 7,352 | false | **This problem is equivalent to finding the longest subarray whose sum is == totalSum - x**\n\n```\npublic class Solution\n{\n public int MinOperations(int[] nums, int x)\n {\n int sum = nums.Sum() - x;\n if(sum < 0) return -1;\n if(sum == 0) return nums.Length;\n int start = 0, cur = ... | 137 | 1 | [] | 17 |
minimum-operations-to-reduce-x-to-zero | Two Sum vs. Hash Map | two-sum-vs-hash-map-by-votrubac-z7hf | Started solving this problem using DP, but then realized that the constraints are too large.\n\nThen, I realized that, to reach x, we will take l numbers from t | votrubac | NORMAL | 2020-11-15T04:04:47.654907+00:00 | 2022-06-11T21:51:33.324811+00:00 | 13,154 | false | Started solving this problem using DP, but then realized that the constraints are too large.\n\nThen, I realized that, to reach `x`, we will take `l` numbers from the left, and `r` numbers from the right. We just need to find `min(l + r)`.\n\n#### Approach 1: Hash Map\nAs you count the sum of the first `l` numbers, you... | 121 | 4 | ['C'] | 21 |
minimum-operations-to-reduce-x-to-zero | [Python] O(n) solution, using cumulative sums | python-on-solution-using-cumulative-sums-1lfl | We can reformulate this problem: we need to choose several values from the beginning and several values from end, such that sum of this numbers is equal to x. I | dbabichev | NORMAL | 2021-01-14T08:47:21.066085+00:00 | 2021-02-26T12:46:17.645169+00:00 | 5,850 | false | We can reformulate this problem: we need to choose several values from the beginning and several values from end, such that sum of this numbers is equal to `x`. It is equivalent to finding some contiguous subarray, such that it has sum of elements equal to `sum(nums) - x`, which has the biggest length. In this way prob... | 103 | 17 | [] | 6 |
minimum-operations-to-reduce-x-to-zero | C++ | Sliding Window | Easy Solution | c-sliding-window-easy-solution-by-_panka-yitw | If helpful do upvote\nThanks\n\n\nclass Solution {\npublic:\n int minOperations(vector<int>& nums, int x) {\n int total=0, n=nums.size();\n for | _pankajMahtolia | NORMAL | 2020-11-17T09:13:35.809490+00:00 | 2020-11-17T09:13:35.809522+00:00 | 5,522 | false | **If helpful do upvote\nThanks**\n\n```\nclass Solution {\npublic:\n int minOperations(vector<int>& nums, int x) {\n int total=0, n=nums.size();\n for(auto i:nums) total+=i;\n if(x>total) return -1;\n int left=0,right=0, curr=0, ans=-1;\n for(; right<n;right++){\n curr+=... | 70 | 8 | [] | 4 |
minimum-operations-to-reduce-x-to-zero | 🚀95.97% || Two Pointers - Sliding Window || Commented Code🚀 | 9597-two-pointers-sliding-window-comment-nkkz | Porblem Description\nGiven an array of integers, nums, and an integer x. Each element in nums can be subtracted from x. The goal is to reduce x to exactly 0 usi | MohamedMamdouh20 | NORMAL | 2023-09-20T00:55:43.092683+00:00 | 2023-09-20T01:23:51.483831+00:00 | 8,033 | false | # Porblem Description\nGiven an array of integers, `nums`, and an integer `x`. Each element in `nums` can be subtracted from x. The **goal** is to reduce `x` to exactly `0` using a **minimum** number of operations.\n\nIn each **operation**, you can choose to **remove** either the `leftmost` or the `rightmost` element f... | 67 | 4 | ['Array', 'Two Pointers', 'Sliding Window', 'Python', 'C++', 'Java', 'Python3'] | 14 |
minimum-operations-to-reduce-x-to-zero | How we think about a solution - O(n) time, O(1) space - Python, JavaScript, Java, C++ | how-we-think-about-a-solution-on-time-o1-ymou | Welcome to my article. Before starting the article, why do I have to get multiple downvotes for a miniute every day? That is obviously deliberate downvotes. I c | niits | NORMAL | 2023-09-20T01:58:22.412434+00:00 | 2023-09-23T16:39:56.080961+00:00 | 2,847 | false | Welcome to my article. Before starting the article, why do I have to get multiple downvotes for a miniute every day? That is obviously deliberate downvotes. I can tell who is doing it. Please show your respect to others! Thanks.\n\n# Intuition\nTry to find target number with sum(nums) - x which is `unnecessary numbers`... | 48 | 6 | ['C++', 'Java', 'Python3', 'JavaScript'] | 5 |
minimum-operations-to-reduce-x-to-zero | C++ | 2 Approaches | c-2-approaches-by-datnguyen2k3-ei91 | Approach 1: Hash map, prefix sum, two sum\n\t- Idea: + create hash map satisfy mapLeft[prefix sum] = i, i is last index in prefix sum\n\t\t\t+ brute force from | datnguyen2k3 | NORMAL | 2022-06-11T01:18:16.295147+00:00 | 2022-10-28T05:46:54.158741+00:00 | 6,989 | false | Approach 1: Hash map, prefix sum, two sum\n\t- Idea: + create hash map satisfy mapLeft[prefix sum] = i, i is last index in prefix sum\n\t\t\t+ brute force from last index in array to first index and create suffix sum, find i satisfy mapLeft[x - suffix sum] > 0 \n\t\t\t => that\'s mean prefix sum + suffix sum = x, a... | 44 | 1 | ['C', 'Sliding Window', 'Prefix Sum'] | 3 |
minimum-operations-to-reduce-x-to-zero | [Python3] O(n) Time O(1) space solution | python3-on-time-o1-space-solution-by-red-19ve | The idea is to find the length of the maximum length subarray that has a sum equal to sum(nums) - x. This makes sense because after removing the optimal element | redsand | NORMAL | 2021-01-14T14:43:00.556612+00:00 | 2021-01-15T14:28:43.900817+00:00 | 2,055 | false | The idea is to find the length of the maximum length subarray that has a sum equal to ```sum(nums) - x```. This makes sense because after removing the optimal elements from the ends, this is what we will be left with. So instead of removing elements, add elements to this subarray in a sliding window fashion and optimiz... | 37 | 0 | [] | 6 |
minimum-operations-to-reduce-x-to-zero | Minimum Operations to Reduce X to Zero | JS | Explanation | beats 100% | minimum-operations-to-reduce-x-to-zero-j-5bfv | Idea:\n\nThis problem is tasking us to essentially find when the sum of the starting and ending subarrays of nums is equal to x. Put another way, it\'s asking u | sgallivan | NORMAL | 2021-01-15T00:26:36.699248+00:00 | 2021-01-15T00:26:36.699277+00:00 | 1,276 | false | ***Idea:***\n\nThis problem is tasking us to essentially find when the sum of the starting and ending subarrays of **nums** is equal to **x**. Put another way, it\'s asking us to find when any subarray of consecutive elements in **nums** is equal to the sum of all elements in **nums** minus **x**.\n```\n ... | 33 | 0 | ['JavaScript'] | 1 |
minimum-operations-to-reduce-x-to-zero | C++| Simple | O(N) | Sliding Window | | c-simple-on-sliding-window-by-persistent-z3iz | Goal is to find a window of longest length which has sum equal to s - x, where s is sum of all the elements of the array.\n When such a window is found all the | persistentBeast | NORMAL | 2022-06-11T07:26:47.147091+00:00 | 2022-06-12T10:27:29.518002+00:00 | 3,589 | false | * Goal is to **find a window of longest length** which has sum equal to `s - x`, where `s` is sum of all the elements of the array.\n* When such a window is found all the elements outside this window will sum to `x` and no of operations will be equal to their count. \n* **TC : O(N) | SC : O(1)**\n\n```\nclass Solution ... | 29 | 0 | ['C', 'Sliding Window'] | 5 |
minimum-operations-to-reduce-x-to-zero | [Java/Python 3] Sliding window: Longest subarray sum to the target = sum(nums) - x. | javapython-3-sliding-window-longest-suba-uev4 | Using sliding window to find the longest subarry that sums to sum(nums) - x.\njava\n public int minOperations(int[] nums, int x) {\n int target = Arra | rock | NORMAL | 2020-11-15T04:42:40.768272+00:00 | 2022-12-03T16:31:11.169729+00:00 | 3,290 | false | Using sliding window to find the longest subarry that sums to `sum(nums) - x`.\n```java\n public int minOperations(int[] nums, int x) {\n int target = Arrays.stream(nums).sum() - x, size = -1, n = nums.length;\n for (int lo = -1, hi = 0, winSum = 0; hi < n; ++hi) {\n winSum += nums[hi];\n ... | 28 | 0 | ['Java', 'Python3'] | 7 |
minimum-operations-to-reduce-x-to-zero | ✅97.65%🔥Sum of Subarray Reduction🔥 | 9765sum-of-subarray-reduction-by-mrake-26e8 | Problem\n\n### This problem involves finding the minimum number of operations to reduce the sum of elements in the array nums to exactly zero, given that you ca | MrAke | NORMAL | 2023-09-20T00:49:33.147774+00:00 | 2023-09-20T00:49:33.147800+00:00 | 2,406 | false | # Problem\n\n### This problem involves finding the minimum number of operations to reduce the sum of elements in the array nums to exactly zero, given that you can perform operations to remove elements from either the left or right end of the array.\n---\n# Solution\n\n##### **1.** Calculate the total sum of all elemen... | 26 | 14 | ['Array', 'Hash Table', 'C', 'Python', 'C++', 'Java', 'Go', 'Python3', 'C#'] | 5 |
minimum-operations-to-reduce-x-to-zero | [Python3] O(N) hash table of prefix | python3-on-hash-table-of-prefix-by-ye15-ljid | Algo \nHere, we maintain a mapping of prefix sum to index mapping. Then, we go over nums inversely to compute suffix sum, and check if a complement exists in th | ye15 | NORMAL | 2020-11-15T04:08:15.092260+00:00 | 2021-01-15T05:11:33.425344+00:00 | 2,109 | false | Algo \nHere, we maintain a mapping of prefix sum to index mapping. Then, we go over `nums` inversely to compute suffix sum, and check if a complement exists in the prefix sum (via hash table). If so, check the length and update `ans`. \n\nImplementation \n```\nclass Solution:\n def minOperations(self, nums: List[int... | 25 | 2 | ['Python3'] | 9 |
minimum-operations-to-reduce-x-to-zero | sliding window [c++] | sliding-window-c-by-vishwasgajawada-vscp | \nclass Solution {\npublic:\n int minOperations(vector<int>& nums, int x) {\n int total=0,n=nums.size();\n for(int num : nums)total+=num;\n | vishwasgajawada | NORMAL | 2021-01-14T09:34:52.724598+00:00 | 2021-10-04T17:38:44.204258+00:00 | 1,706 | false | ```\nclass Solution {\npublic:\n int minOperations(vector<int>& nums, int x) {\n int total=0,n=nums.size();\n for(int num : nums)total+=num;\n int need=total-x,cur=0,longest=0;\n if(need==0)return n;\n int l=0,r=0;\n while(l<=r){\n if(cur<need){\n i... | 19 | 3 | ['C'] | 3 |
minimum-operations-to-reduce-x-to-zero | [Python 3] Two sum problem using HashMap - O(N) - Clean & Concise | python-3-two-sum-problem-using-hashmap-o-b2vb | python\nclass Solution:\n def minOperations(self, nums: List[int], x: int) -> int:\n n = len(nums)\n\n leftMap = dict()\n leftMap[0] = - | hiepit | NORMAL | 2021-05-18T11:43:51.327696+00:00 | 2021-05-18T11:45:14.912315+00:00 | 776 | false | ```python\nclass Solution:\n def minOperations(self, nums: List[int], x: int) -> int:\n n = len(nums)\n\n leftMap = dict()\n leftMap[0] = -1\n left = 0\n for i in range(n):\n left += nums[i]\n if left not in leftMap:\n leftMap[left] = i\n\n ... | 17 | 0 | [] | 0 |
minimum-operations-to-reduce-x-to-zero | C++ || Explained Solution || Best Time and Space Complexity | c-explained-solution-best-time-and-space-4ohc | Please Upvote the solution if helpful --> else comment down the problem and I will try to improve upon that! Thanks for reading .\n```\nclass Solution {\npublic | code-to-learn | NORMAL | 2022-06-11T10:09:08.019965+00:00 | 2022-06-11T10:50:12.255671+00:00 | 1,106 | false | **Please Upvote the solution if helpful --> else comment down the problem and I will try to improve upon that! Thanks for reading .**\n```\nclass Solution {\npublic:\n // O(n) time complexity solution and O(1) space complexity\n // Given Problem (equivalent) => Finding longest subarray with sum=totalSum(nums) - x... | 16 | 0 | [] | 3 |
minimum-operations-to-reduce-x-to-zero | ✅Easiest Solution with explanation and annotations! | easiest-solution-with-explanation-and-an-5r1s | Intuition\nWe can use prefix sum and suffix sum here. \nFor every number in prefix sum we\'ll check if its complement is present in suffix. If it is it means we | parthdharmale008 | NORMAL | 2023-09-20T05:42:48.418526+00:00 | 2023-09-20T05:48:46.071246+00:00 | 2,336 | false | # Intuition\nWe can use prefix sum and suffix sum here. \nFor every number in prefix sum we\'ll check if its complement is present in suffix. If it is it means we can make our `x` if it is not it means we can\'t make `x` with that particular prefix, so move on to the next one.\n\n# Approach\n- Firstly we find our `n` w... | 15 | 0 | ['Array', 'Prefix Sum', 'C++', 'Java'] | 4 |
minimum-operations-to-reduce-x-to-zero | Very Easy C++ Solution with Explanation (Sliding Window) - beats 88.2% submissions | very-easy-c-solution-with-explanation-sl-twpo | Explanation\nThis is a slighty tricky variant of Sliding Window problem.\n\nTrick to make this question easy :\n Total = Sum of all elements of the nums array.\ | vatsalkesarwani | NORMAL | 2021-06-12T22:21:41.261778+00:00 | 2021-06-12T22:21:41.261804+00:00 | 1,271 | false | **Explanation**\nThis is a slighty tricky variant of Sliding Window problem.\n\nTrick to make this question easy :\n* Total = Sum of all elements of the nums array.\n* Now find max operation to find (total-x) in the array\n\nCheck result :\n* If the result comes to 0 then return -1\n* else return size-result\n\nTime Co... | 15 | 1 | ['Two Pointers', 'Greedy', 'C', 'Sliding Window'] | 0 |
minimum-operations-to-reduce-x-to-zero | 💯🔥 1 Method || C++ | Java || Python || JavaScript 💯🔥 | 1-method-c-java-python-javascript-by-use-7c2v | Read Whole article : https://www.nileshblog.tech/minimum-operations-to-reduce-x-to-zero/\n\nExplanation of Problem with Example (HandWritten).\nTime Complexity: | user6845R | NORMAL | 2023-09-20T06:15:23.243780+00:00 | 2023-09-20T06:15:23.243809+00:00 | 1,105 | false | Read Whole article : https://www.nileshblog.tech/minimum-operations-to-reduce-x-to-zero/\n\nExplanation of Problem with Example (HandWritten).\nTime Complexity:\n\n**Prefix Sum + Hashmap (Longest Subarry )Approach : O(N)**\n\nPython :\nJava:\nc++:\nJavaScript:\n\nRead Whole article :https://www.nileshblog.tech/minimum-... | 14 | 1 | ['C', 'Prefix Sum', 'Python', 'Java', 'JavaScript'] | 3 |
minimum-operations-to-reduce-x-to-zero | Java || Greedy + DP + HashMap + BinarySearch + Prefix Sum + Sliding Window || 5 approaches | java-greedy-dp-hashmap-binarysearch-pref-4a1j | \n\t// O(2^n) O(n)\n\t// Recursion\n\t// TLE\n\tpublic int minOperationsRec(int[] nums, int x) {\n\n\t\tint len = nums.length;\n\t\tint count = minOperationsHel | LegendaryCoder | NORMAL | 2021-06-08T02:38:36.745067+00:00 | 2021-06-08T02:39:07.496763+00:00 | 517 | false | \n\t// O(2^n) O(n)\n\t// Recursion\n\t// TLE\n\tpublic int minOperationsRec(int[] nums, int x) {\n\n\t\tint len = nums.length;\n\t\tint count = minOperationsHelper(nums, x, 0, len - 1);\n\t\treturn (count == Integer.MAX_VALUE) ? -1 : count;\n\t}\n\n\t// O(2^n) O(n)\n\t// Recursion\n\tpublic int minOperationsHelper(int[... | 14 | 2 | [] | 1 |
minimum-operations-to-reduce-x-to-zero | C++ | Easy Solution w/ Explanation | Sliding Window ✅ | c-easy-solution-w-explanation-sliding-wi-hfyo | Please upvote the post if you like it :)\n\nIntuition:\nwe can basically think this problem as finding the longest subarray with sum sum - x, since we are delet | prantik0128 | NORMAL | 2022-06-11T04:33:00.684634+00:00 | 2022-06-11T04:36:04.386795+00:00 | 1,095 | false | **Please upvote the post if you like it :)**\n\n**Intuition:**\nwe can basically think this problem as `finding the longest subarray` with **sum** `sum - x`, since we are deleting only from the ends so we will be left out with sum `sum-x` left if we are able to remove elements with sum `x` \nThis is how we come to down... | 13 | 0 | ['C', 'Sliding Window'] | 1 |
minimum-operations-to-reduce-x-to-zero | C++ ☄️ | SLIDING WINDOW | O(1) Space / O(n) Time | c-sliding-window-o1-space-on-time-by-meg-cjud | The problem can be thought of as finding the longest subarray with sum total - x ,because we are deleting elements from the ends only .So there will be a subarr | megamind_ | NORMAL | 2022-06-11T02:01:54.187973+00:00 | 2022-06-11T06:28:29.418926+00:00 | 1,747 | false | The problem can be thought of as finding the longest subarray with sum `total - x` ,because we are deleting elements from the ends only .So there will be a subarray left of sum `total-x` left if we were able to remove elements with sum `x`. \n The approach is to use the concept of the variable-size sliding window using... | 10 | 1 | [] | 1 |
minimum-operations-to-reduce-x-to-zero | [C++] Cumulative Sums Solution Explained, ~100% Time, ~70% Space | c-cumulative-sums-solution-explained-100-vsqh | Okay, I tried a number of approaches, including a 2d matrix for DP and a double-ended BFS, before realising that there was a much easier and performant solution | ajna | NORMAL | 2021-01-14T23:58:00.439707+00:00 | 2021-01-15T00:03:28.978472+00:00 | 1,268 | false | Okay, I tried a number of approaches, including a 2d matrix for DP and a double-ended BFS, before realising that there was a much easier and performant solution; the tips for the problem definitely did not help.\n\nThe core idea is to compute and store all the partial sums from the left together with the indexes in whi... | 10 | 0 | ['C', 'C++'] | 3 |
minimum-operations-to-reduce-x-to-zero | JAVA | Recursion | Sliding Window | 2 Solutions | java-recursion-sliding-window-2-solution-9yu0 | Recursive Solution: (Please note that it will give time limit exceeded (TLE) if submitted )\n\nclass Solution {\n int ans;\n public int minOperations(int[ | deepmehta | NORMAL | 2021-01-14T14:11:48.554505+00:00 | 2021-01-14T14:11:48.554542+00:00 | 1,432 | false | **Recursive Solution:** (Please note that it will give time limit exceeded (TLE) if submitted )\n```\nclass Solution {\n int ans;\n public int minOperations(int[] nums, int x) {\n ans = Integer.MAX_VALUE;\n recur(nums,0,nums.length-1,x,0);\n return ans==Integer.MAX_VALUE?-1:ans;\n }\n p... | 10 | 1 | ['Recursion', 'Sliding Window', 'Java'] | 3 |
minimum-operations-to-reduce-x-to-zero | C++ prefix sums O(n) | c-prefix-sums-on-by-radheshsarma2299-ahp2 | We can observe that we are deleting some part of prefix(possibly none) and some part of suffix(possibly none) from the given array such that sum of prefix and s | radheshsarma2299 | NORMAL | 2020-11-15T11:58:42.393942+00:00 | 2020-11-15T11:58:42.393971+00:00 | 792 | false | We can observe that we are deleting some part of prefix(possibly none) and some part of suffix(possibly none) from the given array such that sum of prefix and suffix deleted is x\nSo we create a map between the suffix sum and the length of suffix\nNow we iterate from left to right,\nlet say we are at index i, then let ... | 10 | 1 | [] | 1 |
minimum-operations-to-reduce-x-to-zero | C++ Prefix Sum & Binary Search/Unordered map/Sliding Window||w Explanation | c-prefix-sum-binary-searchunordered-maps-0v0g | Intuition\n Describe your first thoughts on how to solve this problem. \nThis code efficiently finds the minimum length of a subarray that sums up to x using pr | anwendeng | NORMAL | 2023-09-20T02:24:44.063519+00:00 | 2023-09-20T08:15:24.233524+00:00 | 1,615 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nThis code efficiently finds the minimum length of a subarray that sums up to x using prefix sums and binary search techniques.\n\nSecond approach does not use binary search. Insteadly the hash table unordered_map plays the role! The TC is... | 9 | 0 | ['Hash Table', 'Binary Search', 'Sliding Window', 'Prefix Sum', 'C++'] | 2 |
minimum-operations-to-reduce-x-to-zero | ✅✅JAVA || Beats 99% || 🔥🔥 DP ||🔥🔥 Sliding Window | java-beats-99-dp-sliding-window-by-sumit-8yme | Intuition\n Describe your first thoughts on how to solve this problem. \nDp will give memory limit exceeded.\nso we need to do it by sliding window\n# Approach\ | sumitdarshanala | NORMAL | 2023-09-20T08:52:15.334562+00:00 | 2023-09-20T08:52:15.334581+00:00 | 764 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nDp will give memory limit exceeded.\nso we need to do it by sliding window\n# Approach\n<!-- Describe your approach to solving the problem. -->\nwe need a { subarray of sum } as \ntarget = total sum - x\nthen max subarray sum of target... | 8 | 0 | ['Dynamic Programming', 'Sliding Window', 'Java'] | 3 |
minimum-operations-to-reduce-x-to-zero | [C++/Full Explanation] Simple Sliding Window || Why DP won't work here explanation | cfull-explanation-simple-sliding-window-198pn | class Solution {\npublic:\n \n\t// We can do this question with the help of DP concept of take from front and back and return minimum of the operation but in | Sameer_111 | NORMAL | 2022-06-11T04:56:18.515188+00:00 | 2022-06-11T04:58:21.424655+00:00 | 940 | false | class Solution {\npublic:\n \n\t// We can do this question with the help of DP concept of take from front and back and return minimum of the operation but in that case it will give TLE \n\t// and here we are not able to optimize due to high constraints 1<=x<=10^9.\n\t// That is why sliding window works here, So we f... | 8 | 0 | ['Two Pointers', 'C', 'Sliding Window', 'Prefix Sum', 'C++'] | 1 |
minimum-operations-to-reduce-x-to-zero | Python Sliding Window Using maxLen of continuous subarray concept | python-sliding-window-using-maxlen-of-co-tbjs | Idea is to do using continous subarray with max length size which return continous subarray length. The elements that are not included in the subarray (exterior | aayuskh | NORMAL | 2021-01-14T23:52:47.652390+00:00 | 2021-01-14T23:52:47.652433+00:00 | 459 | false | Idea is to do using continous subarray with max length size which return continous subarray length. The elements that are not included in the subarray (exterior to it) is the ans. So, in this case len(nums) - maxLen of continous subarray.\nNow x is actually sum of exterior, to do continous subarray we need total sum - ... | 8 | 0 | ['Sliding Window', 'Python'] | 1 |
minimum-operations-to-reduce-x-to-zero | Simple solution by finding the longest array with a target in Java | simple-solution-by-finding-the-longest-a-4trx | Essentially, we need to find the longest array with a sum of - (sum(array) - x). If we do find this, this might either be - \n1. Starting from the left end: In | amoghrajesh1999 | NORMAL | 2021-01-14T11:11:30.273381+00:00 | 2021-01-14T11:11:30.273406+00:00 | 870 | false | Essentially, we need to find the longest array with a sum of - (sum(array) - x). If we do find this, this might either be - \n1. Starting from the left end: In this case, We will have found the answer which has minimum length from right\n2. Somewhere in the middle: Will be a valid answer since we can pick elements from... | 8 | 2 | ['Java'] | 1 |
minimum-operations-to-reduce-x-to-zero | C++ | PREFIX SUM | HASHMAP | O(n) TIME COMPLEXITY | c-prefix-sum-hashmap-on-time-complexity-m3k5i | \nclass Solution {\npublic:\n int minOperations(vector<int>& nums, int x) {\n const int n = nums.size();\n int ans = INT_MAX, sum = 0;\n | UnknownOffline | NORMAL | 2022-06-11T03:01:52.357807+00:00 | 2022-06-11T03:01:52.357837+00:00 | 473 | false | ```\nclass Solution {\npublic:\n int minOperations(vector<int>& nums, int x) {\n const int n = nums.size();\n int ans = INT_MAX, sum = 0;\n unordered_map<int,int> m;\n for(int i = n - 1;i >=0;i--) {\n sum += nums[i];\n m[sum] = i + 1;\n if(sum == x) ans = ... | 7 | 0 | ['C', 'Prefix Sum'] | 1 |
minimum-operations-to-reduce-x-to-zero | Python | Prefix Sum & Sliding Window | With Explanation | python-prefix-sum-sliding-window-with-ex-on9w | Detailed Explanation:\n1. Idea: the aim of this question is to find the minimum operations to reduce x to 0, i.e. find the minimum number of elements from the s | Mikey98 | NORMAL | 2022-06-11T01:19:09.316490+00:00 | 2022-06-11T23:25:13.186463+00:00 | 1,545 | false | Detailed Explanation:\n1. Idea: the aim of this question is to find the minimum operations to reduce `x` to 0, i.e. find the minimum number of elements from the start and end that can sum up to `x`. We can transform this problem to: **find the longest subarray in `nums` that sum to `goal`**, where `goal = sum(nums) - x... | 7 | 0 | ['Sliding Window', 'Python', 'Python3'] | 3 |
minimum-operations-to-reduce-x-to-zero | Javascript | Simple Prefix Sum Sliding Window Solution w/ Explanation| beats 100% / 94% | javascript-simple-prefix-sum-sliding-win-avtg | Idea:\n\nThis problem is tasking us to essentially find when the sum of the starting and ending subarrays of nums is equal to x. Put another way, it\'s asking u | sgallivan | NORMAL | 2021-01-14T23:19:15.114947+00:00 | 2021-01-14T23:56:51.173970+00:00 | 418 | false | ***Idea:***\n\nThis problem is tasking us to essentially find when the sum of the starting and ending subarrays of **nums** is equal to **x**. Put another way, it\'s asking us to find when any subarray of consecutive elements in **nums** is equal to the sum of all elements in **nums** minus **x**.\n```\n ... | 7 | 0 | ['JavaScript'] | 0 |
minimum-operations-to-reduce-x-to-zero | Python, find the longest window | python-find-the-longest-window-by-warmr0-n5u8 | Essentially we\'re looking to remove the longest sequence of numbers so that what\'s left is equal to x. \nReformulating the problem this way it becomes a class | warmr0bot | NORMAL | 2021-01-14T10:18:27.317899+00:00 | 2021-01-14T10:18:27.317938+00:00 | 1,046 | false | Essentially we\'re looking to remove the longest sequence of numbers so that what\'s left is equal to `x`. \nReformulating the problem this way it becomes a classic window sliding problem. \nComplexity Time: O(N), Memory: O(1) \n```\ndef minOperations(self, nums: List[int], x: int) -> int:\n\tN = len(nums)\n\ttoremove ... | 7 | 0 | ['Sliding Window', 'Python', 'Python3'] | 0 |
minimum-operations-to-reduce-x-to-zero | [Python] Clear Two Pointers Solution with Video Explanation | python-clear-two-pointers-solution-with-t1nu6 | Video with clear visualization and explanation:\nhttps://youtu.be/kRHZdzYYn50\n\n\nIntuition: Two Pointers\n\n\nCode\n\nclass Solution:\n def minOperations(s | iverson52000 | NORMAL | 2020-11-15T23:26:11.418370+00:00 | 2020-11-15T23:26:11.418407+00:00 | 787 | false | Video with clear visualization and explanation:\nhttps://youtu.be/kRHZdzYYn50\n\n\nIntuition: Two Pointers\n\n\n**Code**\n```\nclass Solution:\n def minOperations(self, nums: List[int], x: int) -> int:\n target = sum(nums)-x\n n = len(nums)\n\n if target < 0: return -1\n if target == 0: r... | 7 | 1 | ['Python'] | 2 |
minimum-operations-to-reduce-x-to-zero | [C++/Java] 2-Pointer O(n) Solutions! | cjava-2-pointer-on-solutions-by-admin007-fzcx | It is easy to observe that in order to find the shortest required operations that sum up to x is finding the longest subarray that sums up to total - x. \n\nApp | admin007 | NORMAL | 2020-11-15T05:15:52.314317+00:00 | 2020-11-15T05:32:20.637861+00:00 | 1,223 | false | It is easy to observe that in order to find the shortest required operations that sum up to *x* is finding the longest subarray that sums up to *total - x*. \n\n**Approach 1 Find the longest subarray sum equals to a target value:**\n\n**C++**\n```\npublic:\n int minOperations(vector<int>& nums, int x) {\n int... | 7 | 0 | ['C', 'Java'] | 1 |
minimum-operations-to-reduce-x-to-zero | Convert to calculate longest subarray with fixed target sum | convert-to-calculate-longest-subarray-wi-para | Just convert to another question:\nThe longest subarray with sum = (sum of nums - x);\n\nThen the question is much easier. Just play the trick of calculating p | nadabao | NORMAL | 2020-11-15T04:07:18.892713+00:00 | 2020-11-15T05:06:00.559015+00:00 | 442 | false | Just convert to another question:\nThe longest subarray with sum = (sum of nums - x);\n\nThen the question is much easier. Just play the trick of calculating presum and store in map. \nFor this type of question, please refer to leetcode 560:\nhttps://leetcode.com/problems/subarray-sum-equals-k/\n\n```\nclass Solution ... | 7 | 1 | [] | 0 |
minimum-operations-to-reduce-x-to-zero | [Javascript] Two pointers O(n) | javascript-two-pointers-on-by-alanchangh-vyv6 | ```\nvar minOperations = function(nums, x) {\n let l = 0,\n r = nums.length-1,\n res = nums.length+1;\n \n let n = x;\n \n while (l | alanchanghsnu | NORMAL | 2020-11-15T04:01:49.122473+00:00 | 2020-11-15T04:13:04.925451+00:00 | 999 | false | ```\nvar minOperations = function(nums, x) {\n let l = 0,\n r = nums.length-1,\n res = nums.length+1;\n \n let n = x;\n \n while (l <= r && n-nums[l] >= 0)\n n -= nums[l++];\n \n if (n === 0)\n res = l;\n \n while (l < r && l >= 0) {\n while (n-nums[r] >= 0)... | 7 | 1 | ['Two Pointers', 'JavaScript'] | 0 |
minimum-operations-to-reduce-x-to-zero | Best O(N) Solution | best-on-solution-by-kumar21ayush03-sj57 | Approach\nSliding Window Technique\n\n# Complexity\n- Time complexity:\nO(n)\n\n- Space complexity:\nO(1)\n\n# Code\n\nclass Solution {\npublic:\n int minOpe | kumar21ayush03 | NORMAL | 2023-09-21T10:32:50.500211+00:00 | 2023-09-21T10:32:50.500238+00:00 | 638 | false | # Approach\nSliding Window Technique\n\n# Complexity\n- Time complexity:\n$$O(n)$$\n\n- Space complexity:\n$$O(1)$$\n\n# Code\n```\nclass Solution {\npublic:\n int minOperations(vector<int>& nums, int x) {\n int n = nums.size();\n int sum = 0;\n for (int i = 0; i < n; i++)\n sum += nu... | 6 | 0 | ['C++'] | 0 |
minimum-operations-to-reduce-x-to-zero | Simple Java Solution | faster than 95.37% | simple-java-solution-faster-than-9537-by-slbr | Finding the minimum number of removing is equal to finding the maximum number of remaining.\nSliding window can help us find the maximum number of remaining eas | angzhanh | NORMAL | 2022-06-11T01:32:21.614252+00:00 | 2022-06-11T01:32:35.768773+00:00 | 1,279 | false | Finding the minimum number of removing is equal to finding the maximum number of remaining.\nSliding window can help us find the maximum number of remaining easily!\n```\nclass Solution {\n public int minOperations(int[] nums, int x) {\n int n = nums.length;\n \n int sum = 0;\n for (int i... | 6 | 0 | ['Sliding Window', 'Java'] | 0 |
minimum-operations-to-reduce-x-to-zero | C++ | O(1) Space | Faster than 97% | Contiguous subarray sum | Unique Approach | c-o1-space-faster-than-97-contiguous-sub-mcip | Explanantion\n Step 1 - Find the SUM of whole array. \n Step 2 - Then, Subtract the given element from the total SUM.\n Step 3 - Now, All we have to do is take | nishsolvesalgo | NORMAL | 2021-09-05T09:35:19.470407+00:00 | 2021-09-05T09:36:30.204806+00:00 | 424 | false | **Explanantion**\n* **Step 1** - Find the ***SUM*** of whole array. \n* **Step 2** - Then, Subtract the given element from the total ***SUM***.\n* **Step 3** - Now, All we have to do is take a left and right pointer traverse through the array and find Contiguous subarrays sum (i.e include all elements from left to righ... | 6 | 0 | ['C', 'Sliding Window'] | 1 |
minimum-operations-to-reduce-x-to-zero | Minimum Operations to Reduce X to Zero || C++ || Easy || 99%beat || Hash map | minimum-operations-to-reduce-x-to-zero-c-mki6 | \nAs we count the sum of the first l numbers, you need to find a complement (x - sum) number formed by numbers on the right. To do so, we can use a hashmap wher | thekalyan001 | NORMAL | 2021-01-14T10:30:01.559669+00:00 | 2021-01-14T10:30:13.194170+00:00 | 520 | false | \nAs we count the sum of the first l numbers, you need to find a complement (x - sum) number formed by numbers on the right. To do so, we can use a hashmap where we will store the rolling sum.\nUpvote if you liked :)\n```\nint minOperations(vector<int>& nums, int x) {\n unordered_map<int, int> left;\n int res = I... | 6 | 1 | ['C'] | 0 |
minimum-operations-to-reduce-x-to-zero | C++ Prefix Sum Solution + Binary Search | c-prefix-sum-solution-binary-search-by-p-2vgx | \nclass Solution {\npublic:\n int minOperations(vector<int>& nums, int x) {\n \n if(nums.size() == 1){\n if(x == nums[0])\n | pikabu | NORMAL | 2020-11-15T04:32:35.604408+00:00 | 2020-11-15T04:32:35.604439+00:00 | 408 | false | ```\nclass Solution {\npublic:\n int minOperations(vector<int>& nums, int x) {\n \n if(nums.size() == 1){\n if(x == nums[0])\n return 1;\n return -1;\n }\n if(nums[0] > x && nums[nums.size() -1] > x)\n return -1;\n \n vector<lo... | 6 | 0 | [] | 0 |
minimum-operations-to-reduce-x-to-zero | JAVA TC: 100% beats , MC: 95% beat, sliding window | java-tc-100-beats-mc-95-beat-sliding-win-e2j1 | sum_total : totla sum of array.\nwe need to find longest lenghted subarray which sums to (sum_total -x)\n\nwe can find the the largest lengthhed subarray using | jerryjohnthomastvm | NORMAL | 2022-06-11T17:06:16.748288+00:00 | 2022-06-11T18:03:38.642017+00:00 | 735 | false | sum_total : totla sum of array.\nwe need to find longest lenghted subarray which sums to (sum_total -x)\n\nwe can find the the largest lengthhed subarray using sliding window,\nmove left pointer when the sum>=(required) otherwise keep on increasing right pointer\n\nAns. total_length - length of largest sub_array with s... | 5 | 0 | ['Sliding Window', 'Java'] | 0 |
minimum-operations-to-reduce-x-to-zero | C++ || EASY TO UNDERSTAND || Prefix Sum + Suffix Sum + Binary Search || 2 Approaches | c-easy-to-understand-prefix-sum-suffix-s-5vmi | Time Complexity O(NlogN)\n\nclass Solution {\npublic:\n int minOperations(vector<int>& nums, int x) {\n int n=nums.size();\n vector<int> pre(n, | aarindey | NORMAL | 2022-06-11T10:29:04.317277+00:00 | 2022-06-11T13:18:41.454643+00:00 | 252 | false | **Time Complexity O(NlogN)**\n```\nclass Solution {\npublic:\n int minOperations(vector<int>& nums, int x) {\n int n=nums.size();\n vector<int> pre(n,0);\n vector<int> suff(n,0);\n pre[0]=nums[0];\n suff[n-1]=nums[n-1];\n for(int i=1;i<n;i++)\n {\n pre[i]=p... | 5 | 0 | [] | 1 |
minimum-operations-to-reduce-x-to-zero | C++ Solution Using Two Pointer | c-solution-using-two-pointer-by-conquist-fnyp | C++ Solution Using Two Pointer\n\nclass Solution {\npublic:\n int minOperations(vector<int>& nums, int x) {\n int n=nums.size();\n int left=0,r | Conquistador17 | NORMAL | 2022-06-11T05:28:57.893469+00:00 | 2022-06-11T05:28:57.893503+00:00 | 1,363 | false | # **C++ Solution Using Two Pointer**\n```\nclass Solution {\npublic:\n int minOperations(vector<int>& nums, int x) {\n int n=nums.size();\n int left=0,right=0;\n int sum=0;\n int ans=-1;\n for(int i=0;i<n;i++){\n sum+=nums[i];\n }\n \n cout<<sum;\n ... | 5 | 0 | ['Two Pointers', 'C', 'C++'] | 1 |
minimum-operations-to-reduce-x-to-zero | ✅ C++ | O(N) | Easy to understand | c-on-easy-to-understand-by-kjs_27-ibj1 | To solve this problem first we need to calculate the maximum length of subarray with sum which is equal to difference of total sum and x. \n\nExample:\nnums = [ | kjs_27 | NORMAL | 2022-06-11T05:20:39.458647+00:00 | 2022-06-18T20:08:29.139476+00:00 | 774 | false | To solve this problem first we need to calculate the maximum length of subarray with sum which is equal to difference of total sum and x. \n\n**Example:**\n**nums** = [3,2,5,4,11,5]\n**x** = 10\n\nTotal sum = 3 + 2 + 5 + 4 + 11 + 5 = 30\nSum - x = 20\n\nLength of maximum subarray with sum = 20 is 3 as shown;\n along with memoization\nBut it gives TLE . The constraints are large for test cases.\n\nTime Comple | nitin23rathod | NORMAL | 2022-06-11T03:47:25.012048+00:00 | 2022-06-11T03:47:25.012085+00:00 | 288 | false | # **Approach 1 :- DP**\nUsed the classic Top Down DP(Knapsack 0/1) along with memoization\n***But it gives TLE*** . The constraints are large for test cases.\n\n**Time Complexity O(N^2).\nSpace Complexity O(N^2).**\n\n```\nclass Solution {\npublic:\n vector<vector<int>>dp;\n int solve(vector<int>&nums,int s,int e... | 5 | 0 | ['C'] | 0 |
minimum-operations-to-reduce-x-to-zero | DP Top Down and Sliding window Solutions in C++ | dp-top-down-and-sliding-window-solutions-dja0 | Approach 1 DP\nUsed the classic Top Down DP(Knapsack 0/1) along with memoization\nBut it gives TLE . The constraints are large for test cases.\n\nTime Complexit | ManthanJoshi | NORMAL | 2021-09-04T11:17:27.573445+00:00 | 2021-09-04T11:19:26.376837+00:00 | 699 | false | **Approach 1** **DP**\nUsed the classic Top Down DP(**Knapsack 0/1**) along with memoization\nBut it gives **TLE** . The constraints are large for test cases.\n\n**Time Complexity O(N^2)\nSpace Complexity O(N^2)**\n```\n// Classic Knapsack 0/1 problem\nclass Solution {\npublic:\n vector<vector<int>>dp;\n int sol... | 5 | 0 | ['Dynamic Programming', 'Sliding Window'] | 0 |
minimum-operations-to-reduce-x-to-zero | Minimum Operations to Reduce X to Zero | C++ | Finding Maximum Sub-Array | 99% beat | minimum-operations-to-reduce-x-to-zero-c-afk4 | As stated by the first hint: "Think in reverse; instead of finding the minimum prefix + suffix, find the maximum subarray." \n\nThis code uses a sliding window | yongweic | NORMAL | 2021-01-15T02:08:24.505841+00:00 | 2021-01-15T02:08:24.505872+00:00 | 186 | false | As stated by the first hint: "Think in reverse; instead of finding the minimum prefix + suffix, find the maximum subarray." \n\nThis code uses a sliding window to find the maximum subarray.\n```\nclass Solution {\npublic:\n int minOperations(vector<int>& nums, int x) {\n int l = 0;\n int r = 0;\n ... | 5 | 2 | [] | 1 |
minimum-operations-to-reduce-x-to-zero | C++ Simple and Easy Solution | c-simple-and-easy-solution-by-yehudisk-bzwu | \nclass Solution {\npublic:\n int findMaxSub(vector<int>& nums, int k) {\n int sum = 0, maxi = 0;\n unordered_map<int, int> m;\n \n | yehudisk | NORMAL | 2021-01-14T12:48:50.437026+00:00 | 2021-01-14T12:48:50.437063+00:00 | 601 | false | ```\nclass Solution {\npublic:\n int findMaxSub(vector<int>& nums, int k) {\n int sum = 0, maxi = 0;\n unordered_map<int, int> m;\n \n for (int i=0; i<nums.size(); i++) {\n sum += nums[i];\n \n if (sum == k)\n maxi = i+1;\n \n ... | 5 | 1 | ['C'] | 0 |
minimum-operations-to-reduce-x-to-zero | 🔥2 Approaches💯 || 🌟O(1) SC || 💯Clean & Best Code✅ | 2-approaches-o1-sc-clean-best-code-by-ad-16ug | Connect with me on LinkedIN : https://www.linkedin.com/in/aditya-jhunjhunwala-51b586195/\n# Complexity\n\n- Time complexity:\nO(n)\n\n- Space complexity:\nO(n) | aDish_21 | NORMAL | 2023-09-20T06:34:59.105751+00:00 | 2023-09-20T14:21:37.252110+00:00 | 411 | false | ### Connect with me on LinkedIN : https://www.linkedin.com/in/aditya-jhunjhunwala-51b586195/\n# Complexity\n```\n- Time complexity:\nO(n)\n\n- Space complexity:\nO(n) for Approach 1\nO(1) for Approach 2\n```\n\n# Code\n## Please Upvote if u found it useful\uD83E\uDD17\n# 1st Approach(Using PrefixSum + HashMap):-\n```\n... | 4 | 0 | ['Array', 'Hash Table', 'Sliding Window', 'Prefix Sum', 'C++'] | 1 |
minimum-operations-to-reduce-x-to-zero | EASY SOLUTION || EXPLANATION || C++ || Beat 98.9 % || | easy-solution-explanation-c-beat-989-by-oxs8m | \n# Approach\n Describe your approach to solving the problem. \n##### Try to find maximum subarray that have the sum == total_sum of array - x\n\n# Code\n\nclas | amol_2004 | NORMAL | 2023-09-20T05:08:43.772745+00:00 | 2023-09-20T05:08:43.772774+00:00 | 327 | false | \n# Approach\n<!-- Describe your approach to solving the problem. -->\n##### Try to find maximum subarray that have the `sum == total_sum of array - x`\n\n# Code\n```\nclass Solution {\npublic:\n int minOperations(vector<int>& nums, int x) {\n int total_sum = 0;\n int n = nums.size();\n for(int ... | 4 | 0 | ['Two Pointers', 'Sliding Window', 'C++'] | 1 |
minimum-operations-to-reduce-x-to-zero | 🔥98.80% Acceptance📈 rate | Using Sliding Window🧭 | Explained in details✅ | 9880-acceptance-rate-using-sliding-windo-3jwg | Intuition\n Describe your first thoughts on how to solve this problem. \nThe intuition behind this code is to find a subarray with the maximum possible sum (equ | nikkipriya_78 | NORMAL | 2023-09-20T04:00:01.834799+00:00 | 2023-10-07T13:55:41.988135+00:00 | 292 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nThe intuition behind this code is to find a subarray with the maximum possible sum (equal to targetSum) within the given nums array. The goal is to minimize the number of operations to reduce x to 0. By calculating the maximum subarray le... | 4 | 0 | ['Array', 'Hash Table', 'Binary Search', 'Sliding Window', 'Prefix Sum', 'Python', 'C++', 'Java', 'Python3', 'JavaScript'] | 0 |
minimum-operations-to-reduce-x-to-zero | very easy to understand solution O(N) time, O(1) SPACE | very-easy-to-understand-solution-on-time-9o1g | Intuition\nwe can take only elements which are present at corner i.e in starting and in last so we try to find window of (total sum of array - k) \nso , n-k win | abhinav_raj531 | NORMAL | 2023-09-20T02:39:31.367671+00:00 | 2023-09-20T02:47:53.560308+00:00 | 134 | false | # Intuition\nwe can take only elements which are present at corner i.e in starting and in last so we try to find window of (total sum of array - k) \nso , n-k window, will contain sum == k. \n\n# Approach\nhere, we are using sliding window approch to find the largest window of elments that contains (total sum of array ... | 4 | 0 | ['Sliding Window', 'C++'] | 0 |
minimum-operations-to-reduce-x-to-zero | C++ | Minimum Operations to Reduce X to Zero | 2 Approaches with Explanation | c-minimum-operations-to-reduce-x-to-zero-o2a2 | Approach 1\n\t1. Maintain two cumulative sum left and right . Also check the edge cases (if x is present in the left or right)\n\t2. Go from start to end of lef | badhansen | NORMAL | 2022-06-11T09:59:41.682010+00:00 | 2022-06-11T10:53:53.392667+00:00 | 189 | false | **Approach 1**\n\t1. Maintain two cumulative sum `left` and `right` . Also check the edge cases (if x is present in the `left` or `right`)\n\t2. Go from start to end of `left` and check if remaining is present or not in the `right` one.\n\t3. `ans` will be sum of two indices.\n\n**Time:** `O(N log N)`\n**Space:** `O(N... | 4 | 0 | ['Binary Search', 'Sliding Window', 'Binary Tree'] | 1 |
minimum-operations-to-reduce-x-to-zero | Kotlin || O(n) time || O(1) space | kotlin-on-time-o1-space-by-yezhizhen-9w5q | \n fun minOperations(nums: IntArray, x: Int): Int {\n //find longest subarray sum to total_sum - x\n var target = nums.sum() - x; var sum = 0\n | yezhizhen | NORMAL | 2022-06-11T07:31:04.772806+00:00 | 2022-06-11T07:31:04.772869+00:00 | 99 | false | ```\n fun minOperations(nums: IntArray, x: Int): Int {\n //find longest subarray sum to total_sum - x\n var target = nums.sum() - x; var sum = 0\n var l = 0; var ans = -1;\n for(r in 0 until nums.size)\n {\n sum += nums[r]\n while(l <= r && sum > target) //mov... | 4 | 0 | ['Kotlin'] | 1 |
minimum-operations-to-reduce-x-to-zero | Binary Search | c++ | O(nlogn) | binary-search-c-onlogn-by-vivek_9680-f3s2 | First, I thought it\'s dp but when I saw the constraint then i came to know that it\'s not dp.\nWhen we see closely, see the prefix[i] + suffix[j]==x then i + ( | vivek_9680 | NORMAL | 2022-06-11T07:01:44.353174+00:00 | 2022-06-11T07:45:41.794587+00:00 | 198 | false | First, I thought it\'s dp but when I saw the constraint then i came to know that it\'s not dp.\nWhen we see closely, see the prefix[i] + suffix[j]==x then i + (n-j) is our answer.\nI am giving the example:\n[1,1,4,2,3] and x = 5\npre = [0,1,2,6,8,11] //add 0 for simplicity\nsuff = [11,10,9,5,3,0] \n\nIn this scenario t... | 4 | 0 | ['C', 'Binary Tree'] | 3 |
minimum-operations-to-reduce-x-to-zero | C++ || Sliding window || 1658. Minimum Operations to Reduce X to Zero || Easy | c-sliding-window-1658-minimum-operations-llf8 | \t// Here ans is the window size we have to maximize the window size\n\t// because more we maximize window size lesser the elements willl be remainig in nums\n\ | anubhavsingh11 | NORMAL | 2022-06-11T06:40:26.363171+00:00 | 2022-06-11T06:41:00.556899+00:00 | 193 | false | \t// Here ans is the window size we have to maximize the window size\n\t// because more we maximize window size lesser the elements willl be remainig in nums\n\t// ans hence our no of operation will be reduced\n\t// Here we are checking for window of max size for sum=totalsum-x \n\t//so that sum of remainig elements b... | 4 | 0 | ['Two Pointers', 'C', 'Sliding Window'] | 0 |
minimum-operations-to-reduce-x-to-zero | Python | reverse thinking | prefix sum + hash map | with comments | python-reverse-thinking-prefix-sum-hash-j7wd3 | When I first read this problem, I think I have seen it elsewhere. Then I realize that it\'s pretty similar to Leetcode 325 Maximum Size Subarray Sum Equals k. \ | BoluoUp | NORMAL | 2022-06-11T05:09:37.422625+00:00 | 2022-06-11T05:09:37.422667+00:00 | 633 | false | When I first read this problem, I think I have seen it elsewhere. Then I realize that it\'s pretty similar to *Leetcode 325 Maximum Size Subarray Sum Equals k*. \n\nTo get the minimum operations to reduce x to 0, we can get the longest length of subarray sum equals "new" target ```sum(nums) - x```. The answer will be `... | 4 | 0 | ['Prefix Sum', 'Python'] | 0 |
minimum-operations-to-reduce-x-to-zero | C++ || Easy Solution || Sliding window || Explained Commented solution | c-easy-solution-sliding-window-explained-6xnv | In this problem you have to find the sum of end points is equal to x and we have to minimize the number of endpoints\n\nSo sum of end points = sum of all elemen | rahulrohilla0204 | NORMAL | 2022-06-11T04:32:47.016439+00:00 | 2022-06-11T04:32:47.016471+00:00 | 483 | false | In this problem you have to find the sum of end points is equal to x and we have to minimize the number of endpoints\n\nSo sum of end points = sum of all elements - sum of subarray i.e. sum of window\nand if we maximize the window the number of end points will be minimum\nso this is the basic idea\n\n| PLEASE UPVOTE I... | 4 | 0 | ['C', 'Sliding Window'] | 1 |
minimum-operations-to-reduce-x-to-zero | JAVA 20 lines of code | O(n) time and constant space | Sliding window | Continuous subarray | java-20-lines-of-code-on-time-and-consta-r6mk | The idea here is to find maximun length continuous subarray which can give sum equals to Target. Here, target is defined as:\nTarget = (Sum of array elements - | arjun8900 | NORMAL | 2021-04-03T10:27:12.547010+00:00 | 2021-04-07T16:39:23.081438+00:00 | 178 | false | The idea here is to find **maximun length continuous subarray** which can give sum equals to **Target**. Here, target is defined as:\n**Target = (Sum of array elements - x)**\n\n```\npublic int minOperations(int[] nums, int x) {\n int sum = 0;\n for(int i=0; i<nums.length; i++){\n sum += nums[... | 4 | 0 | [] | 0 |
minimum-operations-to-reduce-x-to-zero | Python 95% - Kadane's Algorithm | python-95-kadanes-algorithm-by-vvvirenyu-md0m | template similar to LC 209 Minimum Size Subarray Sum\n\n\tdef minOperations(self, nums: List[int], x: int) -> int:\n \n l, curr_sum, n, target, a | vvvirenyu | NORMAL | 2021-01-15T03:44:43.978732+00:00 | 2021-01-15T03:45:39.337103+00:00 | 119 | false | template similar to LC 209 [Minimum Size Subarray Sum](https://leetcode.com/problems/minimum-size-subarray-sum/)\n\n\tdef minOperations(self, nums: List[int], x: int) -> int:\n \n l, curr_sum, n, target, ans = 0, 0, len(nums), sum(nums)-x, -1\n\n if target < 0: return -1\n if target == 0: r... | 4 | 0 | [] | 2 |
minimum-operations-to-reduce-x-to-zero | [Java] Prefix sum with Map and Sliding window beats 100% | java-prefix-sum-with-map-and-sliding-win-t7pp | We can only pick some elements from prefix of array and some elements from suffix of array that add up to x and we need the count of such elements to be minimum | shk10 | NORMAL | 2021-01-14T18:12:15.050941+00:00 | 2021-01-14T18:12:15.050977+00:00 | 343 | false | We can only pick some elements from prefix of array and some elements from suffix of array that add up to ```x``` and we need the count of such elements to be minimum. Another way to look at this problem is that we need to find the maximum length subarray that add up to ```sum(nums) - x``` because this is the middle ar... | 4 | 0 | ['Java'] | 0 |
minimum-operations-to-reduce-x-to-zero | C++ Solution | O(n) time, O(1) memory | c-solution-on-time-o1-memory-by-sgarg367-3k95 | Logic:\nWe need to pick l elements from left and r elements from right such that the sum is equal to x subject to minimizing (l+r)\n1) First of all pick element | sgarg367 | NORMAL | 2021-01-14T16:28:27.063877+00:00 | 2021-01-15T06:29:07.770349+00:00 | 379 | false | **Logic:**\nWe need to pick l elements from left and r elements from right such that the sum is equal to x subject to minimizing (l+r)\n1) First of all pick elements from left till sum is less than x\n2) Then keep on subtracting element from left and add element from right one by one based on condition:\n```\nif(sum >=... | 4 | 1 | [] | 2 |
minimum-operations-to-reduce-x-to-zero | JavaScript Clean Sliding Window Solution | javascript-clean-sliding-window-solution-0k2n | Approach:\nFind the longest subarray where: sum of subarray = sum of the array - x\n\nTime complexity: O(N)\nSpace complexity: O(1)\njavascript\nvar minOperatio | control_the_narrative | NORMAL | 2021-01-14T15:42:18.351349+00:00 | 2021-01-14T15:42:18.351382+00:00 | 459 | false | Approach:\nFind the longest subarray where: sum of subarray = sum of the array - x\n\nTime complexity: O(N)\nSpace complexity: O(1)\n```javascript\nvar minOperations = function(nums, x) {\n const sum = nums.reduce((acc, cur) => acc + cur, 0);\n const target = sum - x;\n \n if(target < 0) return -1;\n if(... | 4 | 1 | ['Sliding Window', 'JavaScript'] | 1 |
minimum-operations-to-reduce-x-to-zero | Java O(n) solution with prefix sum and hashmap | java-on-solution-with-prefix-sum-and-has-txi5 | \nclass Solution {\n public int minOperations(int[] nums, int x) {\n int len = nums.length;\n // only left;\n int min = Integer.MAX_VALU | yzhou78 | NORMAL | 2021-01-14T09:22:44.979064+00:00 | 2021-01-14T09:22:44.979096+00:00 | 371 | false | ```\nclass Solution {\n public int minOperations(int[] nums, int x) {\n int len = nums.length;\n // only left;\n int min = Integer.MAX_VALUE;\n Map<Integer, Integer> map = new HashMap<>(); \n int[] left_sum = new int[len];\n for (int i = 0; i < len; i++) {\n if (i... | 4 | 1 | [] | 0 |
minimum-operations-to-reduce-x-to-zero | c++ solution using map | c-solution-using-map-by-sanjeev1709912-pmdi | In the first look this question give me the wibes of DP problem in which a person can use standered recurssion and memorization to get answer but the problem he | sanjeev1709912 | NORMAL | 2020-11-15T05:18:54.766406+00:00 | 2020-11-15T05:48:43.803392+00:00 | 188 | false | In the first look this question give me the wibes of DP problem in which a person can use standered recurssion and memorization to get answer but the problem here is constrains which does not allow us to use recurssion with memorization.\n\nSo what come next so my next approch is to use prefix sum by using to arrays on... | 4 | 0 | [] | 3 |
minimum-operations-to-reduce-x-to-zero | Python - O(n) time, O(1) memory, two pointers, no extra tools | python-on-time-o1-memory-two-pointers-no-zcjk | First of all, the order of operations does not matter: we just need to find left and right parts of the list that form x. Their cumulative length is the number | ceay | NORMAL | 2020-11-15T04:18:19.913836+00:00 | 2020-11-16T04:22:29.821399+00:00 | 412 | false | First of all, the order of operations does not matter: we just need to find left and right parts of the list that form x. Their cumulative length is the number of operations.\n\nTwo pointers - left and right. The left pointer points at the end of the left part of the list that is included in x, and the right pointer po... | 4 | 0 | [] | 2 |
minimum-operations-to-reduce-x-to-zero | Python O(n) Sliding Window Passed, DP O(n^2) TLE | python-on-sliding-window-passed-dp-on2-t-29tt | The first time I saw this problem, DP jumped into my mind and I spent 30min+ debugging big test cases. However it\'s a O(n^2) solution/aka 10^10 in this case an | kop_fyf | NORMAL | 2020-11-15T04:07:54.037507+00:00 | 2020-11-15T16:50:16.192635+00:00 | 271 | false | The first time I saw this problem, DP jumped into my mind and I spent 30min+ debugging big test cases. However it\'s a O(n^2) solution/aka 10^10 in this case and it got TLE. On the other hand, sliding window just takes O(n). **The problem can be translated into: find the longest subarray with sum = sum(nums) - x**\n``... | 4 | 0 | [] | 1 |
minimum-operations-to-reduce-x-to-zero | EASY SLIDING WINDOW TECHNIQUE | easy-sliding-window-technique-by-2005115-vwdf | PLEASE UPVOTE MY SOLUTION IF YOU LIKE IT\n# CONNECT WITH ME\n### https://www.linkedin.com/in/pratay-nandy-9ba57b229/\n### https://www.instagram.com/pratay_nandy | 2005115 | NORMAL | 2023-09-20T02:51:40.777581+00:00 | 2023-09-22T09:54:56.286953+00:00 | 712 | false | # **PLEASE UPVOTE MY SOLUTION IF YOU LIKE IT**\n# **CONNECT WITH ME**\n### **[https://www.linkedin.com/in/pratay-nandy-9ba57b229/]()**\n### **[https://www.instagram.com/pratay_nandy/]()**\n# Approach\n`minOperations` that takes a vector of integers nums and an integer x as input. The goal of this function is to find th... | 3 | 0 | ['Array', 'Sliding Window', 'C++'] | 2 |
minimum-operations-to-reduce-x-to-zero | Python3 Solution | python3-solution-by-motaharozzaman1996-4sht | \n\nclass Solution:\n def minOperations(self, nums: List[int], x: int) -> int:\n n=len(nums)\n target=sum(nums)-x\n dp=[0]\n res= | Motaharozzaman1996 | NORMAL | 2023-09-20T00:11:18.485542+00:00 | 2023-09-20T00:11:18.485574+00:00 | 448 | false | \n```\nclass Solution:\n def minOperations(self, nums: List[int], x: int) -> int:\n n=len(nums)\n target=sum(nums)-x\n dp=[0]\n res=0\n for num in nums:\n res+=num\n dp.append(res)\n seen={v:i for i,v in enumerate(dp)}\n ans=-1\n for l_val... | 3 | 0 | ['Python', 'Python3'] | 1 |
minimum-operations-to-reduce-x-to-zero | CPP || Variation || Max_len window with target | cpp-variation-max_len-window-with-target-awt2 | \nint solve(vector<int>& nums, int target)\n{\n // Initializing variables\n long long i = 0, j = 0, sum = 0, ans = INT_MIN, n = nums.size();\n\n // Usi | rai-Tarunesh | NORMAL | 2023-07-06T06:58:51.505362+00:00 | 2023-07-06T06:58:51.505397+00:00 | 283 | false | ```\nint solve(vector<int>& nums, int target)\n{\n // Initializing variables\n long long i = 0, j = 0, sum = 0, ans = INT_MIN, n = nums.size();\n\n // Using a sliding window approach\n while (j < n)\n {\n sum += nums[j]; // Add the current element to the sum\n\n if (sum == target) {\n ... | 3 | 0 | ['Sliding Window', 'C++'] | 0 |
minimum-operations-to-reduce-x-to-zero | ✅C++ || Easy Sliding Window Solution || Logical | c-easy-sliding-window-solution-logical-b-jjz2 | \nclass Solution {\npublic:\n int minOperations(vector<int>& nums, int x) {\n \n int sum = accumulate(nums.begin(), nums.end(), 0);\n \n | rishit_30g | NORMAL | 2022-06-30T02:39:33.198215+00:00 | 2022-06-30T02:39:33.198258+00:00 | 354 | false | ```\nclass Solution {\npublic:\n int minOperations(vector<int>& nums, int x) {\n \n int sum = accumulate(nums.begin(), nums.end(), 0);\n \n if(sum < x){\n return -1; \n }\n \n if(sum == x){\n return nums.size();\n }\n \n int ... | 3 | 0 | ['Two Pointers', 'C', 'Sliding Window', 'Prefix Sum', 'C++'] | 0 |
minimum-operations-to-reduce-x-to-zero | C# | Prefix Sum | Dictionary | c-prefix-sum-dictionary-by-yogesh_bodhe-73dt | \npublic class Solution {\n public int MinOperations(int[] nums, int x) {\n long sum = 0;\n int max = -1;\n \n foreach (int num i | yogesh_bodhe | NORMAL | 2022-06-11T06:43:39.149061+00:00 | 2022-06-11T06:43:39.149107+00:00 | 127 | false | ```\npublic class Solution {\n public int MinOperations(int[] nums, int x) {\n long sum = 0;\n int max = -1;\n \n foreach (int num in nums) {\n sum += num;\n }\n \n long target = sum - x;\n Dictionary<long, int> dict = new Dictionary<long, int>();\n ... | 3 | 0 | ['Hash Table', 'Prefix Sum'] | 1 |
minimum-operations-to-reduce-x-to-zero | Can anybody help me in this code🙂? Why it is failing ? | can-anybody-help-me-in-this-code-why-it-5r4u5 | It is giving wrong answer in this test case\n[5207,5594,477,6938,8010,7606,2356,6349,3970,751,5997,6114,9903,3859,6900,7722,2378,1996,8902,228,4461,90,7321,7893 | ritikakhanduri01 | NORMAL | 2022-06-11T06:32:50.048495+00:00 | 2022-06-11T06:35:21.602976+00:00 | 161 | false | **It is giving wrong answer in this test case**\n[5207,5594,477,6938,8010,7606,2356,6349,3970,751,5997,6114,9903,3859,6900,7722,2378,1996,8902,228,4461,90,7321,7893,4879,9987,1146,8177,1073,7254,5088,402,4266,6443,3084,1403,5357,2565,3470,3639,9468,8932,3119,5839,8008,2712,2735,825,4236,3703,2711,530,9630,1521,2174,502... | 3 | 0 | ['Two Pointers', 'C', 'C++'] | 2 |
minimum-operations-to-reduce-x-to-zero | JavaScript (JS) solution 🔥🔥🔥 Beats 100% 💯💯💯 | javascript-js-solution-beats-100-by-abas-8upg | JavaScript solution\n\n\nvar minOperations = function(nums, x) {\n const n = nums.length;\n const sum = nums.reduce((r, n) => r + n, 0);\n const target | abashev | NORMAL | 2022-06-11T05:28:35.260205+00:00 | 2022-06-11T17:42:43.763163+00:00 | 362 | false | JavaScript solution\n\n```\nvar minOperations = function(nums, x) {\n const n = nums.length;\n const sum = nums.reduce((r, n) => r + n, 0);\n const target = sum - x;\n\n let current = 0;\n let ans = -1;\n \n for (let l = 0, r = 0; r < n; r++) {\n current += nums[r];\n \n while ... | 3 | 0 | ['Two Pointers', 'Sliding Window', 'Prefix Sum', 'JavaScript'] | 2 |
minimum-operations-to-reduce-x-to-zero | Easy CPP Prefix sum , Brute Force(Partial execute),Top-down(TimeLimit Exceeded) Explained | easy-cpp-prefix-sum-brute-forcepartial-e-1lao | Brute Force ( Which is not working for all cases ) using two pointer Approach\n\n\nInitially I tought lets check from endings that by comparing two pointers lef | PradeepMission23May8 | NORMAL | 2022-05-22T06:22:17.018695+00:00 | 2022-05-22T08:10:13.954099+00:00 | 467 | false | **Brute Force ( Which is not working for all cases ) using two pointer Approach**\n\n```\nInitially I tought lets check from endings that by comparing two pointers left and right\nbuy taking Maximum number and moving the pointer. It falis in the case of \n\n[3,1,1,4,4,1] k=5 here we can see as per logic it goes to sele... | 3 | 0 | ['Dynamic Programming', 'Prefix Sum', 'C++'] | 0 |
minimum-operations-to-reduce-x-to-zero | Simple C++ code ||O(N) time || O(1) space | simple-c-code-on-time-o1-space-by-prosen-0cq4 | If you like the implementation then Please help me by increasing my reputation. By clicking the up arrow on the left of my image.\n\nclass Solution {\npublic:\n | _pros_ | NORMAL | 2022-05-16T14:49:33.546629+00:00 | 2022-05-16T14:49:33.546675+00:00 | 133 | false | # **If you like the implementation then Please help me by increasing my reputation. By clicking the up arrow on the left of my image.**\n```\nclass Solution {\npublic:\n int minOperations(vector<int>& nums, int x) {\n int n = nums.size();\n int left = -1, right = n-1, tmp = x, store = INT_MAX;\n ... | 3 | 0 | ['Sliding Window'] | 0 |
minimum-operations-to-reduce-x-to-zero | Python easy to read and understand | python-easy-to-read-and-understand-by-sa-tbdx | Recursion:\n\nclass Solution:\n def solve(self, nums, x, cnt):\n if x == 0:\n return cnt\n if len(nums) == 0:\n return fl | sanial2001 | NORMAL | 2022-03-03T07:16:23.635807+00:00 | 2022-03-03T07:16:23.635837+00:00 | 349 | false | * Recursion:\n```\nclass Solution:\n def solve(self, nums, x, cnt):\n if x == 0:\n return cnt\n if len(nums) == 0:\n return float("inf")\n x1 = self.solve(nums[1:], x-nums[0], cnt+1)\n x2 = self.solve(nums[:-1], x-nums[-1], cnt+1)\n return min(x1, x2)\n \n ... | 3 | 0 | ['Recursion', 'Memoization', 'Python'] | 2 |
minimum-operations-to-reduce-x-to-zero | Two python solutions | two-python-solutions-by-flyingspa-nx72 | sliding window\n\nclass Solution:\n\n def minOperations(self, nums: List[int], x: int) -> int:\n s = sum(nums)-x\n ans =float("inf")\n l | flyingspa | NORMAL | 2021-04-06T16:26:11.570432+00:00 | 2021-04-06T16:26:11.570466+00:00 | 506 | false | * sliding window\n\nclass Solution:\n\n def minOperations(self, nums: List[int], x: int) -> int:\n s = sum(nums)-x\n ans =float("inf")\n left = right = 0\n curr = 0\n while(right<len(nums)): # for every left index, check whether there is any subarray that it sum is s\n c... | 3 | 0 | ['Python'] | 0 |
minimum-operations-to-reduce-x-to-zero | DFS Just for your curiosity (TLE) | dfs-just-for-your-curiosity-tle-by-mike1-e6zv | Just as the title mentioned. We can use the dfs approach, \n\n\nclass Solution {\n Integer[][][] dp;\n public int minOperations(int[] nums, int x) {\n | mike1029 | NORMAL | 2021-02-10T05:57:17.291909+00:00 | 2021-02-10T05:58:01.120048+00:00 | 125 | false | Just as the title mentioned. We can use the dfs approach, \n\n```\nclass Solution {\n Integer[][][] dp;\n public int minOperations(int[] nums, int x) {\n dp = new Integer[nums.length][nums.length][x+1];\n int res =dfs(nums, x, 0, nums.length-1);\n return res==Integer.MAX_VALUE? -1: res;\n ... | 3 | 0 | [] | 1 |
minimum-operations-to-reduce-x-to-zero | a few solutions | a-few-solutions-by-claytonjwong-a2s2 | Daily Solutions 2022-06-11:\n\nUse inversion. Find the maximum length subarray i..j sliding window sum equal to the inverted target T, ie. the sum of the input | claytonjwong | NORMAL | 2021-01-14T17:00:18.022165+00:00 | 2022-06-11T13:50:46.155077+00:00 | 152 | false | **Daily Solutions 2022-06-11:**\n\nUse inversion. Find the maximum length subarray `i..j` sliding window sum equal to the inverted target `T`, ie. the sum of the input array `A` minus `K` as the `best` candidate `cand`.\n\n*Kotlin*\n```\nclass Solution {\n fun minOperations(A: IntArray, K: Int): Int {\n var ... | 3 | 0 | [] | 0 |
minimum-operations-to-reduce-x-to-zero | JavaScript Prefix-Sum with HashMap | javascript-prefix-sum-with-hashmap-by-co-p4or | Time: O(N)\nSpace: O(N)\njavascript\nvar minOperations = function(nums, x) {\n const sum = nums.reduce((acc, cur) => acc + cur, 0);\n const target = sum - | control_the_narrative | NORMAL | 2021-01-14T16:18:32.323669+00:00 | 2021-01-14T16:18:32.323717+00:00 | 376 | false | Time: O(N)\nSpace: O(N)\n```javascript\nvar minOperations = function(nums, x) {\n const sum = nums.reduce((acc, cur) => acc + cur, 0);\n const target = sum - x;\n const map = new Map([[0, -1]]);\n \n if(target < 0) return -1;\n if(!target) return nums.length;\n \n let prefixSum = 0, maxLen = -In... | 3 | 0 | ['Prefix Sum', 'JavaScript'] | 0 |
minimum-operations-to-reduce-x-to-zero | [Rust] O(n) cumulative sums + HashMap, Iterators | rust-on-cumulative-sums-hashmap-iterator-i4qr | rust\nuse std::iter::once;\nuse std::collections::HashMap;\n\nimpl Solution {\n pub fn min_operations(nums: Vec<i32>, x: i32) -> i32 {\n let forward: | nizhnik | NORMAL | 2021-01-14T09:34:43.638506+00:00 | 2021-01-14T09:39:35.007533+00:00 | 73 | false | ```rust\nuse std::iter::once;\nuse std::collections::HashMap;\n\nimpl Solution {\n pub fn min_operations(nums: Vec<i32>, x: i32) -> i32 {\n let forward: HashMap<_, _> = sums(nums.iter()).enumerate().map(|(i, s)| (s, i)).collect();\n let n = nums.len() as i32;\n\n sums(nums.iter().rev()).enumerat... | 3 | 0 | ['Iterator'] | 0 |
minimum-operations-to-reduce-x-to-zero | simple hashmap solution in c++ (WITH EXPLANATION) | simple-hashmap-solution-in-c-with-explan-5idz | //here we can use the concept of maximum length subarray of sum equals to K \n// EXPLANATION: if the sum of our array is lets say Z then if we find the the long | anshul_190601 | NORMAL | 2020-11-20T08:22:46.330015+00:00 | 2020-11-21T17:10:51.893581+00:00 | 215 | false | //here we can use the concept of maximum length subarray of sum equals to K \n// EXPLANATION: if the sum of our array is lets say Z then if we find the the longest subarray of sum Z-X then we \n// left with sum equals to X with minimum no of elements and that is the required answer :)\nclass Solution {\npublic:\n in... | 3 | 1 | [] | 1 |
minimum-operations-to-reduce-x-to-zero | Short Python sliding window | short-python-sliding-window-by-user0571r-7zen | We want the largest window so that the sum of the numbers left out is \'x\'\n\n def minOperations(self, nums: List[int], x: int) -> int:\n S = sum(num | user0571rf | NORMAL | 2020-11-15T04:13:45.537585+00:00 | 2020-11-15T04:18:08.166706+00:00 | 275 | false | We want the largest window so that the sum of the numbers left out is \'x\'\n```\n def minOperations(self, nums: List[int], x: int) -> int:\n S = sum(nums)\n left = right = curr = 0\n ans = -1\n while right<len(nums):\n curr += nums[right]\n right+=1\n whi... | 3 | 1 | ['Sliding Window', 'Python3'] | 1 |
minimum-operations-to-reduce-x-to-zero | Still getting TLE | recursion + memorization | Solved using two-sum ( JAVA) | still-getting-tle-recursion-memorization-8yvj | Still getting TLE on this approach: \n\nclass Solution {\n Map<String, Integer> map = new HashMap<>();\n public int minOperations(int[] nums, int x) {\n | syanjo | NORMAL | 2020-11-15T04:07:58.838539+00:00 | 2020-11-15T05:27:48.466323+00:00 | 364 | false | Still getting TLE on this approach: \n```\nclass Solution {\n Map<String, Integer> map = new HashMap<>();\n public int minOperations(int[] nums, int x) {\n int min = recur(nums,x,0,nums.length-1);\n if(min==Integer.MAX_VALUE) return -1;\n else return min;\n }\n public int recur(int[] ar... | 3 | 0 | [] | 3 |
minimum-operations-to-reduce-x-to-zero | Beats 99%, 3 approaches, Beginner friendly. | beats-99-3-approaches-beginner-friendly-22ai4 | Intuition\n Describe your first thoughts on how to solve this problem. \nGet the minimum operations to reduce x to 0.\n\n# Approach\n Describe your approach to | Sarvamayvibhanjana | NORMAL | 2023-09-23T21:57:24.511144+00:00 | 2023-09-23T22:03:53.623655+00:00 | 134 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nGet the minimum operations to reduce x to 0.\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n1. **Memoization:** Take the element from first half or take from second half if target is met return minimum.\n2. **Pref... | 2 | 0 | ['Hash Table', 'Two Pointers', 'Dynamic Programming', 'Memoization', 'Sliding Window', 'Prefix Sum', 'C++'] | 0 |
minimum-operations-to-reduce-x-to-zero | My java solution | my-java-solution-by-raghavrathore7415-24zt | 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 | raghavrathore7415 | NORMAL | 2023-09-20T17:10:49.977108+00:00 | 2023-09-20T17:10:49.977135+00:00 | 211 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ --... | 2 | 0 | ['Java'] | 0 |
minimum-operations-to-reduce-x-to-zero | Video Solution | Explanation With Drawings | In Depth | Java | C++ | video-solution-explanation-with-drawings-ug9q | Intuition, approach and complexity discussed in detail in video solution\nhttps://youtu.be/QsRT1-t0wPQ\n\n# Code\nC++\n\nclass Solution {\npublic:\n int minO | Fly_ing__Rhi_no | NORMAL | 2023-09-20T16:51:25.877966+00:00 | 2023-09-20T17:02:25.271882+00:00 | 37 | false | # Intuition, approach and complexity discussed in detail in video solution\nhttps://youtu.be/QsRT1-t0wPQ\n\n# Code\nC++\n```\nclass Solution {\npublic:\n int minOperations(vector<int>& nums, int x) {\n int sz = nums.size();\n vector<int> prefSum(sz);\n \n for(int indx = 0; indx < sz; indx... | 2 | 0 | ['C++'] | 1 |
minimum-operations-to-reduce-x-to-zero | Java || using a sliding window approach|| | java-using-a-sliding-window-approach-by-93zxj | \n\n# Code\n\nclass Solution {\n public int minOperations(int[] nums, int x) {\n int n = nums.length;\n int targetSum = 0;\n \n | AdityaPandey1 | NORMAL | 2023-09-20T15:56:14.598996+00:00 | 2023-09-20T15:56:14.599015+00:00 | 219 | false | \n\n# Code\n```\nclass Solution {\n public int minOperations(int[] nums, int x) {\n int n = nums.length;\n int targetSum = 0;\n \n for (int num : nums) {\n targetSum += num;\n }\n \n int windowSum = targetSum - x;\n \n if (windowSum < 0) {\n ... | 2 | 0 | ['Java'] | 0 |
minimum-operations-to-reduce-x-to-zero | Easy | Python | Two Pointers | Minimum Operations to Reduce X to Zero | easy-python-two-pointers-minimum-operati-b4eb | see the successfully Accepted Submition\n\n\nclass Solution(object):\n def minOperations(self, nums, x):\n \n total_sum = sum(nums)\n ma | Khosiyat | NORMAL | 2023-09-20T15:39:03.714829+00:00 | 2023-09-20T15:39:26.649529+00:00 | 186 | false | [see the successfully Accepted Submition](https://leetcode.com/submissions/detail/1054631962/)\n\n```\nclass Solution(object):\n def minOperations(self, nums, x):\n \n total_sum = sum(nums)\n max_length = -1\n curr_sum = 0\n count_left = 0\n\n for right_pointer in range(len(... | 2 | 0 | ['Two Pointers', 'Python'] | 0 |
minimum-operations-to-reduce-x-to-zero | Most easiest solution in java | Minimum Operations to Reduce X to Zero | most-easiest-solution-in-java-minimum-op-oiqr | \n\n# Complexity\n- Time complexity:\nO(N)\n\n- Space complexity:\nO(x)\n\n# Code\n\nclass Solution {\n public int minOperations(int[] nums, int x) {\n | Hritiksharma5214 | NORMAL | 2023-09-20T14:56:37.912738+00:00 | 2023-09-20T14:56:37.912773+00:00 | 619 | false | \n\n# Complexity\n- Time complexity:\nO(N)\n\n- Space complexity:\nO(x)\n\n# Code\n```\nclass Solution {\n public int minOperations(int[] nums, int x) {\n int n = nums.length;\n int sum = 0;\n HashMap<Integer,Integer> map = new HashMap<>();\n map.put(0,-1);\n for(int i=0;i<n;i++)\n... | 2 | 0 | ['Java'] | 0 |
minimize-the-maximum-adjacent-element-difference | Binary Search | binary-search-by-votrubac-4wc3 | I did not like this problem. The intuition is not hard, but the input and edge cases are convoluted.\n\nWe compute max_gap between present adjacent elements. We | votrubac | NORMAL | 2024-11-17T07:42:54.352621+00:00 | 2024-11-19T05:28:05.368413+00:00 | 1,698 | false | I did not like this problem. The intuition is not hard, but the input and edge cases are convoluted.\n\nWe compute `max_gap` between *present* adjacent elements. We also track `min_n` and `max_n` of elements adjacent to a *missing* element. \n\nIf we can use only one element as a replacement, we would pick `x = (max_n ... | 25 | 0 | [] | 5 |
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