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values | created_at stringdate 2013-10-25 17:32:12 2025-04-12 09:38:24 | updated_at stringdate 2013-10-25 17:32:12 2025-04-12 09:38:24 | hit_count int64 0 10.6M | has_video bool 2
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minimum-operations-to-make-all-array-elements-equal | C++ | 95% Fast | With best explanation | c-95-fast-with-best-explanation-by-vaibh-m3hi | Intuition\nDivide nums into 2 subarray\nsuppose we have nums = [2,6,8] and we need to find operations for query [5]. \nJust find elements that are greater than | VAIBHAV_SHORAN | NORMAL | 2023-03-30T18:19:11.555074+00:00 | 2023-03-30T18:20:57.070796+00:00 | 190 | false | # Intuition\nDivide nums into 2 subarray\nsuppose we have nums = [2,6,8] and we need to find operations for query [5]. \nJust find elements that are greater than 5 and that are lower than it. Here, [6,8] are greater and [2] is lower.\n\nFor greater, sum of these elements is 6+8 = 14. We need to bring these 2 elements e... | 2 | 0 | ['C++'] | 0 |
minimum-operations-to-make-all-array-elements-equal | Optimized Java solution || Binary Search || Prefix Sum | optimized-java-solution-binary-search-pr-wnt6 | 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 | i_am_sumon | NORMAL | 2023-03-28T16:12:44.863708+00:00 | 2023-04-01T22:27:50.725304+00:00 | 675 | 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$$O(q*log(n))$$\n- Space complexity:\n<!-- Add your space complexity here, e... | 2 | 0 | ['Array', 'Binary Search', 'Sorting', 'Prefix Sum', 'Java'] | 2 |
minimum-operations-to-make-all-array-elements-equal | Easy solution in python-3 | easy-solution-in-python-3-by-pradyumna14-liy4 | Intuition\nIt can be solved using naive approach in O(n^2) time complexity, which includes one outer for loop for iterating in queries list and the inner outer | PRADYUMNA14 | NORMAL | 2023-03-28T09:49:53.145033+00:00 | 2023-03-28T09:50:59.725846+00:00 | 404 | false | # Intuition\nIt can be solved using naive approach in O(n^2) time complexity, which includes one outer for loop for iterating in queries list and the inner outer loop for calculating the sum (as shown) in \'nums\' list. Since, this solution is not optimal, we have to reduce the time complexity further.\n# Naive approac... | 2 | 0 | ['Binary Search', 'Sorting', 'Prefix Sum', 'Python3'] | 0 |
minimum-operations-to-make-all-array-elements-equal | [Entire thought Process + Intuition, Binary Search + Prefix Sum Approach] | entire-thought-process-intuition-binary-2jzvf | Intuition\n\n * Let\'s try, to to make some observations:\n\n Input: nums = [3,1,6,8], queries = [1,5]\n Output: [14, 10]\n \n queires[0] = 1, an | _LearnToCode_ | NORMAL | 2023-03-27T07:44:48.970661+00:00 | 2023-03-27T07:44:48.970693+00:00 | 122 | false | # Intuition\n\n * Let\'s try, to to make some observations:\n\n Input: nums = [3,1,6,8], queries = [1,5]\n Output: [14, 10]\n \n queires[0] = 1, ans = 14, \n answer 14 can be easily compute as (total_sum_of_nums - nums_size * queries[0]) -> (18 - 4 * 1) -> 14\n\n queries[1] = 5, ans = 10, \n but... | 2 | 0 | ['Binary Search', 'Prefix Sum', 'Java'] | 1 |
minimum-operations-to-make-all-array-elements-equal | 📌📌 C++ || Sorting || Prefix Sum || Binary Search || Faster || Easy To Understand 🤷♂️🤷♂️ | c-sorting-prefix-sum-binary-search-faste-fg8x | Using Sorting && Binary Search\n\n Time Complexity :- O(NlogN)\n\n Space Complexity :- O(N)\n\n\nclass Solution {\npublic:\n \n vector<long long> minOpera | __KR_SHANU_IITG | NORMAL | 2023-03-27T06:07:54.976390+00:00 | 2023-03-27T06:07:54.976435+00:00 | 266 | false | * ***Using Sorting && Binary Search***\n\n* ***Time Complexity :- O(NlogN)***\n\n* ***Space Complexity :- O(N)***\n\n```\nclass Solution {\npublic:\n \n vector<long long> minOperations(vector<int>& nums, vector<int>& queries) {\n \n int n = nums.size();\n \n int m = queries.size();\n ... | 2 | 0 | ['Binary Search', 'C', 'Sorting', 'Prefix Sum', 'C++'] | 0 |
minimum-operations-to-make-all-array-elements-equal | 💯🔥Best Code in C++ || BinarySearch🔥 || Sorting✅ || PrefixSum💯 | best-code-in-c-binarysearch-sorting-pref-27bj | Complexity\n- Time complexity:\nO(mlogn)\n\n- Space complexity:\nO(n)\n\n# Code\n### Please Upvote if u liked my Solution\uD83E\uDD17\n\nclass Solution {\npubli | aDish_21 | NORMAL | 2023-03-26T08:40:21.776404+00:00 | 2023-03-26T08:40:21.776444+00:00 | 122 | false | # Complexity\n- Time complexity:\nO(mlogn)\n\n- Space complexity:\nO(n)\n\n# Code\n### Please Upvote if u liked my Solution\uD83E\uDD17\n```\nclass Solution {\npublic:\n vector<long long> minOperations(vector<int>& nums, vector<int>& queries) {\n int n=nums.size(),m=queries.size();\n long long pSum=0;\... | 2 | 0 | ['Binary Search', 'Sorting', 'Prefix Sum', 'C++'] | 0 |
minimum-operations-to-make-all-array-elements-equal | python3 Solution | python3-solution-by-motaharozzaman1996-1gik | \n\nclass Solution:\n def minOperations(self, nums: List[int], queries: List[int]) -> List[int]:\n nums.sort()\n n=len(nums)\n prefix=[0 | Motaharozzaman1996 | NORMAL | 2023-03-26T06:39:49.813491+00:00 | 2023-03-26T06:39:49.813535+00:00 | 1,274 | false | \n```\nclass Solution:\n def minOperations(self, nums: List[int], queries: List[int]) -> List[int]:\n nums.sort()\n n=len(nums)\n prefix=[0]+list(accumulate(nums))\n ans=[]\n for x in queries:\n i=bisect_left(nums,x)\n ans.append(x*(2*i-n)+prefix[n]-2*prefix[i... | 2 | 0 | ['Python', 'Python3'] | 0 |
minimum-operations-to-make-all-array-elements-equal | Java || Sorting + Prefix Sum + Binary Search | java-sorting-prefix-sum-binary-search-by-pdyv | Complexity\n- Time complexity: O((m+n)logn)\n\n- Space complexity: O(n)\n\n## Code\njava []\nclass Solution {\n public List<Long> minOperations(int[] nums, i | nishant7372 | NORMAL | 2023-03-26T05:34:19.890228+00:00 | 2023-03-26T05:46:43.010329+00:00 | 400 | false | ## Complexity\n- Time complexity: O((m+n)logn)\n\n- Space complexity: O(n)\n\n## Code\n``` java []\nclass Solution {\n public List<Long> minOperations(int[] nums, int[] queries) {\n Arrays.sort(nums);\n List<Long> list = new ArrayList<>();\n long[] prefix = new long[nums.length+2]; // taking pre... | 2 | 0 | ['Binary Search', 'Sorting', 'Prefix Sum', 'Java'] | 0 |
minimum-operations-to-make-all-array-elements-equal | [Javascript] Sorting, Binary Search & Prefix Sum | javascript-sorting-binary-search-prefix-t2qez | Solution: Sorting, Binary Search & Prefix Sum\n\n1. Sort nums in asc order.\n2. Get the prefix sum from the left and right of nums.\n3. For each query, binary s | anna-hcj | NORMAL | 2023-03-26T04:00:46.081719+00:00 | 2023-03-26T04:00:46.081764+00:00 | 325 | false | **Solution: Sorting, Binary Search & Prefix Sum**\n\n1. Sort nums in asc order.\n2. Get the prefix sum from the left and right of nums.\n3. For each query, binary search for the split point - the first element where `nums[i] >= query`\n From there, we can calculate the absolute difference of the two segments of nums.\... | 2 | 0 | ['JavaScript'] | 1 |
minimum-operations-to-make-all-array-elements-equal | Easy to Understand ! Beginners Friendly | easy-to-understand-beginners-friendly-by-gt43 | IntuitionApproachFor Each query find the last index of that element in the nums .
The Elements left side of that index are smaller than Query[i] they should be | jayprakash00012 | NORMAL | 2025-03-26T09:00:31.491373+00:00 | 2025-03-26T09:00:31.491373+00:00 | 91 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
# Approach
<!-- Describe your approach to solving the problem. -->
For Each query find the last index of that element in the nums .
The Elements left side of that index are smaller than Query[i] they should be increased
The Elements right ... | 1 | 0 | ['Array', 'Binary Search', 'Sorting', 'Prefix Sum', 'C++'] | 0 |
minimum-operations-to-make-all-array-elements-equal | prefix sum | prefix-sum-by-remedydev-98a5 | Approach\nWe build a prefix sum of differences so we can tell in constant time how many operations needed for sub array for a query\n \nexample: [1,3,4, 6, 8]\ | remedydev | NORMAL | 2023-11-16T03:43:55.336109+00:00 | 2023-11-16T03:43:55.336138+00:00 | 35 | false | # Approach\nWe build a prefix sum of differences so we can tell in constant time how many operations needed for sub array for a query\n ```\nexample: [1,3,4, 6, 8]\n prefix: 0 2 5 10 17\n```\nwe build posfix sum difference as well in order to tell in O(1) time amount of operations needed from left to right indexes... | 1 | 0 | ['JavaScript'] | 0 |
minimum-operations-to-make-all-array-elements-equal | Rust, Python. (n + m) log n. Detailed explanation with math | rust-python-n-m-log-n-detailed-explanati-6b3g | Intuition\nLets look how to address this for one query and the value is $v$.\n\n- If the numbers are below $v$. Then the results are $v - a_1 + v - a_2 + ... v | salvadordali | NORMAL | 2023-05-02T00:55:15.330741+00:00 | 2023-05-02T00:55:15.330776+00:00 | 61 | false | # Intuition\nLets look how to address this for one query and the value is $v$.\n\n- If the numbers are below $v$. Then the results are $v - a_1 + v - a_2 + ... v - a_n = v \\cdot n - \\sum_{i=1}^{n} a_i $\n- If the numbers are above $v$. With similar logic, results are $\\sum_{i=1}^{n} a_i - v \\cdot n$\n\nSo we need ... | 1 | 0 | ['Math', 'Python', 'Rust'] | 0 |
minimum-operations-to-make-all-array-elements-equal | Swift | Prefix Sum + Binary Search | swift-prefix-sum-binary-search-by-upvote-n1vi | Prefix Sum + Binary Search (accepted answer)\n\nclass Solution {\n func minOperations(_ nums: [Int], _ queries: [Int]) -> [Int] {\n let nums = nums.so | UpvoteThisPls | NORMAL | 2023-04-08T23:45:53.447729+00:00 | 2023-04-08T23:45:53.447773+00:00 | 25 | false | **Prefix Sum + Binary Search (accepted answer)**\n```\nclass Solution {\n func minOperations(_ nums: [Int], _ queries: [Int]) -> [Int] {\n let nums = nums.sorted()\n let prefixSums = nums.reduce(into: [0]) { $0.append($1 + $0.last!) }\n \n return queries.map { q in \n var (left... | 1 | 0 | ['Swift'] | 0 |
minimum-operations-to-make-all-array-elements-equal | Magical Formula lol | O(nlogn) | Binary Search | Prefix Sum | magical-formula-lol-onlogn-binary-search-u04l | // Intuition behind this is to isolate numbers less than queries[i] and number greater or equal to queries[i] and then perform this magical formula\n // ans | aryan_sri123 | NORMAL | 2023-03-27T10:13:55.621697+00:00 | 2023-03-27T10:14:42.916414+00:00 | 16 | false | // Intuition behind this is to isolate numbers less than queries[i] and number greater or equal to queries[i] and then perform this magical formula\n // ans[i] = (idx*x-res[idx-1]) + res[n-1]-res[idx-1]-x*n-idx --> key formula I derived (all terms are used in code)\n\t// 1,3,6,8 \n // 1,4,10,18 -> prefix sum... | 1 | 0 | ['Binary Search', 'C', 'Prefix Sum'] | 0 |
minimum-operations-to-make-all-array-elements-equal | Golang 156 ms 10.9 MB | golang-156-ms-109-mb-by-maxhero90-rmip | Complexity\n- Time complexity: O((n+m)*logn)\n- Space complexity: O(n+m)\n# Code\n\nfunc minOperations(nums []int, queries []int) []int64 {\n\tn := len(nums)\n\ | maxhero90 | NORMAL | 2023-03-27T09:22:15.111101+00:00 | 2023-03-27T09:23:41.505039+00:00 | 55 | false | # Complexity\n- Time complexity: $$O((n+m)*logn)$$\n- Space complexity: $$O(n+m)$$\n# Code\n```\nfunc minOperations(nums []int, queries []int) []int64 {\n\tn := len(nums)\n\tsort.Ints(nums)\n\tprefixSums := make([]int, n)\n\tprefixSums[0] = nums[0]\n\tfor i := 1; i < n; i++ {\n\t\tprefixSums[i] = prefixSums[i-1] + nums... | 1 | 0 | ['Go'] | 0 |
minimum-operations-to-make-all-array-elements-equal | C++, prefix sum | c-prefix-sum-by-gyokuro123-2krx | \n# Code\n\nclass Solution {\npublic:\n vector<long long> minOperations(vector<int>& nums, vector<int>& queries) {\n sort(nums.begin(), nums.end());\n | gyokuro123 | NORMAL | 2023-03-27T03:26:46.286732+00:00 | 2023-03-27T03:26:46.286769+00:00 | 30 | false | \n# Code\n```\nclass Solution {\npublic:\n vector<long long> minOperations(vector<int>& nums, vector<int>& queries) {\n sort(nums.begin(), nums.end());\n vector<long long> sums(nums.size(), 0);\n sums[0] = nums[0];\n for(int i = 1; i < nums.size(); i++) sums[i] = nums[i] + sums[i-1];\n ... | 1 | 0 | ['C++'] | 0 |
minimum-operations-to-make-all-array-elements-equal | Sort | Binary Search | sort-binary-search-by-ajayc1007-zhvb | Intuition\nBrute force will give TLE. Look for a pivot from where either we only need to increment or decrement.\n\n# Approach\nCarefully note that the answer d | ajayc1007 | NORMAL | 2023-03-26T19:04:19.974647+00:00 | 2023-03-26T19:04:19.974677+00:00 | 346 | false | # Intuition\nBrute force will give TLE. Look for a pivot from where either we only need to increment or decrement.\n\n# Approach\nCarefully note that the answer does not depend on the order of the elements. In such a case, we can utilize binary search to identify a pivot 1) before which all elements are smaller 2) afte... | 1 | 0 | ['Binary Search', 'Sorting', 'C++'] | 0 |
minimum-operations-to-make-all-array-elements-equal | Sorting+Prefix Sum + Binary Search | sortingprefix-sum-binary-search-by-pande-bczc | Intuition\nIntuition is we need to make all element equal to given target and there is time constraint to we need to think like number of elements that are goin | Pandeyji_1999 | NORMAL | 2023-03-26T14:13:49.904887+00:00 | 2023-03-26T14:13:49.904923+00:00 | 306 | false | # Intuition\nIntuition is we need to make all element equal to given target and there is time constraint to we need to think like number of elements that are going to be increase i.e number which is less than target and decrease the number which is greater than the target.\n\n# Approach\nAs going on each index will lea... | 1 | 0 | ['Binary Search', 'Sorting', 'Prefix Sum'] | 0 |
minimum-operations-to-make-all-array-elements-equal | Easy Solution using Prefix Sum + lower bound + upper bound | easy-solution-using-prefix-sum-lower-bou-fbm8 | Code\n\nclass Solution {\npublic:\n vector<long long> minOperations(vector<int>& nums, vector<int>& queries) {\n \n vector<long long> ans;\n | baquer | NORMAL | 2023-03-26T08:30:08.368056+00:00 | 2023-03-26T08:30:08.368102+00:00 | 201 | false | # Code\n```\nclass Solution {\npublic:\n vector<long long> minOperations(vector<int>& nums, vector<int>& queries) {\n \n vector<long long> ans;\n int n = nums.size();\n sort(nums.begin(), nums.end());\n\n // Ye brute force hai.\n\n // for(int i = 0; i < queries.size(); i++) ... | 1 | 0 | ['Binary Search', 'Prefix Sum', 'C++'] | 0 |
minimum-operations-to-make-all-array-elements-equal | c++ solution | c-solution-by-dilipsuthar60-fkxv | \nclass Solution {\npublic:\n vector<long long> minOperations(vector<int>& nums, vector<int>& q) \n {\n sort(nums.begin(),nums.end());\n int | dilipsuthar17 | NORMAL | 2023-03-26T07:02:39.413723+00:00 | 2023-03-26T07:02:39.413762+00:00 | 110 | false | ```\nclass Solution {\npublic:\n vector<long long> minOperations(vector<int>& nums, vector<int>& q) \n {\n sort(nums.begin(),nums.end());\n int n=nums.size();\n vector<long long>ans;\n long long total=accumulate(nums.begin(),nums.end(),0ll);\n vector<long long>temp(n+1);\n ... | 1 | 0 | ['C', 'Binary Tree', 'C++'] | 0 |
minimum-operations-to-make-all-array-elements-equal | Easy C++ | easy-c-by-avatsal38-6s1d | 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 | avatsal38 | NORMAL | 2023-03-26T06:15:47.163663+00:00 | 2023-03-26T06:15:47.163711+00:00 | 146 | 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. $$O(n)$$ --... | 1 | 0 | ['C++'] | 0 |
minimum-operations-to-make-all-array-elements-equal | Sorting +Prefix Sum + BinarySearch, JAVA, attention the overflow | sorting-prefix-sum-binarysearch-java-att-g0p3 | 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 | tendencymilk | NORMAL | 2023-03-26T05:04:33.080055+00:00 | 2023-03-26T05:04:33.080097+00:00 | 208 | 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. $$O(n)$$ --... | 1 | 0 | ['Java'] | 0 |
minimum-operations-to-make-all-array-elements-equal | C++ Easy Solution | Binary Search + Sorting | O(NlongN) | c-easy-solution-binary-search-sorting-on-vp9j | 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 | manishk4514 | NORMAL | 2023-03-26T04:53:45.368695+00:00 | 2023-03-26T04:53:45.368726+00:00 | 155 | 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. $$O(n)$$ --... | 1 | 0 | ['C++'] | 0 |
minimum-operations-to-make-all-array-elements-equal | Runtime 235 ms Beats 100% | runtime-235-ms-beats-100-by-akshat0610-kmds | Code\n\n#pragma GCC optimize("Ofast","inline","fast-math","unroll-loops","no-stack-protector")\n#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx, | akshat0610 | NORMAL | 2023-03-26T04:43:46.474363+00:00 | 2023-03-26T04:43:46.474427+00:00 | 126 | false | # Code\n```\n#pragma GCC optimize("Ofast","inline","fast-math","unroll-loops","no-stack-protector")\n#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,avx2,tune=native","f16c")\n\nstatic int fast_io = []() \n{ \n std::ios::sync_with_stdio(false); \n cin.tie(nullptr); \n cout.tie(nullptr); \n r... | 1 | 0 | ['C++'] | 0 |
minimum-operations-to-make-all-array-elements-equal | Binary Search + Prefix Sum | binary-search-prefix-sum-by-rishabhsingh-wrk8 | \nclass Solution {\npublic:\n vector<long long> minOperations(vector<int>& nums, vector<int>& queries) {\n \n int n = size(nums);\n sort | Rishabhsinghal12 | NORMAL | 2023-03-26T04:10:57.639772+00:00 | 2023-03-26T04:10:57.639819+00:00 | 87 | false | ```\nclass Solution {\npublic:\n vector<long long> minOperations(vector<int>& nums, vector<int>& queries) {\n \n int n = size(nums);\n sort(begin(nums), end(nums));\n \n vector<long long> pre(n);\n \n pre[0] = nums[0];\n \n for(int i = 1; i < n; i++) {\n... | 1 | 0 | ['Binary Search', 'C', 'Sorting', 'Prefix Sum', 'C++'] | 0 |
minimum-operations-to-make-all-array-elements-equal | NLogN | Easy Solution | nlogn-easy-solution-by-yigitozcelep-gem6 | For each query target in queries, the solution performs binary search on the sorted array nums to find the index i such that nums[i] is the smallest element gre | Yigitozcelep | NORMAL | 2023-03-26T04:01:16.528644+00:00 | 2023-03-26T04:02:08.145454+00:00 | 213 | false | For each query target in queries, the solution performs binary search on the sorted array nums to find the index i such that nums[i] is the smallest element greater than or equal to the target. It then calculates the sum of the following two parts:\n\nThe total number of operations required to make the first i elements... | 1 | 0 | ['Python3'] | 0 |
minimum-operations-to-make-all-array-elements-equal | Java solution | java-solution-by-yashhi_01-adfu | Code | yashhi_01 | NORMAL | 2025-04-09T13:23:52.632860+00:00 | 2025-04-09T13:23:52.632860+00:00 | 2 | false |
# Code
```java []
class Solution {
public int binarySearch(int[] nums, int k){
int left = 0, right = nums.length;
while (left < right) {
int mid = left + (right - left) / 2;
if (nums[mid] < k) left = mid + 1;
else right = mid;
}
return left;
... | 0 | 0 | ['Java'] | 0 |
minimum-operations-to-make-all-array-elements-equal | Java prefixSum + binarySearch | java-prefixsum-binarysearch-by-54v-ag8t | IntuitionIf j numbers in nums that are smaller than query[i] then number of increments = query[i] * j - sum of smaller numbersIf k numbers in nums thare are gre | 54v | NORMAL | 2025-04-08T23:54:52.760057+00:00 | 2025-04-08T23:54:52.760057+00:00 | 2 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
If j numbers in nums that are smaller than query[i] then number of increments = query[i] * j - sum of smaller numbers
If k numbers in nums thare are greater than query[i] then number of decrements = sum of larger nums - query[i] * k
sum ... | 0 | 0 | ['Java'] | 0 |
minimum-operations-to-make-all-array-elements-equal | solution using prefix sum | solution-using-prefix-sum-by-binod24-7wec | IntuitionApproachComplexity
Time complexity:
Space complexity:
Code | binod24 | NORMAL | 2025-03-30T12:06:04.873863+00:00 | 2025-03-30T12:06:04.873863+00:00 | 2 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
# Approach
<!-- Describe your approach to solving the problem. -->
# Complexity
- Time complexity:
<!-- Add your time complexity here, e.g. $$O(n)$$ -->
- Space complexity:
<!-- Add your space complexity here, e.g. $$O(n)$$ -->
# Code
`... | 0 | 0 | ['C++'] | 0 |
minimum-operations-to-make-all-array-elements-equal | Beats 90% of Coders || Best Solution || Pre Sum technique used | beats-90-of-coders-best-solution-pre-sum-f3ts | Code | Deepanshu_Rathore | NORMAL | 2025-03-28T04:24:14.031189+00:00 | 2025-03-28T04:24:14.031189+00:00 | 2 | false | # Code
```cpp []
class Solution {
public:
vector<long long> minOperations(vector<int>& nums, vector<int>& queries) {
sort(nums.begin(), nums.end());
int n = nums.size();
vector<long long> preSum(n+1, 0);
for(int i=0; i<n; ++i) {
preSum[i+1] = preSum[i] + nums[i];
... | 0 | 0 | ['C++'] | 0 |
minimum-operations-to-make-all-array-elements-equal | 100% || C++ | 100-c-by-giri_69-anjt | Intuitionfor a query q, there are some elements smaller than q denoted as leftsum rest are called rightsum. Find teh sum of both of the valuesApproachComplexity | giri_69 | NORMAL | 2025-03-23T08:00:27.754316+00:00 | 2025-03-23T08:00:27.754316+00:00 | 3 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
for a query `q`, there are some elements smaller than `q` denoted as `leftsum` rest are called `rightsum`. Find teh sum of both of the values
# Approach
<!-- Describe your approach to solving the problem. -->
# Complexity
- Time complexit... | 0 | 0 | ['C++'] | 0 |
minimum-operations-to-make-all-array-elements-equal | intuitive bisect solution | intuitive-bisect-solution-by-owenwu4-eats | Intuitionuse binary search to sum up the elements that are not equal to the current - you know bisect will give you right half elements are greater than current | owenwu4 | NORMAL | 2025-02-26T00:16:37.969510+00:00 | 2025-02-26T00:16:37.969510+00:00 | 4 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
use binary search to sum up the elements that are not equal to the current - you know bisect will give you right half elements are greater than current and left are less than, so you can leverage this
# Approach
<!-- Describe your approach... | 0 | 0 | ['Python3'] | 0 |
minimum-operations-to-make-all-array-elements-equal | Ugly looking but easy to understand C++ Binary search solution | ugly-looking-but-easy-to-understand-c-bi-bqgp | Code | amogh007 | NORMAL | 2025-02-22T01:20:45.849390+00:00 | 2025-02-22T01:20:45.849390+00:00 | 5 | false |
# Code
```cpp []
class Solution {
public:
vector<long long> minOperations(vector<int>& nums, vector<int>& queries) {
vector<long long>prefix(nums.size(), 0);
vector<long long>result;
sort(nums.begin(), nums.end());
for (int i = 1; i < nums.size(); i++) {
prefix[i] = p... | 0 | 0 | ['C++'] | 0 |
minimum-operations-to-make-all-array-elements-equal | prefix sum and binary search | prefix-sum-and-binary-search-by-johnchen-34b6 | Intuitionsort the nums, then do prefix sum.find the position of each query, the left part would be query * number of left - sum of left, the right part would be | johnchen | NORMAL | 2025-02-20T15:48:16.537962+00:00 | 2025-02-20T15:48:16.537962+00:00 | 4 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
sort the nums, then do prefix sum.
find the position of each query, the left part would be query * number of left - sum of left, the right part would be sum of right - query * number of right
# Approach
<!-- Describe your approach to solv... | 0 | 0 | ['C++'] | 0 |
minimum-operations-to-make-all-array-elements-equal | Presum + BinarySearch | presum-binarysearch-by-wesleyzhouwj-99oy | IntuitionUsing Presum and Binary searchApproachComplexity
Time complexity: NlogN + MlogN (N is length of nums, M is length of queries)
Space complexity:
O(N)
C | wesleyzhouwj | NORMAL | 2025-02-07T05:16:24.029290+00:00 | 2025-02-07T05:16:24.029290+00:00 | 7 | false | # Intuition
Using Presum and Binary search
# Approach
<!-- Describe your approach to solving the problem. -->
# Complexity
- Time complexity: NlogN + MlogN (N is length of nums, M is length of queries)
<!-- Add your time complexity here, e.g. $$O(n)$$ -->
- Space complexity:
O(N)
# Code
```java []
class Solution {... | 0 | 0 | ['Java'] | 0 |
minimum-operations-to-make-all-array-elements-equal | You shall never get it | you-shall-never-get-it-by-sidsredhar27-5nda | Code | sidsredhar27 | NORMAL | 2025-02-04T11:42:29.991887+00:00 | 2025-02-04T11:42:29.991887+00:00 | 12 | false | # Code
```python3 []
from typing import List
class Solution:
def minOperations(self, nums: List[int], queries: List[int]) -> List[int]:
nums.sort()
res = []
prefix = [0]
for i in nums:
prefix.append(prefix[-1] + i)
def getindex(num):
left, right = 0... | 0 | 0 | ['Python3'] | 0 |
minimum-operations-to-make-all-array-elements-equal | Simple math method | simple-math-method-by-sidsredhar27-zv2i | Code | sidsredhar27 | NORMAL | 2025-02-04T11:41:44.435541+00:00 | 2025-02-04T11:41:44.435541+00:00 | 8 | false | # Code
```python3 []
from typing import List
class Solution:
def minOperations(self, nums: List[int], queries: List[int]) -> List[int]:
nums.sort()
res = []
prefix = [0]
for i in nums:
prefix.append(prefix[-1] + i)
def getindex(num):
left, right = 0... | 0 | 0 | ['Python3'] | 0 |
minimum-operations-to-make-all-array-elements-equal | 2602. Minimum Operations to Make All Array Elements Equal | 2602-minimum-operations-to-make-all-arra-jhck | IntuitionApproachComplexity
Time complexity:
Space complexity:
Code | G8xd0QPqTy | NORMAL | 2025-01-19T04:50:39.267141+00:00 | 2025-01-19T04:50:39.267141+00:00 | 8 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
# Approach
<!-- Describe your approach to solving the problem. -->
# Complexity
- Time complexity:
<!-- Add your time complexity here, e.g. $$O(n)$$ -->
- Space complexity:
<!-- Add your space complexity here, e.g. $$O(n)$$ -->
# Code
`... | 0 | 0 | ['C++'] | 0 |
minimum-operations-to-make-all-array-elements-equal | Simple PrefixSum and BinarySearch | simple-prefixsum-and-binarysearch-by-sri-2mcg | IntuitionThe problem involves calculating the minimum number of operations needed to make all elements in the array nums equal to each query value in the querie | srirammente | NORMAL | 2025-01-14T10:23:11.149540+00:00 | 2025-01-14T10:23:11.149540+00:00 | 12 | false | # Intuition
The problem involves calculating the minimum number of operations needed to make all elements in the array nums equal to each query value in the queries array. Since operations involve incrementing or decrementing elements by 1, we can simplify the calculation by pre-sorting the array and using a prefix sum... | 0 | 0 | ['Binary Search', 'Sorting', 'Prefix Sum', 'Java'] | 0 |
minimum-operations-to-make-all-array-elements-equal | Sort + Prefix Sum + Binary Search | sort-prefix-sum-binary-search-by-up41guy-xzsw | null | Cx1z0 | NORMAL | 2025-01-04T04:08:58.474837+00:00 | 2025-01-04T04:17:17.604891+00:00 | 7 | false | ```javascript []
var minOperations = function (nums, queries) {
const sorted = nums.toSorted((a, b) => a - b);
const n = sorted.length;
// Build prefix sums
const sums = [0];
for (let i = 0; i < n; i++) sums.push(sums[i] + sorted[i]);
// Find position using binary search
const findPos = (target) => {
... | 0 | 0 | ['Array', 'Binary Search', 'Sorting', 'Prefix Sum', 'JavaScript'] | 0 |
minimum-operations-to-make-all-array-elements-equal | BEATS 63% using prefixSumm & binarySearch | beats-63-using-prefixsumm-binarysearch-b-lj5r | Code | alexander_sergeev | NORMAL | 2024-12-26T20:25:26.065808+00:00 | 2024-12-26T20:25:26.065808+00:00 | 9 | false | # Code
```java []
class Solution {
public List<Long> minOperations(int[] nums, int[] queries) {
int n = nums.length;
Arrays.sort(nums);
long[] prefixArr = new long[n + 1];
for (int i = 0; i < n; i++) {
prefixArr[i + 1] = nums[i] + prefixArr[i];
}
List<Lon... | 0 | 0 | ['Java'] | 0 |
minimum-operations-to-make-all-array-elements-equal | Easy | easy-by-rifatt-oq78 | IntuitionApproachComplexity
Time complexity:
Space complexity:
Code | Rifatt | NORMAL | 2024-12-22T20:40:14.788938+00:00 | 2024-12-22T20:40:14.788938+00:00 | 4 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
# Approach
<!-- Describe your approach to solving the problem. -->
# Complexity
- Time complexity:
<!-- Add your time complexity here, e.g. $$O(n)$$ -->
- Space complexity:
<!-- Add your space complexity here, e.g. $$O(n)$$ -->
# Code
`... | 0 | 0 | ['C++'] | 0 |
minimum-operations-to-make-all-array-elements-equal | Absolute PreBinaryS | absolute-prebinarys-by-tonitannoury01-lxl4 | IntuitionApproachComplexity
Time complexity:
O(nlog(n))
Space complexity:
O(n)
Code | tonitannoury01 | NORMAL | 2024-12-22T14:46:43.195436+00:00 | 2024-12-22T14:46:43.195436+00:00 | 5 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
# Approach
<!-- Describe your approach to solving the problem. -->
# Complexity
- Time complexity:
O(nlog(n))
- Space complexity:
O(n)
# Code
```javascript []
/**
* @param {number[]} nums
* @param {number[]} queries
* @return {number... | 0 | 0 | ['JavaScript'] | 0 |
minimum-operations-to-make-all-array-elements-equal | Easy solution using Binary Search | easy-solution-using-binary-search-by-_jy-gqvk | Code\njava []\nclass Solution {\n public List<Long> minOperations(int[] nums, int[] q) {\n Arrays.sort(nums);\n List<Long> ans = new ArrayList< | _jyoti_geek | NORMAL | 2024-11-28T14:51:50.851292+00:00 | 2024-11-28T14:51:50.851320+00:00 | 7 | false | # Code\n```java []\nclass Solution {\n public List<Long> minOperations(int[] nums, int[] q) {\n Arrays.sort(nums);\n List<Long> ans = new ArrayList<>();\n int n = nums.length;\n int m = q.length;\n long[] pre = new long[n];\n pre[0] = nums[0];\n for(int i=1;i<n;i++){\... | 0 | 0 | ['Array', 'Binary Search', 'Sorting', 'Prefix Sum', 'Java'] | 0 |
minimum-operations-to-make-all-array-elements-equal | scala solution | scala-solution-by-vititov-rrul | scala []\nobject Solution {\n def minOperations(nums: Array[Int], queries: Array[Int]): List[Long] = {\n val sorted = nums.to(Vector).map(_.toLong).sorted\n | vititov | NORMAL | 2024-10-14T15:06:17.711713+00:00 | 2024-10-14T15:06:17.711737+00:00 | 0 | false | ```scala []\nobject Solution {\n def minOperations(nums: Array[Int], queries: Array[Int]): List[Long] = {\n val sorted = nums.to(Vector).map(_.toLong).sorted\n val pfx = sorted.iterator.drop(1).scanLeft((sorted.head,0L,1)){case ((last,sum,cnt), num) =>\n (num,sum+(num-last).abs*cnt,cnt+1)\n }.to(Vector)\... | 0 | 0 | ['Array', 'Binary Search', 'Sorting', 'Prefix Sum', 'Scala'] | 0 |
minimum-operations-to-make-all-array-elements-equal | Python3 Combine { sorting, binary search, prefix sums } O(nlgn) + O(qlgn) time O(N) Space Iterative | python3-combine-sorting-binary-search-pr-3bqu | 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 | 2018hsridhar | NORMAL | 2024-10-11T01:50:12.104522+00:00 | 2024-10-11T01:50:12.104553+00:00 | 7 | 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. $$O(n)$$ --... | 0 | 0 | ['Array', 'Binary Search', 'Sorting', 'Prefix Sum', 'Python3'] | 0 |
minimum-operations-to-make-all-array-elements-equal | [Java] Prefix Sums + Binary Search | easy | java-prefix-sums-binary-search-easy-by-j-cip5 | Code\njava []\nclass Solution {\n private int n;\n public List<Long> minOperations(int[] nums, int[] queries) {\n List<Long> res = new ArrayList<>( | jsanchez78 | NORMAL | 2024-10-08T15:55:33.573058+00:00 | 2024-10-08T15:55:33.573087+00:00 | 7 | false | # Code\n```java []\nclass Solution {\n private int n;\n public List<Long> minOperations(int[] nums, int[] queries) {\n List<Long> res = new ArrayList<>();\n Arrays.sort(nums);\n // Init prefixSums\n this.n = nums.length;\n long[] prefix = new long[n + 1];\n for(int i = 1;... | 0 | 0 | ['Array', 'Binary Search', 'Sorting', 'Prefix Sum', 'Java'] | 0 |
minimum-operations-to-make-all-array-elements-equal | sorting, prefix sum, binary search Python solution | sorting-prefix-sum-binary-search-python-kh0bn | Intuition\nOnce nums is sorted, we can find a dividing point for each querie using binary search. Then, the left side of nums is less than the querie and the ri | jbak | NORMAL | 2024-10-04T23:01:16.753433+00:00 | 2024-10-04T23:01:16.753454+00:00 | 24 | false | # Intuition\nOnce `nums` is sorted, we can find a dividing point for each querie using binary search. Then, the left side of `nums` is less than the querie and the right side of `nums` is greater than the querie. Using prefix sum hashmap, we can calculate the difference easily.\n<!-- Describe your first thoughts on how... | 0 | 0 | ['Python3'] | 0 |
minimum-operations-to-make-all-array-elements-equal | C++ Solution || Sorting || Lower Bound || Prefix Sum | c-solution-sorting-lower-bound-prefix-su-oqlf | \ncpp []\nclass Solution {\npublic:\n vector<long long> minOperations(vector<int>& nums, vector<int>& queries) {\n // Sort the array\n sort(num | Vaibhav_Arya007 | NORMAL | 2024-09-29T19:08:17.209648+00:00 | 2024-09-29T19:08:17.209676+00:00 | 5 | false | \n```cpp []\nclass Solution {\npublic:\n vector<long long> minOperations(vector<int>& nums, vector<int>& queries) {\n // Sort the array\n sort(nums.begin(), nums.end());\n \n // Precompute the prefix sum\n int n = nums.size();\n vector<long long> prefixSum(n + 1, 0); // Use ... | 0 | 0 | ['C++'] | 0 |
minimum-operations-to-make-all-array-elements-equal | O(mlogn) solution C++|| beats 85% | omlogn-solution-c-beats-85-by-krishnakan-yjpi | Intuition\n Describe your first thoughts on how to solve this problem. \nintuition was to find a pattern and generalize it.\n\n# Approach\n Describe your approa | krishnakant_1 | NORMAL | 2024-09-22T20:28:23.303012+00:00 | 2024-09-22T20:28:23.303050+00:00 | 3 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nintuition was to find a pattern and generalize it.\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\nI find a number just greater than the query element and calculate the left side contribution, that is difference we... | 0 | 0 | ['C++'] | 0 |
minimum-operations-to-make-all-array-elements-equal | easy line by line cpp code | easy-line-by-line-cpp-code-by-varun_b_v-wqrb | \n# Code\ncpp []\nclass Solution {\npublic:\n vector<long long> minOperations(vector<int>& nums, vector<int>& queries) {\n sort(begin(nums), end(nums) | Varun_B_V | NORMAL | 2024-09-22T12:26:04.929264+00:00 | 2024-09-22T12:26:04.929302+00:00 | 3 | false | \n# Code\n```cpp []\nclass Solution {\npublic:\n vector<long long> minOperations(vector<int>& nums, vector<int>& queries) {\n sort(begin(nums), end(nums)); // Sort nums\n int n = nums.size();\n\n // Create prefix sum array\n vector<long long> ps(n, 0);\n ps[0] = nums[0];\n ... | 0 | 0 | ['Binary Search', 'Prefix Sum', 'C++'] | 0 |
minimum-operations-to-make-all-array-elements-equal | c++ solution | c-solution-by-dilipsuthar60-l0re | \nclass Solution {\npublic:\n int getLowerBound(vector<int>&nums, int value){\n int size=nums.size();\n int left=0,right=size-1,mid=0,index=siz | dilipsuthar17 | NORMAL | 2024-09-13T18:32:12.846664+00:00 | 2024-09-13T18:32:12.846699+00:00 | 1 | false | ```\nclass Solution {\npublic:\n int getLowerBound(vector<int>&nums, int value){\n int size=nums.size();\n int left=0,right=size-1,mid=0,index=size;\n while(left<=right){\n mid=(left+right)/2;\n if(nums[mid]>value){\n index=mid;\n right=mid-1;\... | 0 | 0 | ['C', 'C++'] | 0 |
minimum-operations-to-make-all-array-elements-equal | Simple binary search | simple-binary-search-by-risabhuchiha-njoi | \njava []\nclass Solution {\n public List<Long> minOperations(int[] nums, int[] queries) {\n List<Long>ans=new ArrayList<>();\n Arrays.sort(nums) | risabhuchiha | NORMAL | 2024-09-12T21:41:56.308704+00:00 | 2024-09-12T21:41:56.308751+00:00 | 4 | false | \n```java []\nclass Solution {\n public List<Long> minOperations(int[] nums, int[] queries) {\n List<Long>ans=new ArrayList<>();\n Arrays.sort(nums);\n long[]pre=new long[nums.length];\n pre[0]=nums[0];\n for(int i=1;i<pre.length;i++){\n pre[i]=pre[i-1]+nums[i];\n } \n ... | 0 | 0 | ['Java'] | 0 |
minimum-operations-to-make-all-array-elements-equal | Prefix sum + Binary search. 6 lines | prefix-sum-binary-search-6-lines-by-xxxx-fqsf | python3 []\nclass Solution:\n def minOperations(self, nums: List[int], queries: List[int]) -> List[int]:\n nums.sort()\n pref = [*accumulate(nu | xxxxkav | NORMAL | 2024-09-11T17:36:51.763182+00:00 | 2024-09-11T17:37:05.672518+00:00 | 37 | false | ```python3 []\nclass Solution:\n def minOperations(self, nums: List[int], queries: List[int]) -> List[int]:\n nums.sort()\n pref = [*accumulate(nums, initial = 0)]\n ans, n = [], len(nums)\n for q in queries:\n i = bisect_left(nums, q)\n ans.append((q*i - pref[i])*2 ... | 0 | 0 | ['Binary Search', 'Prefix Sum', 'Python3'] | 0 |
minimum-operations-to-make-all-array-elements-equal | python dynamic programming using left and right sums | python-dynamic-programming-using-left-an-xg83 | Intuition\nLet n = size of nums and m = size of queries. Brute force is O(n*m) but it is too slow to pass the tests. If we sort nums the problem is easier to gr | ie334 | NORMAL | 2024-08-20T04:11:30.918351+00:00 | 2024-08-20T04:11:30.918382+00:00 | 4 | false | # Intuition\nLet n = size of nums and m = size of queries. Brute force is O(n*m) but it is too slow to pass the tests. If we sort nums the problem is easier to grasp. Now for any q in queries we know things bigger than q need to be reduced to q. So use binary search on the sorted nums to find how many such elements are... | 0 | 0 | ['Python3'] | 0 |
minimum-operations-to-make-all-array-elements-equal | Prefix Sum + Binary Search | Detailed Explanation | Python3 | prefix-sum-binary-search-detailed-explan-3vgc | Intuition\nWe have to equalize left and right sums based on where the desired value of all elements should be. This equalization point is a dividing index which | wavejumper45 | NORMAL | 2024-08-09T11:25:29.181635+00:00 | 2024-08-09T11:25:29.181660+00:00 | 36 | false | # Intuition\nWe have to equalize left and right sums based on where the desired value of all elements should be. This equalization point is a dividing index which can be rapidly found if the nums list is pre-sorted, using binary search. The accumulated sums (prefix sum) can be stored for all queries so it does not have... | 0 | 0 | ['Python3'] | 0 |
minimum-operations-to-make-all-array-elements-equal | Easy, Beginner-Friendly - Beats 100% | easy-beginner-friendly-beats-100-by-rich-50ta | \nclass Solution {\n public List<Long> minOperations(int[] nums, int[] queries) {\n int n = nums.length;\n Arrays.sort(nums);\n long[] s | RichardLeee | NORMAL | 2024-08-01T10:56:04.689685+00:00 | 2024-08-01T10:56:04.689713+00:00 | 25 | false | ```\nclass Solution {\n public List<Long> minOperations(int[] nums, int[] queries) {\n int n = nums.length;\n Arrays.sort(nums);\n long[] sum = new long[n + 1];\n for (int i = 0; i < n; i++) {\n sum[i + 1] = sum[i] + nums[i];\n }\n \n List<Long> list = new ... | 0 | 0 | ['Binary Search', 'Prefix Sum', 'Java'] | 0 |
minimum-operations-to-make-all-array-elements-equal | Kotlin beats 100% | kotlin-beats-100-by-usernamectrlc-rwr1 | Intuition\n Describe your first thoughts on how to solve this problem. \nsort nums in order to break sort nums into 2 part. \nfirst part that every element is l | usernamectrlc | NORMAL | 2024-07-21T21:38:40.939876+00:00 | 2024-07-21T21:38:40.939907+00:00 | 4 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nsort nums in order to break sort nums into 2 part. \nfirst part that every element is less than current query \nsecond part that every element is equal or more than current querry\n\nuse binary search to find the first index that element ... | 0 | 0 | ['Binary Search', 'Sorting', 'Prefix Sum', 'Kotlin'] | 0 |
minimum-operations-to-make-all-array-elements-equal | C++ solution | c-solution-by-siddhantnigam-0z4t | Code\n\ntypedef long long ll;\n\nclass Solution {\npublic:\n vector<long long> minOperations(vector<int>& nums, vector<int>& queries) {\n sort(nums.be | siddhantnigam | NORMAL | 2024-07-15T12:57:19.661921+00:00 | 2024-07-15T12:57:19.661954+00:00 | 2 | false | # Code\n```\ntypedef long long ll;\n\nclass Solution {\npublic:\n vector<long long> minOperations(vector<int>& nums, vector<int>& queries) {\n sort(nums.begin(), nums.end());\n\n ll n = nums.size();\n vector<ll> pref(n);\n pref[0] = nums[0];\n for(int i=1; i<n; i++)\n pr... | 0 | 0 | ['C++'] | 0 |
minimum-operations-to-make-all-array-elements-equal | C++ || SORTING || LOWER BOUND || Binary Search || Prefix Sum + BRUTE FORCE | c-sorting-lower-bound-binary-search-pref-y3v8 | \n\n# Complexity\n- Time complexity:\nO(m)+O(nlogN)+O(mlogN)\nFor brute force : O(mn)\n\n- Space complexity:\nO(m)\n\n# Code\n\nclass Solution {\npublic:\n#defi | ayush089 | NORMAL | 2024-07-02T11:52:09.079312+00:00 | 2024-07-02T11:52:09.079337+00:00 | 8 | false | \n\n# Complexity\n- Time complexity:\nO(m)+O(nlogN)+O(mlogN)\nFor brute force : O(mn)\n\n- Space complexity:\nO(m)\n\n# Code\n```\nclass Solution {\npublic:\n#define ll long long\n vector<long long> minOperations(vector<int>& nums, vector<int>& queries) {\n // BRUTE\n // ll n = nums.size();\n /... | 0 | 0 | ['C++'] | 0 |
minimum-operations-to-make-all-array-elements-equal | 🐱👓Simple and easy to understandable approach ✨ || Full Explanation with example | simple-and-easy-to-understandable-approa-95qp | Intuition\n Describe your first thoughts on how to solve this problem. \n\nOur basic intituion is, if we have to reach to y from x (knowing x <= y) considering | ravi_thakur_ | NORMAL | 2024-06-29T14:19:11.331810+00:00 | 2024-06-29T14:19:11.331850+00:00 | 69 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\nOur basic intituion is, if we have to reach to `y` from `x` (knowing x <= y) considering doing one operation at a time. Then, total operation will be `y-x`. and from `x` and `z` to `y` (knowing x, z <= y), the total operation will be `y... | 0 | 0 | ['Array', 'Binary Search', 'Sorting', 'Prefix Sum', 'Python3'] | 0 |
minimum-operations-to-make-all-array-elements-equal | C++ | Prefix Sum | Binary Search | Sorting | Math | c-prefix-sum-binary-search-sorting-math-gdl5c | Code\n\nclass Solution {\npublic:\n vector<long long> minOperations(vector<int>& nums, vector<int>& queries) {\n sort(nums.begin(),nums.end());\n | MrChromatic | NORMAL | 2024-06-27T02:57:23.687182+00:00 | 2024-06-27T02:57:23.687230+00:00 | 3 | false | # Code\n```\nclass Solution {\npublic:\n vector<long long> minOperations(vector<int>& nums, vector<int>& queries) {\n sort(nums.begin(),nums.end());\n int n=nums.size();\n vector<long long> pref(n,0);\n pref[0]=nums[0];\n for(int i=1;i<n;i++)\n pref[i]=nums[i]+pref[i-1];... | 0 | 0 | ['Math', 'Binary Search', 'Prefix Sum', 'C++'] | 0 |
minimum-operations-to-make-all-array-elements-equal | Python3 solution using prefix sum and binary search | python3-solution-using-prefix-sum-and-bi-iaon | 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 | ryderwang | NORMAL | 2024-06-13T00:30:47.873877+00:00 | 2024-06-13T00:30:47.873894+00:00 | 38 | 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. $$O(n)$$ --... | 0 | 0 | ['Python3'] | 0 |
minimum-operations-to-make-all-array-elements-equal | This solution is very beginner friendly. | this-solution-is-very-beginner-friendly-xznwj | 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 | HellishAbhi | NORMAL | 2024-06-11T10:49:14.483036+00:00 | 2024-06-11T10:49:14.483074+00:00 | 4 | 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. $$O(n)$$ --... | 0 | 0 | ['C++'] | 0 |
minimum-operations-to-make-all-array-elements-equal | C# Binary Search | c-binary-search-by-ilya-a-f-cuw1 | csharp\npublic class Solution\n{\n public IList<long> MinOperations(int[] nums, int[] queries)\n {\n Array.Sort(nums);\n\n var sum = 0L;\n | ilya-a-f | NORMAL | 2024-06-03T19:57:31.064929+00:00 | 2024-06-03T19:57:31.064947+00:00 | 22 | false | ```csharp\npublic class Solution\n{\n public IList<long> MinOperations(int[] nums, int[] queries)\n {\n Array.Sort(nums);\n\n var sum = 0L;\n var prefix = Array.ConvertAll(nums, num => sum += num);\n\n return Array.ConvertAll(queries, query =>\n {\n var index = Array.... | 0 | 0 | ['Binary Search', 'C#'] | 0 |
minimum-operations-to-make-all-array-elements-equal | Sorting || Prefix sum|| (nlogn) | sorting-prefix-sum-nlogn-by-kishansingh1-hupy | 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 | kishansingh1 | NORMAL | 2024-06-03T18:18:51.607624+00:00 | 2024-06-03T18:18:51.607655+00:00 | 2 | 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:$$O(nlog( n))$$ \n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, ... | 0 | 0 | ['C++'] | 0 |
minimum-operations-to-make-all-array-elements-equal | c++ code | c-code-by-vikasverma_12-h8fh | 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 | Vikasverma_12 | NORMAL | 2024-05-30T09:00:57.813386+00:00 | 2024-05-30T09:00:57.813415+00:00 | 7 | 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. $$O(n)$$ --... | 0 | 0 | ['C++'] | 0 |
minimum-operations-to-make-all-array-elements-equal | Ez solution | ez-solution-by-ayushprakash-ehqf | 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 | AyushPrakash_ | NORMAL | 2024-05-29T16:39:18.470512+00:00 | 2024-05-29T16:39:18.470530+00:00 | 0 | 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:M*log(N) \n- no of queries == M and length of array is N\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity... | 0 | 0 | ['C++'] | 0 |
minimum-operations-to-make-all-array-elements-equal | JAVA Solution | java-solution-by-anurag0608-oxa2 | Intuition\nThere are already good solutions posted under solutions section. But I felt there are fewer good java solution. So posting it here.\n\n# Complexity\n | anurag0608 | NORMAL | 2024-05-28T21:28:16.726382+00:00 | 2024-05-28T21:28:16.726399+00:00 | 34 | false | # Intuition\nThere are already good solutions posted under solutions section. But I felt there are fewer good java solution. So posting it here.\n\n# Complexity\n- Time complexity: $$O(mlogn)$$, where m is query length and n is the size of array nums.\n\n- Space complexity: $$O(n)$$\n\n# Code\n```\nclass Solution {\n ... | 0 | 0 | ['Binary Search', 'Sorting', 'Prefix Sum', 'Java'] | 1 |
minimum-operations-to-make-all-array-elements-equal | beats 90% at speed, 42ms, O(nlogn + n + m) time | beats-90-at-speed-42ms-onlogn-n-m-time-b-s0iu | 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 | SingSongZepf | NORMAL | 2024-05-22T10:05:19.116687+00:00 | 2024-05-22T10:05:19.116720+00:00 | 1 | 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. $$O(n)$$ --... | 0 | 0 | ['Rust'] | 0 |
minimum-operations-to-make-all-array-elements-equal | Python3 O(N + MlogN) Solution (no bisect tricks) | python3-on-mlogn-solution-no-bisect-tric-3v3w | Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n\n# Complexity\n- Time complexity: O(N +MlogN)\n- Space complexity: O(N)\ | ihateinterviews | NORMAL | 2024-05-20T03:22:36.084441+00:00 | 2024-05-20T03:22:36.084457+00:00 | 47 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n\n# Complexity\n- Time complexity: O(N +MlogN)\n- Space complexity: O(N)\n\n# Code\n```\nclass Solution:\n def minOperations(self, nums: List[int], queries: List[int]) -> List[int]:\n answer = []\n nums.sort... | 0 | 0 | ['Python3'] | 0 |
minimum-operations-to-make-all-array-elements-equal | Easy sol^n | easy-soln-by-singhgolu933600-p61w | 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 | singhgolu933600 | NORMAL | 2024-05-09T11:00:40.461066+00:00 | 2024-05-09T11:00:40.461099+00:00 | 0 | 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. $$O(n)$$ --... | 0 | 0 | ['Array', 'Binary Search', 'Sorting', 'Prefix Sum', 'C++'] | 0 |
minimum-operations-to-make-all-array-elements-equal | beat 100% memory | beat-100-memory-by-lordemilio-kycp | 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 | LordEmilio | NORMAL | 2024-04-18T16:26:29.485662+00:00 | 2024-04-18T16:26:29.485692+00:00 | 33 | 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. $$O(n)$$ --... | 0 | 0 | ['Python'] | 0 |
minimum-operations-to-make-all-array-elements-equal | Prefix Sum + Binary Search || C++ Solution | prefix-sum-binary-search-c-solution-by-l-g5m1 | 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 | Last_Of_UsOO | NORMAL | 2024-04-09T05:27:22.642936+00:00 | 2024-04-09T05:27:22.642985+00:00 | 2 | 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 Time Complexity : O(Nlog(N))\n\n- Space complexity:\n<!-- Add your space... | 0 | 0 | ['Binary Search', 'Sorting', 'Prefix Sum', 'C++'] | 0 |
maximum-difference-score-in-a-grid | [Java/C++/Python] DP, Minimum on Top-Left | javacpython-dp-minimum-on-top-left-by-le-7zk7 | Intuition\nMove from c1 to ck with path c1,c2,..,ck,\nres = (c2 - c1) + (c3 - c2) + .... = ck - c1\n\n\n# Explanation\nTo calculate the best score ending at a c | lee215 | NORMAL | 2024-05-12T04:05:12.446881+00:00 | 2024-05-12T05:16:21.905611+00:00 | 6,521 | false | # **Intuition**\nMove from `c1` to `ck` with path `c1,c2,..,ck`,\n`res = (c2 - c1) + (c3 - c2) + .... = ck - c1`\n<br>\n\n# **Explanation**\nTo calculate the best score ending at a cell `ck`,\njust need to find out the minimum `c1` on its top left area.\n\nTo do this, we can define `dp[i][j]` as the minimum in `A[i][j]... | 101 | 1 | ['C', 'Python', 'Java'] | 15 |
maximum-difference-score-in-a-grid | Explained - Find min upto that cell and take diff | explained-find-min-upto-that-cell-and-ta-iq4k | Intuition\n- Need to find the min max, such that min item cell should be towards left and top of the max item cell.\n\n# Approach\n- Keep tracking the smallest | kreakEmp | NORMAL | 2024-05-12T04:01:07.787783+00:00 | 2024-05-12T05:28:49.745028+00:00 | 3,024 | false | # Intuition\n- Need to find the min max, such that min item cell should be towards left and top of the max item cell.\n\n# Approach\n- Keep tracking the smallest number till the current cell torards its left and towards its top.\n- Consider the current cell as the max item and take diff. Keep tracking the max diff\n\n#... | 39 | 4 | ['C++'] | 6 |
maximum-difference-score-in-a-grid | [C++] with Picture Explanation, DP | c-with-picture-explanation-dp-by-subbuff-mrj9 | Intuition\nThe short simple answer to this problem is that the answer is almost equal to the max value in the matrix minus the minimum value in the matrix.\nHow | SubBuffer | NORMAL | 2024-05-12T04:02:09.286396+00:00 | 2024-07-15T20:24:09.531493+00:00 | 2,652 | false | # Intuition\nThe short simple answer to this problem is that the answer is almost equal to the max value in the matrix minus the minimum value in the matrix.\nHowever, obviously there is more to this.\nLet\'s show why the above is true to a good extent.\nImagine the value in grid[i+k][j+t] (such that t,k>0) is larger t... | 18 | 0 | ['C++'] | 2 |
maximum-difference-score-in-a-grid | Simple java solution | simple-java-solution-by-siddhant_1602-s09e | Complexity\n- Time complexity: O(mn)\n\n- Space complexity: O(mn)\n\n# Code\n\nclass Solution {\n public int maxScore(List<List<Integer>> grid) {\n in | Siddhant_1602 | NORMAL | 2024-05-12T04:04:46.545459+00:00 | 2024-05-12T04:06:47.096760+00:00 | 806 | false | # Complexity\n- Time complexity: $$O(m*n)$$\n\n- Space complexity: $$O(m*n)$$\n\n# Code\n```\nclass Solution {\n public int maxScore(List<List<Integer>> grid) {\n int m = grid.size(), n = grid.get(0).size();\n int ans[][] = new int[m][n];\n ans[m-1][n-1] = grid.get(m-1).get(n-1);\n for(in... | 17 | 1 | ['Java'] | 1 |
maximum-difference-score-in-a-grid | C++ || NO DP || NO RECURSION || NEXT GREATER ELEMENT MATRIX || Beginner Friendly | c-no-dp-no-recursion-next-greater-elemen-q0r4 | First read the question 2-3 times(iff necessary) to get the basic understanding of question.\n\nHint: try to look different ways to get the answer ( I personall | int_float_double | NORMAL | 2024-05-16T17:41:54.448801+00:00 | 2024-05-16T18:23:46.364413+00:00 | 261 | false | ## **First read the question 2-3 times(iff necessary) to get the basic understanding of question.**\n\n**Hint:** try to look different ways to get the answer ( I personally suggest using pen and paper )\n\n\n\n**Explaination:** lets say you are at position (0,1) with value "5" and you want to find the max difference by... | 14 | 0 | ['C', 'Matrix'] | 0 |
maximum-difference-score-in-a-grid | Track Previous Min | track-previous-min-by-votrubac-7sz9 | The path in the grid does not matter; the result is the difference between the first and last element.\n\nSo, we track the previously seen minimum element (to t | votrubac | NORMAL | 2024-05-12T05:07:51.693056+00:00 | 2024-05-12T05:07:51.693082+00:00 | 566 | false | The path in the grid does not matter; the result is the difference between the first and last element.\n\nSo, we track the previously seen minimum element (to the left or above) of the current cell `g[i][j]` in `prev_min[j]`. \n\n**C++**\n```cpp\nint maxScore(vector<vector<int>>& g) {\n int m = g.size(), n = g[0].si... | 13 | 1 | ['C'] | 2 |
maximum-difference-score-in-a-grid | Simple || DP || Tabulation 🔥🔥 | simple-dp-tabulation-by-harsh_padsala_26-36ci | Intuition\nThis is the first ever DP problem i could solved in live contest so i feel i should share to everyone\n\nIntuition was so simple we want to start fro | Harsh_Padsala_265 | NORMAL | 2024-05-12T05:40:43.687511+00:00 | 2024-05-16T03:42:13.579006+00:00 | 748 | false | # Intuition\nThis is the first ever DP problem i could solved in live contest so i feel i should share to everyone\n\nIntuition was so simple we want to start from cell from where we can maximise output. after starting from one cell , we only have three choice either right or down or none of them.\n\nSo i found that st... | 10 | 0 | ['Dynamic Programming', 'Python', 'C++', 'Java'] | 1 |
maximum-difference-score-in-a-grid | Python 3 || 8 lines, padded grid || T/S: 86% / 61% | python-3-8-lines-padded-grid-ts-86-61-by-aa6b | Here\'s the plan:\n1. We pad grid with a top-row and a left-column of inf to avoid checking for edge cases. (Illustrated below with input from Example 1)\n\n | Spaulding_ | NORMAL | 2024-05-13T17:44:36.708274+00:00 | 2024-05-24T01:57:15.979309+00:00 | 222 | false | Here\'s the plan:\n1. We pad `grid` with a top-row and a left-column of `inf` to avoid checking for edge cases. (Illustrated below with input from *Example 1*)\n```\n [inf, inf, inf, inf, inf]\n [inf, 9, 5, 7, 3 ]\n [inf, 8, 9, 6, 1 ]\n [inf, 6, 7, 14, 3 ]\n [inf, ... | 9 | 0 | ['Python3'] | 1 |
maximum-difference-score-in-a-grid | Video Explanation [Optimising N*M*(N+M) solution step by step to N*M] | video-explanation-optimising-nmnm-soluti-dfj8 | Explanation\n\nClick here for the video\n\n# Code\n\ntypedef long long int ll;\n\nconst ll INF = 1e18;\n\nclass Solution {\n vector<vector<ll>> dp;\n vect | codingmohan | NORMAL | 2024-05-12T06:21:22.975026+00:00 | 2024-05-12T06:21:22.975054+00:00 | 283 | false | # Explanation\n\n[Click here for the video](https://youtu.be/pF3XT6cDHpU)\n\n# Code\n```\ntypedef long long int ll;\n\nconst ll INF = 1e18;\n\nclass Solution {\n vector<vector<ll>> dp;\n vector<vector<int>> grid;\n int rows, cols;\n \n /*\n ll MaxScore (int r, int c) {\n if (r == rows-1 && c ==... | 7 | 0 | ['C++'] | 0 |
maximum-difference-score-in-a-grid | 🧐Begginner and Advanced versions 🛀 Clean Code 🪟 Clear explanation ⁉️ Q&A👊 Beat 100% 🥁 | begginner-and-advanced-versions-clean-co-ch1l | \n\n\n# What\'s the short version of your solution?\n\nFor each number in the matrix, you just need to compare it with only one number: the smallest number for | navid | NORMAL | 2024-05-12T04:01:48.011171+00:00 | 2024-05-12T06:27:23.722789+00:00 | 567 | false | \n\n\n# What\'s the short version of your solution?\n\nFor each number in the matrix, you just need to compare it with only one number: the smallest number for which a move could be possible. Then you can s... | 6 | 1 | ['Java'] | 1 |
maximum-difference-score-in-a-grid | 🔥Clean Code💯 with Best Explanation🔥||⚡BottomUP DP🌟 | clean-code-with-best-explanationbottomup-tl7q | Here if u will observe carefully then u will find that for each starting cell there is no need to take choice for all the below rows as well as right columns be | aDish_21 | NORMAL | 2024-05-12T15:30:30.634667+00:00 | 2024-05-12T15:36:38.436137+00:00 | 190 | false | ### Here if u will observe carefully then u will find that for each starting cell there is no need to take choice for all the below rows as well as right columns because the best score for a particular starting cell will always be equal to the maximum difference between the final ending cell & the starting cell as beca... | 5 | 0 | ['Dynamic Programming', 'Greedy', 'C++'] | 0 |
maximum-difference-score-in-a-grid | Easy memoization approach || C++ 🔥🔥 | easy-memoization-approach-c-by-abhinav_s-ttfo | Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n1. solvemaxScore Func | Abhinav_Shakya | NORMAL | 2024-05-12T05:22:25.223168+00:00 | 2024-05-12T06:08:59.980953+00:00 | 402 | 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\n1. **solvemaxScore Function**: This function is a helper function that uses recursion and dynamic programming to solve the problem.\n\n - It takes four parameters... | 5 | 0 | ['C++'] | 0 |
maximum-difference-score-in-a-grid | C++ || Recursion + Memoization || Beginner Friendly 🔥✅💯🤩 | c-recursion-memoization-beginner-friendl-c0m5 | \n\n# Code\n\nclass Solution {\npublic:\n int solve(int i, int j, vector<vector<int>>& grid, vector<vector<int>>& dp) {\n if (i >= grid.size() || j >= | _kvschandra_2234 | NORMAL | 2024-05-12T05:00:28.688685+00:00 | 2024-05-12T07:38:18.623688+00:00 | 1,608 | false | \n\n# Code\n```\nclass Solution {\npublic:\n int solve(int i, int j, vector<vector<int>>& grid, vector<vector<int>>& dp) {\n if (i >= grid.size() || j >= grid[0].size()) return 0;\n if (dp[i][j] != -1) return dp[i][j];\n\n int pick1 = INT_MIN, pick2 = INT_MIN;\n if (i + 1 < grid.size()) p... | 5 | 1 | ['Dynamic Programming', 'Recursion', 'Memoization', 'C++'] | 1 |
maximum-difference-score-in-a-grid | Easiest Python Solution 100% Beat - O(m*n) Time and O(1) Extra Space | easiest-python-solution-100-beat-omn-tim-qgrf | Intuition\nThe first thing to notice is that the maximum score can always come from a single move. Consider the case we have a maximum score with multiple moves | rohanvedula | NORMAL | 2024-05-15T14:38:42.955640+00:00 | 2024-05-15T14:38:42.955691+00:00 | 156 | false | # Intuition\nThe first thing to notice is that the maximum score can always come from a single move. Consider the case we have a maximum score with multiple moves. Let\'s assume that it start\'s with a value $$v_1$$ and ends with a value $$v_n$$. We know that the score of the path is $$v_1 - v_n$$ and as a part of goin... | 4 | 0 | ['Dynamic Programming', 'Matrix', 'Python', 'Python3'] | 1 |
maximum-difference-score-in-a-grid | C++ Solution | Dp Recursion + Memoization and Tabulation bottom up | c-solution-dp-recursion-memoization-and-31d7t | Tried solving the question using recursion and memoization in contest but two test cases were giving tle. Anyone have any suggetion to optimize the recusive dp | Abhishek_499 | NORMAL | 2024-05-12T05:12:38.489741+00:00 | 2024-05-12T05:12:38.489770+00:00 | 185 | false | Tried solving the question using recursion and memoization in contest but two test cases were giving tle. Anyone have any suggetion to optimize the recusive dp code please suggest.\n\n\nTried solving using tabulation bottom up dp and it passed all the test cases\n\n\nCode 1 ----> Recursion + Memoization\nCode 2 -----> ... | 4 | 0 | ['C++'] | 1 |
maximum-difference-score-in-a-grid | Java - Extended Brute Force Solution | java-extended-brute-force-solution-by-ia-sdnm | \nclass Solution {\n private Integer dp[][];\n public int maxScore(List<List<Integer>> grid) {\n int row = grid.size();\n int col = grid.get | iamvineettiwari | NORMAL | 2024-05-12T04:24:18.640413+00:00 | 2024-05-12T04:56:51.338529+00:00 | 316 | false | ```\nclass Solution {\n private Integer dp[][];\n public int maxScore(List<List<Integer>> grid) {\n int row = grid.size();\n int col = grid.get(0).size();\n dp = new Integer[row][col];\n\n int max = Integer.MIN_VALUE;\n\n for (int i = 0; i < row; i++) {\n for (int j =... | 4 | 0 | ['Recursion', 'Memoization', 'Java'] | 3 |
maximum-difference-score-in-a-grid | Cleanest Memoization JAVA code you will ever see || CLEAN CODE || | cleanest-memoization-java-code-you-will-5pbh4 | 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 | Akshatjain_ | NORMAL | 2024-06-06T11:16:33.289194+00:00 | 2024-06-06T11:16:33.289226+00:00 | 105 | 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. $$O(n)$$ --... | 3 | 0 | ['Dynamic Programming', 'Recursion', 'Memoization', 'Java'] | 2 |
maximum-difference-score-in-a-grid | Python (Simple DP) | python-simple-dp-by-rnotappl-yda9 | 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 | rnotappl | NORMAL | 2024-05-12T09:18:39.103860+00:00 | 2024-05-12T09:18:39.103912+00:00 | 226 | 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. $$O(n)$$ --... | 3 | 0 | ['Python3'] | 1 |
maximum-difference-score-in-a-grid | Easiest Python solution | easiest-python-solution-by-edwards310-wpq0 | 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\n# Complexity\n- T | Edwards310 | NORMAL | 2024-05-12T04:07:57.199623+00:00 | 2024-05-12T04:07:57.199642+00:00 | 234 | false | # Intuition\n\n\n<!-- Describe your first thoughts on how to... | 3 | 1 | ['Python3'] | 4 |
maximum-difference-score-in-a-grid | ✅✅Beats 100% Users ✅ Easy Solution ✅ Best Approaches✅ | beats-100-users-easy-solution-best-appro-sz4s | Intuition\nThe problem involves finding the maximum score possible while moving from the top-left corner to the bottom-right corner of the grid. You can move ei | arslanarsal | NORMAL | 2024-05-12T04:04:03.504243+00:00 | 2024-05-12T04:04:03.504279+00:00 | 243 | false | # Intuition\nThe problem involves finding the maximum score possible while moving from the top-left corner to the bottom-right corner of the grid. You can move either right or down at each step. The score of the path is determined by the difference between the values of the cells along the path. \n\n# Approach\n1. **Dy... | 3 | 0 | ['C++'] | 1 |
maximum-difference-score-in-a-grid | 2D-DP,Memoization, Easy to understand C++ | 2d-dpmemoization-easy-to-understand-c-by-kh32 | \n\n# Complexity\n- Time complexity: O(mn)\n\n- Space complexity: O(mn)\n\n\n# Code\n\nclass Solution {\npublic:\n int f(vector<vector<int>>&dp,vector<vector | raj_g1729 | NORMAL | 2024-05-19T08:53:55.812223+00:00 | 2024-05-19T08:55:40.159552+00:00 | 75 | false | \n\n# Complexity\n- Time complexity: O(m*n)\n\n- Space complexity: O(m*n)\n\n\n# Code\n```\nclass Solution {\npublic:\n int f(vector<vector<int>>&dp,vector<vector<int>>&grid,int i,int j)\n {\n if(i>=grid.size() || j>=grid[0].size()) return 0;\n if(dp[i][j]!=-1) return dp[i][j];\n int rowTake=... | 2 | 0 | ['C++'] | 0 |
maximum-difference-score-in-a-grid | Memoized Version of Lee's Solution | memoized-version-of-lees-solution-by-nik-ffm4 | Intuition\n Describe your first thoughts on how to solve this problem. \nFor those whose intuition works by recursion and memoization, this solution is the simp | nik_07 | NORMAL | 2024-05-14T13:34:19.260344+00:00 | 2024-05-14T13:35:22.683562+00:00 | 144 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nFor those whose intuition works by recursion and memoization, this solution is the simplest way to solve this problem.\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\nLee\'s solution explains the intuition as follo... | 2 | 0 | ['Dynamic Programming', 'Recursion', 'Memoization', 'C++'] | 0 |
maximum-difference-score-in-a-grid | Simple recursive solution with Memoization | simple-recursive-solution-with-memoizati-o643 | Intuition\nRecursively go in the bottom and right direction making a note of max. Use memoization to optimize.\n\n# Approach\n- Start from top left (0, 0)\n- Fi | sadanandpai | NORMAL | 2024-05-13T19:13:00.351527+00:00 | 2024-05-13T19:13:00.351552+00:00 | 225 | false | # Intuition\nRecursively go in the bottom and right direction making a note of max. Use memoization to optimize.\n\n# Approach\n- Start from top left (0, 0)\n- Find the max of bottomMax and rightMax\n- Update if max at the current cell is greater\n- Update the memoized cell and return\n\n# Complexity\n- Time complexity... | 2 | 0 | ['Dynamic Programming', 'Recursion', 'Memoization', 'JavaScript'] | 0 |
maximum-difference-score-in-a-grid | O(N*M) Time Complexity, Simple DP Solution | onm-time-complexity-simple-dp-solution-b-rfo7 | The key idea is to find the minimum element in the top left region and then take the maximum among the difference of grid and the minimum array constructed.\n\n | N0V1C3 | NORMAL | 2024-05-13T08:29:40.421184+00:00 | 2024-05-13T08:39:23.588296+00:00 | 62 | false | The key idea is to find the minimum element in the top left region and then take the maximum among the difference of grid and the minimum array constructed.\n\nThis is because going from a->b->c gives the answer as (b-a)+(c-b) which is essentially (c-a).\n\n\n**Python:**\n\n```\nclass Solution:\n def maxScore(self, ... | 2 | 0 | ['Dynamic Programming', 'Python'] | 3 |
maximum-difference-score-in-a-grid | Easy C++ Solution || (Recursion + Memoization) ✅✅ | easy-c-solution-recursion-memoization-by-1ggm | Code\n\nclass Solution {\npublic:\n int helper(vector<vector<int>>& grid,int i,int j,int n,int m,vector<vector<int>> &dp){\n if(i==(n-1) && j==(m-1)){ | Abhi242 | NORMAL | 2024-05-13T08:18:37.155373+00:00 | 2024-05-13T08:18:37.155394+00:00 | 309 | false | # Code\n```\nclass Solution {\npublic:\n int helper(vector<vector<int>>& grid,int i,int j,int n,int m,vector<vector<int>> &dp){\n if(i==(n-1) && j==(m-1)){\n return dp[i][j]=0;\n }\n int ans=-1e6;\n if(dp[i][j]!=-1e6){\n return dp[i][j];\n }\n if((i+1)<... | 2 | 0 | ['Dynamic Programming', 'Recursion', 'Memoization', 'C++'] | 0 |
maximum-difference-score-in-a-grid | [python3] easy dp solution + explanation/contest comentary | python3-easy-dp-solution-explanationcont-gv7e | Intuition and Thoughts\n you\'re just looking for the maximum cell that\'s further to the right or further down from the cell you\'re looking at. \n Gonna be si | horseshoecrab | NORMAL | 2024-05-12T22:31:33.888666+00:00 | 2024-05-12T22:42:39.533686+00:00 | 9 | false | #### Intuition and Thoughts\n* you\'re just looking for the maximum cell that\'s further to the right or further down from the cell you\'re looking at. \n* Gonna be size of O(grid) time for each cell if we iterate to get max cell (box with top left at (i,j) and bottom right at (m-1,n-1)), that sucks. \n* If we know the... | 2 | 0 | [] | 0 |
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