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russian-doll-envelopes | Simple Dp LIS Box Stacking | simple-dp-lis-box-stacking-by-chaharnish-24rf | It is a simple application of Longest Increasing subsequence. \nYou Just have to increase the sequence with a smaller element. \nThere is also a NlogN solution | chaharnishant | NORMAL | 2021-03-30T18:38:52.175839+00:00 | 2021-03-30T18:39:42.661102+00:00 | 654 | false | It is a simple application of Longest Increasing subsequence. \nYou Just have to increase the sequence with a smaller element. \nThere is also a NlogN solution do check it out!\n\n```\n\tstatic bool compare(pair<int,pair<int,int>>& a,pair<int,pair<int,int>>& b){\n return a.first>b.first;\n }\n \n int ma... | 4 | 0 | ['Dynamic Programming'] | 0 |
russian-doll-envelopes | Java DP solution with comments for explanation | java-dp-solution-with-comments-for-expla-7jp0 | java\nclass Solution {\n public int maxEnvelopes(int[][] envelopes) {\n // sort the envelopes in ascending order based on width\n Arrays.sort(e | viet-quocnguyen | NORMAL | 2021-03-30T15:12:15.746741+00:00 | 2021-03-30T15:12:15.746787+00:00 | 180 | false | ```java\nclass Solution {\n public int maxEnvelopes(int[][] envelopes) {\n // sort the envelopes in ascending order based on width\n Arrays.sort(envelopes, (a, b) -> a[0] - b[0]); \n \n // keep track of max envelopes could be Russian doll for each envelop.\n int[] dp = new int[env... | 4 | 1 | [] | 1 |
russian-doll-envelopes | My Java Solution using Sort and Longest Increasing Subsequence concept | my-java-solution-using-sort-and-longest-3xibh | \nclass Solution {\n public int maxEnvelopes(int[][] envelopes) {\n // sort as \n // if widths are equal, sort based on the decreeasing height\ | vrohith | NORMAL | 2021-03-30T09:10:50.396348+00:00 | 2021-03-30T09:15:19.983172+00:00 | 315 | false | ```\nclass Solution {\n public int maxEnvelopes(int[][] envelopes) {\n // sort as \n // if widths are equal, sort based on the decreeasing height\n // else carry on with width on increasing\n Arrays.sort(envelopes, (a, b) -> a[0] == b[0] ? b[1] - a[1] : a[0] - b[0]);\n // now just ... | 4 | 1 | ['Sorting', 'Binary Tree', 'Java'] | 0 |
russian-doll-envelopes | C++ | Variation of LIS | DP | c-variation-of-lis-dp-by-wh0ami-akzw | \nclass Solution {\n static bool compare(vector<int>v1, vector<int>v2) {\n return v1[0] < v2[0] || (v1[0] == v2[0] && v1[1] < v2[1]);\n }\npublic:\ | wh0ami | NORMAL | 2020-05-09T15:21:31.186380+00:00 | 2020-05-09T15:21:31.186411+00:00 | 249 | false | ```\nclass Solution {\n static bool compare(vector<int>v1, vector<int>v2) {\n return v1[0] < v2[0] || (v1[0] == v2[0] && v1[1] < v2[1]);\n }\npublic:\n int maxEnvelopes(vector<vector<int>>& envelopes) {\n \n int n = envelopes.size();\n if (n == 0) return 0;\n \n sort(e... | 4 | 1 | [] | 0 |
russian-doll-envelopes | Accepted C# (Sort + LIS) solution: Easy to understand | accepted-c-sort-lis-solution-easy-to-und-yso6 | \npublic class Solution {\n public int MaxEnvelopes(int[][] envelopes)\n {\n if (envelopes == null || envelopes.Length == 0)\n return 0; | pantigalt | NORMAL | 2020-03-06T09:20:38.917909+00:00 | 2020-03-06T09:20:38.917943+00:00 | 128 | false | ```\npublic class Solution {\n public int MaxEnvelopes(int[][] envelopes)\n {\n if (envelopes == null || envelopes.Length == 0)\n return 0;\n \n Array.Sort(envelopes, (a, b) => a[0] == b[0] ? b[1].CompareTo(a[1]) : a[0].CompareTo(b[0]));\n return LengthOfLIS(envelopes.Select... | 4 | 0 | [] | 0 |
russian-doll-envelopes | Simple C# Solution | simple-c-solution-by-maxpushkarev-j87v | \n public class Solution\n {\n public int MaxEnvelopes(int[][] envelopes)\n {\n if (envelopes.Length == 0)\n {\n | maxpushkarev | NORMAL | 2020-03-05T03:07:00.101704+00:00 | 2020-03-05T03:07:00.101750+00:00 | 392 | false | ```\n public class Solution\n {\n public int MaxEnvelopes(int[][] envelopes)\n {\n if (envelopes.Length == 0)\n {\n return 0;\n }\n\n if (envelopes.Length == 1)\n {\n return 1;\n }\n\n Array.So... | 4 | 2 | ['Dynamic Programming'] | 2 |
russian-doll-envelopes | DFS or topological sorting | dfs-or-topological-sorting-by-guagua_2-sp8d | The solution I propose may not be as efficient as many other solutions posted. But I find topological sorting is very intuitive here and is quite interesting. C | guagua_2 | NORMAL | 2019-12-30T06:30:52.439296+00:00 | 2019-12-30T06:30:52.439485+00:00 | 433 | false | The solution I propose may not be as efficient as many other solutions posted. But I find topological sorting is very intuitive here and is quite interesting. Consider each doll is a node in a graph. Two nodes are connected if one can contain another. It is easy to come up with a O(n^2) solution with DFS:\n\n```\nclass... | 4 | 1 | [] | 0 |
russian-doll-envelopes | Simple Java Graph based solution (bad performance) | simple-java-graph-based-solution-bad-per-thyp | Came up with this different approach to the problem. I figured we could model these envelopes as a graph where an edge from n1 -> n2 would mean that n1 can enve | dominoanty | NORMAL | 2019-09-03T06:44:38.929009+00:00 | 2019-09-03T06:44:38.929047+00:00 | 241 | false | Came up with this different approach to the problem. I figured we could model these envelopes as a graph where an edge from n1 -> n2 would mean that n1 can envelope n2 [n1.width > n2.width && n1.height > n2.height]. Now that we have the graph, we can find the longest path and the length of that would give us the answer... | 4 | 0 | ['Graph'] | 0 |
russian-doll-envelopes | JAVA dp solution with comments | java-dp-solution-with-comments-by-siddha-6m3a | \nclass Solution {\n \n int[] dp;\n \n public int lengthOfRussianDoll(int index, int[][] envelopes){\n \n\t\t// terminating condition aka bas | siddhantiitbmittal3 | NORMAL | 2019-05-17T12:55:51.591421+00:00 | 2019-05-17T12:55:51.591486+00:00 | 661 | false | ```\nclass Solution {\n \n int[] dp;\n \n public int lengthOfRussianDoll(int index, int[][] envelopes){\n \n\t\t// terminating condition aka base case\n if(index > envelopes.length-1)\n return 0;\n \n if(dp[index] > 0)\n return dp[index];\n\n int maxC... | 4 | 0 | [] | 2 |
russian-doll-envelopes | Longest path in DAG. DP solution. With follow up solution. | longest-path-in-dag-dp-solution-with-fol-xpla | dp[i] = max(0, dp[j] | if rectangle j can be embedded to rectangle i) + 1\n\n public class Solution {\n \n boolean canFit(int[] a, int[] b) { | zzz1322 | NORMAL | 2016-06-07T02:09:14+00:00 | 2016-06-07T02:09:14+00:00 | 2,246 | false | **dp[i] = max(0, dp[j]** | if rectangle j can be embedded to rectangle i) **+ 1**\n\n public class Solution {\n \n boolean canFit(int[] a, int[] b) { // Rectangle a can fit into rectangle b. \n return (a[0] < b[0] && a[1] < b[1]);\n }\n \n public int maxEnvelopes(int[][... | 4 | 1 | [] | 1 |
russian-doll-envelopes | My Three C++ Solutions: DP, Binary Search and Lower_bound | my-three-c-solutions-dp-binary-search-an-akei | Solution One:\n\n class Solution {\n public:\n int maxEnvelopes(vector>& envelopes) {\n int res = 0, n = envelopes.size();\n \t\tvect | grandyang | NORMAL | 2016-06-07T16:36:16+00:00 | 2016-06-07T16:36:16+00:00 | 869 | false | Solution One:\n\n class Solution {\n public:\n int maxEnvelopes(vector<pair<int, int>>& envelopes) {\n int res = 0, n = envelopes.size();\n \t\tvector<int> dp(n, 1);\n \t\tsort(envelopes.begin(), envelopes.end());\n \t\tfor (int i = 0; i < n; ++i) {\n \t\t\tfor (int j = 0; j < i; ++j... | 4 | 1 | [] | 0 |
russian-doll-envelopes | C++ DP Solution with O(n^2) Time complexity O(n) Space complexity | c-dp-solution-with-on2-time-complexity-o-91zo | class Solution {\n public:\n int maxEnvelopes(vector<pair<int, int>>& envelopes) {\n int ans = 0;\n vector<int> dp;\n | yular | NORMAL | 2016-06-09T07:48:57+00:00 | 2016-06-09T07:48:57+00:00 | 906 | false | class Solution {\n public:\n int maxEnvelopes(vector<pair<int, int>>& envelopes) {\n int ans = 0;\n vector<int> dp;\n if(!envelopes.size())\n return ans;\n sort(envelopes.begin(), envelopes.end(), cmpfunc);\n dp.resize(envelopes.size())... | 4 | 1 | ['Dynamic Programming', 'C++'] | 2 |
russian-doll-envelopes | Clean C++ 11 Implementation with Explaination refered from @kamyu104 | clean-c-11-implementation-with-explainat-nqmq | It is easy to relate this problem with the previous LIS problem. But how to solve it under the 2 dimensional cases. The key ideas lay at that how we deal with t | rainbowsecret | NORMAL | 2016-06-12T02:38:58+00:00 | 2018-10-20T11:37:12.659773+00:00 | 984 | false | It is easy to relate this problem with the previous LIS problem. But how to solve it under the 2 dimensional cases. The key ideas lay at that how we deal with the equal width but different hight cases. \n\nA clever solution is to sort the pair<int,int> array according the width, if the width is equal, just sort by the... | 4 | 1 | [] | 0 |
russian-doll-envelopes | Longest increasing Subsequence || O(Nlogn) || Easy CPP Code | longest-increasing-subsequence-onlogn-ea-vuow | 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 | ayushrakwal94 | NORMAL | 2024-03-07T08:14:11.934568+00:00 | 2024-03-07T08:14:11.934591+00:00 | 756 | 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 | ['Binary Search', 'Dynamic Programming', 'Sorting', 'C++'] | 0 |
russian-doll-envelopes | C++ Solution | Stepwise Explained | Hard problem made easy | Binary Search | Sorting | DP | c-solution-stepwise-explained-hard-probl-c4jr | Intuition and Approach\n\nThe goal is to find the maximum number of envelopes that can be nested inside each other. The sorting is done based on the width, and | darkaadityaa | NORMAL | 2024-02-06T19:04:33.063616+00:00 | 2024-02-06T19:16:47.416988+00:00 | 277 | false | # Intuition and Approach\n\nThe goal is to find the maximum number of envelopes that can be nested inside each other. The sorting is done based on the width, and then the problem is reduced to finding the **Longest Increasing Subsequence (LIS)** based on the height.\n\n1) **Sort Based on Width:**\n-- Sort the envelopes... | 3 | 0 | ['Binary Search', 'Dynamic Programming', 'Sorting', 'C++'] | 0 |
russian-doll-envelopes | Easy || Relatable approach|| learn from 300. | easy-relatable-approach-learn-from-300-b-2ogf | Intuition\n Describe your first thoughts on how to solve this problem. \nSee this problem is similar to 300. Longest Increasing Subsequence , I have explained i | Abhishekkant135 | NORMAL | 2024-01-06T07:50:51.854454+00:00 | 2024-01-06T07:50:51.854489+00:00 | 1,030 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nSee this problem is similar to 300. Longest Increasing Subsequence , I have explained in detailed how patience sort works read it here https://leetcode.com/problems/longest-increasing-subsequence/solutions/4514634/easy-patience-sorting-o-... | 3 | 0 | ['Java'] | 1 |
russian-doll-envelopes | Most Detailed and BEGINNER FRIENDLY Explanation || Binary Search and DP (Memoization) | most-detailed-and-beginner-friendly-expl-kqpo | Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \nv = [[5,4],[6,4],[6,7], | BihaniManan | NORMAL | 2023-07-19T16:11:15.224564+00:00 | 2024-09-21T10:39:47.232370+00:00 | 375 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\nv = [[5,4],[6,4],[6,7],[2,3]]\n\nSort this vector in increasing order wrt to the first element.\nIf v[i].first == v[i + 1].first, in this case sort in decreasing order... | 3 | 1 | ['Binary Search', 'Dynamic Programming', 'Recursion', 'Memoization', 'C++'] | 0 |
russian-doll-envelopes | 4 APPROCH WITH EXPLAIN->SAME AS LIS | 4-approch-with-explain-same-as-lis-by-pa-1qfk | Intuition\n Describe your first thoughts on how to solve this problem. \nLONGEST INCREASING SUBSEQUENCE SAME\n# Approach\n Describe your approach to solving the | parth_gujral_ | NORMAL | 2023-03-23T13:22:13.169139+00:00 | 2023-03-23T13:22:13.169178+00:00 | 1,528 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nLONGEST INCREASING SUBSEQUENCE SAME\n# Approach\n<!-- Describe your approach to solving the problem. -->\nRussian dolls hai unko envelops me dalna hai to sort kardia width wise but agar width same hui kisi ki toh kya kre?\n\nto hmne HIEGH... | 3 | 0 | ['C++'] | 1 |
russian-doll-envelopes | What i understood from all lis solution | what-i-understood-from-all-lis-solution-rmnom | \n// this is nlogn lis variant\nwe need to sort the weights in increasing order but if they are equal then\n heights should be in decreasing order , this is d | pranavrocksharma | NORMAL | 2022-06-20T16:03:10.944465+00:00 | 2022-06-20T16:04:52.583596+00:00 | 160 | false | \n// this is nlogn lis variant\nwe need to sort the weights in increasing order but if they are equal then\n heights should be in decreasing order , this is done to ensure that when we\n do lis on the height the lower one do not gets picked up. eg\n \n [[5,4],[6,5],[6,7],[2,3]]\n \n after sorting in inc wid... | 3 | 0 | ['Dynamic Programming'] | 0 |
russian-doll-envelopes | Russian Doll Envelopes || O(NlogN) || LIS || Binary Search | russian-doll-envelopes-onlogn-lis-binary-a3b2 | First, we sort the Envelopes. width in ascending order and height in decending order. Then apply LIS (Longest Increasing Subsequence) on height to find the requ | tanwar02 | NORMAL | 2022-05-25T21:19:19.779468+00:00 | 2022-05-27T12:08:59.863381+00:00 | 450 | false | First, we sort the Envelopes. width in ascending order and height in decending order. Then apply LIS (Longest Increasing Subsequence) on height to find the required length.\n\nWhy height in decending order?\nIf we sort the height in ascending order and If the width of two envelopes are same and the height of first enve... | 3 | 0 | ['Binary Tree', 'Java'] | 0 |
russian-doll-envelopes | ✅C++ | ✅2 Approaches well Explained O(N^2) & O(NlogN) | ✅Easy & clean Code | c-2-approaches-well-explained-on2-onlogn-c74a | Approach-1: O(N^2) Approach | passed 85/87 testcases\nstep-1 sort the ds\n step-2: Solve it like LIS problem (https://leetcode.com/problems/longest-increasing-s | shm_47 | NORMAL | 2022-05-25T11:20:48.763845+00:00 | 2022-05-25T11:21:49.953061+00:00 | 335 | false | **Approach-1:** O(N^2) Approach | passed 85/87 testcases\nstep-1 sort the ds\n step-2: Solve it like LIS problem (https://leetcode.com/problems/longest-increasing-subsequence/discuss/2072338/c-on2-solution-explained-through-comments) \n\n```\nclass Solution {\npublic:\n int maxEnvelopes(vector<vector<int>>& envelope... | 3 | 0 | ['C'] | 1 |
russian-doll-envelopes | C++ | 4 Approaches | Recursion -> Memoization -> Tabulation -> Binary Search | c-4-approaches-recursion-memoization-tab-it88 | 1. Recursion\nCODE:\n\n // Recursion *** Will Give TLE \n int solve(int idx, int prev, vector> &arr, int n) {\n if(idx == n)\n return 0 | Kajal_39 | NORMAL | 2022-05-25T07:25:29.183951+00:00 | 2022-05-25T07:51:41.724872+00:00 | 264 | false | **1. Recursion\nCODE:**\n\n // Recursion *** Will Give TLE ***\n int solve(int idx, int prev, vector<vector<int>> &arr, int n) {\n if(idx == n)\n return 0;\n int take = 0, untake = 0;\n untake = solve(idx+1, prev, arr, n);\n if((prev == -1) || (arr[idx][0] > arr[prev][0] &&... | 3 | 0 | ['Binary Search', 'Dynamic Programming'] | 0 |
russian-doll-envelopes | ✅ JAVA | SORTING + LIS | TLE | MLE | O(N^2) | O(N LOGN) | 👍🏼 | | java-sorting-lis-tle-mle-on2-on-logn-by-xcd9f | COUPLE OF SOLUTIONS IN JAVA\n Below is Naive approach which throws TLE.\n \nclass Solution {\n \n private int MAX_VALUE = 100002;\n \n public in | bharathkalyans | NORMAL | 2022-05-25T07:20:20.398042+00:00 | 2022-05-25T07:22:32.491276+00:00 | 531 | false | __COUPLE OF SOLUTIONS IN JAVA__\n* Below is Naive approach which throws TLE.\n ```\nclass Solution {\n \n private int MAX_VALUE = 100002;\n \n public int maxEnvelopes(int[][] envelopes) {\n Arrays.sort(envelopes, (a, b) -> a[0] - b[0]);\n int[] prev = new int[]{MAX_VALUE, MAX_VALUE};\n ... | 3 | 0 | ['Dynamic Programming', 'Binary Tree', 'Java'] | 3 |
russian-doll-envelopes | c++ | Russian Doll Envelopes || Simple Solution | c-russian-doll-envelopes-simple-solution-vgrp | Why we need to sort?\n\nIn these types of problem when we are dealing with two dimensions, we need to reduce the problem from two-dimensional array into a one-d | babasaheb256 | NORMAL | 2022-05-25T06:04:05.513829+00:00 | 2022-05-25T06:14:23.307342+00:00 | 357 | false | Why we need to sort?\n\nIn these types of problem when we are dealing with two dimensions, we need to reduce the problem from two-dimensional array into a one-dimensional array in order to improve time complexity.\n\t\n"**Sort first when things are undecided**", sorting can make the data orderly, reduce the degree of c... | 3 | 0 | ['C', 'Sorting', 'C++'] | 0 |
russian-doll-envelopes | [C++] Explaination why sort like that works? || LIS + Binary search | c-explaination-why-sort-like-that-works-ts8vj | The question can be confusing even though you know the solution of classic lis using binary search.I recommend learning first the Binary search solution of clas | iShubhamRana | NORMAL | 2022-05-25T03:23:44.363235+00:00 | 2022-05-25T03:41:03.852356+00:00 | 491 | false | The question can be confusing even though you know the solution of classic lis using binary search.I recommend learning first the Binary search solution of classic LIS. \nThe sorting part caused a lot of confusion to me also ;)\n\n**This post focuses on why sorting a particular way yields the solution**\n\nFor a envelo... | 3 | 0 | [] | 0 |
minimum-operations-to-make-the-array-increasing | [Java/Python 3] Straight Forward codes. | javapython-3-straight-forward-codes-by-r-xurf | Use a dummy value of 0 to initialize the prev; \n2. Traverse the input array, compare each element cur with the modified previous element prev; If cur > prev, u | rock | NORMAL | 2021-04-17T16:04:48.515866+00:00 | 2021-04-17T17:59:36.118667+00:00 | 7,935 | false | 1. Use a dummy value of `0` to initialize the `prev`; \n2. Traverse the input array, compare each element `cur` with the **modified** previous element `prev`; If `cur > prev`, update `prev` to the value of `cur` for the prospective comparison in the next iteration; otherwise, we need to increase `cur` to at least `prev... | 67 | 3 | [] | 10 |
minimum-operations-to-make-the-array-increasing | C++ Track Last | c-track-last-by-votrubac-q11e | cpp\nint minOperations(vector<int>& nums) {\n int res = 0, last = 0;\n for (auto n : nums) {\n res += max(0, last - n + 1);\n last = max(n, | votrubac | NORMAL | 2021-04-17T16:48:46.466909+00:00 | 2021-04-17T16:48:46.467025+00:00 | 5,681 | false | ```cpp\nint minOperations(vector<int>& nums) {\n int res = 0, last = 0;\n for (auto n : nums) {\n res += max(0, last - n + 1);\n last = max(n, last + 1);\n }\n return res;\n}\n``` | 62 | 2 | [] | 11 |
minimum-operations-to-make-the-array-increasing | Easy C++ Solution | easy-c-solution-by-anshika_28-201t | Do upvote if useful!\n\nclass Solution {\npublic:\n int minOperations(vector<int>& nums) {\n int output=0;\n for(int i=0;i<nums.size()-1;i++){\ | anshika_28 | NORMAL | 2021-05-02T08:19:27.465721+00:00 | 2021-05-02T08:19:27.465752+00:00 | 3,895 | false | **Do upvote if useful!**\n```\nclass Solution {\npublic:\n int minOperations(vector<int>& nums) {\n int output=0;\n for(int i=0;i<nums.size()-1;i++){\n if(nums[i]<nums[i+1])\n continue;\n else{\n output=output+(nums[i]+1-nums[i+1]);\n n... | 47 | 3 | ['C', 'C++'] | 5 |
minimum-operations-to-make-the-array-increasing | 100% 1ms Java Solution | 100-1ms-java-solution-by-rentx-v5i4 | Using num as the latest value to compare with next nums[i], instead of increasing nums[i] and using it to compare with nums[i+1]. Changing the value of nums[i] | rentx | NORMAL | 2021-05-26T23:02:02.469589+00:00 | 2021-05-26T23:02:02.469622+00:00 | 2,839 | false | Using num as the latest value to compare with next nums[i], instead of increasing nums[i] and using it to compare with nums[i+1]. Changing the value of nums[i] costs more time than changing the value of the variable num. \n\n```\nclass Solution {\n public int minOperations(int[] nums) {\n if (nums.length <= 1... | 21 | 0 | ['Java'] | 6 |
minimum-operations-to-make-the-array-increasing | Python3 simple solution beats 90% users | python3-simple-solution-beats-90-users-b-ai7s | \nclass Solution:\n def minOperations(self, nums: List[int]) -> int:\n count = 0\n for i in range(1,len(nums)):\n if nums[i] <= nums | EklavyaJoshi | NORMAL | 2021-04-27T03:09:05.660999+00:00 | 2021-04-27T03:09:05.661030+00:00 | 2,065 | false | ```\nclass Solution:\n def minOperations(self, nums: List[int]) -> int:\n count = 0\n for i in range(1,len(nums)):\n if nums[i] <= nums[i-1]:\n x = nums[i]\n nums[i] += (nums[i-1] - nums[i]) + 1\n count += nums[i] - x\n return count\n```\n*... | 18 | 2 | ['Python3'] | 1 |
minimum-operations-to-make-the-array-increasing | ✅ Short & Easy Solution w/ Explanation | 4 Lines of Code! | short-easy-solution-w-explanation-4-line-wqhf | \u2714\uFE0F Solution - I\n\nThe problem can be solved as -\n\n1. Just iterate over the array.\n2. If at any point, nums[i] <= nums[i - 1], then we need to incr | archit91 | NORMAL | 2021-04-17T16:03:56.108367+00:00 | 2021-04-17T18:39:27.315639+00:00 | 1,997 | false | \u2714\uFE0F ***Solution - I***\n\nThe problem can be solved as -\n\n1. Just iterate over the array.\n2. If at any point, `nums[i] <= nums[i - 1]`, then we need to increment `nums[i]` to make the array strictly increasing. The number of increments needed is given by - `nums[i - 1] + nums[i] + 1`. Basically, it is the n... | 17 | 1 | ['C'] | 5 |
minimum-operations-to-make-the-array-increasing | Python O(n) simple and short solution | python-on-simple-and-short-solution-by-t-o13f | Python :\n\n\ndef minOperations(self, nums: List[int]) -> int:\n\tans = 0\n\n\tfor i in range(1, len(nums)):\n\t\tif nums[i] <= nums[i - 1]:\n\t\t\tans += (nums | TovAm | NORMAL | 2021-11-07T15:42:03.331855+00:00 | 2021-11-07T15:42:03.331896+00:00 | 1,672 | false | **Python :**\n\n```\ndef minOperations(self, nums: List[int]) -> int:\n\tans = 0\n\n\tfor i in range(1, len(nums)):\n\t\tif nums[i] <= nums[i - 1]:\n\t\t\tans += (nums[i - 1] - nums[i] + 1)\n\t\t\tnums[i] = (nums[i - 1] + 1)\n\n\treturn ans\n```\n\n**Like it ? please upvote !** | 13 | 1 | ['Python', 'Python3'] | 1 |
minimum-operations-to-make-the-array-increasing | Easiest Java Solution [ 100% ] [ 2ms ] | easiest-java-solution-100-2ms-by-rajarsh-t260 | Approach\n1. Initialization:\n - Get the length of the array nums and store it in len.\n - Create a new array arr of the same length as nums.\n - Initiali | RajarshiMitra | NORMAL | 2024-06-18T07:01:57.922048+00:00 | 2024-06-18T07:08:03.456238+00:00 | 506 | false | # Approach\n1. **Initialization**:\n - Get the length of the array `nums` and store it in `len`.\n - Create a new array `arr` of the same length as `nums`.\n - Initialize a variable `count` to keep track of the total number of operations.\n\n2. **Copying the Original Array**:\n - Copy each element of `nums` int... | 12 | 0 | ['Java'] | 0 |
minimum-operations-to-make-the-array-increasing | [Java] - Simple One Pass Solution | java-simple-one-pass-solution-by-makubex-s5ye | Check if the next number (nums[i]), is smaller than current number (nums[i - 1]). If it is smaller, get the difference between current number nums[i - 1] and ne | makubex74 | NORMAL | 2021-04-19T05:53:07.871391+00:00 | 2021-04-20T05:26:43.788207+00:00 | 765 | false | * Check if the next number (```nums[i]```), is smaller than current number (```nums[i - 1]```). If it is smaller, get the difference between current number ```nums[i - 1]``` and next number ```nums[i]``` and add a 1 to it to make the next number strictly higher than current number.\n* Add the computed difference value ... | 9 | 2 | [] | 0 |
minimum-operations-to-make-the-array-increasing | Python [one pass solution] | python-one-pass-solution-by-it_bilim-j8t5 | \nclass Solution(object):\n def minOperations(self, nums):\n res = 0\n for i in range(1,len(nums)):\n if nums[i]<=nums[i-1]:\n | it_bilim | NORMAL | 2021-04-17T17:08:52.059753+00:00 | 2021-04-17T17:08:52.059785+00:00 | 657 | false | ```\nclass Solution(object):\n def minOperations(self, nums):\n res = 0\n for i in range(1,len(nums)):\n if nums[i]<=nums[i-1]:\n diff = nums[i-1]-nums[i]+1\n res += diff\n nums[i] = nums[i-1]+1\n return res\n``` | 9 | 3 | [] | 0 |
minimum-operations-to-make-the-array-increasing | Simple Javascript Solution | simple-javascript-solution-by-nileshsain-v53j | \nvar minOperations = function(nums) {\n if(nums.length < 2) return 0;\n let count = 0;\n for(let i = 1; i<nums.length; i++) {\n if(nums[i] <= nu | nileshsaini_99 | NORMAL | 2021-09-23T12:47:55.930629+00:00 | 2021-09-23T12:47:55.930660+00:00 | 931 | false | ```\nvar minOperations = function(nums) {\n if(nums.length < 2) return 0;\n let count = 0;\n for(let i = 1; i<nums.length; i++) {\n if(nums[i] <= nums[i-1]) {\n let change = nums[i-1] - nums[i] + 1;\n count += change;\n nums[i] += change;\n }\n }\n \n return c... | 8 | 0 | ['JavaScript'] | 0 |
minimum-operations-to-make-the-array-increasing | [Java] Easy Solution with explanation beats 100% Time: O(n), Space O(1) | java-easy-solution-with-explanation-beat-krno | Approach :\n Start from index 1\n if the current element (nums[i]) is less than the previous element(nums[i-1]) then have the difference count to store the di | vinsinin | NORMAL | 2021-04-17T16:14:43.005909+00:00 | 2021-04-17T16:22:26.801419+00:00 | 1,137 | false | **Approach :**\n* Start from `index 1`\n* if the current element (`nums[i]`) is less than the previous element(`nums[i-1]`) then have the difference `count` to store the difference, also `+1` to make it increasing\n* Add the difference to the result `count += diff`\n* Add the difference to the current element `nums... | 7 | 0 | ['Java'] | 2 |
minimum-operations-to-make-the-array-increasing | ✅Easiest Basic C++ solution | Full Explanations with Dry Run | No advance Logiv✅ | easiest-basic-c-solution-full-explanatio-12yz | Very Easy solution..just imagine elements of array as I proceed\n\nSolution:\n\nclass Solution {\npublic:\n int minOperations(vector<int>& nums) {\n i | dev_yash_ | NORMAL | 2024-02-18T08:09:29.189034+00:00 | 2024-02-18T08:09:29.189058+00:00 | 388 | false | Very Easy solution..just imagine elements of array as I proceed\n\n**Solution:**\n```\nclass Solution {\npublic:\n int minOperations(vector<int>& nums) {\n int operations=0; //cant keep 1 and do operations++ as for single ele array its inv\n for(size_t i {1};i<nums.size();i++) //keep i=1 not 0 as if c... | 6 | 0 | ['Array', 'Greedy', 'C', 'C++'] | 0 |
minimum-operations-to-make-the-array-increasing | [Python3] Easy Solution without Modifying Array | python3-easy-solution-without-modifying-3ssj5 | Modifying mutable input data is bad practice in Python for functions like this, because when using it, the user may not be aware of the changes, and this will l | rgalyeon | NORMAL | 2021-04-17T22:51:59.657645+00:00 | 2021-04-17T22:51:59.657676+00:00 | 725 | false | Modifying mutable input data is bad practice in Python for functions like this, because when using it, the user may not be aware of the changes, and this will lead to subsequent errors.\nWe can solve this problem without modifying a list by using an additional variable (`max_elem`).\n\nAlgo:\n* Initialize variables: ... | 6 | 0 | ['Python'] | 1 |
minimum-operations-to-make-the-array-increasing | Python Easy and Efficient Solution . | python-easy-and-efficient-solution-by-as-613p | Intuition\n Describe your first thoughts on how to solve this problem. \nWe have to Return the minimum number of operations needed to make nums strictly increas | Aswin_Bharath | NORMAL | 2023-11-01T05:05:10.815480+00:00 | 2023-11-01T05:05:10.815508+00:00 | 898 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nWe have to Return the minimum number of operations needed to make nums strictly increasing.\n# Approach\n<!-- Describe your approach to solving the problem. -->\nTaking the previous value and comparing with the current value and increasin... | 5 | 0 | ['Python3'] | 0 |
minimum-operations-to-make-the-array-increasing | 5 LINE JAVA SOLUTION || Easy peasy lemon squeezy😊 || SIMPLE | 5-line-java-solution-easy-peasy-lemon-sq-hqjv | \n\n# Code\n\nclass Solution {\n public int minOperations(int[] nums) {\n int minimum_operations = 0;\n for(int i=1;i<nums.length;i++){\n | Sauravmehta | NORMAL | 2023-02-13T20:56:58.978116+00:00 | 2023-02-13T20:56:58.978161+00:00 | 841 | false | \n\n# Code\n```\nclass Solution {\n public int minOperations(int[] nums) {\n int minimum_operations = 0;\n for(int i=1;i<nums.length;i++){\n minimum_operations += Math.max(nums[i-1]+1,nums[i]) - nums[i];\n nums[i] = Math.max(nums[i-1]+1,nums[i]);\n }\n return minimum... | 5 | 0 | ['Java'] | 1 |
minimum-operations-to-make-the-array-increasing | Java Clean Solution & explanation | java-clean-solution-explanation-by-aswin-fh7m | Please \uD83D\uDD3C upvote this post if you find the answer useful & do comment about your thoughts \uD83D\uDCAC\n\n## Explanation\n\nThis is the most clean sol | aswinb | NORMAL | 2021-04-17T16:19:42.097912+00:00 | 2022-05-19T07:08:50.572107+00:00 | 264 | false | **Please** \uD83D\uDD3C **upvote this post if you find the answer useful & do comment about your thoughts** \uD83D\uDCAC\n\n## Explanation\n\nThis is the most clean solution in Java which calculates number of operations along with array updation of the operations to be made.\n\n- At first we check whether the given arr... | 5 | 1 | ['Java'] | 0 |
minimum-operations-to-make-the-array-increasing | Python Simple Solution | python-simple-solution-by-lokeshsk1-fxvr | \nclass Solution:\n def minOperations(self, nums: List[int]) -> int:\n c=0\n for i in range(len(nums)-1):\n if nums[i+1] <= nums[i]: | lokeshsk1 | NORMAL | 2021-04-17T16:09:12.549316+00:00 | 2021-04-18T03:45:34.735632+00:00 | 396 | false | ```\nclass Solution:\n def minOperations(self, nums: List[int]) -> int:\n c=0\n for i in range(len(nums)-1):\n if nums[i+1] <= nums[i]:\n d = (nums[i] + 1) - nums[i+1]\n nums[i+1] += d\n c += d\n return c\n``` | 5 | 1 | [] | 1 |
minimum-operations-to-make-the-array-increasing | Two Solutions ✅|| Equal Complexity || Varied Runtimes ⏱️🚀 || JAVA ☕ | two-solutions-equal-complexity-varied-ru-agma | Intuition\nThe goal is to ensure each element in the array is strictly greater than the previous element. If an element is not greater than the previous one, we | Megha_Mathur18 | NORMAL | 2024-06-14T05:05:32.245752+00:00 | 2024-06-14T05:05:32.245785+00:00 | 345 | false | # Intuition\nThe goal is to ensure each element in the array is strictly greater than the previous element. If an element is not greater than the previous one, we need to increase it to be at least one more than the previous element.\n\n# Approach\n1. Initialize a counter count to keep track of the total number of oper... | 4 | 0 | ['Java'] | 0 |
minimum-operations-to-make-the-array-increasing | simple and easy C++ solution 😍❤️🔥 | simple-and-easy-c-solution-by-shishirrsi-3nvv | if it\'s help, please up \u2B06 vote! \u2764\uFE0F\n\n\n# Code\n\nclass Solution {\npublic:\n int minOperations(vector<int>& nums) \n {\n int ans = | shishirRsiam | NORMAL | 2024-06-05T07:17:47.612288+00:00 | 2024-06-05T07:17:47.612307+00:00 | 408 | false | # if it\'s help, please up \u2B06 vote! \u2764\uFE0F\n\n\n# Code\n```\nclass Solution {\npublic:\n int minOperations(vector<int>& nums) \n {\n int ans = 0, n = nums.size();\n vector<int>temp = nums;\n for(int i=0;i<n-1;i++)\n {\n if(temp[i] >= temp[i+1])\n tem... | 4 | 0 | ['Array', 'Greedy', 'C++'] | 3 |
minimum-operations-to-make-the-array-increasing | Beginner-friendly || Simple solution in Python3/TypeScript | beginner-friendly-simple-solution-in-pyt-iukh | Intuition\nThe problem description is the following:\n- there\'s a list of nums\n- our goal is to make nums strongly increasing\n\nThere\'s no need in sorting. | subscriber6436 | NORMAL | 2023-09-26T22:37:43.547317+00:00 | 2023-09-26T22:37:43.547342+00:00 | 292 | false | # Intuition\nThe problem description is the following:\n- there\'s a list of `nums`\n- our goal is to make `nums` **strongly increasing**\n\nThere\'s no need in sorting. The approach is straightforward and requires to calculate difference between adjacent integers.\n\n# Approach\n1. initialize `ans` to store the answer... | 4 | 0 | ['Array', 'Greedy', 'Python3'] | 1 |
minimum-operations-to-make-the-array-increasing | Full Accuracy python3 | full-accuracy-python3-by-ganjinaveen-spzs | \n\n# Find difference between two elements and add 1\n\nclass Solution(object):\n def minOperations(self, nums):\n count = current= 0\n for n i | GANJINAVEEN | NORMAL | 2023-02-08T18:23:48.687612+00:00 | 2023-03-20T17:48:26.895032+00:00 | 260 | false | \n\n# Find difference between two elements and add 1\n```\nclass Solution(object):\n def minOperations(self, nums):\n count = current= 0\n for n in nums:\n if n <= current:\n current += 1\n count += current - n\n else:\n current = n\n ... | 4 | 0 | ['Python'] | 0 |
minimum-operations-to-make-the-array-increasing | Java solution 2ms | java-solution-2ms-by-saha_souvik-jg5q | Intuition\nFor every element, which is less than or equal to its predecessor, perform the required no. of make it strictly increasing\n\n# Approach\nTraverse th | Saha_Souvik | NORMAL | 2023-01-25T13:44:48.780944+00:00 | 2023-01-25T13:44:48.780997+00:00 | 761 | false | # Intuition\nFor every element, which is less than or equal to its predecessor, perform the required no. of make it strictly increasing\n\n# Approach\nTraverse the array from left to right, for every element, ```if nums[i] <= nums[i-1] ```, then we increment the i-th element and add the required no.of steps to the resu... | 4 | 0 | ['Java'] | 0 |
minimum-operations-to-make-the-array-increasing | JavaScript faster than 85% | javascript-faster-than-85-by-seredulichk-e35p | \nconst minOperations = nums => {\n let prevEl = nums[0];\n let counter = 0;\n \n for (let i = 0; i < nums.length - 1; i += 1) {\n if (nums[i | seredulichka | NORMAL | 2022-08-25T12:04:02.956745+00:00 | 2022-08-25T12:04:02.956778+00:00 | 562 | false | ```\nconst minOperations = nums => {\n let prevEl = nums[0];\n let counter = 0;\n \n for (let i = 0; i < nums.length - 1; i += 1) {\n if (nums[i + 1] <= prevEl) {\n const diff = prevEl + 1 - nums[i+1] \n counter += diff\n prevEl += 1\n } else {\n pre... | 4 | 0 | ['JavaScript'] | 0 |
minimum-operations-to-make-the-array-increasing | 100% Faster and 100% Less Space. Easy Solution in Python | 100-faster-and-100-less-space-easy-solut-6sjl | ```class Solution:\n def minOperations(self, nums: List[int]) -> int:\n base = nums[0]\n count = 0\n for i in range(1,len(nums)):\n | shivamsingh99 | NORMAL | 2021-04-18T14:08:52.754126+00:00 | 2021-04-18T19:56:07.217995+00:00 | 598 | false | ```class Solution:\n def minOperations(self, nums: List[int]) -> int:\n base = nums[0]\n count = 0\n for i in range(1,len(nums)):\n if nums[i] < base:\n x = base - nums[i] + 1\n count += x\n nums[i] += x\n base = nums[i]\n ... | 4 | 1 | ['Python'] | 3 |
minimum-operations-to-make-the-array-increasing | python 3 solution beats 99.8% TIME , O(1) SPACE used | python-3-solution-beats-998-time-o1-spac-b6a8 | stats\n Describe your first thoughts on how to solve this problem. \n\n\n\n# Approach\n Describe your approach to solving the problem. \nwe can have a temporary | Sumukha_g | NORMAL | 2023-08-10T12:59:19.808820+00:00 | 2023-08-10T12:59:19.808844+00:00 | 471 | false | # stats\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\nwe can have a temporary... | 3 | 0 | ['Array', 'Python3'] | 2 |
minimum-operations-to-make-the-array-increasing | Best approach in JAVA,C++,PYTHON,GO with 0ms runtime!! | best-approach-in-javacpythongo-with-0ms-urhkn | \n\n# Approach\nThe given code defines a class named "Solution" with a member function named "minOperations". This function takes a vector of integers "nums" as | madhavbsnl013 | NORMAL | 2023-07-02T09:13:27.691516+00:00 | 2023-07-02T09:14:15.364711+00:00 | 596 | false | \n\n# Approach\nThe given code defines a class named "Solution" with a member function named "minOperations". This function takes a vector of integers "nums" as input and returns an integer.\n\nThe purpose of the function is to calculate the minimum number of operations required to make the elements in the input vector... | 3 | 0 | ['Greedy', 'Python', 'C++', 'Java', 'Go'] | 0 |
minimum-operations-to-make-the-array-increasing | Easy solution | easy-solution-by-wtfcoder-41i8 | 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 | wtfcoder | NORMAL | 2023-06-23T14:24:28.103258+00:00 | 2023-06-23T14:24:28.103289+00:00 | 343 | 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(N)\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:O(1)\n<!-- Add your space complexity here, e.g. $$... | 3 | 0 | ['C++'] | 0 |
minimum-operations-to-make-the-array-increasing | FASTEST way with JAVA(2ms) | fastest-way-with-java2ms-by-amin_aziz-h20u | Code\n\nclass Solution {\n public int minOperations(int[] nums) {\n int count = 0;\n int dep = nums[0];\n for(int a = 0; a < nums.leng | amin_aziz | NORMAL | 2023-04-13T06:56:56.681525+00:00 | 2023-04-13T06:56:56.681563+00:00 | 621 | false | # Code\n```\nclass Solution {\n public int minOperations(int[] nums) {\n int count = 0;\n int dep = nums[0];\n for(int a = 0; a < nums.length-1; a++){\n if(dep>=nums[a+1]){\n dep++;\n count += dep-nums[a+1];\n }else{\n dep = nu... | 3 | 0 | ['Java'] | 0 |
minimum-operations-to-make-the-array-increasing | 8 lines solution using python3 simplest of all | 8-lines-solution-using-python3-simplest-khs9x | Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time | Mohammad_tanveer | NORMAL | 2023-03-20T17:13:55.132520+00:00 | 2023-03-20T17:13:55.132562+00:00 | 668 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ --... | 3 | 0 | ['Python3'] | 0 |
minimum-operations-to-make-the-array-increasing | Java | Array | Easy Approach | java-array-easy-approach-by-divyansh__26-nz6i | \nclass Solution {\n public int minOperations(int[] nums) {\n int count=0;\n for(int i=1;i<nums.length;i++){\n if(nums[i]<=nums[i-1] | Divyansh__26 | NORMAL | 2022-09-14T15:45:48.037494+00:00 | 2022-09-14T15:45:48.037528+00:00 | 645 | false | ```\nclass Solution {\n public int minOperations(int[] nums) {\n int count=0;\n for(int i=1;i<nums.length;i++){\n if(nums[i]<=nums[i-1]){\n count=count+nums[i-1]-nums[i]+1;\n nums[i]=nums[i-1]+1;\n }\n }\n return count;\n }\n}\n```\nK... | 3 | 0 | ['Array', 'Java'] | 0 |
minimum-operations-to-make-the-array-increasing | easy python code | easy-python-code-by-dakash682-wj5h | \nclass Solution:\n def minOperations(self, nums: List[int]) -> int:\n count = 0\n for i in range(len(nums)-1):\n if nums[i] >= nums | dakash682 | NORMAL | 2022-04-09T03:26:57.696625+00:00 | 2022-04-09T03:26:57.696671+00:00 | 409 | false | ```\nclass Solution:\n def minOperations(self, nums: List[int]) -> int:\n count = 0\n for i in range(len(nums)-1):\n if nums[i] >= nums[i+1]:\n count += (nums[i]+1)-nums[i+1]\n nums[i+1] = nums[i]+1\n return count\n```\nhope it helped you,\nif it did, plz... | 3 | 0 | ['Python', 'Python3'] | 0 |
minimum-operations-to-make-the-array-increasing | C++ Easy to understand 4 Lines only | c-easy-to-understand-4-lines-only-by-nex-a5ak | \nclass Solution {\npublic:\n int minOperations(vector<int>& nums) {\n int res = 0, last = 0;\n for (auto n : nums) {\n res += max(0, last - | NextThread | NORMAL | 2022-01-04T15:14:46.273550+00:00 | 2022-01-04T15:14:46.273585+00:00 | 226 | false | ```\nclass Solution {\npublic:\n int minOperations(vector<int>& nums) {\n int res = 0, last = 0;\n for (auto n : nums) {\n res += max(0, last - n + 1);\n last = max(n, last + 1);\n }\n return res;\n } \n};\n``` | 3 | 0 | ['C'] | 1 |
minimum-operations-to-make-the-array-increasing | Easy Readable Solution | easy-readable-solution-by-himanshuchhika-c1gk | Explanation:\ngoal is to find the minimum number of operation so if we are going to perform operation i.e if case is nums[i-1]>=nums[i] then make make nums[i]=n | himanshuchhikara | NORMAL | 2021-04-17T16:13:40.724320+00:00 | 2021-04-17T16:41:01.820694+00:00 | 181 | false | **Explanation:**\ngoal is to find the minimum number of operation so if we are going to perform operation i.e if case is nums[i-1]>=nums[i] then make make nums[i]=nums[i-1]+1. So cost to make nums[i] equal to (nums[i-1]+1) is nums[i-1] - nums[i] + 1.\n\n**CODE:**\n```\n public int minOperations(int[] nums) {\n ... | 3 | 1 | ['Java'] | 1 |
minimum-operations-to-make-the-array-increasing | Solution | solution-by-ftm_v20-ncnm | 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 | ftm_v20 | NORMAL | 2024-07-02T14:51:07.001079+00:00 | 2024-07-02T14:51:07.001108+00:00 | 250 | 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)$$ --... | 2 | 0 | ['Python3'] | 0 |
minimum-operations-to-make-the-array-increasing | Rust || O(n) || 0ms | rust-on-0ms-by-user7454af-4iwg | Complexity\n- Time complexity: O(n)\n\n- Space complexity: O(1)\n\n# Code\n\nimpl Solution {\n pub fn min_operations(mut nums: Vec<i32>) -> i32 {\n le | user7454af | NORMAL | 2024-06-06T02:37:37.205904+00:00 | 2024-06-06T02:37:37.205937+00:00 | 206 | false | # Complexity\n- Time complexity: $$O(n)$$\n\n- Space complexity: $$O(1)$$\n\n# Code\n```\nimpl Solution {\n pub fn min_operations(mut nums: Vec<i32>) -> i32 {\n let mut ans = 0;\n for i in 1..nums.len() {\n if nums[i] <= nums[i-1] {\n let change = nums[i-1] - nums[i] + 1;\n ... | 2 | 0 | ['Rust'] | 0 |
minimum-operations-to-make-the-array-increasing | Easy C++ Solution | With Explaination | easy-c-solution-with-explaination-by-vai-fjnv | \nclass Solution {\npublic:\nint minOperations(vector<int>& nums) {\n int count=0;\n\t\t//Loop through the vector starting from the second element\n | vaibhavS_07 | NORMAL | 2024-03-06T22:26:49.259036+00:00 | 2024-03-06T22:28:20.017163+00:00 | 595 | false | ```\nclass Solution {\npublic:\nint minOperations(vector<int>& nums) {\n int count=0;\n\t\t//Loop through the vector starting from the second element\n for(int i=1;i<nums.size();i++){\n\t\t //Check if the current element is less than or equal to the previous one\n if(nums[i]<=nums[i-1]){\n\t... | 2 | 0 | ['C'] | 0 |
minimum-operations-to-make-the-array-increasing | Super Easy || upvote if u like :) | super-easy-upvote-if-u-like-by-hrugved00-mzwk | \n\n# Code\n\nclass Solution {\n public int minOperations(int[] nums) {\n int step=0;\n for(int i=1;i<nums.length;i++){\n if(nums[i] | hrugved001 | NORMAL | 2023-06-16T06:42:21.871038+00:00 | 2023-06-16T06:42:21.871066+00:00 | 260 | false | \n\n# Code\n```\nclass Solution {\n public int minOperations(int[] nums) {\n int step=0;\n for(int i=1;i<nums.length;i++){\n if(nums[i]<=nums[i-1]){\n step+=Math.abs(nums[i]-nums[i-1])+1;\n nums[i]=nums[i-1]+1;\n }\n }\n return step;\n }\n}\n... | 2 | 0 | ['Java'] | 0 |
minimum-operations-to-make-the-array-increasing | Simple JAVA Solution for beginners. 2ms. Beats 95.3%. | simple-java-solution-for-beginners-2ms-b-4m98 | 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 | sohaebAhmed | NORMAL | 2023-05-09T03:29:14.941866+00:00 | 2023-05-09T03:29:14.941903+00:00 | 307 | 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)$$ --... | 2 | 0 | ['Java'] | 0 |
minimum-operations-to-make-the-array-increasing | C++ Solution || Simple Approach | c-solution-simple-approach-by-asad_sarwa-3k10 | Complexity\n- Time complexity: O(n)\n Add your time complexity here, e.g. O(n) \n\n- Space complexity: O(1)\n Add your space complexity here, e.g. O(n) \n\n# Co | Asad_Sarwar | NORMAL | 2023-03-24T11:52:10.114206+00:00 | 2023-03-24T11:52:10.114238+00:00 | 1,283 | false | # Complexity\n- Time complexity: O(n)\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity: O(1)\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\nclass Solution {\npublic:\n int minOperations(vector<int>& nums) {\n\n if(nums.size()==1)\n return 0;\n int n=n... | 2 | 0 | ['C++'] | 0 |
minimum-operations-to-make-the-array-increasing | Simple C++ code || beats 94.88% and 84.15% ✌ | simple-c-code-beats-9488-and-8415-by-kra-ozol | \n# Approach\n Describe your approach to solving the problem. \niterates through index =1, checks if the element in the current index is smaller than the elemen | Kraken_02 | NORMAL | 2023-02-07T04:48:07.598352+00:00 | 2023-02-07T04:48:07.598399+00:00 | 715 | false | \n# Approach\n<!-- Describe your approach to solving the problem. -->\niterates through index =1, checks if the element in the current index is smaller than the element in the previous one, if it is then adds value equal to previousElement - currentElement + 1 , also stores the increased number to a counter to return l... | 2 | 0 | ['C++'] | 1 |
minimum-operations-to-make-the-array-increasing | C++ Solution | c-solution-by-mayuri_goswami-3e39 | Intuition\nFor every element, which is less than or equal to its predecessor, we will perform the required number of operations to make it strictly increasing.\ | Mayuri_Goswami | NORMAL | 2023-01-28T03:07:45.922674+00:00 | 2023-01-28T03:07:45.922717+00:00 | 1,271 | false | # Intuition\nFor every element, which is less than or equal to its predecessor, we will perform the required number of operations to make it strictly increasing.\n\n# Approach\nFirst we will declare a variable m, and will assign the nums[0] value to it. Now, traverse the array from left to right, for every element, if ... | 2 | 0 | ['C++'] | 1 |
minimum-operations-to-make-the-array-increasing | Java&Javascript Solution (JW) | javajavascript-solution-jw-by-specter01w-7d07 | 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 | specter01wj | NORMAL | 2023-01-26T15:49:27.023462+00:00 | 2023-01-26T15:49:27.023510+00:00 | 193 | 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', 'JavaScript'] | 0 |
minimum-operations-to-make-the-array-increasing | C++ easy solution TC:O(n) SC:O(n) | c-easy-solution-tcon-scon-by-om_limbhare-ttpz | \nclass Solution {\npublic:\n int minOperations(vector<int>& nums) {\n if(nums.size()==1) return 0;\n int count=0,sum=0;\n for(int i=1;i | om_limbhare | NORMAL | 2022-11-24T22:25:48.756143+00:00 | 2022-11-24T22:25:48.756190+00:00 | 353 | false | ```\nclass Solution {\npublic:\n int minOperations(vector<int>& nums) {\n if(nums.size()==1) return 0;\n int count=0,sum=0;\n for(int i=1;i<nums.size();i++){\n if(nums[i]<=nums[i-1]){\n count+=nums[i-1]+1;\n sum+=count-nums[i];\n nums[i]=co... | 2 | 0 | ['C', 'C++'] | 0 |
minimum-operations-to-make-the-array-increasing | [ C++ ] - Runtime : 8 ms - 99.66 % faster || Easy Solution | c-runtime-8-ms-9966-faster-easy-solution-ahnw | Code\n\nclass Solution {\npublic:\n int minOperations(vector<int>& nums) {\n int sum=0;\n for(int i=0;i<nums.size()-1;i++){\n if(num | ideepakchauhan7 | NORMAL | 2022-11-13T14:30:24.482245+00:00 | 2022-11-13T17:21:03.608449+00:00 | 575 | false | # Code\n```\nclass Solution {\npublic:\n int minOperations(vector<int>& nums) {\n int sum=0;\n for(int i=0;i<nums.size()-1;i++){\n if(nums[i]>nums[i+1]||nums[i]==nums[i+1]){\n sum+=nums[i]-nums[i+1]+1;\n nums[i+1]+=nums[i]-nums[i+1]+1;\n }\n }\... | 2 | 0 | ['Array', 'Greedy', 'C++'] | 0 |
minimum-operations-to-make-the-array-increasing | BEGINNER FRIENDLY || EASY TO UNDERSTAND [JAVA] | beginner-friendly-easy-to-understand-jav-a7ko | \nclass Solution {\n public int minOperations(int[] nums) {\n int count=0;\n \n for(int i=1;i<nums.length;i++) {\n if(nums[i]<= | priyankan_23 | NORMAL | 2022-08-31T16:48:19.218986+00:00 | 2022-08-31T16:48:19.219025+00:00 | 271 | false | ```\nclass Solution {\n public int minOperations(int[] nums) {\n int count=0;\n \n for(int i=1;i<nums.length;i++) {\n if(nums[i]<=nums[i-1]){\n count+=nums[i-1]-nums[i]+1;\n nums[i]+=nums[i-1]-nums[i]+1;\n }\n \n }\n return co... | 2 | 0 | [] | 0 |
minimum-operations-to-make-the-array-increasing | Java || Easy Fastest Solution || 0ms | java-easy-fastest-solution-0ms-by-ranjit-ehlw | \n int res=0;\n if(nums.length==1){\n return res;\n }\n int val=0;\n for(int i=1;i<nums.length;i++){\n \n if | Ranjit-Kumar-Nayak | NORMAL | 2022-06-16T02:30:57.778511+00:00 | 2022-06-16T02:30:57.778554+00:00 | 216 | false | ```\n int res=0;\n if(nums.length==1){\n return res;\n }\n int val=0;\n for(int i=1;i<nums.length;i++){\n \n if(nums[i-1]>=nums[i]){\n val+=(nums[i-1]- nums[i])+1;\n nums[i]=1+nums[i-1];\n }\n }\n return val;\n``` | 2 | 0 | ['Greedy', 'Java'] | 1 |
minimum-operations-to-make-the-array-increasing | C++ Simple Greedy Solution | O(n) | One-Pass | c-simple-greedy-solution-on-one-pass-by-a1igm | Every element should be atmost 1 greater than the previous element.\nSo, iterate the array & make every element nums[i] = max(nums[i], nums[i-1] + 1) and count | Mythri_Kaulwar | NORMAL | 2022-02-13T15:51:59.122661+00:00 | 2022-02-13T15:51:59.122689+00:00 | 299 | false | Every element should be atmost 1 greater than the previous element.\nSo, iterate the array & make every element ```nums[i] = max(nums[i], nums[i-1] + 1)``` and count the no. of ```1```s to be added to make this change.\n**Code :**\n```\nclass Solution {\npublic:\n int minOperations(vector<int>& nums) {\n int ... | 2 | 0 | ['Greedy', 'C', 'C++'] | 0 |
minimum-operations-to-make-the-array-increasing | [ Java ] Beats 100% Very simple to understand solution with explanation | java-beats-100-very-simple-to-understand-g02f | Concept : Since we want a purely increasing array, an element at a[i] if smaller or equal to a[i-1] has to be made at least a[i-1]+1 in order to satisfy our req | KapProDes | NORMAL | 2021-12-09T02:25:29.256184+00:00 | 2021-12-09T02:25:29.256230+00:00 | 205 | false | __Concept__ : Since we want a purely increasing array, an element at __a[i]__ if smaller or equal to __a[i-1]__ has to be made at least __a[i-1]+1__ in order to satisfy our requirements.\n\n__Steps__ : \n1. check __if a[i] <= a[i-1]__. This is the only case where we need to make an operations.\n2. If we find a[i] <= a[... | 2 | 0 | ['Java'] | 0 |
minimum-operations-to-make-the-array-increasing | WEEB DOES PYTHON | weeb-does-python-by-skywalker5423-12e9 | \n\tclass Solution:\n\t\tdef minOperations(self, nums: List[int]) -> int:\n\t\t\tcount = 0\n\n\t\t\tfor i in range(1,len(nums)):\n\t\t\t\tif nums[i] <= nums[i-1 | Skywalker5423 | NORMAL | 2021-12-03T07:24:57.471241+00:00 | 2021-12-03T07:24:57.471285+00:00 | 164 | false | \n\tclass Solution:\n\t\tdef minOperations(self, nums: List[int]) -> int:\n\t\t\tcount = 0\n\n\t\t\tfor i in range(1,len(nums)):\n\t\t\t\tif nums[i] <= nums[i-1]:\n\t\t\t\t\tinitial = nums[i] \n\t\t\t\t\tnums[i] = nums[i-1] + 1\n\t\t\t\t\tcount += nums[i] - initial\n\n\t\t\treturn count\n\nAlright leetcoders, take a br... | 2 | 0 | ['Python', 'Python3'] | 1 |
minimum-operations-to-make-the-array-increasing | C++ solution without updating the vector nums | c-solution-without-updating-the-vector-n-6ac7 | \nclass Solution {\npublic:\n int minOperations(vector<int>& nums) {\n if(nums.size()==0)return 0;\n int op=0,count=0;\n for(int | codedguy | NORMAL | 2021-09-26T10:43:25.177949+00:00 | 2021-09-26T10:43:25.177980+00:00 | 37 | false | ```\nclass Solution {\npublic:\n int minOperations(vector<int>& nums) {\n if(nums.size()==0)return 0;\n int op=0,count=0;\n for(int i=1;i<nums.size();i++){\n if(nums[i]<=nums[i-1]+count){\n count+=nums[i-1]-nums[i]+1; \n op+=... | 2 | 0 | [] | 0 |
minimum-operations-to-make-the-array-increasing | easy python solution | easy-python-solution-by-bhupatjangid-vrae | \nclass Solution:\n def minOperations(self, nums: List[int]) -> int:\n ans=0\n for i in range(1,len(nums)):\n ans+=(max(0,nums[i-1]+ | bhupatjangid | NORMAL | 2021-08-16T15:16:31.222038+00:00 | 2021-08-16T15:16:31.222086+00:00 | 81 | false | ```\nclass Solution:\n def minOperations(self, nums: List[int]) -> int:\n ans=0\n for i in range(1,len(nums)):\n ans+=(max(0,nums[i-1]+1-nums[i]))\n nums[i]=max(nums[i],nums[i-1]+1)\n return ans\n``` | 2 | 0 | [] | 0 |
minimum-operations-to-make-the-array-increasing | [JAVA] s'mple short O(n) solution | java-smple-short-on-solution-by-zeldox-r1mh | if you like it pls upvote\n\nJAVA\n\nclass Solution {\n public int minOperations(int[] nums) {\n int res = 0;\n for(int i = 1;i<nums.length;i++ | zeldox | NORMAL | 2021-08-14T21:48:17.594328+00:00 | 2021-08-14T21:48:17.594356+00:00 | 110 | false | if you like it pls upvote\n\nJAVA\n```\nclass Solution {\n public int minOperations(int[] nums) {\n int res = 0;\n for(int i = 1;i<nums.length;i++){\n if(nums[i]<=nums[i-1]){\n res+= nums[i-1]-nums[i]+1;\n nums[i] = nums[i-1]+1;\n }\n }\n ... | 2 | 0 | ['Java'] | 0 |
minimum-operations-to-make-the-array-increasing | JavaScript reduce solution | javascript-reduce-solution-by-kkchengaf-bq56 | every time increase by (max - cur)\n\nvar minOperations = function(nums) {\n var max = 0;\n return nums.reduce((acc, cur) => { \n max = Math.max(cu | kkchengaf | NORMAL | 2021-06-09T03:52:51.979139+00:00 | 2021-06-09T03:52:51.979179+00:00 | 180 | false | every time increase by (max - cur)\n```\nvar minOperations = function(nums) {\n var max = 0;\n return nums.reduce((acc, cur) => { \n max = Math.max(cur, ++max);\n return acc + max - cur;\n }, 0)\n};\n``` | 2 | 0 | ['JavaScript'] | 2 |
minimum-operations-to-make-the-array-increasing | python | beats 99.8% | easy | python-beats-998-easy-by-abhishen99-c44v | \nclass Solution:\n def minOperations(self, nums: List[int]) -> int:\n \n \n pre,ans=0,0\n \n for i in nums:\n | Abhishen99 | NORMAL | 2021-05-21T18:12:15.271143+00:00 | 2021-05-21T18:12:46.564590+00:00 | 252 | false | ```\nclass Solution:\n def minOperations(self, nums: List[int]) -> int:\n \n \n pre,ans=0,0\n \n for i in nums:\n if pre < i:\n pre=i\n else:\n pre+=1\n ans+=pre-i\n return ans\n``` | 2 | 0 | ['Python', 'Python3'] | 2 |
minimum-operations-to-make-the-array-increasing | Python easy solution | python-easy-solution-by-iamkshitij77-u7dh | ```\nclass Solution:\n def minOperations(self, nums: List[int]) -> int:\n ans = 0\n for i in range(1,len(nums)):\n if nums[i-1]>=num | iamkshitij77 | NORMAL | 2021-05-06T20:57:10.484988+00:00 | 2021-05-06T20:57:10.485017+00:00 | 71 | false | ```\nclass Solution:\n def minOperations(self, nums: List[int]) -> int:\n ans = 0\n for i in range(1,len(nums)):\n if nums[i-1]>=nums[i]:\n ans+=nums[i-1]-nums[i]+1\n nums[i] = nums[i-1] + 1\n return ans\n | 2 | 0 | [] | 0 |
minimum-operations-to-make-the-array-increasing | Simple Javascript solution beats 95% | simple-javascript-solution-beats-95-by-s-c1it | \n/**\n * @param {number[]} nums\n * @return {number}\n */\nvar minOperations = function(nums) {\n let ops = 0, prevNum = nums[0]\n for(let i = 1; i < num | saynn | NORMAL | 2021-04-27T07:17:33.799652+00:00 | 2021-04-27T07:17:33.799702+00:00 | 318 | false | ```\n/**\n * @param {number[]} nums\n * @return {number}\n */\nvar minOperations = function(nums) {\n let ops = 0, prevNum = nums[0]\n for(let i = 1; i < nums.length; i++){\n if(nums[i] <= prevNum){\n ops += prevNum + 1 - nums[i]\n nums[i] = prevNum + 1\n }\n prevNum = n... | 2 | 0 | ['JavaScript'] | 0 |
minimum-operations-to-make-the-array-increasing | Go golang solution | go-golang-solution-by-leaf_peng-kpl6 | Runtime: 16 ms, faster than 47.56% of Go online submissions for Minimum Operations to Make the Array Increasing.\nMemory Usage: 6.2 MB, less than 6.10% of Go on | leaf_peng | NORMAL | 2021-04-25T03:19:47.199021+00:00 | 2021-04-25T03:19:47.199057+00:00 | 56 | false | >Runtime: 16 ms, faster than 47.56% of Go online submissions for Minimum Operations to Make the Array Increasing.\nMemory Usage: 6.2 MB, less than 6.10% of Go online submissions for Minimum Operations to Make the Array Increasing.\n\n```go\nfunc minOperations(nums []int) int {\n ans := 0\n for i := 1; i < len(num... | 2 | 0 | [] | 0 |
minimum-operations-to-make-the-array-increasing | A greedy solution for a greedy problem (C++) | a-greedy-solution-for-a-greedy-problem-c-l2lz | The resultant array after performing all of the increment operations has to be strictly increasing (e.g., [1,2,3,4,5]).For a strictly increasing array each elem | vatsss | NORMAL | 2021-04-20T17:38:35.869198+00:00 | 2021-04-20T17:38:35.869246+00:00 | 277 | false | The resultant array after performing all of the increment operations has to be strictly increasing (e.g., [1,2,3,4,5]).For a strictly increasing array each element has to be greater than its previous element by atleast 1.So, we can take a greedy approach such that whenever we find an element that is not greater than th... | 2 | 0 | ['Greedy', 'C'] | 0 |
minimum-operations-to-make-the-array-increasing | Simple C++ and Java Solution Time O(N) and Space O(1) | simple-c-and-java-solution-time-on-and-s-otuk | C++ solution: \n\n\n\nclass Solution {\npublic:\n int minOperations(vector<int>& nums) {\n if(nums.size()<1)return 0;\n int cost=0;\n fo | millenniumdart09 | NORMAL | 2021-04-18T10:39:22.676687+00:00 | 2021-05-25T19:13:30.571739+00:00 | 186 | false | **C++ solution:** \n```\n\n\nclass Solution {\npublic:\n int minOperations(vector<int>& nums) {\n if(nums.size()<1)return 0;\n int cost=0;\n for(int i=1;i<nums.size();i++)\n {\n if(nums[i]<=nums[i-1])\n {\n cost+=nums[i-1]-nums[i]+1;\n n... | 2 | 0 | ['C', 'Java'] | 0 |
minimum-operations-to-make-the-array-increasing | Rust Simple Solution | rust-simple-solution-by-kohbis-fm5x | rust\nimpl Solution {\n pub fn min_operations(nums: Vec<i32>) -> i32 {\n let mut count: i32 = 0;\n let mut prev: i32 = 0;\n for num in n | kohbis | NORMAL | 2021-04-18T01:47:40.611447+00:00 | 2021-04-18T01:47:40.611478+00:00 | 77 | false | ```rust\nimpl Solution {\n pub fn min_operations(nums: Vec<i32>) -> i32 {\n let mut count: i32 = 0;\n let mut prev: i32 = 0;\n for num in nums {\n if prev >= num {\n count += (prev + 1) - num;\n prev += 1\n } else {\n prev = num\... | 2 | 0 | ['Rust'] | 0 |
minimum-operations-to-make-the-array-increasing | python easy solution | python-easy-solution-by-akaghosting-ckdz | \tclass Solution:\n\t\tdef minOperations(self, nums: List[int]) -> int:\n\t\t\tres = 0\n\t\t\tfor i in range(1, len(nums)):\n\t\t\t\tif nums[i] <= nums[i - 1]:\ | akaghosting | NORMAL | 2021-04-17T16:16:29.266753+00:00 | 2021-04-17T16:16:29.266789+00:00 | 128 | false | \tclass Solution:\n\t\tdef minOperations(self, nums: List[int]) -> int:\n\t\t\tres = 0\n\t\t\tfor i in range(1, len(nums)):\n\t\t\t\tif nums[i] <= nums[i - 1]:\n\t\t\t\t\tres += nums[i - 1] - nums[i] + 1\n\t\t\t\t\tnums[i] = nums[i - 1] + 1\n\t\t\treturn res | 2 | 0 | [] | 0 |
minimum-operations-to-make-the-array-increasing | [Python3] sweep | python3-sweep-by-ye15-jtyk | \nclass Solution:\n def minOperations(self, nums: List[int]) -> int:\n ans = 0\n for i in range(1, len(nums)):\n if nums[i-1] >= num | ye15 | NORMAL | 2021-04-17T16:09:35.818346+00:00 | 2021-04-17T16:09:35.818385+00:00 | 136 | false | ```\nclass Solution:\n def minOperations(self, nums: List[int]) -> int:\n ans = 0\n for i in range(1, len(nums)):\n if nums[i-1] >= nums[i]: \n ans += 1 + nums[i-1] - nums[i]\n nums[i] = 1 + nums[i-1]\n return ans \n``` | 2 | 0 | ['Python3'] | 0 |
minimum-operations-to-make-the-array-increasing | [C++] One pass solution (100% time & 100% space) | c-one-pass-solution-100-time-100-space-b-vxfl | \nclass Solution {\npublic:\n int minOperations(vector<int>& nums) {\n int result = 0;\n \n for (int i = 1; i < nums.size(); ++i) {\n | bhaviksheth | NORMAL | 2021-04-17T16:03:49.791097+00:00 | 2021-04-17T16:03:49.791130+00:00 | 235 | false | ```\nclass Solution {\npublic:\n int minOperations(vector<int>& nums) {\n int result = 0;\n \n for (int i = 1; i < nums.size(); ++i) {\n if (nums[i] <= nums[i - 1]) {\n result += nums[i - 1] + 1 - nums[i];\n nums[i] = nums[i - 1] + 1;\n }\n ... | 2 | 0 | [] | 0 |
minimum-operations-to-make-the-array-increasing | Easy and just count it | easy-and-just-count-it-by-sairangineeni-a1ye | ApproachCreate a count variable to track the number of operations.Loop through the array from index 0 to n - 2 (we compare each element with the next).For each | Sairangineeni | NORMAL | 2025-03-31T06:05:56.507022+00:00 | 2025-03-31T06:05:56.507022+00:00 | 127 | false | # Approach
Create a count variable to track the number of operations.
Loop through the array from index 0 to n - 2 (we compare each element with the next).
For each pair nums[i] and nums[i+1]:
If nums[i] >= nums[i+1], it means we need to increase nums[i+1].
Calculate how much to increase it by: diff = nums[i] - num... | 1 | 0 | ['Java'] | 0 |
minimum-operations-to-make-the-array-increasing | Greedy Incremental Adjustment to Enforce Strict Monotonicity | greedy-incremental-adjustment-to-enforce-bezz | IntuitionThe goal is to transform the input array into a strictly increasing sequence using the minimum number of increment operations. To achieve this, we scan | expert07 | NORMAL | 2025-03-28T12:29:50.002714+00:00 | 2025-03-28T12:29:50.002714+00:00 | 127 | false | # Intuition
The goal is to transform the input array into a strictly increasing sequence using the minimum number of increment operations. To achieve this, we scan the array from left to right. If the current element is not strictly less than the next, we increment the next element until it becomes greater. This greedy... | 1 | 0 | ['Array', 'Greedy', 'Simulation', 'C++'] | 0 |
minimum-operations-to-make-the-array-increasing | T.C-> O(n) and S.C-> O(1) | cpp 🤩 | tc-on-and-sc-o1-cpp-by-varuntyagig-gjz1 | Complexity
Time complexity:
O(n)
Space complexity:
O(1)
Code | varuntyagig | NORMAL | 2025-03-27T21:33:40.093547+00:00 | 2025-03-27T21:33:40.093547+00:00 | 61 | false | # Complexity
- Time complexity:
$$O(n)$$
- Space complexity:
$$O(1)$$
# Code
```cpp []
class Solution {
public:
int minOperations(vector<int>& nums) {
int steps = 0;
for (int i = 0; i < nums.size() - 1; ++i) {
if (nums[i] > nums[i + 1]) {
steps += ((nums[i] - nums[i + ... | 1 | 0 | ['Array', 'C++'] | 0 |
minimum-operations-to-make-the-array-increasing | <<EASY THAN YOUR EXPECTATIONS>> | easy-than-your-expectations-by-dakshesh_-wob3 | PLEASE UPVOTE MECode | Dakshesh_vyas123 | NORMAL | 2025-03-20T10:41:09.631710+00:00 | 2025-03-20T10:41:09.631710+00:00 | 98 | false | # PLEASE UPVOTE ME
# Code
```cpp []
class Solution {
public:
int minOperations(vector<int>& nums) {
int a=0;
for(int i=0;i<nums.size()-1;i++){
if(nums[i+1]<=nums[i]) {
a+=(nums[i]-nums[i+1])+1;
nums[i+1]=nums[i]+1;}
}
return a;
}
};
``` | 1 | 0 | ['C++'] | 0 |
minimum-operations-to-make-the-array-increasing | C++ Solution ||n100% Beats | c-solution-n100-beats-by-jeetgajera-upuq | Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time | Jeetgajera | NORMAL | 2024-09-12T03:44:20.384949+00:00 | 2024-09-12T03:44:20.384981+00:00 | 18 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity : O(N)\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity : O(1)\n<!-- Add your space complexity here, e.g.... | 1 | 0 | ['Greedy', 'C++'] | 0 |
minimum-operations-to-make-the-array-increasing | Easy 100% | easy-100-by-oybek_0005-cvtg | Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time | oybek_0005 | NORMAL | 2024-08-12T14:43:20.171526+00:00 | 2024-08-12T14:43:20.171571+00:00 | 255 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ --... | 1 | 0 | ['Java'] | 1 |
minimum-operations-to-make-the-array-increasing | simple C++ sol | simple-c-sol-by-vivek_0104-4hf2 | Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time | Vivek_0104 | NORMAL | 2024-03-02T07:28:27.325401+00:00 | 2024-03-02T07:28:27.325423+00:00 | 6 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\nO(N)\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n... | 1 | 0 | ['C++'] | 0 |
minimum-operations-to-make-the-array-increasing | Simple java code 2 ms beats 99 % | simple-java-code-2-ms-beats-99-by-arobh-zxya | \n# Complexity\n- \n\n# Code\n\nclass Solution {\n public int minOperations(int[] nums) {\n int max = nums[0];\n int sum = 0;\n for(int | Arobh | NORMAL | 2024-01-14T03:41:49.603307+00:00 | 2024-01-14T03:41:49.603331+00:00 | 18 | false | \n# Complexity\n- \n\n# Code\n```\nclass Solution {\n public int minOperations(int[] nums) {\n int max = nums[0];\n int sum = 0;\n for(int i=1; i<nums.length; i++) {\n if(... | 1 | 0 | ['Java'] | 0 |
minimum-operations-to-make-the-array-increasing | Easy C++ solution || Greedy approach | easy-c-solution-greedy-approach-by-bhara-h30f | \n\n# Code\n\nclass Solution {\npublic:\n int minOperations(vector<int>& nums) {\n int n = nums.size(), count = 0;\n if(n == 1){\n r | bharathgowda29 | NORMAL | 2024-01-07T14:08:51.579170+00:00 | 2024-01-07T14:08:51.579195+00:00 | 450 | false | \n\n# Code\n```\nclass Solution {\npublic:\n int minOperations(vector<int>& nums) {\n int n = nums.size(), count = 0;\n if(n == 1){\n return 0;\n }\n for(int i=1; i<n; i++){\n if(nums[i] > nums[i-1]){\n continue;\n }\n else{\n ... | 1 | 0 | ['Array', 'Greedy', 'C++'] | 0 |
minimum-operations-to-make-the-array-increasing | simplest java sol | simplest-java-sol-by-anoopchaudhary1-ncpl | \n\n# Code\n\nclass Solution {\n public int minOperations(int[] nums) {\n int[] arr = new int[nums.length];\n int count =0;\n for(int i | Anoopchaudhary1 | NORMAL | 2023-12-14T08:31:52.227610+00:00 | 2023-12-14T08:31:52.227644+00:00 | 140 | false | \n\n# Code\n```\nclass Solution {\n public int minOperations(int[] nums) {\n int[] arr = new int[nums.length];\n int count =0;\n for(int i = 0 ; i<nums.length ; i++){\n arr[i] = nums[i];\n }\n for(int i =0 ; i<arr.length-1 ; i++){\n if(arr[i+1] <=arr[i]){\n ... | 1 | 0 | ['Java'] | 0 |
minimum-operations-to-make-the-array-increasing | using adjacent difference. | using-adjacent-difference-by-akhilaaa-mimj | 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 | Akhilaaa | NORMAL | 2023-11-17T11:44:18.420638+00:00 | 2023-11-17T11:44:18.420667+00:00 | 86 | 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 |
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