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count-prefixes-of-a-given-string | Count Prefixes of a Given String | count-prefixes-of-a-given-string-by-akas-owqm | \nint count=0;\n for(auto it:words){\n string str=s.substr(0,it.size());\n if(str==it)\n count++;\n }\n \n | Akash_Pandey | NORMAL | 2023-01-29T18:00:54.336854+00:00 | 2023-01-29T18:00:54.336890+00:00 | 707 | false | ```\nint count=0;\n for(auto it:words){\n string str=s.substr(0,it.size());\n if(str==it)\n count++;\n }\n \n return count;\n\t\t\n``` | 1 | 0 | ['C', 'Iterator'] | 0 |
count-prefixes-of-a-given-string | Simple || CPP || (Datta Bayo) | simple-cpp-datta-bayo-by-shreyas_6379-j30j | Nothing To explain Brut force approach.\n\n# Code\n\nclass Solution {\npublic:\n int countPrefixes(vector<string>& words, string s) {\n int count =0;\ | shreyas_6379 | NORMAL | 2023-01-29T12:01:33.552857+00:00 | 2023-01-29T12:01:33.552902+00:00 | 10 | false | Nothing To explain Brut force approach.\n\n# Code\n```\nclass Solution {\npublic:\n int countPrefixes(vector<string>& words, string s) {\n int count =0;\n for(int i=0;i<=s.length();i++)\n {\n string check = s.substr(0,i);\n for(int j=0;j<words.size();j++)\n {\n ... | 1 | 0 | ['C++'] | 1 |
count-prefixes-of-a-given-string | Efficient one-liner solution with startsWith and reduce | efficient-one-liner-solution-with-starts-5x53 | Code\n\n/**\n * @param {string[]} words\n * @param {string} s\n * @return {number}\n */\nvar countPrefixes = function(words, s) {\n return words.reduce((acc, | ods967 | NORMAL | 2023-01-27T23:30:40.870114+00:00 | 2023-01-27T23:30:55.480121+00:00 | 55 | false | # Code\n```\n/**\n * @param {string[]} words\n * @param {string} s\n * @return {number}\n */\nvar countPrefixes = function(words, s) {\n return words.reduce((acc, word) => s.startsWith(word) ? acc + 1 : acc, 0);\n};\n``` | 1 | 1 | ['JavaScript'] | 0 |
count-prefixes-of-a-given-string | C# Linq 1-liner | 87ms 95% | c-linq-1-liner-87ms-95-by-legon2k-tdg4 | Code\n\npublic class Solution {\n public int CountPrefixes(string[] words, string s) \n => words.Count(s.StartsWith);\n}\n | Legon2k | NORMAL | 2023-01-17T10:01:48.292996+00:00 | 2023-01-17T10:03:22.570950+00:00 | 50 | false | # Code\n```\npublic class Solution {\n public int CountPrefixes(string[] words, string s) \n => words.Count(s.StartsWith);\n}\n``` | 1 | 0 | ['C#'] | 0 |
count-prefixes-of-a-given-string | 2255. Count Prefixes of a Given String | 2255-count-prefixes-of-a-given-string-by-xj1v | \n# Code\n\n/**\n * @param {string[]} words\n * @param {string} s\n * @return {number}\n */\nvar countPrefixes = function(words, s) {\n return words.reduce(( | YusupovJaloliddin | NORMAL | 2022-12-20T18:25:52.148102+00:00 | 2022-12-20T18:25:52.148153+00:00 | 49 | false | \n# Code\n```\n/**\n * @param {string[]} words\n * @param {string} s\n * @return {number}\n */\nvar countPrefixes = function(words, s) {\n return words.reduce((acc, val) => acc + s.startsWith(val), 0);\n};\n``` | 1 | 0 | ['JavaScript'] | 0 |
count-prefixes-of-a-given-string | Count Prefixes of a Given String Solution Java | count-prefixes-of-a-given-string-solutio-btmi | class Solution {\n public int countPrefixes(String[] words, String s) {\n return (int) Arrays.stream(words).filter(word -> s.startsWith(word)).count();\n } | bhupendra786 | NORMAL | 2022-11-11T17:29:34.802072+00:00 | 2022-11-11T17:29:34.802112+00:00 | 60 | false | class Solution {\n public int countPrefixes(String[] words, String s) {\n return (int) Arrays.stream(words).filter(word -> s.startsWith(word)).count();\n }\n}\n | 1 | 0 | ['Array', 'String'] | 0 |
count-prefixes-of-a-given-string | JAVA solution | startsWith() or indexOf() ✅ | java-solution-startswith-or-indexof-by-s-ox2b | Please Upvote :D\n\nclass Solution {\n public int countPrefixes(String[] words, String s) {\n int count = 0;\n \n for (String str : word | sourin_bruh | NORMAL | 2022-10-07T12:42:10.630976+00:00 | 2022-10-07T12:42:10.631015+00:00 | 103 | false | ### Please Upvote :D\n```\nclass Solution {\n public int countPrefixes(String[] words, String s) {\n int count = 0;\n \n for (String str : words) {\n // if (s.indexOf(str) == 0) count++;\n if (s.startsWith(str)) count++;\n }\n \n return count;\n }\n}... | 1 | 0 | ['Java'] | 0 |
count-prefixes-of-a-given-string | c++ | short | 97% faster than all | c-short-97-faster-than-all-by-venomhighs-5y8r | \n# Code\n\nclass Solution {\npublic:\n int countPrefixes(vector<string>& words, string s) {\n int count = 0;\n for(auto word:words){\n | venomhighs7 | NORMAL | 2022-10-02T16:17:37.173033+00:00 | 2022-10-02T16:17:37.173073+00:00 | 867 | false | \n# Code\n```\nclass Solution {\npublic:\n int countPrefixes(vector<string>& words, string s) {\n int count = 0;\n for(auto word:words){\n bool isPrefix = true;\n for(int i=0; i<word.length(); i++){\n if(word[i] != s[i]){\n isPrefix = false;\n ... | 1 | 0 | ['C++'] | 1 |
count-prefixes-of-a-given-string | JAVA SOLN | java-soln-by-jatin0287-r4ga | class Solution {\n public int countPrefixes(String[] words, String s) {\n \n int count =0;\n for(int i=0; i<words.length; i++){\n | Jatin0287 | NORMAL | 2022-09-23T05:37:54.403387+00:00 | 2022-09-23T05:37:54.403430+00:00 | 95 | false | class Solution {\n public int countPrefixes(String[] words, String s) {\n \n int count =0;\n for(int i=0; i<words.length; i++){\n if(s.startsWith(words[i])){\n count++;\n }\n } \n return count;\n }\n} | 1 | 0 | [] | 0 |
count-prefixes-of-a-given-string | Javascript easy solution | javascript-easy-solution-by-yorkinjon-iz0f | \n/**\n * @param {string[]} words\n * @param {string} s\n * @return {number}\n */\nvar countPrefixes = function(words, s) {\n let count = 0;\n for(let i=0 | Yorkinjon | NORMAL | 2022-09-20T13:48:42.847921+00:00 | 2022-09-20T13:48:42.847959+00:00 | 24 | false | ```\n/**\n * @param {string[]} words\n * @param {string} s\n * @return {number}\n */\nvar countPrefixes = function(words, s) {\n let count = 0;\n for(let i=0; i<words.length; i++) {\n if(words[i] === s.slice(0, words[i].length)) {\n count++;\n }\n }\n return count;\n};\n``` | 1 | 0 | ['JavaScript'] | 0 |
count-prefixes-of-a-given-string | C++ solution | c-solution-by-lit2021036_iiitl-grnd | \nclass Solution {\npublic:\n int countPrefixes(vector<string>& words, string s) {\n int ct=0;\n for(string&word:words){\n bool flag=t | lit2021036_iiitl | NORMAL | 2022-09-11T08:09:21.934477+00:00 | 2022-09-11T08:09:21.934515+00:00 | 88 | false | ```\nclass Solution {\npublic:\n int countPrefixes(vector<string>& words, string s) {\n int ct=0;\n for(string&word:words){\n bool flag=true;\n if(word.length()<=s.length()){\n for(int i=0; i<min(word.length(),s.length()); ++i){\n if(word[i]!=s[i]){\n ... | 1 | 0 | [] | 0 |
count-prefixes-of-a-given-string | Count Prefixes of a Given String | count-prefixes-of-a-given-string-by-dhan-j0ms | python3 sol\n\nclass Solution:\n def countPrefixes(self, words: List[str], s: str) -> int:\n count= 0 \n for i in words:\n if i ==s: | dhananjayaduttmishra | NORMAL | 2022-09-07T16:43:30.887949+00:00 | 2022-09-07T16:43:30.887990+00:00 | 170 | false | python3 sol\n```\nclass Solution:\n def countPrefixes(self, words: List[str], s: str) -> int:\n count= 0 \n for i in words:\n if i ==s:\n count+=1\n else:\n if i ==s[:len(i)]:\n count+=1\n return count\n``` | 1 | 0 | ['Python'] | 0 |
count-prefixes-of-a-given-string | easy fast short solution | easy-fast-short-solution-by-hurshidbek-6rij | \nPLEASE UPVOTE IF YOU LIKE\n\n```\n int count = 0;\n for(String temp: words)\n if (s.indexOf(temp) == 0)\n count++;\n | Hurshidbek | NORMAL | 2022-08-30T03:27:18.354705+00:00 | 2022-08-30T03:27:18.354743+00:00 | 108 | false | ```\nPLEASE UPVOTE IF YOU LIKE\n```\n```\n int count = 0;\n for(String temp: words)\n if (s.indexOf(temp) == 0)\n count++;\n return count; | 1 | 0 | ['Java'] | 0 |
minimum-operations-to-make-the-array-k-increasing | [C++/Python] Longest Non-Decreasing Subsequence - Clean & Concise | cpython-longest-non-decreasing-subsequen-2pt3 | Idea\n- If k = 1, we need to find the minimum number of operations to make the whole array non-decreasing.\n- If k = 2, we need to make:\n\t- newArr1: Elements | hiepit | NORMAL | 2021-12-19T04:01:35.409106+00:00 | 2021-12-20T20:30:22.045560+00:00 | 13,371 | false | **Idea**\n- If `k = 1`, we need to find the minimum number of operations to make the whole array non-decreasing.\n- If `k = 2`, we need to make:\n\t- newArr1: Elements in index [0, 2, 4, 6...] are non-decreasing.\n\t- newArr2: Elements in index [1, 3, 5, 7...] are non-decreasing.\n- If `k = 3`, we need to make:\n - ... | 216 | 8 | [] | 22 |
minimum-operations-to-make-the-array-k-increasing | LIS in Disguise | lis-in-disguise-by-votrubac-a6xs | We have k independent subarrays in our array, and we need to make those subarrays non-decreasing.\n\nFor each subarray, we need to find the longest non-decreasi | votrubac | NORMAL | 2021-12-19T04:00:39.889036+00:00 | 2021-12-20T00:56:37.716095+00:00 | 5,667 | false | We have `k` independent subarrays in our array, and we need to make those subarrays non-decreasing.\n\nFor each subarray, we need to find the longest non-decreasing subsequence. We keep numbers in that subsequence as-is, and change the rest. We can use the approach from [300. Longest Increasing Subsequence](https://lee... | 99 | 3 | ['C', 'Java'] | 6 |
minimum-operations-to-make-the-array-k-increasing | [Python] Short LIS, explained | python-short-lis-explained-by-dbabichev-pbsk | There are two observations we need to make.\n1. It is independent to consider subproblems on 0, k, 2k, ..., 1, k+1, 2k+1, ... and so on.\n2. For each subproblem | dbabichev | NORMAL | 2021-12-19T04:00:49.129799+00:00 | 2021-12-19T04:00:49.129829+00:00 | 1,258 | false | There are two observations we need to make.\n1. It is independent to consider subproblems on `0, k, 2k, ...`, `1, k+1, 2k+1, ...` and so on.\n2. For each subproblem it is enough to solve LIS (longest increasing(not strictly) subsequence) problem. Indeed, if we have LIS of length `t`, than we need to change not more tha... | 35 | 8 | [] | 5 |
minimum-operations-to-make-the-array-k-increasing | [Python] Explanation with pictures, LIS | python-explanation-with-pictures-lis-by-njmrf | Given k = 3 for example. Let\'s stay that we start with index = 0, the first condition is A[0] <= A[3], \n\n\nWe also need to guarantee A[3] <= A[6], A[6] <= | Bakerston | NORMAL | 2021-12-19T04:02:00.303575+00:00 | 2021-12-19T04:02:48.704231+00:00 | 1,396 | false | Given k = 3 for example. Let\'s stay that we start with index = 0, the first condition is **A[0] <= A[3]**, \n\n\nWe also need to guarantee **A[3] <= A[6]**, **A[6] <= A[9]**, etc, if such larger indexs exist... | 34 | 2 | [] | 2 |
minimum-operations-to-make-the-array-k-increasing | Longest increasing subsequence | Explanation + Code | longest-increasing-subsequence-explanati-aj34 | There will be K independent arrays each having max length n/k\n Independently find lis of all K independent arrays (using binary search/BIT)\n Minimum change = | Priyansh_34 | NORMAL | 2021-12-19T04:01:04.989380+00:00 | 2021-12-19T05:11:34.766392+00:00 | 1,787 | false | * There will be K **independent** arrays each having max length n/k\n* Independently find **lis** of all K independent arrays (using binary search/BIT)\n* **Minimum change = length of the array - lis of the array**\n* Sum all the minimum change needed for all K arrays\n\n**CODE** (CPP)\n```\nclass Solution {\npubli... | 29 | 2 | ['C'] | 5 |
minimum-operations-to-make-the-array-k-increasing | Java + Intuition + LIS + Binary Search | java-intuition-lis-binary-search-by-lear-w1sf | Intuition:\nThe intuition behind this solution is that we extract all k-separated subsequences and find out the longest increasing subsequence (LIS) for each k- | learn4Fun | NORMAL | 2021-12-19T05:57:19.468309+00:00 | 2021-12-22T22:16:45.979534+00:00 | 1,770 | false | **Intuition:**\nThe intuition behind this solution is that we extract all k-separated subsequences and find out the longest increasing subsequence (LIS) for each k-separated subsequences and then for each of these extracted subsequences we calculate the min operations (delete/update) to make it sorted/increasing.\n\n``... | 28 | 0 | ['Binary Tree', 'Java'] | 7 |
minimum-operations-to-make-the-array-k-increasing | Java | Longest non-decreasing subsequence k times | java-longest-non-decreasing-subsequence-bgx23 | Based on 300. Longest Increasing Subsequence, but allowing duplicates in sequence (hence non-decreasing).\nTo solve the problem:\n- iterate k subsequences of th | prezes | NORMAL | 2021-12-19T04:16:41.033833+00:00 | 2021-12-19T17:23:20.697316+00:00 | 1,018 | false | Based on [300. Longest Increasing Subsequence](https://leetcode.com/problems/longest-increasing-subsequence/), but allowing duplicates in sequence (hence non-decreasing).\nTo solve the problem:\n- iterate k subsequences of the original sequence (elements separated by k steps belong to the same subsequence),\n- calculat... | 11 | 0 | [] | 2 |
minimum-operations-to-make-the-array-k-increasing | Beats 100% on runtime [EXPLAINED] | beats-100-on-runtime-explained-by-r9n-cqcx | Intuition\nModify an array so that certain subsequences are non-decreasing. By minimizing the number of changes needed, we can achieve a K-increasing array wher | r9n | NORMAL | 2024-11-04T05:29:10.232001+00:00 | 2024-11-04T05:29:10.232024+00:00 | 43 | false | # Intuition\nModify an array so that certain subsequences are non-decreasing. By minimizing the number of changes needed, we can achieve a K-increasing array where elements spaced k apart are in the correct order.\n\n# Approach\nDivide the array into k subsequences based on their indices, find the longest non-decreasin... | 7 | 0 | ['Array', 'Binary Search', 'TypeScript'] | 0 |
minimum-operations-to-make-the-array-k-increasing | C++ Solution, group and LIS. | c-solution-group-and-lis-by-11235813-zjg8 | \n\n1. group arr into k groups.\n2. get LIS for each group, accumulate group.size - LIS.\n\nclass Solution {\npublic:\n int LIS(vector<int> &arr) {\n | 11235813 | NORMAL | 2021-12-19T04:00:31.558819+00:00 | 2021-12-19T04:00:31.558863+00:00 | 902 | false | \n\n1. group arr into k groups.\n2. get LIS for each group, accumulate `group.size - LIS`.\n```\nclass Solution {\npublic:\n int LIS(vector<int> &arr) {\n vector<int> lis(arr.size(), INT_MAX);\n int ans = 0;\n for(int i = 0; i < arr.size(); i++) {\n int idx = upper_bound(lis.begin(), ... | 6 | 0 | [] | 1 |
minimum-operations-to-make-the-array-k-increasing | ✅ [Python/C++/Java] LIS || 100% Faster || Detailed Explanation || Clean and Easy to Understand | pythoncjava-lis-100-faster-detailed-expl-v18g | First divide arr into k groups and consider each group individually.\n In each group, we need to find out the length of the longest increasing sequence, because | linfq | NORMAL | 2021-12-19T13:57:37.806914+00:00 | 2021-12-19T15:31:00.363974+00:00 | 860 | false | * First divide arr into k groups and consider each group individually.\n* In each group, we need to find out the length of the longest increasing sequence, because the longest increasing sequence does not need to be operated, and the remaining positions can be set to the value of the previous or next in the longest inc... | 5 | 0 | ['C', 'Python', 'C++', 'Java'] | 0 |
minimum-operations-to-make-the-array-k-increasing | [Python3] almost LIS | python3-almost-lis-by-ye15-d0pi | Please check out this commit for solutions of weekly 272. \n\n\nclass Solution:\n def kIncreasing(self, arr: List[int], k: int) -> int:\n \n de | ye15 | NORMAL | 2021-12-19T04:06:32.314805+00:00 | 2021-12-19T18:08:29.123111+00:00 | 338 | false | Please check out this [commit](https://github.com/gaosanyong/leetcode/commit/55c6a88797eef9ac745a3dbbff821a2aac735a70) for solutions of weekly 272. \n\n```\nclass Solution:\n def kIncreasing(self, arr: List[int], k: int) -> int:\n \n def fn(sub): \n """Return ops to make sub non-decreasing."... | 4 | 1 | ['Python3'] | 0 |
minimum-operations-to-make-the-array-k-increasing | C++ | With explanation | longest non-decreasing subsequence | c-with-explanation-longest-non-decreasin-4bra | Explanation:-\n1. if we see carefully then we will find that we have to make these subsequences non-decreasing:-\n arr[i]<=arr[i+k]<=arr[i+2k]... so on\n he | aman282571 | NORMAL | 2021-12-19T04:01:04.986872+00:00 | 2021-12-19T04:01:04.986901+00:00 | 1,649 | false | **Explanation:-**\n1. if we see carefully then we will find that we have to make these subsequences non-decreasing:-\n ```arr[i]<=arr[i+k]<=arr[i+2k]... so on```\n here ```i``` wil start from ```0``` and go upto ```k-1```\n2. For making these non-decreasing we have to find ```longest non-decreasing subsequence``` i... | 4 | 1 | ['C'] | 1 |
minimum-operations-to-make-the-array-k-increasing | C++ Longest Increasing Subsequence | c-longest-increasing-subsequence-by-lzl1-ee3e | \nSee my latest update in repo LeetCode\n\n## Solution 1. Longest Increasing Subsequence (LIS)\n\nSplit the numbers in A into k groups: [0, k, 2k, 3k, ...], [1, | lzl124631x | NORMAL | 2021-12-19T04:00:55.075327+00:00 | 2021-12-19T04:00:55.075371+00:00 | 971 | false | \nSee my latest update in repo [LeetCode](https://github.com/lzl124631x/LeetCode)\n\n## Solution 1. Longest Increasing Subsequence (LIS)\n\nSplit the numbers in `A` into `k` groups: `[0, k, 2k, 3k, ...]`, `[1, 1+k, 1+2k, 1+3k, ...]`, ...\n\nCompute the minimum operations needed to make a group non-decreasing. Assume th... | 4 | 0 | [] | 1 |
minimum-operations-to-make-the-array-k-increasing | C++ || Longest increasing subsequence variation | c-longest-increasing-subsequence-variati-7u7r | Goal : to find the minimum numbers of operations to make this array k-increasing.\nApproach: find the minimum numbers of operation of each increasing subsequenc | VineetKumar2023 | NORMAL | 2021-12-19T05:45:23.901809+00:00 | 2021-12-19T07:35:07.073030+00:00 | 409 | false | Goal : to find the minimum numbers of operations to make this array k-increasing.\nApproach: find the minimum numbers of operation of each increasing subsequence( formed by arr[i] -> arr[i+k] -> arr[i+2k] -> ... where 0<=i<k ) and add them up.\n\nFinding the minimum operations: if we find the longest non-decreasing su... | 3 | 0 | ['C'] | 0 |
minimum-operations-to-make-the-array-k-increasing | C++ Simple LIS Simple and Short Solution Explained | c-simple-lis-simple-and-short-solution-e-t2wx | \n//function to calculate LIS in O(n*logn) using binary search and DP\n//NOTE : O(n*n) method of finding LIS will give TLE\nint helper(vector<int>&nums){\n | rishabh_devbanshi | NORMAL | 2021-12-19T04:12:30.656493+00:00 | 2021-12-19T04:12:30.656521+00:00 | 652 | false | ```\n//function to calculate LIS in O(n*logn) using binary search and DP\n//NOTE : O(n*n) method of finding LIS will give TLE\nint helper(vector<int>&nums){\n vector<int> lis;\n for (int i = 0; i < nums.size(); ++i) {\n int x = nums[i];\n if (lis.empty() || lis[lis.size() - 1] <= x... | 3 | 0 | ['Dynamic Programming', 'Binary Tree'] | 1 |
minimum-operations-to-make-the-array-k-increasing | 🔥🔥🔥C++ | super easy | clean code | LIS | easy to grasp🔥🔥🔥 | c-super-easy-clean-code-lis-easy-to-gras-jt0t | 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 | Algo-Messihas | NORMAL | 2023-08-16T16:03:23.368265+00:00 | 2023-08-16T16:03:23.368294+00:00 | 175 | 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 | ['Array', 'Binary Search', 'C++'] | 0 |
minimum-operations-to-make-the-array-k-increasing | C++| O(Nlog(N)) | c-onlogn-by-blue_tiger-nyxr | // If u observe carefully you will get to know that the problem is similar to LIS(Longest Increasing Subsequence) problem. In LIS we have only one set of data t | Blue_tiger | NORMAL | 2022-05-30T08:51:03.392198+00:00 | 2022-05-30T08:51:03.392228+00:00 | 287 | false | // If u observe carefully you will get to know that the problem is similar to LIS(Longest Increasing Subsequence) problem. In LIS we have only one set of data to find, but in this problem we have to find in k different set. If u dry run the test case u will find that all the set are independent to each other so we can ... | 2 | 0 | [] | 0 |
minimum-operations-to-make-the-array-k-increasing | LIS || CPP || Easy to understand | lis-cpp-easy-to-understand-by-riki05-8vrd | \nclass Solution {\npublic:\nint lengthOfLIS(vector<int>& nums) {\n \n //LIS dp approach will give TLE\n int n=nums.size();\n vector<int> v; | riki05 | NORMAL | 2022-02-03T09:04:43.251918+00:00 | 2022-02-03T09:04:43.251950+00:00 | 384 | false | ```\nclass Solution {\npublic:\nint lengthOfLIS(vector<int>& nums) {\n \n //LIS dp approach will give TLE\n int n=nums.size();\n vector<int> v;v.push_back(nums[0]);\n for(int i=1;i<n;i++)\n {\n if(nums[i]>=v.back()) v.push_back(nums[i]); \n else{\n ... | 2 | 0 | ['C', 'C++'] | 1 |
minimum-operations-to-make-the-array-k-increasing | C++ | LIS(MODIFIED) | IMPLEMENTATION | c-lismodified-implementation-by-chikzz-0tty | PLEASE UPVOTE IF U LIKE MY SOLUTION AND EXPLANATION\n\nclass Solution {\npublic:\n int kIncreasing(vector<int>& arr, int k) {\n \n int n=arr.si | chikzz | NORMAL | 2021-12-21T05:24:35.747840+00:00 | 2021-12-21T05:24:35.747875+00:00 | 158 | false | **PLEASE UPVOTE IF U LIKE MY SOLUTION AND EXPLANATION**\n```\nclass Solution {\npublic:\n int kIncreasing(vector<int>& arr, int k) {\n \n int n=arr.size();\n \n //we consider a visited array so let say for i we traverse i+k,i+2*k,i+3*k,...\n //we do not need to consider doing opera... | 2 | 0 | [] | 0 |
minimum-operations-to-make-the-array-k-increasing | O(nlogn/k) | Detailed Explanation | Longest Non Decreasing Subsequence | onlognk-detailed-explanation-longest-non-iu2s | We need to find minumum number of operations to make following arrays non-decreasing. \n[arr[0], arr[k], arr[2k], ....]\n[arr[1], arr[k + 1], arr[2k + 1], ...]\ | ratio123 | NORMAL | 2021-12-19T06:54:12.783125+00:00 | 2021-12-19T07:01:38.247809+00:00 | 881 | false | We need to find minumum number of operations to make following arrays non-decreasing. \n[arr[0], arr[k], arr[2k], ....]\n[arr[1], arr[k + 1], arr[2k + 1], ...]\n....\n[arr[k - 1], arr[2k - 1], ...]\n\nClearly, each arrary listed above is independent of others, and we just need to find the minumum operation number for e... | 2 | 0 | ['C', 'Binary Tree'] | 0 |
minimum-operations-to-make-the-array-k-increasing | Java | O(NlogN) | Binaray-Search for LIS| Detailed Explanation | java-onlogn-binaray-search-for-lis-detai-91oj | The problem is asking us to find a longest non-decreasing subsequence\n1. case when k == 1:\nwe are trying to minimize the change operation -> if we find a long | zunyiliu | NORMAL | 2021-12-19T04:24:36.237110+00:00 | 2021-12-19T04:24:36.237136+00:00 | 173 | false | The problem is asking us to find a longest non-decreasing subsequence\n**1. case when k == 1:**\nwe are trying to minimize the change operation -> if we find a longest non-decreasing subsequence for arr, then \nthe result would be **arr.length - length of LIS**\n\n**2. how to find a LIS ?**\nwe use an ArrayList to reco... | 2 | 0 | ['Binary Tree'] | 1 |
minimum-operations-to-make-the-array-k-increasing | C# - O(N Log N) Based on Longest Increasing Subsequence | c-on-log-n-based-on-longest-increasing-s-fyrv | csharp\npublic int KIncreasing(int[] arr, int k)\n{\n\tint count = 0;\n\tList<int>[] current = new List<int>[k];\n\tfor (int i = 0; i < k; i++)\n\t{\n\t\tcurren | christris | NORMAL | 2021-12-19T04:01:39.822269+00:00 | 2021-12-19T04:01:39.822307+00:00 | 193 | false | ```csharp\npublic int KIncreasing(int[] arr, int k)\n{\n\tint count = 0;\n\tList<int>[] current = new List<int>[k];\n\tfor (int i = 0; i < k; i++)\n\t{\n\t\tcurrent[i] = new List<int>();\n\t}\n\n\tfor (int i = 0; i < arr.Length; i++)\n\t{\n\t\tint index = i % k;\n\t\tcurrent[index].Add(arr[i]);\n\t}\n\n\tforeach (var l... | 2 | 0 | [] | 0 |
minimum-operations-to-make-the-array-k-increasing | C++ || LIS || easy to understand. | c-lis-easy-to-understand-by-abortive02-fhrw | Intuition\n Describe your first thoughts on how to solve this problem. \nLIS DP\n\n# Approach\n Describe your approach to solving the problem. \nu all know abou | abortive02 | NORMAL | 2023-06-29T13:16:59.672818+00:00 | 2023-06-29T13:16:59.672837+00:00 | 70 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nLIS DP\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\nu all know about how to solve LIS using binary search then it will be simple for u to understand.\n\njust forget about k : problem is broken down in to samller... | 1 | 0 | ['C++'] | 0 |
minimum-operations-to-make-the-array-k-increasing | Java Simple Solution || Binary Search || O(Nlogn) | java-simple-solution-binary-search-onlog-rxvn | \nclass Solution {\n public int kIncreasing(int[] arr, int k) {\n int n=arr.length;\n int ans=0;\n for(int i=0; i<k; i++)\n {\n | saurabhtrigunayat | NORMAL | 2022-10-01T12:09:25.341698+00:00 | 2022-10-01T12:09:25.341732+00:00 | 48 | false | ```\nclass Solution {\n public int kIncreasing(int[] arr, int k) {\n int n=arr.length;\n int ans=0;\n for(int i=0; i<k; i++)\n {\n ArrayList<Integer> list=new ArrayList();\n for(int j=i; j<n; j=j+k)\n list.add(arr[j]);\n ans+=list.size()-LIS... | 1 | 0 | [] | 0 |
minimum-operations-to-make-the-array-k-increasing | Simple Solution in Java || Easy to Understand | simple-solution-in-java-easy-to-understa-3rev | \nclass Solution {\n public int kIncreasing(int[] arr, int k) {\n \n int ans = 0;\n for(int i=0; i<k; i++){\n List<Integer> l | PRANAV_KUMAR99 | NORMAL | 2022-04-12T04:57:32.932914+00:00 | 2022-04-12T04:57:32.932948+00:00 | 134 | false | ```\nclass Solution {\n public int kIncreasing(int[] arr, int k) {\n \n int ans = 0;\n for(int i=0; i<k; i++){\n List<Integer> li = new ArrayList<>();\n \n for(int j=i; j<arr.length; j += k){\n li.add(arr[j]);\n }\n \n ... | 1 | 0 | [] | 0 |
minimum-operations-to-make-the-array-k-increasing | C++ | c-by-jwang777-f5ei | \nclass Solution {\npublic:\n int kIncreasing(vector<int>& arr, int k) {\n int n = arr.size();\n \n int result = arr.size();\n fo | jwang777 | NORMAL | 2022-01-16T08:35:21.618689+00:00 | 2022-01-16T08:35:21.618725+00:00 | 115 | false | ```\nclass Solution {\npublic:\n int kIncreasing(vector<int>& arr, int k) {\n int n = arr.size();\n \n int result = arr.size();\n for (int i=0; i<k; ++i) {\n vector<int> seq = {arr[i]};\n for (int j=i+k; j<n; j+=k) {\n if (arr[j] >= seq.back()) {\n ... | 1 | 0 | [] | 0 |
minimum-operations-to-make-the-array-k-increasing | LIS | Multiset | lis-multiset-by-ghost_tsushima-whln | \nclass Solution {\npublic:\n \n int findAns(vector<int> &v) {\n multiset<int>s;\n \n for(int i=0; i<v.size(); i++) {\n if | ghost_tsushima | NORMAL | 2021-12-25T16:59:38.493827+00:00 | 2021-12-25T16:59:38.493862+00:00 | 123 | false | ```\nclass Solution {\npublic:\n \n int findAns(vector<int> &v) {\n multiset<int>s;\n \n for(int i=0; i<v.size(); i++) {\n if (s.empty()) {\n s.insert(v[i]);\n continue;\n }\n s.insert(v[i]);\n multiset<int>::iterator i... | 1 | 0 | ['C'] | 0 |
minimum-operations-to-make-the-array-k-increasing | LIS | lis-by-abhi_nav2011-koim | \nclass Solution {\npublic:\n int helper(vector<int> v){\n //find lis\n vector<int> dp;\n dp.push_back(v[0]);\n \n for(int | abhi_nav2011 | NORMAL | 2021-12-25T10:42:44.218883+00:00 | 2021-12-25T10:42:44.218911+00:00 | 124 | false | ```\nclass Solution {\npublic:\n int helper(vector<int> v){\n //find lis\n vector<int> dp;\n dp.push_back(v[0]);\n \n for(int i=1;i<v.size();++i){\n if(v[i] < dp.back()){\n int index = upper_bound(dp.begin(),dp.end(),v[i]) - dp.begin();\n dp... | 1 | 0 | ['C'] | 0 |
minimum-operations-to-make-the-array-k-increasing | a few solutions | a-few-solutions-by-claytonjwong-ajd4 | This problem is 300. Longest Increasing Subsequence in disguise, ie. for each kth "bucket" from 0..K-1 inclusive, we accumulate the LIS, then subtract the accum | claytonjwong | NORMAL | 2021-12-23T20:05:47.622561+00:00 | 2021-12-24T14:36:47.904712+00:00 | 92 | false | This problem is [300. Longest Increasing Subsequence](https://leetcode.com/problems/longest-increasing-subsequence/discuss/385203/The-ART-of-Dynamic-Programming) in disguise, ie. for each `k`<sup>th</sup> "bucket" from `0..K-1` inclusive, we accumulate the [LIS](https://leetcode.com/problems/longest-increasing-subseque... | 1 | 0 | [] | 0 |
minimum-operations-to-make-the-array-k-increasing | 2111. Minimum Operations to Make the Array K-Increasing | 2111-minimum-operations-to-make-the-arra-ld0s | LNDS (Longest Non-Decreasing Subarray) Approach\n\nclass Solution {\n public int upperBound(ArrayList<Integer> a, int val){\n int l=0;int r=a.size()-1 | kalinga | NORMAL | 2021-12-22T08:38:46.293614+00:00 | 2021-12-22T08:38:46.293643+00:00 | 808 | false | LNDS (Longest Non-Decreasing Subarray) Approach\n```\nclass Solution {\n public int upperBound(ArrayList<Integer> a, int val){\n int l=0;int r=a.size()-1;\n while(l<r){\n int m=(l+r)/2;\n if(a.get(m)<=val){\n l=m+1;\n }\n else{\n ... | 1 | 0 | ['Array', 'Binary Tree', 'Java'] | 1 |
minimum-operations-to-make-the-array-k-increasing | ✅ [c++] || Longest Increasing Subsequence | c-longest-increasing-subsequence-by-xor0-l9u3 | \nclass Solution {\npublic:\n int LIS(vector<int>& vec){\n int n=vec.size();\n \n vector<int> lis; //keep track of all the element which | xor09 | NORMAL | 2021-12-20T04:20:04.346926+00:00 | 2021-12-20T04:20:04.346970+00:00 | 155 | false | ```\nclass Solution {\npublic:\n int LIS(vector<int>& vec){\n int n=vec.size();\n \n vector<int> lis; //keep track of all the element which takes part in LIS\n lis.push_back(vec[0]);\n \n for(int i=1; i<n; ++i){\n if(vec[i] >= lis.back()) lis.push_back(vec[i]);\n ... | 1 | 0 | ['C', 'C++'] | 0 |
minimum-operations-to-make-the-array-k-increasing | Using Longest Increasing Subsequence || C++ Code || Upper_bound | using-longest-increasing-subsequence-c-c-ars9 | \nclass Solution {\n int lengthOfLIS(vector<int>& nums) {\n \n int n = nums.size();\n vector<int>seq;\n seq.push_back(nums[0]);\n | kirtanprajapati | NORMAL | 2021-12-19T17:44:00.368739+00:00 | 2021-12-19T17:44:00.368772+00:00 | 122 | false | ```\nclass Solution {\n int lengthOfLIS(vector<int>& nums) {\n \n int n = nums.size();\n vector<int>seq;\n seq.push_back(nums[0]);\n \n for(int i=1;i<n;i++){\n if(seq.back() <= nums[i]){\n seq.push_back(nums[i]);\n }\n else{\n ... | 1 | 0 | ['C', 'Binary Tree'] | 1 |
minimum-operations-to-make-the-array-k-increasing | [Rust] LIS | rust-lis-by-bovinovain-r0tw | Partition_pointis supported since 1.52.0 in Rust. But it is not yet supported on leetcode. As a work around, we can manually implement the binary search partiti | bovinovain | NORMAL | 2021-12-19T16:32:36.870824+00:00 | 2021-12-19T16:32:36.870856+00:00 | 48 | false | `Partition_point`is supported since 1.52.0 in Rust. But it is not yet supported on leetcode. As a work around, we can manually implement the binary search partition.\n``` rust\nimpl Solution {\n pub fn k_increasing(arr: Vec<i32>, k: i32) -> i32 {\n let mut s = 0;\n for i in 0..k{\n let now:V... | 1 | 0 | [] | 0 |
minimum-operations-to-make-the-array-k-increasing | Find by longest Non Decreasing Sub sequence solution clean | find-by-longest-non-decreasing-sub-seque-3gkt | \njavascript\nvar kIncreasing = function (arr, k) {\n let res = 0,\n n = arr.length;\n for (let i = 0; i < k; i++) {\n const newArr = [];\n for (let | articFz | NORMAL | 2021-12-19T09:25:30.261535+00:00 | 2021-12-19T09:25:30.261565+00:00 | 143 | false | \n```javascript\nvar kIncreasing = function (arr, k) {\n let res = 0,\n n = arr.length;\n for (let i = 0; i < k; i++) {\n const newArr = [];\n for (let j = i; j < n; j += k) {\n newArr.push(arr[j]);\n }\n res += newArr.length - longestNonDecreasingSub(newArr);\n }\n return res;\n};\n\nfunction l... | 1 | 0 | ['JavaScript'] | 0 |
minimum-operations-to-make-the-array-k-increasing | Python [Longest non-decreasing subsequence] | python-longest-non-decreasing-subsequenc-g5bg | Suppose that you have a function lnds(sub) that gives you the longest non-decreasing subsequence of a subarray. For example, sub = [1, 2, 3, 3, 2, 2, 2] would r | gsan | NORMAL | 2021-12-19T06:33:15.742717+00:00 | 2021-12-19T06:33:15.742747+00:00 | 174 | false | Suppose that you have a function `lnds(sub)` that gives you the longest non-decreasing subsequence of a subarray. For example, `sub = [1, 2, 3, 3, 2, 2, 2]` would return `5` which corresponds to the subsequence `[1, 2, 2, 2, 2]`.\n\nThe longest strictly increasing subsequence is actually Problem 300. Compared to that a... | 1 | 0 | [] | 0 |
minimum-operations-to-make-the-array-k-increasing | Swift LIS (but actually Longest Non-Decreasing Sequence) | swift-lis-but-actually-longest-non-decre-fk5y | Inspired from @hiepit\nhttps://leetcode.com/problems/minimum-operations-to-make-the-array-k-increasing/discuss/1635013/C%2B%2BPython-Longest-Non-Decreasing-Subs | chung1991 | NORMAL | 2021-12-19T04:39:06.419166+00:00 | 2021-12-19T05:19:12.374602+00:00 | 103 | false | Inspired from @hiepit\nhttps://leetcode.com/problems/minimum-operations-to-make-the-array-k-increasing/discuss/1635013/C%2B%2BPython-Longest-Non-Decreasing-Subsequence-Clean-and-Concise\nThe description from above article is very nice details.\n\n```\nclass Solution {\n // Time: O(nlogn)\n // Space: O(1)\n fun... | 1 | 0 | [] | 0 |
minimum-operations-to-make-the-array-k-increasing | (C++) 2111. Minimum Operations to Make the Array K-Increasing | c-2111-minimum-operations-to-make-the-ar-sm48 | \n\nclass Solution {\npublic:\n int kIncreasing(vector<int>& arr, int k) {\n \n auto fn = [](vector<int>& sub) {\n vector<int> vals; | qeetcode | NORMAL | 2021-12-19T04:11:16.837692+00:00 | 2021-12-19T17:59:37.095373+00:00 | 92 | false | \n```\nclass Solution {\npublic:\n int kIncreasing(vector<int>& arr, int k) {\n \n auto fn = [](vector<int>& sub) {\n vector<int> vals; \n for (auto& x : sub) {\n auto it = upper_bound(vals.begin(), vals.end(), x); \n if (it == vals.end()) vals.push_b... | 1 | 0 | ['C'] | 0 |
minimum-operations-to-make-the-array-k-increasing | Java Longest Increasing Subsequence Solution | java-longest-increasing-subsequence-solu-amk1 | This question is quite similar to 300. longest increasing subsequence, the only difference is that we are looping through the array by k interval, and that we a | kevinyu0506 | NORMAL | 2021-12-19T04:09:18.497327+00:00 | 2021-12-19T04:09:18.497367+00:00 | 180 | false | This question is quite similar to [300. longest increasing subsequence](https://leetcode.com/problems/longest-increasing-subsequence/), the only difference is that we are looping through the array by `k` interval, and that we are not restricting the condition to `strict increase` . To give a more relax definition of `i... | 1 | 0 | [] | 1 |
minimum-operations-to-make-the-array-k-increasing | Java | Longest non-decreasing subsequence | java-longest-non-decreasing-subsequence-ncc8n | Run longest non-decreasing subsequence each chain (i, i+k, i+2k ... )\nAnswer will be total length - sum (subsequence length)\n\n\n\nclass Solution {\n publi | myih | NORMAL | 2021-12-19T04:01:19.924766+00:00 | 2021-12-19T05:15:24.147158+00:00 | 286 | false | Run longest non-decreasing subsequence each chain (i, i+k, i+2k ... )\nAnswer will be total length - sum (subsequence length)\n\n\n```\nclass Solution {\n public int kIncreasing(int[] arr, int k) {\n int count = 0;\n for(int i=0; i<k; i++) {\n count += longestSub(arr, k, i);\n }\n ... | 1 | 0 | [] | 0 |
minimum-operations-to-make-the-array-k-increasing | Longest Non-Decreasing Subsequence | longest-non-decreasing-subsequence-by-da-o2m5 | \nIf you understand the idea of 300. Longest Increasing Subsequence problem, this solution is quite straitforward\n\n Split array to k sub_arrays \n Find minimu | danghuybk | NORMAL | 2021-12-19T04:00:44.938573+00:00 | 2021-12-19T04:05:42.076352+00:00 | 1,006 | false | \nIf you understand the idea of [300. Longest Increasing Subsequence](https://leetcode.com/problems/longest-increasing-subsequence/) problem, this solution is quite straitforward\n\n* Split array to k sub_arrays \n* Find minimum operation to make each sub_array non-decreasing, and it equal length subarray - longest non... | 1 | 0 | ['Binary Tree'] | 0 |
minimum-operations-to-make-the-array-k-increasing | Python, LIS solution with explanation | python-lis-solution-with-explanation-by-99cgd | IntuitionIn arr, we have k groups.
In a group, its element's index % k is the same.Approachwe just find each group's length of LIS,
and len(arr) - sum of length | shun6096tw | NORMAL | 2025-02-08T02:40:34.619036+00:00 | 2025-02-08T02:40:34.619036+00:00 | 11 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
In ```arr```, we have ```k``` groups.
In a group, its element's ```index % k``` is the same.
# Approach
<!-- Describe your approach to solving the problem. -->
we just find each group's length of LIS,
and len(arr) - sum of length of each ... | 0 | 0 | ['Binary Search', 'Python3'] | 0 |
minimum-operations-to-make-the-array-k-increasing | 2111. Minimum Operations to Make the Array K-Increasing | 2111-minimum-operations-to-make-the-arra-06qa | IntuitionApproachComplexity
Time complexity:
Space complexity:
Code | G8xd0QPqTy | NORMAL | 2025-01-18T02:41:59.766832+00:00 | 2025-01-18T02:41:59.766832+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 | ['Python3'] | 0 |
minimum-operations-to-make-the-array-k-increasing | 🐲 Another One Liner 🐉 | another-one-liner-by-jhzxgqngzm-kpew | Complexity
Time complexity:
O(nlog(n))Codeor over multibe lines | jHZXGQNgZM | NORMAL | 2025-01-09T10:33:32.620711+00:00 | 2025-01-09T10:33:32.620711+00:00 | 8 | false | # Complexity
- Time complexity:
<!-- Add your time complexity here, e.g. $$O(n)$$ -->
$$O(nlog(n))$$
# Code
```python3
class Solution:
def kIncreasing(self, arr: List[int], k: int) -> int:
return len(arr)-sum([(d:=[1e10]),any(d.__setitem__(bisect_right(d,i),i)for i in arr[m::k] if i<d[-1] or d.append(i) )... | 0 | 0 | ['Python3'] | 0 |
minimum-operations-to-make-the-array-k-increasing | Short and Fast T̶h̶a̶t̶’s̶ w̶h̶a̶t̶ s̶h̶e̶ s̶a̶i̶d̶ | short-and-fast-thats-what-she-said-by-jh-eh3m | Approachfind the longest non monoton increasing subsequence for each sublist, and subtract the length of the subsequences from the total length.Complexity
Time | jHZXGQNgZM | NORMAL | 2025-01-09T10:03:05.345083+00:00 | 2025-01-09T10:03:05.345083+00:00 | 7 | false | # Approach
<!-- Describe your approach to solving the problem. -->
find the longest non monoton increasing subsequence for each sublist, and subtract the length of the subsequences from the total length.
# Complexity
- Time complexity:
<!-- Add your time complexity here, e.g. $$O(n)$$ -->
$$O(n log(n))$$
- Space comple... | 0 | 0 | ['Python3'] | 0 |
minimum-operations-to-make-the-array-k-increasing | LIS concept | lis-concept-by-vats_lc-y9jl | Code | vats_lc | NORMAL | 2025-01-09T05:28:30.570357+00:00 | 2025-01-09T05:28:30.570357+00:00 | 4 | false | # Code
```cpp []
class Solution {
public:
int getAns(vector<int>& a) {
vector<int> temp;
int n = a.size();
for (int i = 0; i < n; i++) {
if (temp.size() == 0)
temp.push_back(a[i]);
else {
int lb =
upper_bound(temp.be... | 0 | 0 | ['C++'] | 0 |
minimum-operations-to-make-the-array-k-increasing | Minimum Operations to Make the Array K-Increassing | minimum-operations-to-make-the-array-k-i-ykz6 | Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \nif we see carefully the | Naeem_ABD | NORMAL | 2024-11-30T16:04:53.328923+00:00 | 2024-11-30T16:04:53.328967+00:00 | 3 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\nif we see carefully then we will find that we have to make these subsequences non-decreasing:-\narr[i]<=arr[i+k]<=arr[i+2k]... so on\nhere i wil start from 0 and go up... | 0 | 0 | ['C++'] | 0 |
minimum-operations-to-make-the-array-k-increasing | 2111. Minimum Operations to Make the Array K-Increasing.cpp | 2111-minimum-operations-to-make-the-arra-ft2v | Code\n\nclass Solution {\npublic:\n int kIncreasing(vector<int>& arr, int k) {\n \n auto fn = [](vector<int>& sub) {\n vector<int> v | 202021ganesh | NORMAL | 2024-11-01T10:48:32.167774+00:00 | 2024-11-01T10:48:32.167800+00:00 | 1 | false | **Code**\n```\nclass Solution {\npublic:\n int kIncreasing(vector<int>& arr, int k) {\n \n auto fn = [](vector<int>& sub) {\n vector<int> vals; \n for (auto& x : sub) {\n auto it = upper_bound(vals.begin(), vals.end(), x); \n if (it == vals.end()) val... | 0 | 0 | ['C'] | 0 |
minimum-operations-to-make-the-array-k-increasing | LIS | lis-by-maxorgus-uf89 | Code\npython3 []\nclass Solution:\n def lengthOfLIS(self, nums):\n a = []\n for n in nums:\n if len(a) == 0 or a[-1] <= n:\n | MaxOrgus | NORMAL | 2024-09-04T02:34:17.172018+00:00 | 2024-09-04T02:34:31.278726+00:00 | 7 | false | # Code\n```python3 []\nclass Solution:\n def lengthOfLIS(self, nums):\n a = []\n for n in nums:\n if len(a) == 0 or a[-1] <= n:\n a.append(n)\n else:\n i = bisect.bisect_right(a, n)\n a[i] = n\n \n return len(a)\n def k... | 0 | 0 | ['Python3'] | 0 |
minimum-operations-to-make-the-array-k-increasing | [Rust / Elixir] patience sort | rust-elixir-patience-sort-by-minamikaze3-z4cy | Approach\nUse patience sort to find the longest non-decreasing subsequence for each:\narr.iter().skip(i).step_by(k) or\nEnum.drop(i) |> Enum.take_every(k),\nwhe | Minamikaze392 | NORMAL | 2024-07-12T07:37:37.537982+00:00 | 2024-07-12T10:01:20.916183+00:00 | 2 | false | # Approach\nUse patience sort to find the longest non-decreasing subsequence for each:\n`arr.iter().skip(i).step_by(k)` or\n`Enum.drop(i) |> Enum.take_every(k)`,\nwhere `0 <= i < k`.\n\n# Complexity\n- Time complexity: $$O(n*log(n / k))$$\n\n- Space complexity: $$O(n)$$\n\n# Code\n```Rust []\nimpl Solution {\n pub f... | 0 | 0 | ['Binary Search', 'Rust', 'Elixir'] | 0 |
minimum-operations-to-make-the-array-k-increasing | Python Hard | python-hard-by-lucasschnee-ura2 | \nclass Solution:\n def kIncreasing(self, arr: List[int], k: int) -> int:\n N = len(arr)\n lookup = defaultdict(list)\n for i in range(k | lucasschnee | NORMAL | 2024-07-02T14:54:41.636915+00:00 | 2024-07-02T14:54:41.636934+00:00 | 3 | false | ```\nclass Solution:\n def kIncreasing(self, arr: List[int], k: int) -> int:\n N = len(arr)\n lookup = defaultdict(list)\n for i in range(k):\n j = i\n while j < N:\n lookup[j % k].append(arr[j])\n j += k\n\n total = 0\n for key i... | 0 | 0 | ['Python3'] | 0 |
minimum-operations-to-make-the-array-k-increasing | EASY Longest Increasing Subsequence Solution | easy-longest-increasing-subsequence-solu-gs01 | Intuition\nobviously ,we can divide the array in almost k independent parts\nnow final answer is just sum of mininum operation in all these parts\n# Approach\n | debargha_nath | NORMAL | 2024-07-01T12:02:18.253795+00:00 | 2024-07-01T12:02:18.253831+00:00 | 8 | false | # Intuition\nobviously ,we can divide the array in almost k independent parts\nnow final answer is just sum of mininum operation in all these parts\n# Approach\n<!-- Describe your approach to solving the problem. -->\nminimum operation is just the LIS in that part\nand we are done....\n# Complexity\n- Time complexity:\... | 0 | 0 | ['Binary Search', 'Dynamic Programming', 'C++'] | 0 |
minimum-operations-to-make-the-array-k-increasing | C++ | LIS | Beats 70.80% | c-lis-beats-7080-by-cosmoctopus-9xgh | \n# Code\n\nclass Solution {\npublic:\n /* longest increasing subsequence */\n int LIS(vector<int> &arr)\n {\n vector<int> lis(arr.size(), INT_M | cosmoctopus | NORMAL | 2024-06-30T14:49:55.886642+00:00 | 2024-06-30T14:49:55.886682+00:00 | 1 | false | \n# Code\n```\nclass Solution {\npublic:\n /* longest increasing subsequence */\n int LIS(vector<int> &arr)\n {\n vector<int> lis(arr.size(), INT_MAX);\n int ans = 0;\n for(int i = 0; i < arr.size(); i++)\n {\n int idx = upper_bound(lis.begin(), lis.end(), arr[i]) - lis.b... | 0 | 0 | ['C++'] | 0 |
minimum-operations-to-make-the-array-k-increasing | O(nlogn), max size subsequence | onlogn-max-size-subsequence-by-szoro-8nky | \n\n# Code\n\nclass Solution {\npublic:\n int kIncreasing(vector<int>& a, int k) {\n int n = a.size(),ans=0;\n if(k==n)return 0;\n for(i | szoro | NORMAL | 2024-06-29T11:42:41.493646+00:00 | 2024-06-29T11:42:41.493684+00:00 | 2 | false | \n\n# Code\n```\nclass Solution {\npublic:\n int kIncreasing(vector<int>& a, int k) {\n int n = a.size(),ans=0;\n if(k==n)return 0;\n for(int i=0;i<k;i++){\n vector<int> v;\n int j=i;\n for(j=i;j<n;j+=k){\n if(v.size()==0||v.back()<=a[j])v.push_bac... | 0 | 0 | ['C++'] | 0 |
minimum-operations-to-make-the-array-k-increasing | Easy solution using longest non-decreasing subsequence | easy-solution-using-longest-non-decreasi-84y7 | Intuition\nThe intutition is simple to take all subarrays of the defined types. ie. subarrays (0,k,2k ...), (1,k+1,2k+1...) and so on.\nFor all this subarrays w | kumarajesh | NORMAL | 2024-05-20T17:55:21.468215+00:00 | 2024-05-20T17:55:21.468277+00:00 | 0 | false | # Intuition\nThe intutition is simple to take all subarrays of the defined types. ie. subarrays (0,k,2*k ...), (1,k+1,2*k+1...) and so on.\nFor all this subarrays we will find how many elements we need to update so that the subarrays become non-decreasing.\n\n# Approach\nAll the subarrays that are formed are independan... | 0 | 0 | ['C++'] | 0 |
minimum-operations-to-make-the-array-k-increasing | [C++] Easy LIS based solution | c-easy-lis-based-solution-by-ashigup-t19x | Code\n\nclass Solution {\npublic:\n int kIncreasing(vector<int>& arr, int k) {\n int ans = 0;\n for(int i=0;i<k;i++){\n vector<int> | princegup678 | NORMAL | 2024-03-21T02:51:14.843760+00:00 | 2024-03-21T02:51:14.843794+00:00 | 1 | false | # Code\n```\nclass Solution {\npublic:\n int kIncreasing(vector<int>& arr, int k) {\n int ans = 0;\n for(int i=0;i<k;i++){\n vector<int> sorted;\n sorted.push_back(arr[i]);\n int size = 1;\n for(int j=i+k;j<arr.size();j+=k){\n size++;\n ... | 0 | 0 | ['C++'] | 0 |
minimum-operations-to-make-the-array-k-increasing | [C++] LIS in nlogn solution | c-lis-in-nlogn-solution-by-ashigup-g0ym | Code\n\nclass Solution {\npublic:\n int LIS(vector<int> &arr){\n vector<int> sorted;\n\n sorted.push_back(arr[0]);\n\n for(int i=1;i<arr | princegup678 | NORMAL | 2024-03-21T02:47:38.732181+00:00 | 2024-03-21T02:47:38.732217+00:00 | 3 | false | # Code\n```\nclass Solution {\npublic:\n int LIS(vector<int> &arr){\n vector<int> sorted;\n\n sorted.push_back(arr[0]);\n\n for(int i=1;i<arr.size();i++){\n if(arr[i] >= sorted.back()){\n sorted.push_back(arr[i]);\n }\n else{\n int i... | 0 | 0 | ['C++'] | 0 |
minimum-operations-to-make-the-array-k-increasing | LIS | Binary-Search | TC - O(nlogn) | lis-binary-search-tc-onlogn-by-aavvikk-k3y3 | IDEA - For each group find the LIS. By finding LIS we will get the max len which has non-decreasing element, we can replace the remaing elements with elements o | aavvikk__ | NORMAL | 2024-03-07T14:10:11.366288+00:00 | 2024-03-07T14:10:11.366310+00:00 | 1 | false | **IDEA -** For each group find the LIS. By finding LIS we will get the max len which has non-decreasing element, we can replace the remaing elements with elements of LIS. we only need the count, so we subtract LIS length from the len of each group.\n\n```\nclass Solution {\npublic:\n int kIncreasing(vector<int>& arr... | 0 | 0 | ['C', 'Binary Tree'] | 0 |
minimum-operations-to-make-the-array-k-increasing | Binary Search, C++ | binary-search-c-by-goku_2022-snbq | 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 | goku_2022 | NORMAL | 2024-02-21T09:10:40.350887+00:00 | 2024-02-21T09:10:40.350905+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-the-array-k-increasing | using the concept of longest increasing subsequence | using-the-concept-of-longest-increasing-4pcdb | 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 | satya9546 | NORMAL | 2024-02-03T20:32:12.894788+00:00 | 2024-02-03T20:32:12.894815+00:00 | 3 | 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-the-array-k-increasing | Javascript dp | javascript-dp-by-igorjin-3u0g | 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 | igorjin | NORMAL | 2024-01-16T20:40:48.600546+00:00 | 2024-01-16T20:40:48.600576+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\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ --... | 0 | 0 | ['JavaScript'] | 0 |
minimum-operations-to-make-the-array-k-increasing | CPP solution!! | cpp-solution-by-grindingninja-cqx2 | \n\n# Complexity\n- Time complexity:\nO(nlogn)\n\n- Space complexity:\n Add your space complexity here, e.g. O(n) \n\n# Code\n\nclass Solution {\nprivate:\n | grindingninja | NORMAL | 2024-01-12T00:19:59.174350+00:00 | 2024-01-12T00:19:59.174371+00:00 | 3 | false | \n\n# Complexity\n- Time complexity:\nO(nlogn)\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\nclass Solution {\nprivate:\n int findIndex(vector<int>& arr, int val) {\n int l = 0;\n int r = arr.size();\n\n while(l < r) {\n int m = l + (r-... | 0 | 0 | ['C++'] | 0 |
minimum-operations-to-make-the-array-k-increasing | LIS of K arrays | lis-of-k-arrays-by-theabbie-pacc | \nimport bisect\n\nclass Solution:\n def kIncreasing(self, arr: List[int], k: int) -> int:\n n = len(arr)\n lis = [[] for _ in range(k)]\n | theabbie | NORMAL | 2024-01-05T05:04:42.967975+00:00 | 2024-01-05T05:06:04.207455+00:00 | 2 | false | ```\nimport bisect\n\nclass Solution:\n def kIncreasing(self, arr: List[int], k: int) -> int:\n n = len(arr)\n lis = [[] for _ in range(k)]\n res = n\n for i in range(n):\n x = bisect.bisect_left(lis[i % k], (arr[i], i))\n if x < len(lis[i % k]):\n lis... | 0 | 0 | [] | 0 |
minimum-operations-to-make-the-array-k-increasing | C++ | c-by-tinachien-n4yz | \nclass Solution {\npublic:\n int kIncreasing(vector<int>& arr, int k) {\n int ret = 0;\n for(int i = 0; i < k; i++){\n vector<int>g | TinaChien | NORMAL | 2023-12-19T23:59:30.320024+00:00 | 2023-12-19T23:59:30.320047+00:00 | 0 | false | ```\nclass Solution {\npublic:\n int kIncreasing(vector<int>& arr, int k) {\n int ret = 0;\n for(int i = 0; i < k; i++){\n vector<int>group;\n for(int j = i; j < arr.size(); j+=k){\n group.push_back(arr[j]);\n }\n ret += LIS(group);\n }\... | 0 | 0 | [] | 0 |
minimum-operations-to-make-the-array-k-increasing | Longest non-decreasing subsequence - Binary search | longest-non-decreasing-subsequence-binar-8rmz | Intuition\nThe question can be simplified by just solving for "minimum number of elements to change to make the array non decreasing". The reason that is so is | sisoj | NORMAL | 2023-10-27T17:26:59.175136+00:00 | 2023-10-27T17:27:21.496420+00:00 | 1 | false | # Intuition\nThe question can be simplified by just solving for "minimum number of elements to change to make the array non decreasing". The reason that is so is because the question is composed of multiple arrays (k arrays). So if we solve the question for any given array, we can solve it for k arrays.\nLets say K=3 a... | 0 | 0 | ['C++'] | 0 |
minimum-operations-to-make-the-array-k-increasing | Rust | LIS | rust-lis-by-soyflourbread-myrv | Intuition\n\nSeparate the vector into k subvectors that have no businees with each other.\n\n# Approach\n\nSee Longest Increasing Subsequence. Tho this question | soyflourbread | NORMAL | 2023-08-09T02:38:59.088334+00:00 | 2023-08-09T02:38:59.088371+00:00 | 5 | false | # Intuition\n\nSeparate the vector into `k` subvectors that have no businees with each other.\n\n# Approach\n\nSee [Longest Increasing Subsequence](https://leetcode.com/problems/longest-increasing-subsequence/). Tho this question requires longest nondecreasing subsequence, so a bit of modification is needed.\n\n# Compl... | 0 | 0 | ['Binary Search', 'Monotonic Stack', 'Rust'] | 0 |
minimum-operations-to-make-the-array-k-increasing | Python (Simple LIS) | python-simple-lis-by-rnotappl-08m6 | 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 | 2023-04-19T18:48:03.004981+00:00 | 2023-04-19T18:48:03.005012+00:00 | 84 | 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-the-array-k-increasing | Minimum Operations To Make The Array K - Increasing, C++ Explained Solution | minimum-operations-to-make-the-array-k-i-9p84 | Do Upvote If Found Helpful !!!\n\n# Approach\n Describe your approach to solving the problem. \nThe problem here is quite interesting and requires a bit of an o | ShuklaAmit1311 | NORMAL | 2023-03-21T19:27:14.262132+00:00 | 2023-03-21T19:27:14.262156+00:00 | 72 | false | **Do Upvote If Found Helpful !!!**\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\nThe problem here is quite interesting and requires a bit of an observation. Basically for each set of numbers present at a distance of **K**, we need to find minimum operations to make that set of numbers a non-de... | 0 | 0 | ['Math', 'Dynamic Programming', 'C++'] | 0 |
minimum-operations-to-make-the-array-k-increasing | Short Easy C++ | short-easy-c-by-devilabhipro-2kio | \n\nclass Solution {\npublic:\n int kIncreasing(vector<int>& arr, int k) {\n int n = arr.size();\n int ans = 0;\n for(int i=0;i<k;i++){\ | devilabhipro | NORMAL | 2023-03-06T09:50:42.853178+00:00 | 2023-03-06T09:50:42.853219+00:00 | 13 | false | ```\n\nclass Solution {\npublic:\n int kIncreasing(vector<int>& arr, int k) {\n int n = arr.size();\n int ans = 0;\n for(int i=0;i<k;i++){\n vector<int> lis;\n int len = 0;\n for(int j=i;j<n;j+=k){\n if(lis.empty() || lis.back() <= arr[j]){\n ... | 0 | 0 | [] | 0 |
minimum-operations-to-make-the-array-k-increasing | Just a runnable solution | just-a-runnable-solution-by-ssrlive-q8g3 | Code\n\nimpl Solution {\n pub fn k_increasing(arr: Vec<i32>, k: i32) -> i32 {\n pub fn upper_bound<T: Ord>(vec: &Vec<T>, x: &T) -> Result<usize, usize | ssrlive | NORMAL | 2023-03-03T16:20:21.744211+00:00 | 2023-03-03T16:20:21.744247+00:00 | 15 | false | # Code\n```\nimpl Solution {\n pub fn k_increasing(arr: Vec<i32>, k: i32) -> i32 {\n pub fn upper_bound<T: Ord>(vec: &Vec<T>, x: &T) -> Result<usize, usize> {\n vec.iter().position(|y| y > x).ok_or(vec.len())\n }\n\n let mut longest = 0;\n for i in 0..k {\n let mut m... | 0 | 0 | ['Rust'] | 0 |
minimum-operations-to-make-the-array-k-increasing | short and concise solution (c++),NlogN/k Intutive | short-and-concise-solution-cnlognk-intut-pjdn | Intuition\n Describe your first thoughts on how to solve this problem. \nWe need to keep in mind the = sign mentioned in the question .\nThat is the biggest hin | rokon123 | NORMAL | 2023-02-19T08:22:48.022603+00:00 | 2023-02-19T08:23:15.326572+00:00 | 20 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nWe need to keep in mind the = sign mentioned in the question .\nThat is the biggest hint.\n\nThe K arrays which are formed starting from index 0...K and every next element at a k distance will be independent .\n\nThose arrays should be so... | 0 | 0 | ['C++'] | 0 |
minimum-operations-to-make-the-array-k-increasing | C++ || EXPLAINED || CLEAN CODE || | c-explained-clean-code-by-karma_kar-ohet | 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 | d4Nf3Bf4 | NORMAL | 2023-02-04T23:33:40.793353+00:00 | 2023-02-04T23:34:15.068021+00:00 | 45 | 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-the-array-k-increasing | C++ || LIS Pattern || Easy Implementation || Binary Search | c-lis-pattern-easy-implementation-binary-gqcm | \nclass Solution {\npublic:\n int LIS(vector<int>&nums){\n if(nums.size() == 0) return 0;\n vector<int> ans;\n ans.push_back(nums[0]);\n | aman_2_0_2_3 | NORMAL | 2023-01-21T07:34:09.866274+00:00 | 2023-01-21T07:34:09.866316+00:00 | 42 | false | ```\nclass Solution {\npublic:\n int LIS(vector<int>&nums){\n if(nums.size() == 0) return 0;\n vector<int> ans;\n ans.push_back(nums[0]);\n for(int i=1;i<nums.size();i++){\n if(nums[i] >= ans.back()){\n ans.push_back(nums[i]);\n }\n else{\n ... | 0 | 0 | ['Binary Search', 'C'] | 0 |
minimum-operations-to-make-the-array-k-increasing | Fastest code | fastest-code-by-ajayluhach7-6xbz | Intuition\n Describe your first thoughts on how to solve this problem. \nI dont know more than basic java so don\'t blame for cheating in last 20 cases\n# Appro | ajayluhach7 | NORMAL | 2023-01-19T21:35:50.266282+00:00 | 2023-01-19T21:35:50.266318+00:00 | 70 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nI dont know more than basic java so don\'t blame for cheating in last 20 cases\n# Approach\n<!-- Describe your approach to solving the problem. -->\nI saw they wrote related topic as binary search and array so people using array list are ... | 0 | 0 | ['Array', 'Binary Search', 'Java'] | 0 |
minimum-operations-to-make-the-array-k-increasing | Beats 100% LIS Solution Explained Python | beats-100-lis-solution-explained-python-7lkze | \n\nHow to find the min no. of operations needed to make a sequence not decreasing?\n\nYou find the longest increasing subsequence (LIS) of the sequence.\nAll t | sarthakBhandari | NORMAL | 2023-01-11T18:09:12.786110+00:00 | 2023-01-11T18:10:22.084751+00:00 | 82 | false | \n\n**How to find the min no. of operations needed to make a sequence not decreasing?**\n\nYou find the longest increasing subsequence (LIS) of the sequence.\nAll the elements not in LIS needed to modified.... | 0 | 0 | ['Dynamic Programming', 'Python'] | 0 |
minimum-operations-to-make-the-array-k-increasing | C++ || LIS | c-lis-by-adityasingh011011110-xv5s | \nclass Solution {\npublic:\n int kIncreasing(vector<int>& arr, int k) {\n int ans = 0,prev,s,n = arr.size();\n for(int i=0;i<k;i++){\n | adityasingh011011110 | NORMAL | 2022-10-07T06:24:10.838113+00:00 | 2022-10-07T06:24:10.838151+00:00 | 38 | false | ```\nclass Solution {\npublic:\n int kIncreasing(vector<int>& arr, int k) {\n int ans = 0,prev,s,n = arr.size();\n for(int i=0;i<k;i++){\n prev=arr[i];\n s = 0;\n vector<int> lis;\n vector<int>:: iterator it;\n for(int j=i;j<n;j=j+k){\n ... | 0 | 0 | ['C'] | 0 |
minimum-operations-to-make-the-array-k-increasing | python solution (faster 92%) | python-solution-faster-92-by-dugu0607-s3no | \tclass Solution:\n\t\tdef longestNonDecreasingSubsequence(self, arr):\n\t\t\tsub = []\n\t\t\tfor i, x in enumerate(arr):\n\t\t\t\tif len(sub) == 0 or sub[-1] < | Dugu0607 | NORMAL | 2022-10-05T07:13:46.892031+00:00 | 2022-10-05T07:13:46.892072+00:00 | 27 | false | \tclass Solution:\n\t\tdef longestNonDecreasingSubsequence(self, arr):\n\t\t\tsub = []\n\t\t\tfor i, x in enumerate(arr):\n\t\t\t\tif len(sub) == 0 or sub[-1] <= x: # Append to LIS if new element is >= last element in LIS\n\t\t\t\t\tsub.append(x)\n\t\t\t\telse:\n\t\t\t\t\tidx = bisect_right(sub, x) # Find the index o... | 0 | 0 | [] | 0 |
minimum-operations-to-make-the-array-k-increasing | C++ || Simple LIS solution | c-simple-lis-solution-by-omik_07-diem | \nint kIncreasing(vector<int>& arr, int k) {\n \n int mx=0; \n for(int i=0;i<k;i++)\n {\n vector<int>v;\n fo | Omik_07 | NORMAL | 2022-09-27T13:42:42.778883+00:00 | 2022-09-27T13:42:42.778922+00:00 | 18 | false | ```\nint kIncreasing(vector<int>& arr, int k) {\n \n int mx=0; \n for(int i=0;i<k;i++)\n {\n vector<int>v;\n for(int j=i;j<arr.size();j+=k)\n {\n if(v.empty() || v.back()<=arr[j])\n {\n v.push_back(arr[j]);\n... | 0 | 0 | ['C'] | 0 |
minimum-operations-to-make-the-array-k-increasing | Short and easy LIS solution in Python with explanation | short-and-easy-lis-solution-in-python-wi-10dl | The solution is to compute the longest non-decreasing subsequence for the k sequences. \nThe sequence can be very long, thus the effient O(n log n) solution to | metaphysicalist | NORMAL | 2022-09-25T07:20:08.017761+00:00 | 2022-09-25T07:20:08.017804+00:00 | 83 | false | The solution is to compute the longest non-decreasing subsequence for the k sequences. \nThe sequence can be very long, thus the effient O(n log n) solution to longest non-decreasing subsequence is required. \n\n```\nclass Solution:\n def lis(self, start):\n results = []\n total = 0\n for i in r... | 0 | 0 | ['Binary Search', 'Python3'] | 0 |
minimum-operations-to-make-the-array-k-increasing | Python3: Almost correct; But, how do I "debug"? | python3-almost-correct-but-how-do-i-debu-8tfb | \nclass Solution:\n def kIncreasing(self, arr: List[int], k: int) -> int:\n count = 0\n for i in range(k, len(arr)):\n if arr[i - k] | prasadnr606 | NORMAL | 2022-09-04T06:24:12.159655+00:00 | 2022-09-04T06:24:12.159694+00:00 | 30 | false | ```\nclass Solution:\n def kIncreasing(self, arr: List[int], k: int) -> int:\n count = 0\n for i in range(k, len(arr)):\n if arr[i - k] > arr[i]:\n arr[i] = arr[i - k] + 1\n count += 1\n \n return count\n``` | 0 | 0 | [] | 0 |
minimum-operations-to-make-the-array-k-increasing | LIS Python | lis-python-by-2019csb1077-3w12 | \nfrom bisect import bisect_right as bl\nclass Solution:\n def lengthOfLIS(self, nums: List[int]) -> int:\n l = []\n for i in nums:\n | 2019csb1077 | NORMAL | 2022-09-01T05:13:38.668892+00:00 | 2022-09-01T05:13:38.668931+00:00 | 38 | false | ```\nfrom bisect import bisect_right as bl\nclass Solution:\n def lengthOfLIS(self, nums: List[int]) -> int:\n l = []\n for i in nums:\n if not l:\n l.append(i)\n continue\n idx = bl(l,i)\n if idx==len(l):\n l.append(i)\n ... | 0 | 0 | [] | 0 |
minimum-operations-to-make-the-array-k-increasing | [C++] Longest Non-Decreasing Subsequence technique. O(k*n*log(n)) | c-longest-non-decreasing-subsequence-tec-lf7b | This approach uses the upper bound function in C++.\n\nUpper Bound return an iterator to the next smallest greater number than a key in a sorted array.\nEg: if | vipulbhardwaj1011 | NORMAL | 2022-08-14T07:07:59.255959+00:00 | 2022-08-14T07:07:59.256001+00:00 | 103 | false | This approach uses the **upper bound** function in C++.\n\nUpper Bound return an iterator to the **next smallest greater number** than a **key** in a **sorted array**.\n*Eg: if sorted arr = [1, 2, 3, 4, 5], and key = 4, then **upper_bound(arr.begin, arr.end(), key)** would return an iterator to the **first occurance of... | 0 | 0 | ['C', 'Binary Tree'] | 0 |
minimum-operations-to-make-the-array-k-increasing | c++ lis | c-lis-by-me_abhi_2511-b5x0 | class Solution {\npublic:\n\n int lis(vectornums)\n {\n int n=nums.size();\n vectordp(n+1,INT_MAX);\n dp[0]=INT_MIN;\n for(int | me_abhi_2511 | NORMAL | 2022-08-11T18:53:04.370643+00:00 | 2022-08-11T18:53:04.370677+00:00 | 58 | false | class Solution {\npublic:\n\n int lis(vector<int>nums)\n {\n int n=nums.size();\n vector<int>dp(n+1,INT_MAX);\n dp[0]=INT_MIN;\n for(int i=0;i<n;i++)\n {\n int idx=upper_bound(dp.begin(),dp.end(),nums[i])-dp.begin();\n if(dp[idx]>nums[i]&&dp[idx-1]<=nums[i]... | 0 | 0 | ['C'] | 0 |
minimum-operations-to-make-the-array-k-increasing | C++ 243ms beats 99% run LIS k times | c-243ms-beats-99-run-lis-k-times-by-ccch-ht40 | Separate arr into k subsequences. \nNumber of operations on each subsequence would be total elements count minus LIS length. \nFor example, 1 subsequence might | ccchan0109 | NORMAL | 2022-08-08T10:24:36.675710+00:00 | 2022-08-08T10:26:15.212730+00:00 | 59 | false | Separate arr into k subsequences. \nNumber of operations on each subsequence would be total elements count minus LIS length. \nFor example, 1 subsequence might be [1,4,3] generated from arr=[1,2,4,5,3] when k=2. Then, total elements count is 3 and LIS length would be 2, ie. [1,3]. Thus the operations count would be 3-2... | 0 | 0 | ['C'] | 0 |
minimum-operations-to-make-the-array-k-increasing | Python from scratch | python-from-scratch-by-mrpython-tzet | Python without any library: Complete implementation of bisect right\nTime Complexity - k * nlogn\nSpace - O(k)\n\n\nclass Solution:\n def kIncreasing(self, a | mrPython | NORMAL | 2022-08-08T05:13:01.775337+00:00 | 2022-08-08T05:13:01.775369+00:00 | 147 | false | Python without any library: Complete implementation of bisect right\nTime Complexity - k * nlogn\nSpace - O(k)\n\n```\nclass Solution:\n def kIncreasing(self, arr: List[int], k: int) -> int:\n \n def LIS(nums):\n res = []\n for i in nums:\n low,high = 0, len(res)\n ... | 0 | 0 | ['Binary Tree', 'Python', 'Python3'] | 0 |
minimum-operations-to-make-the-array-k-increasing | C++ LIS variant | c-lis-variant-by-fzh-bt4m | \n\n\n\n// Ideas / Approach: an variant of LISS ie the Longest Increasing SubSeq ~~\n// because this problem is about increasing sequence. BTW, if it\'s about d | fzh | NORMAL | 2022-07-24T20:04:08.132949+00:00 | 2022-07-24T20:04:08.132982+00:00 | 76 | false | \n\n\n```\n// Ideas / Approach: an variant of LISS ie the Longest Increasing SubSeq ~~\n// because this problem is about increasing sequence. BTW, if it\'s about decreasing sequence, the\n// LISS can apply as well.\n// * If k == 1, then the answer is A.size() - LIS(A), where A is the input array.\n// * If k > 1, then t... | 0 | 0 | ['C'] | 0 |
minimum-operations-to-make-the-array-k-increasing | C++ || LIS | c-lis-by-rahulmahotra-thrt | \nint lis(vector<int>& v){\n \n vector<int> seq;\n \n for(int i=0; i<v.size(); i++){\n \n if(seq.empty() or se | rahulMahotra | NORMAL | 2022-07-17T13:18:08.622146+00:00 | 2022-07-17T13:18:08.622235+00:00 | 86 | false | ```\nint lis(vector<int>& v){\n \n vector<int> seq;\n \n for(int i=0; i<v.size(); i++){\n \n if(seq.empty() or seq.back() <= v[i]) seq.push_back(v[i]); \n else{\n int idx = upper_bound(seq.begin(), seq.end(), v[i]) \n ... | 0 | 0 | [] | 1 |
minimum-operations-to-make-the-array-k-increasing | binary search easy | binary-search-easy-by-abhi_009-bgri | \nclass Solution:\n \n def LIS(self, arr):\n n = len(arr)\n dp = [\'None\']*n\n dp[0] = arr[0]\n j= 0 \n for i in range | Abhi_009 | NORMAL | 2022-07-14T09:21:41.962259+00:00 | 2022-07-14T09:21:41.962304+00:00 | 128 | false | ```\nclass Solution:\n \n def LIS(self, arr):\n n = len(arr)\n dp = [\'None\']*n\n dp[0] = arr[0]\n j= 0 \n for i in range(1,n):\n if arr[i]>=dp[j]:\n dp[j+1] = arr[i]\n j+=1\n else:\n idx = bisect.bisect(dp,arr[... | 0 | 0 | ['Python', 'Python3'] | 0 |
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