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minimum-number-of-operations-to-make-array-continuous | C++ sorting + Binary Search | c-sorting-binary-search-by-vinyr-4ar3 | \nclass Solution {\npublic:\n int minOperations(vector<int>& A) {\n int n = A.size();\n if(n == 1) return 0;\n sort(A.begin(), A.end()); | vinyr | NORMAL | 2021-09-18T16:27:56.887539+00:00 | 2021-09-19T04:10:26.440961+00:00 | 249 | false | ```\nclass Solution {\npublic:\n int minOperations(vector<int>& A) {\n int n = A.size();\n if(n == 1) return 0;\n sort(A.begin(), A.end());\n \n vector<int> nums = {A[0]};\n for(int i=1; i<n; ++i)\n if(A[i] != A[i-1]) nums.push_back(A[i]); //remove duplicates\n\n ... | 3 | 0 | ['Binary Search', 'C', 'Sorting'] | 0 |
minimum-number-of-operations-to-make-array-continuous | [C++] Sorting + Binary Search || Easy solution with comments | c-sorting-binary-search-easy-solution-wi-vsp3 | \nclass Solution {\npublic:\n int minOperations(vector<int>& nums) {\n sort(nums.begin(),nums.end());\n unordered_map<int,int> mp; // store | manishbishnoi897 | NORMAL | 2021-09-18T16:02:02.214993+00:00 | 2021-09-18T16:42:49.398413+00:00 | 330 | false | ```\nclass Solution {\npublic:\n int minOperations(vector<int>& nums) {\n sort(nums.begin(),nums.end());\n unordered_map<int,int> mp; // stores the frequency of each element\n vector<int> dp(nums.size(),0); // dp[i] represents no of duplicate elements from 0....i\n dp[0]=0;\n m... | 3 | 0 | ['Binary Search', 'C', 'Sorting'] | 0 |
minimum-number-of-operations-to-make-array-continuous | Easy to understand Python Solution using Sliding Window | easy-to-understand-python-solution-using-jv42 | Code | Vinit_K_P | NORMAL | 2025-01-09T19:17:55.261231+00:00 | 2025-01-09T19:17:55.261231+00:00 | 53 | false |
# Code
```python3 []
class Solution:
def minOperations(self, nums: List[int]) -> int:
i = 0
ans = 0
unique_nums = sorted(set(nums))
for j in range(len(unique_nums)):
while unique_nums[j]-unique_nums[i]>=(len(nums)):
i+=1
ans = max(ans, j-i+1... | 2 | 0 | ['Sliding Window', 'Python3'] | 0 |
minimum-number-of-operations-to-make-array-continuous | 10 lines of Code using Binary Search Easy Approach | 10-lines-of-code-using-binary-search-eas-ueqo | Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n\n\n# Complexity\n- Time complexity:\nO(NlogN)\n\n- Space complexity:\nO( | GodX07 | NORMAL | 2023-10-10T19:42:52.801922+00:00 | 2023-10-10T19:42:52.801939+00:00 | 294 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n\n\n# Complexity\n- Time complexity:\nO(NlogN)\n\n- Space complexity:\nO(1)\n\n# Code\n```\nclass Solution {\npublic:\n int minOperations(vector<int>& nums) {\n int n = nums.size(); // Get the number of elements i... | 2 | 0 | ['Binary Search', 'C++'] | 0 |
minimum-number-of-operations-to-make-array-continuous | Java Solution✅||Beats 100%💯||Easy to understand||Detailed Explanation||Sliding Window | java-solutionbeats-100easy-to-understand-l7vx | \n# Approach\n Describe your approach to solving the problem. \n- Sort the array and than run a loop to find the unique elements.\n- When encounters a unique el | Vishu6403 | NORMAL | 2023-10-10T18:41:08.401674+00:00 | 2023-10-10T18:41:08.401698+00:00 | 164 | false | \n# Approach\n<!-- Describe your approach to solving the problem. -->\n- Sort the array and than run a loop to find the unique elements.\n- When encounters a unique element (one that is not equal to the previous element), it stores it at the beginning of the array, effectively reducing the array size to only include un... | 2 | 0 | ['Array', 'Java'] | 0 |
minimum-number-of-operations-to-make-array-continuous | Easy Solution || C++ | easy-solution-c-by-prakas_h-on84 | 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 | prakas_h | NORMAL | 2023-10-10T16:50:17.546028+00:00 | 2023-10-10T16:50:17.546053+00:00 | 65 | 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++'] | 1 |
minimum-number-of-operations-to-make-array-continuous | Easy java Solution using Binary Search | easy-java-solution-using-binary-search-b-soey | \n\n# Code\n\nclass Solution {\n public int minOperations(int[] nums) {\n int m=nums.length;\n Set<Integer> set=new HashSet<>();\n for(in | SagarRuhela | NORMAL | 2023-10-10T16:24:35.364264+00:00 | 2023-10-10T16:24:35.364282+00:00 | 85 | false | \n\n# Code\n```\nclass Solution {\n public int minOperations(int[] nums) {\n int m=nums.length;\n Set<Integer> set=new HashSet<>();\n for(int i=0;i<m;i++){\n set.add(nums[i]);\n } \n int arr[]=new int[set.size()];\n int idx=0;\n int minAns=Integer.MAX_VALUE;\n ... | 2 | 0 | ['Array', 'Binary Search', 'Java'] | 0 |
minimum-number-of-operations-to-make-array-continuous | Easy c++ solution | easy-c-solution-by-geetanjali-18-ost9 | 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 | geetanjali-18 | NORMAL | 2023-10-10T16:15:42.401131+00:00 | 2023-10-10T16:15:42.401154+00:00 | 20 | 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 | ['C++'] | 0 |
minimum-number-of-operations-to-make-array-continuous | Easy Solution using python | easy-solution-using-python-by-jaya_surya-7xwa | 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 | Jaya_Surya_3 | NORMAL | 2023-10-10T14:39:07.463937+00:00 | 2023-10-10T14:39:07.463966+00:00 | 153 | 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', 'Python3'] | 0 |
minimum-number-of-operations-to-make-array-continuous | ✅☑[C++/C/Java/Python/JavaScript] || O(nlogn) || Sliding Window || EXPLAINED🔥 | ccjavapythonjavascript-onlogn-sliding-wi-e1cx | PLEASE UPVOTE IF IT HELPED\n\n---\n\n\n# Approaches\n(Also explained in the code)\n1. The code aims to find the minimum operations required to make the array "c | MarkSPhilip31 | NORMAL | 2023-10-10T14:30:26.577822+00:00 | 2023-10-12T20:25:17.416718+00:00 | 384 | false | # *PLEASE UPVOTE IF IT HELPED*\n\n---\n\n\n# Approaches\n***(Also explained in the code)***\n1. The code aims to find the minimum operations required to make the array "consecutive," where consecutive elements have a difference of at most `n-1`.\n\n1. We first insert all unique elements from the input array `nums` into... | 2 | 0 | ['Array', 'Hash Table', 'Binary Search', 'C', 'Sliding Window', 'Ordered Set', 'C++', 'Java', 'Python3', 'JavaScript'] | 3 |
minimum-number-of-operations-to-make-array-continuous | Standard approach | standard-approach-by-deepakvrma-43w9 | \n# Code\n\nclass Solution {\npublic:\n int minOperations(std::vector<int>& nums) {\n int n = nums.size();\n std::sort(nums.begin(), nums.end() | DeepakVrma | NORMAL | 2023-10-10T12:15:11.496048+00:00 | 2023-10-10T12:15:11.496092+00:00 | 21 | false | \n# Code\n```\nclass Solution {\npublic:\n int minOperations(std::vector<int>& nums) {\n int n = nums.size();\n std::sort(nums.begin(), nums.end());\n std::vector<int> uniqueNums(nums.begin(), std::unique(nums.begin(), nums.end()));\n int ans = std::numeric_limits<int>::max();\n\n ... | 2 | 0 | ['Sorting', 'C++'] | 0 |
minimum-number-of-operations-to-make-array-continuous | Efficient Approach Using Sliding window and Sorting | efficient-approach-using-sliding-window-ffv8y | Intuition\nThe provided code uses a sliding window technique to efficiently find the minimum number of operations required to \nmake all elements in the input a | gautam309 | NORMAL | 2023-10-10T12:05:49.754401+00:00 | 2023-10-10T12:05:49.754421+00:00 | 34 | false | # Intuition\nThe provided code uses a sliding window technique to efficiently find the minimum number of operations required to \nmake all elements in the input array distinct. It begins by sorting the array and removing duplicates, resulting in \na sorted array with unique elements.\n\nThe sliding window mechanism is ... | 2 | 0 | ['Sliding Window', 'Sorting', 'C++'] | 0 |
minimum-number-of-operations-to-make-array-continuous | Easy solution using set | easy-solution-using-set-by-samarthnag-chae | 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 | samarthnag | NORMAL | 2023-10-10T11:25:07.988675+00:00 | 2023-10-10T11:25:07.988694+00:00 | 163 | 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 | ['C++'] | 0 |
minimum-number-of-operations-to-make-array-continuous | Simple Binery search Implementation | simple-binery-search-implementation-by-s-okgp | Intuition\n Describe your first thoughts on how to solve this problem. \n\n\n\n\n# Approach\n\n1. Sort the input array nums in ascending order.\n2. Create a new | seefat | NORMAL | 2023-10-10T09:21:11.798848+00:00 | 2023-10-10T09:21:11.798870+00:00 | 94 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n\n\n\n# Approach\n\n1. Sort the input array `nums` in ascending order.\n2. Create a new array `uniqueNums` to store the unique elements of `nums` using a Set.\n3. Define a binary search function to find the maximum consecutive length of... | 2 | 0 | ['Two Pointers', 'Binary Search', 'JavaScript'] | 0 |
minimum-number-of-operations-to-make-array-continuous | Java ✅: O(n2) HashSet & O(nlogn) SW approach | java-on2-hashset-onlogn-sw-approach-by-a-yoow | TC - O(n2) - Using HashSet - TLE\n\n // O(n2) approach considering every element as minimum \n public int minOperations(int[] nums){\n int n = nums | Ayush7865 | NORMAL | 2023-08-05T06:36:35.140465+00:00 | 2023-09-18T19:00:42.447902+00:00 | 19 | false | ### TC - O(n2) - Using HashSet - TLE\n```\n // O(n2) approach considering every element as minimum \n public int minOperations(int[] nums){\n int n = nums.length;\n int min = Integer.MAX_VALUE;\n HashSet<Integer> set = new HashSet<>();\n for(int num : nums){\n set.add(num);\... | 2 | 0 | ['Java'] | 0 |
minimum-number-of-operations-to-make-array-continuous | o(nlogn) | onlogn-by-anil-budamakuntla-q88c | 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 | ANIL-BUDAMAKUNTLA | NORMAL | 2023-06-12T02:08:57.589414+00:00 | 2024-06-18T04:49:42.535491+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:0(nlogn)\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O... | 2 | 0 | ['C++'] | 0 |
minimum-number-of-operations-to-make-array-continuous | Java Sliding Windows | java-sliding-windows-by-mra-0qmu | java\npublic int minOperations(int[] nums) {\n\tint[] arr = Arrays.stream(nums).sorted().distinct().toArray();\n\tint res = nums.length;\n\tfor (int i = 0, j = | mra | NORMAL | 2022-10-20T13:39:04.338452+00:00 | 2022-10-20T13:41:10.082442+00:00 | 407 | false | ```java\npublic int minOperations(int[] nums) {\n\tint[] arr = Arrays.stream(nums).sorted().distinct().toArray();\n\tint res = nums.length;\n\tfor (int i = 0, j = 0; j < arr.length; j++) {\n\t\twhile (arr[j] - arr[i] >= nums.length) {\n\t\t\ti++;\n\t\t}\n\n\t\tres = Integer.min(res, nums.length - (j - i + 1));\n\t}\n\t... | 2 | 0 | ['Java'] | 0 |
minimum-number-of-operations-to-make-array-continuous | Simple Solution Explained Python | simple-solution-explained-python-by-sart-kor2 | ending - starting = n - 1\nwe want something like [a, a+1, ......, a+n-1] (sorted representation)\nmax operations = n-1, min operations = 0\neach value in nums | sarthakBhandari | NORMAL | 2022-10-14T14:33:38.519522+00:00 | 2022-10-14T14:49:18.989225+00:00 | 301 | false | ending - starting = n - 1\nwe want something like [a, a+1, ......, a+n-1] (sorted representation)\nmax operations = n-1, min operations = 0\neach value in nums can be a valid starting point\neach value in nums can be a valid ending point\n\nif each value can be a valid starting point, find the number of elements outsid... | 2 | 0 | ['Binary Search', 'Sliding Window', 'Python'] | 0 |
minimum-number-of-operations-to-make-array-continuous | Java | Binary Search | O(nlogn) | java-binary-search-onlogn-by-harshraj998-5qf0 | Inuition:\nConsidering each element as the last element of answer array, we will find how many numbers are present between the element and (element - array.leng | HarshRaj9988 | NORMAL | 2022-10-01T05:32:19.034488+00:00 | 2022-10-01T05:32:19.034533+00:00 | 750 | false | Inuition:\nConsidering each element as the last element of answer array, we will find how many numbers are present between the element and (element - array.length +1), inclusive these two. keep track of maximum number of valid elements. and at last return the difference between total number of element of array and the ... | 2 | 1 | [] | 0 |
minimum-number-of-operations-to-make-array-continuous | C++|| Sliding Window || Sorting || Solution with explaination | c-sliding-window-sorting-solution-with-e-b0jj | The idea is to remove the duplicate elements and sort the array \nduplicate elements are removed as they are definitely not the part of our continuous array \nn | jahnavimahara | NORMAL | 2022-08-21T17:33:01.539844+00:00 | 2022-08-21T17:33:01.539886+00:00 | 516 | false | **The idea is to remove the duplicate elements and sort the array \nduplicate elements are removed as they are definitely not the part of our continuous array \nnow iterate through the array and consider every element as the minimum element of continuous subarray\nmake a sliding window from that element and increase it... | 2 | 0 | ['C', 'Sliding Window', 'Sorting'] | 0 |
minimum-number-of-operations-to-make-array-continuous | Java | O(NLog(N) Time | O(1) Space | Sliding Window with Comments | java-onlogn-time-o1-space-sliding-window-1z8l | \nclass Solution {\n public int minOperations(int[] nums) {\n int n = nums.length;\n Arrays.sort(nums); // Sort in ascending order | hawtsauce-iwnl | NORMAL | 2022-08-03T14:58:38.127101+00:00 | 2022-08-03T14:59:17.820588+00:00 | 347 | false | ```\nclass Solution {\n public int minOperations(int[] nums) {\n int n = nums.length;\n Arrays.sort(nums); // Sort in ascending order, so left points to min element and right to max element in window\n \n int left = 0, right = 0; \n int maxDiff = n - 1; ... | 2 | 0 | ['Sliding Window', 'Java'] | 0 |
minimum-number-of-operations-to-make-array-continuous | C++ | Sets | c-sets-by-__ikigai-bz4k | \nclass Solution {\npublic:\n int minOperations(vector<int>& nums) {\n int len=nums.size()-1;\n int mini=INT_MAX, maxi=INT_MIN;\n set<in | __ikigai__ | NORMAL | 2022-08-01T16:13:06.737510+00:00 | 2022-08-01T16:13:06.737541+00:00 | 103 | false | ```\nclass Solution {\npublic:\n int minOperations(vector<int>& nums) {\n int len=nums.size()-1;\n int mini=INT_MAX, maxi=INT_MIN;\n set<int> count;\n for(auto &it:nums) {\n count.insert(it);\n mini=min(mini, it); \n maxi=max(maxi, it);\n }\n ... | 2 | 0 | ['Ordered Set'] | 0 |
minimum-number-of-operations-to-make-array-continuous | HEAVILY COMMENTED CODE + SIMPLEST EXPLANATION-USING BINARY SEARCH | heavily-commented-code-simplest-explanat-4xrk | \n/*Intution : while pondering at the problem, I was thinking of o(n2) solution wherein, I\'ll deem every element as the minimum element and will see how many e | user2085X | NORMAL | 2022-07-29T06:20:19.195108+00:00 | 2022-07-29T06:20:19.195147+00:00 | 221 | false | `\n/*Intution : while pondering at the problem, I was thinking of o(n2) solution wherein, I\'ll deem every element as the minimum element and will see how many elements are there in the array that together with this element will form a "continuous array" with minimum changes. \nSomething like this :*/ `\n\n```\nfor(int... | 2 | 0 | ['Binary Tree', 'Ordered Set'] | 0 |
minimum-number-of-operations-to-make-array-continuous | Binary Search | binary-search-by-diyora13-l0ka | \nclass Solution {\npublic:\n int minOperations(vector<int>& a) \n {\n map<int,int> mp;\n int n=a.size(),ans=1e9;\n for(int i=0;i<n;i | diyora13 | NORMAL | 2022-06-07T01:45:00.171938+00:00 | 2022-06-07T01:45:00.172005+00:00 | 408 | false | ```\nclass Solution {\npublic:\n int minOperations(vector<int>& a) \n {\n map<int,int> mp;\n int n=a.size(),ans=1e9;\n for(int i=0;i<n;i++)\n mp[a[i]]++;\n a.clear();\n for(auto i:mp)\n a.push_back(i.first);\n sort(a.begin(),a.end());\n for(in... | 2 | 0 | ['Binary Search', 'C', 'C++'] | 0 |
minimum-number-of-operations-to-make-array-continuous | C++ short | c-short-by-abhishekaaru-wykm | \tclass Solution {\n\tpublic:\n\t\tint minOperations(vector& nums) {\n\t\t\tint n = nums.size();\n\t\t\t\n\t\t\tint ans=INT_MAX;\n\t\t\tsort(nums.begin(),nums.e | abhishekaaru | NORMAL | 2022-04-07T11:18:34.581927+00:00 | 2022-04-07T11:18:34.581958+00:00 | 311 | false | \tclass Solution {\n\tpublic:\n\t\tint minOperations(vector<int>& nums) {\n\t\t\tint n = nums.size();\n\t\t\t\n\t\t\tint ans=INT_MAX;\n\t\t\tsort(nums.begin(),nums.end());\n\t\t\tnums.erase(unique(nums.begin(),nums.end()),nums.end());\n\t\t\tint m = nums.size();\n\t\t\t\n\t\t\tint j=0;\n\t\t\tfor(int i=0;i<m;i++){\n\t\... | 2 | 0 | ['C'] | 0 |
minimum-number-of-operations-to-make-array-continuous | JAVA solution (sorting+ treeset) | java-solution-sorting-treeset-by-flyroko-8gqi | \n \n\t class Solution {\n public int minOperations(int[] nums) {\n\t\t//using treeset to store them in sorted fashion and ignoring duplicates\n | flyRoko123 | NORMAL | 2021-12-19T14:50:44.468547+00:00 | 2021-12-19T14:50:44.468586+00:00 | 154 | false | \n \n\t class Solution {\n public int minOperations(int[] nums) {\n\t\t//using treeset to store them in sorted fashion and ignoring duplicates\n \n TreeSet<Integer> s=new TreeSet<>();\n int max=Integer.MIN_VALUE;\n int range=nums.length-1;\n \n //sorting the array to m... | 2 | 0 | [] | 0 |
minimum-number-of-operations-to-make-array-continuous | Pigeonhole Principle | pigeonhole-principle-by-throwaway1973-xei8 | Didn\'t see anyone mention this, but here\'s how I thought of the problem. You can kind of think of this according to the pigeonhole principle used to reason ab | throwaway1973 | NORMAL | 2021-11-19T09:28:31.755324+00:00 | 2021-11-19T09:28:55.305147+00:00 | 227 | false | Didn\'t see anyone mention this, but here\'s how I thought of the problem. You can kind of think of this according to the pigeonhole principle used to reason about hash tables. Think of it this way, I have a range of numbers (buckets) between the maximum and minimum number candidates of the array, and I want to cram th... | 2 | 0 | [] | 0 |
minimum-number-of-operations-to-make-array-continuous | C++ Simple Solution using Set | c-simple-solution-using-set-by-lunatic_p-3afn | Idea is sort the array and erase duplicates if any,\nSo set is bettter compare to erase in a vector.\n\n\nint minOperations(vector<int>& A) {\n int N = | lunatic_peace | NORMAL | 2021-10-05T02:04:13.740575+00:00 | 2021-10-05T02:04:28.797049+00:00 | 201 | false | Idea is sort the array and erase duplicates if any,\nSo set is bettter compare to erase in a vector.\n\n```\nint minOperations(vector<int>& A) {\n int N = A.size(), i = 0, j = 0; \n set<int> st(A.begin(), A.end());\n vector<int> B (st.begin(), st.end());\n \n for (int M = B.size()... | 2 | 0 | ['C'] | 1 |
minimum-number-of-operations-to-make-array-continuous | Sorting + Dedup + Binary Search with Java | sorting-dedup-binary-search-with-java-by-s3kr | LC2009. Minimum Number of Operations to Make Array Continuous\n## Method. Sorting + Dedup + Binary Search\n### Main Idea\nObservations\n1. Since the length of t | leungogogo | NORMAL | 2021-09-24T17:07:42.839694+00:00 | 2021-09-24T17:16:48.600304+00:00 | 206 | false | # LC2009. Minimum Number of Operations to Make Array Continuous\n## Method. Sorting + Dedup + Binary Search\n### Main Idea\n**Observations**\n1. Since the length of the input array is fixed, once the starting point is determined, the end point is also fixed, which is `end = start + n - 1`.\n2. To see how many modificat... | 2 | 0 | [] | 0 |
minimum-number-of-operations-to-make-array-continuous | Simple Java Sliding Window | simple-java-sliding-window-by-fang2018-pbov | \nclass Solution {\n public int minOperations(int[] nums) {\n int n = nums.length, maxNum = 0; \n Arrays.sort(nums);\n Map<Integer, Inte | fang2018 | NORMAL | 2021-09-19T04:35:45.701271+00:00 | 2021-09-19T04:35:45.701294+00:00 | 106 | false | ```\nclass Solution {\n public int minOperations(int[] nums) {\n int n = nums.length, maxNum = 0; \n Arrays.sort(nums);\n Map<Integer, Integer> map = new HashMap<>();\n for (int i = 0, j = 0; i < n; i++) {\n map.put(nums[i], map.getOrDefault(nums[i], 0) + 1);\n while... | 2 | 0 | [] | 0 |
minimum-number-of-operations-to-make-array-continuous | [C++] Upper_bound || Faster than 100% -O[nlogn] with Simple Explanation | c-upper_bound-faster-than-100-onlogn-wit-820l | Conclusion from the question\n1) Final array will be a continious series of unique elements of length nums.size().\n2) Intuitively, It is not very difficult to | Shivu__122 | NORMAL | 2021-09-18T20:59:08.061777+00:00 | 2021-09-18T21:11:38.436814+00:00 | 149 | false | **Conclusion from the question**\n1) Final array will be a continious series of unique elements of length nums.size().\n2) Intuitively, It is not very difficult to see that the resulting series will start from any optimal element which is already present in the array. \n\n3) Now we only need to find from **which elemen... | 2 | 0 | [] | 1 |
minimum-number-of-operations-to-make-array-continuous | Simple solution in c++ | simple-solution-in-c-by-dhruvppatel2506-9trn | Simple solution using order statistics tree.\nRefered this \n\n#include <ext/pb_ds/assoc_container.hpp>\n#include <ext/pb_ds/tree_policy.hpp>\n\n\nusing namespa | dhruvppatel2506 | NORMAL | 2021-09-18T16:13:43.076376+00:00 | 2021-09-18T16:15:32.264422+00:00 | 87 | false | Simple solution using order statistics tree.\nRefered [this](https://codeforces.com/blog/entry/11080) \n```\n#include <ext/pb_ds/assoc_container.hpp>\n#include <ext/pb_ds/tree_policy.hpp>\n\n\nusing namespace __gnu_pbds;\n\ntemplate <class T> using Tree = tree<T, null_type, less<T>, rb_tree_tag,tree_order_statistics_no... | 2 | 1 | [] | 2 |
minimum-number-of-operations-to-make-array-continuous | [Python 3] O(nlogn) solution with visualization | python-3-onlogn-solution-with-visualizat-ocqw | Explanation with visualization\n\nObservation 1:\nWe can consider only unique elements (No duplicates)\n\nObservation 2: \nFor the difference between maximum an | sbh69840 | NORMAL | 2021-09-18T16:01:30.461688+00:00 | 2021-09-19T02:28:17.320199+00:00 | 124 | false | **Explanation with visualization**\n\n**Observation 1:**\nWe can consider only unique elements (No duplicates)\n\n**Observation 2:** \nFor the difference between maximum and minimum number to be equal to len(nums)-1 we want the elements to be a contiguous sequence. \n\n**Solution:**\nSince the final result has to be a ... | 2 | 0 | [] | 0 |
minimum-number-of-operations-to-make-array-continuous | [Kotlin] Simple Short O(nlogn) Solution | kotlin-simple-short-onlogn-solution-by-c-556z | \nclass Solution {\n fun minOperations(nums: IntArray): Int {\n val distinctNums = nums.sorted().distinct()\n var maxCount = 1\n var fir | catcatcute | NORMAL | 2021-09-18T16:01:27.447856+00:00 | 2021-09-18T16:04:12.473335+00:00 | 94 | false | ```\nclass Solution {\n fun minOperations(nums: IntArray): Int {\n val distinctNums = nums.sorted().distinct()\n var maxCount = 1\n var firstNumIndex = 0\n for (i in 1..distinctNums.lastIndex) {\n while (distinctNums[i] - distinctNums[firstNumIndex] >= nums.size) {\n ... | 2 | 0 | ['Kotlin'] | 1 |
minimum-number-of-operations-to-make-array-continuous | Python TC O(n log(n)), SC O(n): Proof that a Best Window Ends with a Value in nums | python-tc-on-logn-sc-on-proof-that-a-bes-h4la | Intuition\n\nThe idea is that we want to find the sliding window of range n-1 that involves the fewest moves to make all elements unique in the window.\n\nSuppo | biggestchungus | NORMAL | 2024-08-18T04:31:28.233005+00:00 | 2024-08-18T04:31:28.233030+00:00 | 10 | false | # Intuition\n\nThe idea is that we want to find the sliding window of range `n-1` that involves the fewest moves to make all elements unique in the window.\n\nSuppose we have `u` unique elements in some window `lo..hi` where `hi-lo+1 == n` as described in the problem statement. Then we have to move the other `n-u` elem... | 1 | 0 | ['Python3'] | 0 |
minimum-number-of-operations-to-make-array-continuous | Using diference array | using-diference-array-by-aditya3435-cezm | \n\n# Code\n\nclass Solution {\npublic:\n int minOperations(vector<int>& nums) {\n int f = nums.size();\n set<int> st(nums.begin(), nums.end()) | aditya3435 | NORMAL | 2023-11-02T14:11:02.862154+00:00 | 2023-11-02T14:11:02.862182+00:00 | 4 | false | \n\n# Code\n```\nclass Solution {\npublic:\n int minOperations(vector<int>& nums) {\n int f = nums.size();\n set<int> st(nums.begin(), nums.end());\n nums.clear();\n for(auto c : st) nums.push_back(c);\n int mn = INT_MIN, mni = 0;\n for(int i=0; i<nums.size(); i++) {\n ... | 1 | 0 | ['C++'] | 0 |
minimum-number-of-operations-to-make-array-continuous | Python | Easy | Queue | Heap | Minimum Number of Operations to Make Array Continuous | python-easy-queue-heap-minimum-number-of-p8u3 | \nsee the Successfully Accepted Submission\nPython\nclass Solution:\n def minOperations(self, nums):\n n = len(nums)\n nums.sort() # Sort the | Khosiyat | NORMAL | 2023-10-12T14:21:30.138380+00:00 | 2023-10-12T14:21:30.138431+00:00 | 27 | false | \n[see the Successfully Accepted Submission](https://leetcode.com/submissions/detail/1071611939/)\n```Python\nclass Solution:\n def minOperations(self, nums):\n n = len(nums)\n nums.sort() # Sort the input nums in non-decreasing order.\n \n if n == 1:\n return 0 # If there\'s... | 1 | 0 | ['Queue', 'Python'] | 0 |
minimum-number-of-operations-to-make-array-continuous | Java | Sliding Window | Full Explanation | java-sliding-window-full-explanation-by-7xpxe | Intuition\nThe problem requires finding the minimum number of operations to make the array continuous. To achieve this, we can use a HashSet to remove duplicate | harshs29 | NORMAL | 2023-10-10T18:12:44.879775+00:00 | 2023-10-10T18:12:44.879805+00:00 | 27 | false | # Intuition\nThe problem requires finding the minimum number of operations to make the array continuous. To achieve this, we can use a HashSet to remove duplicates, then sort the array. Next, we use a sliding window approach to find the smallest subarray containing n unique elements.\n\n\n# Approach\n1. Initialize vari... | 1 | 0 | ['Array', 'Sliding Window', 'Java'] | 0 |
minimum-number-of-operations-to-make-array-continuous | C++ Binary Search , Understandable to Beginners as well | c-binary-search-understandable-to-beginn-wjrr | 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 | Ankita2905 | NORMAL | 2023-10-10T17:10:19.313880+00:00 | 2023-10-10T17:10:19.313902+00:00 | 14 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ --... | 1 | 0 | ['C++'] | 0 |
minimum-number-of-operations-to-make-array-continuous | [Rust] | Idiomatic Iterator Chain | rust-idiomatic-iterator-chain-by-marktan-ozhi | \nimpl Solution {\n pub fn min_operations(mut nums: Vec<i32>) -> i32 {\n let n: usize = nums.len();\n let n: i32 = n as i32;\n\n nums.so | MarkTanashchukID | NORMAL | 2023-10-10T16:57:26.055504+00:00 | 2023-10-10T16:57:26.055535+00:00 | 6 | false | ```\nimpl Solution {\n pub fn min_operations(mut nums: Vec<i32>) -> i32 {\n let n: usize = nums.len();\n let n: i32 = n as i32;\n\n nums.sort();\n nums.dedup();\n\n nums.iter()\n .enumerate()\n .map(|(i, num)| nums.partition_point(|&p| p < num + n) - i)\n ... | 1 | 0 | ['Rust'] | 0 |
minimum-number-of-operations-to-make-array-continuous | Simple Solution || Beginner Friendly || Easy to Understand C++ | simple-solution-beginner-friendly-easy-t-djcv | Intuition\n\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Tim | Ishu-Gupta-9-ISHGPT | NORMAL | 2023-10-10T15:56:08.579298+00:00 | 2023-10-10T15:56:08.579316+00:00 | 12 | false | # Intuition\n\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:O(nlogn)\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:O(n)\n<!-- Add your space complexity here, e.... | 1 | 0 | ['C++'] | 0 |
minimum-number-of-operations-to-make-array-continuous | ||easy solution||O(1) space|| | easy-solutiono1-space-by-sainiankur0508-cst9 | Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time | sainiankur0508 | NORMAL | 2023-10-10T14:27:18.077702+00:00 | 2023-10-10T14:27:18.077733+00:00 | 33 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n$$O(nlogn)$$\n\n- Space complexity:\n$$O(1)$$\n\n# Code\n```\nclass Solution {\npublic:\n int minOperations(vector<int>& nums) {... | 1 | 0 | ['Array', 'Sliding Window', 'Sorting', 'C++'] | 0 |
minimum-number-of-operations-to-make-array-continuous | binary search | binary-search-by-mr_stark-hepv | \nclass Solution {\npublic:\n \n \n int minOperations(vector<int>& nums) {\n int nn = nums.size();\n set<int> st(nums.begin(),nums.end());\n | mr_stark | NORMAL | 2023-10-10T14:11:29.477535+00:00 | 2023-10-10T14:11:48.228397+00:00 | 9 | false | ```\nclass Solution {\npublic:\n \n \n int minOperations(vector<int>& nums) {\n int nn = nums.size();\n set<int> st(nums.begin(),nums.end());\n vector<int> v(st.begin(),st.end());\n int n = v.size();\n int res = INT_MAX;\n for(int i = 0;i<n;i++)\n {\n i... | 1 | 0 | [] | 0 |
minimum-number-of-operations-to-make-array-continuous | Binary Search | O(nlogn) | Problem seems trickier, but it's not. | binary-search-onlogn-problem-seems-trick-plh0 | Intuition\n Describe your first thoughts on how to solve this problem. \nFirst, we need to understand that a solution always exists for this problem. Why? Let\' | Nik28 | NORMAL | 2023-10-10T13:38:43.455472+00:00 | 2023-10-10T13:38:43.455490+00:00 | 75 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nFirst, we need to understand that a solution always exists for this problem. Why? Let\'s see.\n\nIf we\'ve an array `[1,2,4,6]`, any set of 4 consecutive elements is a contiguous array. \n`[1,2,3,4], [5,6,7,8], [100,101,102,103]`, any. \n... | 1 | 0 | ['Binary Search', 'C++'] | 0 |
minimum-number-of-operations-to-make-array-continuous | ✅Accepted Java Code || Beats 100% | accepted-java-code-beats-100-by-thilaknv-7vw1 | Complexity\n- Time complexity : O(nlog(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) | thilaknv | NORMAL | 2023-10-10T12:55:36.229728+00:00 | 2023-10-10T12:55:36.229751+00:00 | 82 | false | # Complexity\n- Time complexity : $$O(nlog(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)$$ -->\n\n# Code\n```\nclass Solution {\npublic int minOperations(int[] nums) {\n Arrays.sort(nums);\n int max = 1, temp = 1;\n ... | 1 | 0 | ['Array', 'Binary Search', 'Java'] | 0 |
minimum-number-of-operations-to-make-array-continuous | Java easy solution 👨💻💕😎 | java-easy-solution-by-ziddiraj-a683 | Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time | ZIDDIRAJ | NORMAL | 2023-10-10T12:39:02.158600+00:00 | 2023-10-10T12:39:02.158624+00:00 | 59 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ --... | 1 | 0 | ['Array', 'Binary Search', 'Java'] | 0 |
minimum-number-of-operations-to-make-array-continuous | Beats 99.02%of users with Java | beats-9902of-users-with-java-by-kushal_5-z1h4 | Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time | kushal_589 | NORMAL | 2023-10-10T11:38:05.346063+00:00 | 2023-10-10T11:38:05.346088+00:00 | 75 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ --... | 1 | 0 | ['Java'] | 1 |
minimum-number-of-operations-to-make-array-continuous | Python super easy solution with intuition. Concept( set & binary search) | python-super-easy-solution-with-intuitio-uhbg | Intuition\n Describe your first thoughts on how to solve this problem. \n If you observe the question carefully you can see that\n1. the frequency of the number | mkrishnendu079 | NORMAL | 2023-10-10T11:11:38.179166+00:00 | 2023-10-10T11:11:38.179183+00:00 | 101 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n If you observe the question carefully you can see that\n1. the frequency of the numbers does not matter the same number comes up once or many times does not matter in the end you just need to find maximum **unique** numbers that are agre... | 1 | 0 | ['Python3'] | 0 |
minimum-number-of-operations-to-make-array-continuous | C solution | c-solution-by-hoouhoou-i4f6 | \nint cmp(const void *a, const void *b){\n return *(int*)a - *(int*)b;\n}\n\nint binarySearch(int target, int *nums, int size){\n int l = 0, r = size, mid | HoouHoou | NORMAL | 2023-10-10T10:00:09.504226+00:00 | 2023-10-10T10:00:09.504242+00:00 | 19 | false | ```\nint cmp(const void *a, const void *b){\n return *(int*)a - *(int*)b;\n}\n\nint binarySearch(int target, int *nums, int size){\n int l = 0, r = size, mid;\n while (l < r){\n mid = (l + r) >> 1;\n if (nums[mid] > target)\n r = mid;\n else\n l = mid + 1;\n }\n ... | 1 | 0 | ['C'] | 0 |
minimum-number-of-operations-to-make-array-continuous | C# Two Pointers With Explanation | c-two-pointers-with-explanation-by-yulin-pf5x | Approach\nFirst, by reading the description, we know array can be sorted and will not affect the result, and not expecting any duplicated element.\nBy these rul | yulinchao | NORMAL | 2023-10-10T09:14:06.335715+00:00 | 2023-10-10T09:14:06.335733+00:00 | 43 | false | # Approach\nFirst, by reading the description, we know array can be sorted and will not affect the result, and not expecting any duplicated element.\nBy these rules, we know the numbers in nums are continous and unique.\n\nTo make the actions as less as possible, we wish the interval formed by left and right(left and r... | 1 | 0 | ['C#'] | 2 |
minimum-number-of-operations-to-make-array-continuous | Fast and Simple 8-line solution | Deque | fast-and-simple-8-line-solution-deque-by-bq9n | Complexity\n- Time complexity: O(NLogN)\n Add your time complexity here, e.g. O(n) \n\n- Space complexity: O(N)\n Add your space complexity here, e.g. O(n) \n\ | swapniltyagi17 | NORMAL | 2023-10-10T08:49:19.568660+00:00 | 2023-10-10T09:02:17.685268+00:00 | 22 | false | # Complexity\n- Time complexity: O(NLogN)\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity: O(N)\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\nclass Solution {\npublic:\n\n int minOperations(vector<int>& nums) {\n sort(nums.begin(),nums.end());\n i... | 1 | 0 | ['C++'] | 0 |
flip-string-to-monotone-increasing | C++ one-pass DP solution, 0ms O(n) | O(1), one line, with explaination. | c-one-pass-dp-solution-0ms-on-o1-one-lin-pslr | This is a typical case of DP. Let's see the sub-question of DP first. Suppose that you have a string s, and the solution to the mono increase question is alread | liamhuang | NORMAL | 2018-11-05T03:45:46.753674+00:00 | 2018-11-05T03:45:46.753717+00:00 | 26,691 | false |
This is a typical case of DP.
Let's see the sub-question of DP first.
Suppose that you have a string `s`, and the solution to the mono increase question is already solved. That is, for string `s`, `counter_flip` flips are required for the string, and there were `counter_one` `'1'`s in the original string `s`.
Let's... | 625 | 8 | [] | 36 |
flip-string-to-monotone-increasing | ✅[C++]Easy solution with explaination[Without Dp]✅ | ceasy-solution-with-explainationwithout-1b88t | \uD83C\uDFA5\uD83D\uDD25 Exciting News! Join my Coding Journey! Subscribe Now! \uD83D\uDD25\uD83C\uDFA5\n\n\uD83D\uDD17 Link in the leetcode profile \n\nNew cod | vishnoi29 | NORMAL | 2023-01-17T00:29:24.675938+00:00 | 2023-06-14T02:09:29.883062+00:00 | 18,013 | false | \uD83C\uDFA5\uD83D\uDD25 Exciting News! Join my Coding Journey! Subscribe Now! \uD83D\uDD25\uD83C\uDFA5\n\n\uD83D\uDD17 Link in the leetcode profile \n\nNew coding channel alert! \uD83D\uDE80\uD83D\uDCBB Subscribe to unlock amazing coding content and tutorials. Help me reach 1K subs to start posting more videos! Join n... | 415 | 3 | ['C++'] | 20 |
flip-string-to-monotone-increasing | Prefix-Suffix Java O(N) One Pass Solution - Space O(1) | prefix-suffix-java-on-one-pass-solution-eghnf | Algorithm:\n1. Skip 0\'s until we encounter the first 1.\n2. Keep track of number of 1\'s in onesCount (Prefix).\n3. Any 0 that comes after we encounter 1 can b | vicky_11 | NORMAL | 2018-10-21T03:26:43.762711+00:00 | 2018-10-24T12:29:41.770616+00:00 | 18,867 | false | Algorithm:\n1. Skip 0\'s until we encounter the first 1.\n2. Keep track of number of 1\'s in onesCount (Prefix).\n3. Any 0 that comes after we encounter 1 can be a potential candidate for flip. Keep track of it in flipCount.\n4. If flipCount exceeds oneCount - (Prefix 1\'s flipped to 0\'s)\n\t\ta. Then we are trying to... | 242 | 5 | [] | 22 |
flip-string-to-monotone-increasing | C++/Java 4 lines O(n) | O(1), DP | cjava-4-lines-on-o1-dp-by-votrubac-t1hv | We need to split the array into left \'0\' and right \'1\' sub-arrays, so that sum of \'1\' -> \'0\' flips (left) and \'0\' -> \'1\' flips (right) is minimal.\ | votrubac | NORMAL | 2018-10-21T03:02:15.588980+00:00 | 2022-01-05T23:45:56.994365+00:00 | 16,762 | false | We need to split the array into left \'0\' and right \'1\' sub-arrays, so that sum of \'1\' -> \'0\' flips (left) and \'0\' -> \'1\' flips (right) is minimal.\n- Count of \'1\' -> \'0\' flips going left to right, and store it in ```f0```.\n- Count of \'0\' -> \'1\' flips going right to left, and store it in ```f1```.\... | 211 | 7 | [] | 21 |
flip-string-to-monotone-increasing | Two concepts, four implementation with explanation for beginners. | two-concepts-four-implementation-with-ex-na9k | The idea:\nLike many question, the description of this question is overly complicated and we can simplify it greatly. One of the best things you can do before s | haoran_xu | NORMAL | 2021-06-13T15:36:35.190962+00:00 | 2021-07-24T13:07:03.557932+00:00 | 6,795 | false | The idea:\nLike many question, the description of this question is overly complicated and we can simplify it greatly. One of the best things you can do before starting to write code is figure out exactly what you want. This might seem obvious but if you miss this step, it can cause a lot of pain down the road. First, I... | 208 | 1 | [] | 25 |
flip-string-to-monotone-increasing | C++ Simple and Short O(n) Solution, With Explanation | c-simple-and-short-on-solution-with-expl-xsxw | idea:\nThis is simple dynamic programming.\nWe loop through the string.\nIf we got a 1, go on. No need to flip. We just increment the 1 counter.\nIf we got a 0, | yehudisk | NORMAL | 2021-08-10T08:08:43.240588+00:00 | 2021-08-10T08:08:43.240613+00:00 | 5,375 | false | **idea:**\nThis is simple dynamic programming.\nWe loop through the string.\nIf we got a 1, go on. No need to flip. We just increment the 1 counter.\nIf we got a 0, we increment the flips counter.\nNow, we have two options. Either to flip the new zero to one or to flip all previous ones to zeros.\nSo we take the min be... | 128 | 6 | ['C'] | 9 |
flip-string-to-monotone-increasing | python O(n) time O(1) space solution with explanation(with extra Chinese explanation) | python-on-time-o1-space-solution-with-ex-ijpe | you can get Chinese explanation in \nhttps://buptwc.github.io/2018/10/21/Leetcode-926-Flip-String-to-Monotone-Increasing/\n\nwe assume the string after flip is | bupt_wc | NORMAL | 2018-10-21T14:39:19.127681+00:00 | 2018-10-24T18:40:56.570397+00:00 | 8,612 | false | you can get Chinese explanation in \nhttps://buptwc.github.io/2018/10/21/Leetcode-926-Flip-String-to-Monotone-Increasing/\n\nwe assume the string after flip is `s = \'0\'*i + \'1\'*j`, the index of first \'1\' is `i`\n\nwe just need to find the `i` in the initial string. We make all `1` before index `i` flip to `0`, a... | 125 | 2 | [] | 20 |
flip-string-to-monotone-increasing | Python 3 || 7 lines, w/ explanation and example || T/S: 96% / 98% | python-3-7-lines-w-explanation-and-examp-x3o2 | Here\'s the plan:\n- We iterate throughs, but for all0before the initial 1, we count nothing. It only becomes interesting once we hit a1. We could rephrase the | Spaulding_ | NORMAL | 2023-01-17T00:57:20.096096+00:00 | 2024-05-31T17:43:05.718139+00:00 | 6,697 | false | Here\'s the plan:\n- We iterate through`s`, but for all`0`before the initial `1`, we count nothing. It only becomes interesting once we hit a`1`. We could rephrase the problem to: "Discard all the initial`0`, then determine which requires fewer flips: changing the remaining string to all`1`, or flip the`1` and start ov... | 99 | 0 | ['Python3'] | 15 |
flip-string-to-monotone-increasing | Simple Explanation With Images || O(n) time, O(1) Space | simple-explanation-with-images-on-time-o-uxb3 | Intuition\nHow about we think about the cost and gain? As we go from left to right, the zeroes we get are our costs and the ones we get are our gains.\n Describ | hridoy100 | NORMAL | 2023-01-17T05:05:18.070117+00:00 | 2023-01-17T05:10:52.072277+00:00 | 4,347 | false | # Intuition\nHow about we think about the cost and gain? As we go from left to right, the zeroes we get are our costs and the ones we get are our gains.\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n\n it can retain its value.\n2) it can be flipped.\n\n- We can have all 0s in the s | shivaye | NORMAL | 2021-08-10T10:13:47.632136+00:00 | 2021-08-10T10:13:47.632172+00:00 | 2,676 | false | **Solution :**\n***\n```\n#observation:\nEvery character in the string has 2 choices,\n1) it can retain its value.\n2) it can be flipped.\n\n- We can have all 0s in the string,\n- We can have all 1s in the string,\n- We can have mixture of 0s and 1s but with the condition that no 0 should occur after 1\n -But the 1s... | 66 | 3 | [] | 4 |
flip-string-to-monotone-increasing | ✅✅C++ ||Most Easy Solution || 💯💯DP Approach || Heavily Commented | c-most-easy-solution-dp-approach-heavily-otki | \u2705\u2705C++ || Easy Solution || \uD83D\uDCAF\uD83D\uDCAFDP Approach || Heavily Commented\n# Please Upvote as it really motivates me\n\nclass Solution {\npub | Conquistador17 | NORMAL | 2023-01-17T04:01:19.234366+00:00 | 2023-01-17T04:01:19.234414+00:00 | 4,986 | false | # **\u2705\u2705C++ || Easy Solution || \uD83D\uDCAF\uD83D\uDCAFDP Approach || Heavily Commented**\n# **Please Upvote as it really motivates me**\n```\nclass Solution {\npublic:\n int func(int idx,string&s,char ch,vector<vector<int>>&dp){\n int n=s.size();\n\t\t//If our index idx is more than n then we will r... | 47 | 0 | ['Dynamic Programming', 'Memoization', 'C', 'C++'] | 7 |
flip-string-to-monotone-increasing | Java DP solution O(n) time O(1) space | java-dp-solution-on-time-o1-space-by-sel-1gyj | ``` / tranverse the string * use two variables to record the minimim number of flip need to make the substring from (0 - i) to be MonoIncr with end with 1 or 0 | self_learner | NORMAL | 2018-10-21T03:04:18.763665+00:00 | 2018-10-26T14:21:20.097984+00:00 | 4,580 | false | ```
/* tranverse the string
* use two variables to record the minimim number of flip need to make the substring from (0 - i) to be MonoIncr with end with 1 or 0;
* (1) for position i + 1, if preceding string is end with 1, the current string can only end with one,
* so cntEndWithOne can only be used to update the... | 47 | 2 | [] | 9 |
flip-string-to-monotone-increasing | Easy Python | Explained | O(n) time, O(1) space | easy-python-explained-on-time-o1-space-b-grvb | Easy Python | Explained | O(n) time, O(1) space\n\nThe Python code below presents a simple solution for this problem, where we track the cost of generating diff | aragorn_ | NORMAL | 2020-07-21T23:33:09.785947+00:00 | 2020-07-21T23:37:53.073099+00:00 | 3,142 | false | **Easy Python | Explained | O(n) time, O(1) space**\n\nThe Python code below presents a simple solution for this problem, where we track the cost of generating different sorted arrays with the form "[0] * a + [1] * b". Initially, we compute the cost of having an array of "ones" exclusively. Then, we start allowing the ... | 44 | 0 | ['Python', 'Python3'] | 7 |
flip-string-to-monotone-increasing | Java Solution with Explanation | java-solution-with-explanation-by-sartha-gmt0 | \n\n# Approach and Explanation\n Describe your approach to solving the problem. \n\n1. This is a solution in Java for the problem of finding the minimum number | Sarthak_Singh_ | NORMAL | 2023-01-17T03:15:09.299673+00:00 | 2023-01-17T03:18:22.141473+00:00 | 5,666 | false | \n\n# Approach and Explanation\n<!-- Describe your approach to solving the problem. -->\n\n1. This is a solution in Java for the problem of finding the minimum number of flips required to make a binary string monotonically increasing. A monotonically increasing string is one in which the characters are in increasing or... | 41 | 1 | ['Java'] | 4 |
flip-string-to-monotone-increasing | [Python] two O(n) solutions: dp and accumulate, explained | python-two-on-solutions-dp-and-accumulat-3idy | Solution 1 - dp\nThe first idea I thought about when I saw this problem is to use dynamic programming. Let us define by dp(i, k) the state where we are on index | dbabichev | NORMAL | 2021-08-10T08:30:08.904300+00:00 | 2021-08-10T18:14:54.995924+00:00 | 2,654 | false | #### Solution 1 - dp\nThe first idea I thought about when I saw this problem is to use dynamic programming. Let us define by `dp(i, k)` the state where we are on index `i` at the moment and we make last element equal to `k`. From what states we can get to this state? It is for sure `dp(i-1, j)` and the question is what... | 39 | 2 | ['Dynamic Programming'] | 4 |
flip-string-to-monotone-increasing | ⛱ Easy Javascript Solution with Detailed Explanation - o(n) time, o(1) space 😎 | easy-javascript-solution-with-detailed-e-6rh9 | Explanation\nWe want our string to be monotonic. \nStrings "0000", "00111" and "1111" are monotonic.\nWhat if string is already monotonic and we add one more ch | avakatushka | NORMAL | 2021-08-11T07:26:33.215367+00:00 | 2021-08-11T07:35:14.186967+00:00 | 1,718 | false | ## Explanation\nWe want our string to be monotonic. \nStrings "0000", "00111" and "1111" are monotonic.\nWhat if string is already monotonic and we add one more character?\n* If this character is \'1\', string will always remain monotonic. No problem here \n\uD83E\uDD17 "0000" + "1", "0000111" + "1" or "1111" + "1".\n*... | 37 | 0 | ['Dynamic Programming', 'JavaScript'] | 6 |
flip-string-to-monotone-increasing | [Python] DP is EASY - From O(N) to O(1) space - Clean & Concise | python-dp-is-easy-from-on-to-o1-space-cl-li02 | \u2714\uFE0F Solution 1: Top down DP\n- Supper easy, with each s[i], we have two choices:\n\t- Don\'t flip s[i].\n\t- Flip s[i]\n- Choose the choice which has t | hiepit | NORMAL | 2021-08-10T07:09:38.910467+00:00 | 2021-08-11T04:48:31.689323+00:00 | 2,466 | false | **\u2714\uFE0F Solution 1: Top down DP**\n- Supper easy, with each `s[i]`, we have two choices:\n\t- Don\'t flip `s[i]`.\n\t- Flip `s[i]`\n- Choose the choice which has the minimum number of flips.\n- Let `dp(i, last)` is the minimum number of flips to make `s[i..n-1]` monotone increasing, where `s[i]` must be >= `last... | 37 | 4 | [] | 7 |
flip-string-to-monotone-increasing | Python/JS/Java/Go/C++ O(n) by pattern analysis [w/ Hint] | pythonjsjavagoc-on-by-pattern-analysis-w-eavk | Observe and analyze two possible cases on binary string pattern\n\nLet S start on empty string "", and add either \'0\' or \'1\' on the tail to make it longer.\ | brianchiang_tw | NORMAL | 2021-08-10T09:06:47.681480+00:00 | 2022-10-22T10:48:19.291772+00:00 | 2,491 | false | Observe and analyze two possible cases on binary string pattern\n\nLet S start on empty string "", and add either \'0\' or \'1\' on the tail to make it longer.\n\n---\nCase #1\nS + \'**1**\' : No need to flip, because \'**1**\' **keeps monotone increasing** always.\n\n---\n\n---\nCase #2\nS + \'**0**\' : Need **flip op... | 33 | 0 | ['Math', 'Dynamic Programming', 'C', 'Python', 'C++', 'Java', 'Go', 'Python3', 'JavaScript'] | 4 |
flip-string-to-monotone-increasing | Count prefix ones and suffix zeros | count-prefix-ones-and-suffix-zeros-by-wa-1owz | leftOne[i] represents count of 1s in S[0...i]\nrightZero[i] represents count of 0s in S[i...n-1]\nThen min = min(leftOne[i] + rightZero[i + 1]), where i = 0, .. | wangzi6147 | NORMAL | 2018-10-21T03:07:10.851135+00:00 | 2018-10-23T02:49:02.594272+00:00 | 2,794 | false | `leftOne[i]` represents count of `1`s in `S[0...i]`\n`rightZero[i]` represents count of `0`s in `S[i...n-1]`\nThen `min = min(leftOne[i] + rightZero[i + 1])`, where `i = 0, ... n-1`\n```\nclass Solution {\n public int minFlipsMonoIncr(String S) {\n int n = S.length();\n int[] leftOne = new int[n];\n ... | 31 | 0 | [] | 8 |
flip-string-to-monotone-increasing | 🔥 [LeetCode The Hard Way] 🔥 Explained Line By Line | leetcode-the-hard-way-explained-line-by-slkps | \uD83D\uDD34 Check out LeetCode The Hard Way for more solution explanations and tutorials. \n\uD83D\uDFE0 Check out our Discord for live discussion.\n\uD83D\uDF | __wkw__ | NORMAL | 2023-01-17T02:12:31.445157+00:00 | 2023-01-17T14:44:12.297524+00:00 | 2,055 | false | \uD83D\uDD34 Check out [LeetCode The Hard Way](https://wingkwong.github.io/leetcode-the-hard-way/) for more solution explanations and tutorials. \n\uD83D\uDFE0 Check out our [Discord](https://discord.gg/Nqm4jJcyBf) for live discussion.\n\uD83D\uDFE2 Give a star on [Github Repository](https://github.com/wingkwong/leetco... | 29 | 0 | ['Python', 'C++', 'Java', 'Rust'] | 3 |
flip-string-to-monotone-increasing | C++ Bruteforce O(n2) -> Optimal (Prefix - Suffix) Easy to understand | c-bruteforce-on2-optimal-prefix-suffix-e-o2ns | Intuition:\n\n\tAll characters in left side of the string should be 0\'s\n \tAll characters in right side of the string should be 1\'s\n\t\n\twe cut string at | madhu20 | NORMAL | 2021-08-11T09:47:59.116531+00:00 | 2021-08-11T09:56:40.085752+00:00 | 1,132 | false | **Intuition:**\n\n\tAll characters in left side of the string should be 0\'s\n \tAll characters in right side of the string should be 1\'s\n\t\n\twe cut string at each index\n\t\tbefore cut all characters should be 0\'s ( so left side we convert 1\'s to 0\'s if present)\n\t\tafter cut all characters should be 1\'s (so... | 26 | 0 | ['C', 'Prefix Sum', 'C++'] | 3 |
flip-string-to-monotone-increasing | Java DP using O(N) time and O(1) space | java-dp-using-on-time-and-o1-space-by-ev-9vw0 | Idea is to keep two variables storing min flips to be done for current digit to be one or to be zero.\n\n\nclass Solution {\n public int minFlipsMonoIncr(Str | evilnearby | NORMAL | 2018-10-21T03:05:13.047130+00:00 | 2018-10-23T02:38:12.284957+00:00 | 2,069 | false | Idea is to keep two variables storing min flips to be done for current digit to be one or to be zero.\n\n```\nclass Solution {\n public int minFlipsMonoIncr(String S) {\n char[] arr = S.toCharArray();\n int one = 0, zero = 0;\n \n for (int i = 0; i < arr.length; i++) {\n if (ar... | 23 | 0 | [] | 5 |
flip-string-to-monotone-increasing | C++ prefix sum ✅ | c-prefix-sum-by-deleted_user-9gxh | Intuition\n Describe your first thoughts on how to solve this problem. \nwe are using prefix sum method to solve the problme\n\n# Approach\n Describe your appro | deleted_user | NORMAL | 2023-01-17T00:29:17.356126+00:00 | 2023-01-17T00:29:17.356173+00:00 | 2,651 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nwe are using prefix sum method to solve the problme\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\nSkip the loop until we find the first \'1\' \nafter finding first occurrence of \'1\' we have to keep a count of n... | 22 | 4 | ['C++'] | 7 |
flip-string-to-monotone-increasing | An Intuitive & Visual Explanation in Detail | an-intuitive-visual-explanation-in-detai-3c05 | THE CORE IDEA\nThe idea is simple. 0...01...1 kind of sequence is beautiful. A bunch of zeros, followed by a bunch of ones. Given a random string, we determine | chaudhary1337 | NORMAL | 2021-08-10T09:17:41.451361+00:00 | 2021-08-10T09:41:45.197328+00:00 | 775 | false | **THE CORE IDEA**\nThe idea is simple. `0...01...1` kind of sequence is beautiful. A bunch of zeros, followed by a bunch of ones. Given a random string, we determine the **breaking point** of that string, to the left will then be *only* 0s and to the right be *only* 1s. \n\nExample: A string like `00110` has breaking p... | 19 | 1 | [] | 2 |
flip-string-to-monotone-increasing | Java Solutions with Full Explanations (B-F, DP x2, Constant Time & Space) | java-solutions-with-full-explanations-b-95nk8 | Reference: LeetCode\nDifficulty: Medium\n\n## Problem\n\n> A string of \'0\'s and \'1\'s is monotone increasing if it consists of some number of \'0\'s (possibl | junhaowanggg | NORMAL | 2019-10-16T06:27:05.676305+00:00 | 2019-10-16T06:27:05.676340+00:00 | 1,385 | false | Reference: [LeetCode](https://leetcode.com/problems/flip-string-to-monotone-increasing)\nDifficulty: <span class="orange">Medium</span>\n\n## Problem\n\n> A string of `\'0\'`s and `\'1\'`s is monotone increasing if it consists of some number of `\'0\'`s (possibly 0), followed by some number of `\'1\'`s (also possibly 0... | 18 | 0 | ['Dynamic Programming', 'Java'] | 4 |
flip-string-to-monotone-increasing | Java DP solution | java-dp-solution-by-lf963-yzts | DP Solution\n\nMantain two variables flipToZero and flipToOne\nflipToZero: how many flips we need so that we can flip current digit to zero.\nflipToOne: how man | lf963 | NORMAL | 2018-10-28T06:35:10.873860+00:00 | 2018-10-28T06:35:10.873904+00:00 | 922 | false | DP Solution\n\nMantain two variables ```flipToZero``` and ```flipToOne```\n```flipToZero```: how many flips we need so that we can flip current digit to zero.\n```flipToOne```: how many flips we need so that we can flip current digit to one.\n\nDP recursion:\n```\nif current digit is zero\n\tflipToZero[i] = 0 + flipToZ... | 18 | 0 | [] | 1 |
flip-string-to-monotone-increasing | C++ || Explanations and Comments || 4 Solutions || Greedy || DP | c-explanations-and-comments-4-solutions-p5f6t | Intuition\nWe want to iterate over the string and for each index calc the best answer on it\n\n# Approach\nTraverse on the string and for each index treat it l | 7oskaa | NORMAL | 2023-01-17T11:02:14.007315+00:00 | 2023-01-17T12:13:31.078657+00:00 | 905 | false | # Intuition\nWe want to iterate over the string and for each index calc the best answer on it\n\n# Approach\nTraverse on the string and for each index treat it like all previous number are `0`\'s and all next numbers are `1`\'s\n\n# Complexity\n- Time complexity:\n$O(n)$ - where $n$ is the number of characters\n\n- Sp... | 17 | 0 | ['Dynamic Programming', 'Greedy', 'Prefix Sum', 'C++'] | 2 |
flip-string-to-monotone-increasing | Python O(N)/O(1), One Pass, Very Simple | python-ono1-one-pass-very-simple-by-dudu-lv75 | This question should be quite straightforward and no need to over think.\nWe check the string from left or right, and keep the substring prior to current positi | dudupipi | NORMAL | 2022-03-04T20:57:07.755691+00:00 | 2022-03-13T06:06:55.162205+00:00 | 690 | false | This question should be quite straightforward and no need to over think.\nWe check the string from left or right, and keep the substring prior to current position monotone increasing.\n\nIf we encounter a 1, do nothing since appending this won\'t break the monotonicity.\nIf we encounter a 0, then two options to keep su... | 17 | 0 | ['Dynamic Programming', 'Python'] | 3 |
flip-string-to-monotone-increasing | Python 3-liner | python-3-liner-by-cenkay-bi2u | We start with assuming "111.." section occupies all string, s.\n Then we update "000.." section as s[:i + 1] and "111.." section as s[i + 1:] during iteration a | cenkay | NORMAL | 2018-10-21T11:00:15.986407+00:00 | 2018-10-21T16:18:26.691852+00:00 | 1,214 | false | * We start with assuming "111.." section occupies all string, s.\n* Then we update "000.." section as s[:i + 1] and "111.." section as s[i + 1:] during iteration as well as the result\n* "zeros" variable counts all misplaced "0"s and "ones" variable counts all misplaced "1"s\n```\nclass Solution:\n def minFlipsMonoI... | 17 | 1 | [] | 1 |
flip-string-to-monotone-increasing | [C++/Java/Python] Real-World Problem Explained ✅ | cjavapython-real-world-problem-explained-nwxy | Real-World Problem:\nDo you know why these problem is quite an easy yet and important problem? Here\'s what i know:\n\nThis problem is inspired by a Real-World | prantik0128 | NORMAL | 2023-01-17T05:10:24.990762+00:00 | 2023-01-17T05:55:35.987187+00:00 | 1,079 | false | **Real-World Problem:**\nDo you know why these problem is quite an easy yet and important problem? Here\'s what i know:\n\nThis problem is inspired by a **Real-World Scenario** of:\n\n\tOptimizing Data Storage & Retrieval In a Computer\'s Memory.\n\nIn this scenario, we might have a` large amount of data` that needs to... | 15 | 0 | ['C', 'Python', 'Java'] | 1 |
flip-string-to-monotone-increasing | 🚀Easy Explanation🚀|| C++ || Python || O(N)✅ | easy-explanation-c-python-on-by-naman_ag-hhok | Consider\uD83D\uDC4D\uD83D\uDC4D\n\n Please Upvote If You Find It Helpful.\n\n# Intuition:\nCounting the no. of 1\'s to flip on left side | naman_ag | NORMAL | 2023-01-17T02:18:47.853771+00:00 | 2023-01-17T05:36:14.316505+00:00 | 1,004 | false | # Consider\uD83D\uDC4D\uD83D\uDC4D\n```\n Please Upvote If You Find It Helpful.\n```\n# Intuition:\nCounting the no. of 1\'s to flip on left side + no. of 0\'s to flip on right side of each index i and then select minimum flips from it.\n\n# Approach:\nStep 1. For each index i, store sum of (no. ... | 15 | 0 | ['String', 'C++', 'Python3'] | 2 |
flip-string-to-monotone-increasing | ✅Flip String to Monotone Increasing || Basic Implementation w/o DP || C++ | Python | Java | flip-string-to-monotone-increasing-basic-qv14 | ALGORITHM:\n\n Set n - length of the string , flipCount = 0 and oneCount = 0\n For i in range 0 to n \u2013 1\n\t If S[i] is 0, then\n\t\t If oneCount = 0, then | Maango16 | NORMAL | 2021-08-10T07:17:14.182927+00:00 | 2021-08-10T07:25:29.986939+00:00 | 722 | false | **ALGORITHM:**\n\n* Set `n - length of the string` , `flipCount = 0` and `oneCount = 0`\n* For i in range `0 to n \u2013 1`\n\t* If `S[i] is 0`, then\n\t\t* If `oneCount = 0`, then skip to the next iteration\n\t\t* Increase the flipCount by 1\n\t* Otherwise `increase oneCount` by 1\n\t* If `oneCount < flipCount`, then ... | 15 | 5 | [] | 5 |
flip-string-to-monotone-increasing | Python greedy solution, count 1s and 0s solution, 1 pass | python-greedy-solution-count-1s-and-0s-s-0k03 | When there are equal 0s following 1s, we flip all the 1s.\npython\nones = zeros = 0\nresult = 0\n\nfor c in s:\n\tif c == "1":\n\t\tones += 1\n\telse:\n\t\n\t\t | FACEPLANT | NORMAL | 2021-08-10T16:07:14.713853+00:00 | 2021-08-10T16:11:53.605299+00:00 | 372 | false | When there are equal 0s following 1s, we flip all the 1s.\n```python\nones = zeros = 0\nresult = 0\n\nfor c in s:\n\tif c == "1":\n\t\tones += 1\n\telse:\n\t\n\t\t# We start to count 0s after 1s\n\t\tif ones:\n\t\t\tzeros += 1\n\t\t\t\n\t\t\t# when 0s are equal to 1s, we flip all 1s, and reset counters\n\t\t\tif zeros ... | 12 | 0 | [] | 1 |
flip-string-to-monotone-increasing | Clean Codes🔥|| Dynamic Programming ✅|| C++|| Java || Python3 | clean-codes-dynamic-programming-c-java-p-rm7c | Request \uD83D\uDE0A :\n- If you find this solution easy to understand and helpful, then Please Upvote\uD83D\uDC4D\uD83D\uDC4D\n# Code [C++| Java| Python3] :\nC | N7_BLACKHAT | NORMAL | 2023-01-17T04:58:17.294179+00:00 | 2023-01-17T05:05:43.927764+00:00 | 1,682 | false | # Request \uD83D\uDE0A :\n- If you find this solution easy to understand and helpful, then Please Upvote\uD83D\uDC4D\uD83D\uDC4D\n# Code [C++| Java| Python3] :\n```C++ []\nclass Solution {\n public:\n int minFlipsMonoIncr(string S) {\n vector<int> dp(2);\n\n for (int i = 0; i < S.length(); ++i) {\n int temp... | 10 | 0 | ['String', 'Dynamic Programming', 'C++', 'Java', 'Python3'] | 0 |
flip-string-to-monotone-increasing | Easy Java Solution | easy-java-solution-by-amanchandna-80h0 | Code\n\nclass Solution {\n public int minFlipsMonoIncr(String s) {\n //intuition : reach first one\n //after reaching first one I can count the | amanchandna | NORMAL | 2023-01-17T01:45:24.511908+00:00 | 2023-01-17T01:45:24.511950+00:00 | 2,750 | false | # Code\n```\nclass Solution {\n public int minFlipsMonoIncr(String s) {\n //intuition : reach first one\n //after reaching first one I can count the number of zeros : those zeros are having possibility to get flipped\n //if count of 0s > count of 1s -> will flip ones\n\n int n = s.length(... | 10 | 0 | ['Prefix Sum', 'Java'] | 2 |
flip-string-to-monotone-increasing | Recursive DP Solution Easy to Understand with Detail Explanation | recursive-dp-solution-easy-to-understand-qysl | In the problem, you have to create a monotonic increasing sequence i.e a sequence of 0\'s followed by 1\'s or a sequence of all 0\'s or perhaps a sequence of al | Arnav_Tiwari | NORMAL | 2021-08-10T19:05:44.733626+00:00 | 2021-08-10T19:09:54.629347+00:00 | 816 | false | In the problem, you have to create a monotonic increasing sequence i.e a sequence of 0\'s followed by 1\'s or a sequence of all 0\'s or perhaps a sequence of all 1\'s whichever might cost the least number of flips.\nThe problem can be solved using the following idea:\n**1. 0 can be followed by either 1 or 0.**\n**2. 1 ... | 10 | 0 | ['String', 'Dynamic Programming', 'Recursion', 'Memoization'] | 2 |
flip-string-to-monotone-increasing | Prefix-Suffix Python O(n) time, Very Easy Explanation. | prefix-suffix-python-on-time-very-easy-e-oq7w | For Monotone Increasing String, there must be an index where all the values on the left hand side are "0" and all the values on the right hand side are "1".\n S | raja_nikshith_katta | NORMAL | 2021-08-10T08:13:00.875483+00:00 | 2021-08-10T08:13:00.875532+00:00 | 394 | false | * For Monotone Increasing String, there must be an index where all the values on the left hand side are "0" and all the values on the right hand side are "1".\n* So now to convert the input string into Monotone Increasing String, We need to choose an index and flip all "1"s to "0" on the left and flip all "0"s to "1".\... | 10 | 0 | ['Suffix Array', 'Python'] | 0 |
flip-string-to-monotone-increasing | ✅Easy C++ solution || ✅Short & Simple || ✅Best Method || ✅Easy-To-Understand | easy-c-solution-short-simple-best-method-rtkx | 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# ComplexityBold\n- T | akajayraj001 | NORMAL | 2023-01-17T04:45:52.526011+00:00 | 2023-01-17T04:45:52.526052+00:00 | 1,508 | 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**Bold**\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... | 9 | 1 | ['C++'] | 6 |
flip-string-to-monotone-increasing | ✅ C++ || Easy Detailed Solution || Beginner friendly || O(n), O(1) | c-easy-detailed-solution-beginner-friend-8mxz | Approach\n##### My entire approach is based on the idea that all the ones to the left (including pointer) of the pointer and all the zeros to the right of the p | underground_coder | NORMAL | 2023-01-17T04:05:53.246395+00:00 | 2023-01-17T04:05:53.246428+00:00 | 1,033 | false | # Approach\n##### My entire approach is based on the idea that all the `ones` to the left (including pointer) of the pointer and all the `zeros` to the right of the pointer must be flipped.\n- Initially I\'ve stored all the zeros of string in `totalZero` variable to initialize my `rightZero` and `ans` variables.\n- As ... | 9 | 0 | ['C++'] | 2 |
flip-string-to-monotone-increasing | One-pass intuitive python solution | one-pass-intuitive-python-solution-by-yi-akas | Very similar to @LiamHuang\'s solution. For case we encounter \'1\', we actually do no action. I wanted to make it more clear:\n\n\nclass Solution:\n def min | yilmazali | NORMAL | 2022-01-05T09:34:01.412827+00:00 | 2022-01-05T09:34:01.412860+00:00 | 607 | false | Very similar to @LiamHuang\'s solution. For case we encounter \'1\', we actually do no action. I wanted to make it more clear:\n\n```\nclass Solution:\n def minFlipsMonoIncr(self, s: str) -> int:\n \n n = len(s)\n min_flip, one_count = 0, 0\n \n for i in range(n):\n char... | 8 | 0 | ['Python'] | 1 |
flip-string-to-monotone-increasing | Python One Liner (Actually good, unexplained) | python-one-liner-actually-good-unexplain-c2y7 | \nclass Solution:\n def minFlipsMonoIncr(self, s: str) -> int:\n return min(0,*accumulate([[-1,1][int(c)]for c in s]))+s.count(\'0\')\n | likey_00 | NORMAL | 2023-01-17T17:59:59.506718+00:00 | 2023-01-17T17:59:59.506761+00:00 | 1,071 | false | ```\nclass Solution:\n def minFlipsMonoIncr(self, s: str) -> int:\n return min(0,*accumulate([[-1,1][int(c)]for c in s]))+s.count(\'0\')\n``` | 7 | 0 | ['Python3'] | 0 |
flip-string-to-monotone-increasing | Picture Explanation || Java Recursion, and Memo and another fast Solution | picture-explanation-java-recursion-and-m-bl3g | \n\nRecursive Solution: (Gives TLE)\nbut forms the base for using dp\n\n\nclass Solution { \n private int solve(String s, int index, int prev){\n if(i | as-15 | NORMAL | 2023-01-17T11:22:56.609462+00:00 | 2023-01-17T11:42:52.102582+00:00 | 335 | false | \n\n**Recursive Solution: (Gives TLE)**\nbut forms the base for using dp\n\n```\nclass Solution { \n private int solve(String s, int index, int prev){\n if(index >= s.length())\n return 0;... | 7 | 0 | ['Dynamic Programming', 'Recursion', 'Memoization', 'Java'] | 1 |
flip-string-to-monotone-increasing | JAVA | Easy solution | Explained ✅ | java-easy-solution-explained-by-sourin_b-om2w | Please Upvote :D\n\njava []\nclass Solution {\n public int minFlipsMonoIncr(String s) {\n // variable \'flips\' will store the number of flips we will | sourin_bruh | NORMAL | 2023-01-17T09:07:14.152626+00:00 | 2023-01-17T13:04:14.543766+00:00 | 242 | false | # Please Upvote :D\n\n``` java []\nclass Solution {\n public int minFlipsMonoIncr(String s) {\n // variable \'flips\' will store the number of flips we will perform\n // variable \'ones\' will store the number of 1s\n // we initially want to flip 0s to 1s\n int flips = 0, ones = 0;\n ... | 7 | 0 | ['String', 'Java'] | 0 |
flip-string-to-monotone-increasing | ✅C++| Simple O(n)|O(1) | 5 lines clear soln '✔✔ | c-simple-ono1-5-lines-clear-soln-by-xaho-bqnk | Intuition\n Describe your first thoughts on how to solve this problem. \nJust check if flipping into one\'s has minimum answer or flipping into zeroes and we al | Xahoor72 | NORMAL | 2023-01-17T06:28:53.410280+00:00 | 2023-01-17T14:04:26.227407+00:00 | 686 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nJust check if flipping into one\'s has minimum answer or flipping into zeroes and we also want first 0\'s then 1\'s ,so we can check like this :\n- If we got a 1 just do nothing only store its occurence i.e do countOne++.\n- If we got a 0... | 7 | 0 | ['C++'] | 3 |
flip-string-to-monotone-increasing | Go Greedy !! | go-greedy-by-yadivyanshu-033w | Approach\n Describe your approach to solving the problem. \nAt any \'i\', if count(1) < count(0), then increase your count(ans) by count(1) as at instance \'i\' | yadivyanshu | NORMAL | 2023-01-17T05:35:49.113473+00:00 | 2023-01-17T05:35:49.113524+00:00 | 65 | false | # Approach\n<!-- Describe your approach to solving the problem. -->\nAt any \'i\', if count(1) < count(0), then increase your count(ans) by count(1) as at instance \'i\' we are making all previous elements 0 and reinitialize the count variables.\n\nAnd at the end we are taking minimum of count(0) and count(1) to flip a... | 7 | 0 | ['C++'] | 0 |
flip-string-to-monotone-increasing | Easiest Recursive Solution C++ | easiest-recursive-solution-c-by-asmitpap-x1qm | \nclass Solution {\npublic:\n \n int t[20005][2] ;\n \n int Solve(string &s , int i , bool ones)\n {\n if(i==s.size())\n return | asmitpapney | NORMAL | 2021-07-15T00:59:08.709764+00:00 | 2021-07-15T00:59:08.709790+00:00 | 281 | false | ```\nclass Solution {\npublic:\n \n int t[20005][2] ;\n \n int Solve(string &s , int i , bool ones)\n {\n if(i==s.size())\n return 0 ;\n \n if(t[i][ones]!=-1) \n return t[i][ones] ;\n \n if(s[i]==\'0\') // check if there is one before , if yes set... | 7 | 0 | [] | 1 |
flip-string-to-monotone-increasing | Java clean solution with explanation | java-clean-solution-with-explanation-by-9fg3x | \nclass Solution {\n public int minFlipsMonoIncr(String s) {\n int flipCount = 0, oneCount = 0;\n for (int i = 0; i < s.length(); i++) {\n | akiramonster | NORMAL | 2020-01-28T16:55:35.298694+00:00 | 2020-01-28T16:55:35.298728+00:00 | 342 | false | ```\nclass Solution {\n public int minFlipsMonoIncr(String s) {\n int flipCount = 0, oneCount = 0;\n for (int i = 0; i < s.length(); i++) {\n char c = s.charAt(i);\n if (c == \'0\') {\n if (oneCount == 0) continue; // if all beginning char is zero, no need to flip\n... | 7 | 0 | [] | 0 |
flip-string-to-monotone-increasing | Java DP easy to understand | java-dp-easy-to-understand-by-grhb-56i3 | \npublic int minFlipsMonoIncr(String S) {\n int l = S.length();\n int[] f = new int[l];// f[i] means flips need for substring S[0..i]\n f[0 | grhb | NORMAL | 2019-02-20T03:48:26.222687+00:00 | 2019-02-20T03:48:26.222727+00:00 | 431 | false | ```\npublic int minFlipsMonoIncr(String S) {\n int l = S.length();\n int[] f = new int[l];// f[i] means flips need for substring S[0..i]\n f[0] = 0;//no matter \'0\' or \'1\', no need to flip\n int numOne = S.charAt(0) == \'1\' ? 1 : 0;\n for(int i = 1; i<l; i++){\n if(S.ch... | 7 | 0 | [] | 1 |
flip-string-to-monotone-increasing | Python 148ms solution beaten by 92% - three passes, but DEAD SIMPLE | python-148ms-solution-beaten-by-92-three-dwia | For each i, find # of 1s to its left. (first pass) For each i, find # of 0s to its right. (second pass) For each i, sum of 1s to its left and 0s to its right is | lawrence7 | NORMAL | 2018-10-24T07:29:54.777596+00:00 | 2018-10-24T14:50:04.555660+00:00 | 493 | false | For each i, find # of 1s to its left. (first pass)
For each i, find # of 0s to its right. (second pass)
For each i, sum of 1s to its left and 0s to its right is the # of flips required. A third pass to find minimum.
```
class Solution:
def minFlipsMonoIncr(self, S):
"""
:type S: str
:rtyp... | 7 | 0 | [] | 2 |
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