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minimum-operations-to-make-the-array-increasing | Type Script Solution | type-script-solution-by-diyoufkv7-e11x | Intuition\nMathematic Expression Approch\n\n# Approach\n- Finding the Difference between the Elements that is smaller than the\nnext element\n\n# Complexity\n- | diyoufkv7 | NORMAL | 2023-11-03T04:14:52.942209+00:00 | 2023-11-03T04:14:52.942240+00:00 | 26 | false | # Intuition\nMathematic Expression Approch\n\n# Approach\n- Finding the Difference between the Elements that is smaller than the\nnext element\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. $$O(n)$$ -->\n\n# Co... | 1 | 0 | ['TypeScript'] | 0 |
minimum-operations-to-make-the-array-increasing | Do this sample code. | do-this-sample-code-by-0varun-5xx4 | 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 | 0varun | NORMAL | 2023-11-01T04:42:35.795915+00:00 | 2023-11-01T04:42:35.795945+00:00 | 152 | 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 | ['JavaScript'] | 2 |
minimum-operations-to-make-the-array-increasing | Simple Solution | simple-solution-by-adwxith-9yfl | 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 | adwxith | NORMAL | 2023-11-01T04:40:55.098722+00:00 | 2023-11-01T04:40:55.098761+00:00 | 257 | 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', 'JavaScript'] | 0 |
minimum-operations-to-make-the-array-increasing | Best Java Solution || Beats 100% | best-java-solution-beats-100-by-ravikuma-z18i | 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 | ravikumar50 | NORMAL | 2023-10-09T17:09:06.853331+00:00 | 2023-10-09T17:09:06.853349+00:00 | 987 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ --... | 1 | 0 | ['Java'] | 0 |
minimum-operations-to-make-the-array-increasing | Solution which beats 96% in C# | solution-which-beats-96-in-c-by-akhundov-l683 | \npublic class Solution {\n public int MinOperations(int[] nums) {\n int currElement = nums[0], count = 0;\n for (int i = 1; i < nums.Len | akhundovaykhan | NORMAL | 2023-09-08T13:24:47.441777+00:00 | 2023-09-08T13:24:47.441842+00:00 | 164 | false | ```\npublic class Solution {\n public int MinOperations(int[] nums) {\n int currElement = nums[0], count = 0;\n for (int i = 1; i < nums.Length; i++)\n {\n if (currElement < nums[i])\n {\n currElement = nums[i];\n }\n ... | 1 | 0 | ['C#'] | 0 |
minimum-operations-to-make-the-array-increasing | Minimum Operations to Make the Array Increasing 🧑💻🧑💻 || JAVA solution code 💁💁... | minimum-operations-to-make-the-array-inc-kzfq | Code\n\nclass Solution {\n public int minOperations(int[] nums) {\n int count = 0;\n\n for(int i=1;i<nums.length;i++){\n\n if(nums[i | Jayakumar__S | NORMAL | 2023-08-04T16:45:22.346913+00:00 | 2023-08-04T16:45:22.346942+00:00 | 526 | false | # Code\n```\nclass Solution {\n public int minOperations(int[] nums) {\n int count = 0;\n\n for(int i=1;i<nums.length;i++){\n\n if(nums[i]<=nums[i-1]){\n count += Math.abs(nums[i]-nums[i-1])+1;\n nums[i] = nums[i-1]+1;\n\n }\n }\n return count;\... | 1 | 0 | ['Java'] | 0 |
minimum-operations-to-make-the-array-increasing | Minimum Operations to Make the Array Increasing Solution in C++ | minimum-operations-to-make-the-array-inc-49sa | 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 | The_Kunal_Singh | NORMAL | 2023-03-04T05:55:20.014738+00:00 | 2023-03-04T05:55:20.014777+00:00 | 35 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\nO(n)\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$... | 1 | 0 | ['C++'] | 0 |
minimum-operations-to-make-the-array-increasing | Easy c++ Solution for Beginners | easy-c-solution-for-beginners-by-sunnyya-mzi0 | Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time | sunnyyad2002 | NORMAL | 2023-02-20T06:05:52.074071+00:00 | 2023-02-20T06:05:52.074105+00:00 | 106 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ --... | 1 | 0 | ['C++'] | 0 |
minimum-operations-to-make-the-array-increasing | c++ || using loop || easy approach | c-using-loop-easy-approach-by-jauliadhik-j6yf | \n# Complexity\n- Time complexity:\nO(N)\n\n- Space complexity:\nO(1)\n\n# Code\n\nclass Solution {\npublic:\n int minOperations(vector<int>& nums) \n {\n | jauliadhikari2012 | NORMAL | 2023-01-17T03:42:46.491597+00:00 | 2023-01-17T03:42:46.491644+00:00 | 796 | false | \n# Complexity\n- Time complexity:\nO(N)\n\n- Space complexity:\nO(1)\n\n# Code\n```\nclass Solution {\npublic:\n int minOperations(vector<int>& nums) \n {\n int count =0;\n for(int i=0;i<nums.size()-1;i++)\n {\n if(nums[i] < nums[i+1])\n {\n continue;\n ... | 1 | 0 | ['C++'] | 1 |
minimum-operations-to-make-the-array-increasing | C++ brute force | c-brute-force-by-mishalalshahari-dybu | \nint minOperations(vector<int>& nums) {\n int count=0;\n for(int i=0;i<nums.size()-1;i++){\n if(nums[i]<nums[i+1]){\n c | mishalalshahari | NORMAL | 2022-10-31T17:45:54.813800+00:00 | 2022-10-31T17:45:54.813858+00:00 | 16 | false | ```\nint minOperations(vector<int>& nums) {\n int count=0;\n for(int i=0;i<nums.size()-1;i++){\n if(nums[i]<nums[i+1]){\n continue;\n }\n else{\n count+=nums[i]+1-nums[i+1];\n nums[i+1]=nums[i]+1;\n }\n }\n ... | 1 | 0 | ['C++'] | 0 |
minimum-operations-to-make-the-array-increasing | 80% SPACE AND TIME BEATS || C++ || EASY || SIMPLE | 80-space-and-time-beats-c-easy-simple-by-b74t | \nclass Solution {\npublic:\n int minOperations(vector<int>& nums) \n {\n int moves=0;\n //first we will sort the array \n //reazon-->we | akshat0610 | NORMAL | 2022-09-26T14:02:25.087213+00:00 | 2022-09-26T14:02:25.087253+00:00 | 535 | false | ```\nclass Solution {\npublic:\n int minOperations(vector<int>& nums) \n {\n int moves=0;\n //first we will sort the array \n //reazon-->we will convert the array into increasign array\n //how->we will increment the curr element and make it bigger than its previous\n\n //fist 0th ind... | 1 | 0 | ['C', 'Sorting', 'C++'] | 0 |
minimum-operations-to-make-the-array-increasing | 80% SPACE AND TIME BEATS || C++ || EASY || SIMPLE | 80-space-and-time-beats-c-easy-simple-by-3yiv | \nclass Solution {\npublic:\n int minOperations(vector<int>& nums) {\n int ans = 0;\n for(int i = 1; i < nums.size(); i++){\n if(num | abhay_12345 | NORMAL | 2022-09-23T19:29:30.099954+00:00 | 2022-09-23T19:29:30.099994+00:00 | 329 | false | ```\nclass Solution {\npublic:\n int minOperations(vector<int>& nums) {\n int ans = 0;\n for(int i = 1; i < nums.size(); i++){\n if(nums[i]==nums[i-1]){\n ans++;\n nums[i]++;\n }\n else if(nums[i]-nums[i-1]<0){\n ans += -nums... | 1 | 0 | ['Greedy', 'C', 'C++'] | 0 |
minimum-operations-to-make-the-array-increasing | Java Solution || 100% Optimized || 100% faster || O(1) Space || O(N) time | java-solution-100-optimized-100-faster-o-dh1t | class Solution {\n public int minOperations(int[] nums) {\n \n int count =0; // Operations Count\n int nextVal = nums[0]+1; // We ch | avadarshverma737 | NORMAL | 2022-09-13T14:01:39.722026+00:00 | 2022-09-13T14:02:14.804148+00:00 | 289 | false | class Solution {\n public int minOperations(int[] nums) {\n \n int count =0; // Operations Count\n int nextVal = nums[0]+1; // We choose the next value to be our successive element\n for(int i=1;i<nums.length;i++){\n \n if(nums[i]<=nums[i-1]){\n co... | 1 | 0 | ['Java'] | 0 |
minimum-operations-to-make-the-array-increasing | C++ EASY SOLUTION | c-easy-solution-by-abhay18e-ffc2 | Would love your feedback\n\n int minOperations(vector<int>& nums) {\n int operationCount = 0;\n \n for(int i=1; i<nums.size(); i++){\n | abhay18e | NORMAL | 2022-08-14T12:01:23.125001+00:00 | 2022-08-14T12:01:23.125045+00:00 | 169 | false | *Would love your feedback*\n```\n int minOperations(vector<int>& nums) {\n int operationCount = 0;\n \n for(int i=1; i<nums.size(); i++){\n int diff = nums[i]-nums[i-1];\n \n if(diff<=0){\n nums[i] += abs(diff) +1;\n operation... | 1 | 0 | [] | 1 |
minimum-operations-to-make-the-array-increasing | Java Easiest Solution | java-easiest-solution-by-aditya001-l8y4 | ```\nclass Solution {\n public int minOperations(int[] nums) {\n int ans = 0;\n for(int i=0;i<nums.length-1;i++){\n if(nums[i+1]<=nu | Aditya001 | NORMAL | 2022-08-05T04:52:33.212971+00:00 | 2022-08-05T04:52:33.213020+00:00 | 306 | false | ```\nclass Solution {\n public int minOperations(int[] nums) {\n int ans = 0;\n for(int i=0;i<nums.length-1;i++){\n if(nums[i+1]<=nums[i]){\n ans += nums[i]-nums[i+1]+1;\n nums[i+1] = nums[i]+1;\n }\n }\n return ans;\n }\n} | 1 | 0 | ['Java'] | 0 |
minimum-operations-to-make-the-array-increasing | C++ | brute force solution | c-brute-force-solution-by-shobhit2205-dfg8 | \nclass Solution {\npublic:\n int minOperations(vector<int>& nums) {\n int count = 0;\n for(int i = 1; i < nums.size(); i++){\n if(n | Shobhit2205 | NORMAL | 2022-08-03T15:15:10.878738+00:00 | 2022-08-03T15:15:10.878806+00:00 | 186 | false | ```\nclass Solution {\npublic:\n int minOperations(vector<int>& nums) {\n int count = 0;\n for(int i = 1; i < nums.size(); i++){\n if(nums[i] <= nums[i - 1]) {\n count += (nums[i-1] - nums[i] + 1);\n nums[i] = nums[i-1] + 1;\n }\n }\n re... | 1 | 0 | ['C'] | 0 |
minimum-operations-to-make-the-array-increasing | Java Easy Solution 2ms | java-easy-solution-2ms-by-ramsingh27-waia | class Solution {\n public int minOperations(int[] nums) {\n int n=nums.length;\n int count=0;\n for(int i=0;i<n-1;i++)\n {\n | Ramsingh27 | NORMAL | 2022-07-30T17:13:08.449013+00:00 | 2022-07-30T17:13:08.449051+00:00 | 54 | false | class Solution {\n public int minOperations(int[] nums) {\n int n=nums.length;\n int count=0;\n for(int i=0;i<n-1;i++)\n {\n if(nums[i+1]<=nums[i])\n {\n int r=nums[i]-nums[i+1]+1;\n count+=r;\n nums[i+1]+=r;\n ... | 1 | 0 | [] | 0 |
minimum-operations-to-make-the-array-increasing | Vasu vro | vasu-vro-by-revanth2311-7gq0 | \'\'\' \nclass Solution {\npublic:\n int minOperations(vector& nums) {\n int k=0,op=0;\n for(int i=1;i=nums[i]){\n k=nums[i-1]-n | revanth2311 | NORMAL | 2022-07-28T08:21:00.727200+00:00 | 2022-07-28T08:21:00.727242+00:00 | 29 | false | \'\'\' \nclass Solution {\npublic:\n int minOperations(vector<int>& nums) {\n int k=0,op=0;\n for(int i=1;i<nums.size();i++){\n k=0;\n if(nums[i-1]>=nums[i]){\n k=nums[i-1]-nums[i]+1;\n nums[i]=nums[i]+k;\n }\n op=op+k;\n ... | 1 | 0 | [] | 0 |
minimum-operations-to-make-the-array-increasing | Easy C++ solution | easy-c-solution-by-mubin86-fikc | \nclass Solution {\npublic:\n int minOperations(vector<int>& nums) {\n int result = 0;\n for(int i = 1; i<nums.size(); i++){\n if(nu | mubin86 | NORMAL | 2022-07-24T17:37:25.522201+00:00 | 2022-07-24T17:37:25.522253+00:00 | 51 | false | ```\nclass Solution {\npublic:\n int minOperations(vector<int>& nums) {\n int result = 0;\n for(int i = 1; i<nums.size(); i++){\n if(nums[i] <= nums[i-1]){ \n int newNum = nums[i-1] + 1;\n result += newNum - nums[i];\n nums[i] = newNum; \n ... | 1 | 0 | [] | 0 |
minimum-operations-to-make-the-array-increasing | Easiest Approach C++ | easiest-approach-c-by-sagarkesharwnnni-21hd | \nclass Solution {\npublic:\n int minOperations(vector<int>& nums) {\n int count=0;\n int temp=0;\n \n for(int i=0;i<nums.size()- | sagarkesharwnnni | NORMAL | 2022-07-20T10:41:55.133058+00:00 | 2022-07-20T10:41:55.133099+00:00 | 65 | false | ```\nclass Solution {\npublic:\n int minOperations(vector<int>& nums) {\n int count=0;\n int temp=0;\n \n for(int i=0;i<nums.size()-1;i++){\n if(nums[i]>=nums[i+1]){\n temp=nums[i]-nums[i+1]+1;\n nums[i+1]+=temp;\n count+=temp; \... | 1 | 0 | ['C'] | 0 |
minimum-operations-to-make-the-array-increasing | c++|O(n)|Simple Logic | consimple-logic-by-deepakhacker21-grza | class Solution {\npublic:\n\n int minOperations(vector& nums) {\n int count = 0;\n for(int i=1;i= nums[i]) {\n count = count + ( | deepakHacker21 | NORMAL | 2022-07-14T18:45:56.130502+00:00 | 2022-07-14T18:45:56.130543+00:00 | 39 | false | class Solution {\npublic:\n\n int minOperations(vector<int>& nums) {\n int count = 0;\n for(int i=1;i<nums.size();i++) {\n if(nums[i-1] >= nums[i]) {\n count = count + ((nums[i-1]+1)-nums[i]);\n nums[i] = nums[i-1] +1;\n }\n }\n return c... | 1 | 0 | ['C'] | 0 |
minimum-operations-to-make-the-array-increasing | 90% fast || c++ || O(N) | 90-fast-c-on-by-sumit_singh0044-9l26 | \n int minOperations(vector<int>& nums) {\n if(nums.size()==1)\n return 0;\n int count=0;\n for(int i=1;i<nums.size();i++)\n | sumit_singh0044 | NORMAL | 2022-07-08T11:15:22.700652+00:00 | 2022-07-08T11:15:22.700675+00:00 | 57 | false | ```\n int minOperations(vector<int>& nums) {\n if(nums.size()==1)\n return 0;\n int count=0;\n for(int i=1;i<nums.size();i++)\n {\n if(nums[i]==nums[i-1])\n {\n nums[i]++;\n count++;\n }\n else if(nums[i]... | 1 | 0 | ['C', 'C++'] | 0 |
minimum-operations-to-make-the-array-increasing | ✅C++ || Short && clean || O(N) && O(1) | c-short-clean-on-o1-by-abhinav_0107-gdq7 | \n\n\n\n\tclass Solution {\n\t\tpublic:\n\t\t\tint minOperations(vector& nums) {\n\t\t\t\tint k=0;\n\t\t\t\tfor(int i=0;i<nums.size()-1;i++){\n\t\t\t\t\tif(nums | abhinav_0107 | NORMAL | 2022-06-28T08:43:24.402103+00:00 | 2022-06-28T08:43:24.402140+00:00 | 149 | false | \n\n\n\n\tclass Solution {\n\t\tpublic:\n\t\t\tint minOperations(vector<int>& nums) {\n\t\t\t\tint k=0;\n\t\t\t\tfor(int i=0;i<nums.size()-1;i++){\n\t\t\t\t\tif(nums[i+1]<=nums[i]){\n\t\t\t\t\t\tk+=(nums[i]-num... | 1 | 0 | ['C', 'C++'] | 0 |
minimum-operations-to-make-the-array-increasing | Java solution. O(n) | java-solution-on-by-pvp-ltcb | \nclass Solution {\n public int minOperations(int[] nums) {\n int ans = 0;\n for (int i = 1; i < nums.length; i++) {\n if (nums[i] < | pvp | NORMAL | 2022-06-27T06:17:07.982618+00:00 | 2022-06-27T06:17:07.982657+00:00 | 40 | false | ```\nclass Solution {\n public int minOperations(int[] nums) {\n int ans = 0;\n for (int i = 1; i < nums.length; i++) {\n if (nums[i] <= nums[i - 1]) {\n ans += nums[i - 1] + 1 - nums[i];\n nums[i] = nums[i - 1] + 1;\n }\n }\n return ans... | 1 | 0 | [] | 0 |
minimum-operations-to-make-the-array-increasing | Python | Easy & Understanding Solution | python-easy-understanding-solution-by-ba-v6km | ```\nclass Solution:\n def minOperations(self, nums: List[int]) -> int:\n n=len(nums)\n if(n==1):\n return 0\n \n ans= | backpropagator | NORMAL | 2022-06-14T18:09:32.572031+00:00 | 2022-06-14T18:09:32.572077+00:00 | 371 | false | ```\nclass Solution:\n def minOperations(self, nums: List[int]) -> int:\n n=len(nums)\n if(n==1):\n return 0\n \n ans=0\n \n for i in range(1,n):\n if(nums[i]<=nums[i-1]):\n ans+=nums[i-1]-nums[i]+1\n nums[i]=nums[i-1]+1\n ... | 1 | 0 | ['Python', 'Python3'] | 0 |
construct-string-with-repeat-limit | Easy understanding C++ code with comments | easy-understanding-c-code-with-comments-v9h2s | We want to form a lexicographically largest string using the characters in a given string.\nBut we have to make sure that a letter is not repeated more than the | venkatasaitanish | NORMAL | 2022-02-20T04:02:43.434365+00:00 | 2022-05-23T06:26:12.675157+00:00 | 7,523 | false | We want to form a lexicographically largest string using the characters in a given string.\nBut we have to make sure that a letter is not repeated more than the given limit in a row (i.e consecutively).\n\nSo we use a ***priority_queue*** to pop the element which has the highest priority. *If it\'s count is less than t... | 115 | 8 | ['Greedy', 'C', 'Heap (Priority Queue)'] | 19 |
construct-string-with-repeat-limit | Java, Priority Queue and stack - Larger String with repeat limit | java-priority-queue-and-stack-larger-str-vh3v | At first on looking into this problem it looked little complex,\n\nBut after thinking for some time, I am able to unwire it little easily..\n\nHere is the appro | Surendaar | NORMAL | 2022-02-20T04:03:39.932617+00:00 | 2022-02-20T04:03:39.932659+00:00 | 5,579 | false | At first on looking into this problem it looked little complex,\n\nBut after thinking for some time, I am able to unwire it little easily..\n\nHere is the approach I took.\n\nIn-order to get a lexicographically larger number we know that it should have all larger alphabets in the starting of string like zzzzzzyyyyyyxxx... | 60 | 6 | ['Stack', 'Heap (Priority Queue)', 'Java'] | 18 |
construct-string-with-repeat-limit | C++ Greedy + Counting O(N) Time O(1) space | c-greedy-counting-on-time-o1-space-by-lz-dzpf | See my latest update in repo LeetCode\n\n\n## Solution 1. Greedy + Counting\n\nStore frequency of characters in int cnt[26].\n\nWe pick characters in batches. I | lzl124631x | NORMAL | 2022-02-20T04:00:54.263930+00:00 | 2022-02-20T04:00:54.263968+00:00 | 5,276 | false | See my latest update in repo [LeetCode](https://github.com/lzl124631x/LeetCode)\n\n\n## Solution 1. Greedy + Counting\n\nStore frequency of characters in `int cnt[26]`.\n\nWe pick characters in batches. In each batch, we pick the first character from `z` to `a` whose `cnt` is positive with the following caveats:\n1. If... | 57 | 3 | [] | 13 |
construct-string-with-repeat-limit | 👉Two Pointer Simple Solution 🧠 | two-pointer-simple-solution-by-sumeet_sh-8ac1 | 🧠 IntuitionWe are tasked with constructing the lexicographically largest string possible from the string s, while ensuring that no character appears more than r | Sumeet_Sharma-1 | NORMAL | 2024-12-17T01:25:33.461938+00:00 | 2024-12-17T05:34:35.395965+00:00 | 20,125 | false | # \uD83E\uDDE0 Intuition\n\nWe are tasked with constructing the lexicographically largest string possible from the string `s`, while ensuring that no character appears more than `repeatLimit` times consecutively.\n\nTo achieve this, we need to:\n- **Maximize lexicographical order**: We prioritize placing the lexicograp... | 50 | 4 | ['Two Pointers', 'C++', 'Java', 'Python3', 'Ruby', 'JavaScript', 'C#'] | 18 |
construct-string-with-repeat-limit | [Python3] priority queue | python3-priority-queue-by-ye15-pyvn | Please pull this commit for solutions of weekly 281. \n\n\nclass Solution:\n def repeatLimitedString(self, s: str, repeatLimit: int) -> str:\n pq = [( | ye15 | NORMAL | 2022-02-20T04:03:24.517116+00:00 | 2022-02-21T22:06:51.329967+00:00 | 3,451 | false | Please pull this [commit](https://github.com/gaosanyong/leetcode/commit/793daa0aab0733bfadd4041fdaa6f8bdd38fe229) for solutions of weekly 281. \n\n```\nclass Solution:\n def repeatLimitedString(self, s: str, repeatLimit: int) -> str:\n pq = [(-ord(k), v) for k, v in Counter(s).items()] \n heapify(pq)\n... | 44 | 7 | ['Python3'] | 9 |
construct-string-with-repeat-limit | Priority Queue Solution ✅✅ | priority-queue-solution-by-arunk_leetcod-99ot | IntuitionThe problem requires constructing a string where no character appears more than k consecutive times while maximizing the lexicographical order. To achi | arunk_leetcode | NORMAL | 2024-12-17T05:33:24.621350+00:00 | 2024-12-17T05:33:24.621350+00:00 | 4,711 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nThe problem requires constructing a string where no character appears more than k consecutive times while maximizing the lexicographical order. To achieve this, the largest characters should be used as much as possible, but when a charact... | 25 | 0 | ['Heap (Priority Queue)', 'Python', 'C++', 'Java', 'Python3'] | 3 |
construct-string-with-repeat-limit | Greedy | greedy-by-votrubac-ayeg | Count all characters in 26 buckets (\'a\' to \'z\'). \n2. While we have characters left:\n\t- Find the largest bucket i that still has characters.\n\t- Take up | votrubac | NORMAL | 2022-02-20T04:01:57.409501+00:00 | 2022-02-21T06:20:17.477404+00:00 | 2,580 | false | 1. Count all characters in 26 buckets (\'a\' to \'z\'). \n2. While we have characters left:\n\t- Find the largest bucket `i` that still has characters.\n\t- Take up to limit characters from bucket `i`.\n\t- Find the second largest bucket `j` that has characters.\n\t- Pick 1 character from bucket `j`.\n\nCaveats:\n- The... | 25 | 1 | [] | 6 |
construct-string-with-repeat-limit | Greedy make_heap+ C-array vs priority_queue | greedy-mak_heap-c-array-by-anwendeng-de3i | IntuitionAgain Greedy heap with k operations where k= # of of diffrent chars.
Similar process is done by used priority_queue. 2 approaches have slight different | anwendeng | NORMAL | 2024-12-17T02:02:31.593668+00:00 | 2024-12-17T06:33:46.871328+00:00 | 3,448 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nAgain Greedy heap with k operations where k= # of of diffrent chars.\nSimilar process is done by used priority_queue. 2 approaches have slight different processes for the 2nd max character.\n# Approach\n<!-- Describe your approach to solv... | 21 | 0 | ['String', 'Greedy', 'Heap (Priority Queue)', 'C++'] | 8 |
construct-string-with-repeat-limit | Python | Straightforward Solution | Counting | python-straightforward-solution-counting-ms2v | We store the frequency of each character of the string in an array(A[26]). Now we start from the last index of the array which will store z as we need to genera | mohd_mustufa | NORMAL | 2022-02-20T04:16:54.244534+00:00 | 2022-02-20T04:16:54.244561+00:00 | 1,221 | false | We store the frequency of each character of the string in an array(A[26]). Now we start from the last index of the array which will store z as we need to generate output in lexiographically increasing order.\n\n```\n def repeatLimitedString(self, s: str, repeatLimit: int) -> str:\n arr = [0] * 26\n a =... | 15 | 4 | [] | 2 |
construct-string-with-repeat-limit | Python | Greedy Algorithm & Priority Queue | python-greedy-algorithm-priority-queue-b-tddx | see the Successfully Accepted SubmissionCodeSteps:
Frequency Count:
Use a Counter to count the frequency of each character in the string.
Max Heap:
Use a max | Khosiyat | NORMAL | 2024-12-17T01:08:00.090981+00:00 | 2024-12-17T01:08:00.090981+00:00 | 1,233 | false | [see the Successfully Accepted Submission](https://leetcode.com/problems/construct-string-with-repeat-limit/submissions/1480683191/?envType=daily-question&envId=2024-12-17)\n\n# Code\n```python3 []\nclass Solution:\n def repeatLimitedString(self, s: str, repeatLimit: int) -> str:\n # Count frequencies of each... | 11 | 0 | ['Python3'] | 0 |
construct-string-with-repeat-limit | Logic explained With Comments || Priority Queue | logic-explained-with-comments-priority-q-m8x1 | As we need lexicographically largest string so start adding largest character available.\nOne thing we need to take care of is we cannot repeat a character more | kamisamaaaa | NORMAL | 2022-02-20T04:22:00.167222+00:00 | 2022-02-21T04:28:43.709618+00:00 | 879 | false | As we **need lexicographically largest string** so start **adding largest character available**.\nOne thing we need to **take care of** is we **cannot repeat a character more than repeatLimit times** so **when this happens** **add 1 next greatest charcter.**\n```\nclass Solution {\npublic:\n string repeatLimitedStri... | 9 | 0 | ['C', 'Heap (Priority Queue)'] | 2 |
construct-string-with-repeat-limit | Python short solution using heap | python-short-solution-using-heap-by-cubi-n101 | \n \n \n The idea is simple, keep a max heap of all the characters and their available counts.\n If the max character element at the t | Cubicon | NORMAL | 2022-02-20T04:01:49.333320+00:00 | 2022-02-20T04:01:49.333355+00:00 | 505 | false | \n \n \n The idea is simple, keep a max heap of all the characters and their available counts.\n If the max character element at the top of the heap is already exceeding the limit interms of count,\n then pop the next one and add to heap. Also add back the unused max character element t... | 9 | 0 | [] | 0 |
construct-string-with-repeat-limit | Solution for LeetCode#2182 | solution-for-leetcode2182-by-samir023041-us3r | IntuitionThe problem requires creating a lexicographically largest string with a limit on consecutive repeating characters. The intuition is to use a greedy app | samir023041 | NORMAL | 2024-12-17T05:18:14.199718+00:00 | 2024-12-17T05:21:22.713648+00:00 | 558 | false | \n\n# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nThe problem requires creating a lexicographically largest string with a limit on consecutive repeating characters. The i... | 8 | 0 | ['Counting', 'Java'] | 1 |
construct-string-with-repeat-limit | Simple solution C++| Java| Python | Javascript | simple-solution-c-java-python-javascript-5sth | Solution in C++, Python, Java, and JavaScriptIntuitionWe need to construct a string that ensures no character appears more than repeatLimit times consecutively. | BijoySingh7 | NORMAL | 2024-12-17T04:50:51.159486+00:00 | 2024-12-17T04:50:51.159486+00:00 | 2,205 | false | \n\n###### *Solution in C++, Python, Java, and JavaScript*\n\n```cpp []\nclass Solution {\npublic:\n string repeatLimitedString(string s, int repeatLimit) {\n vector<int> freq(26, 0);\n for (char c : s) freq[c - \'a\']++;\n priority_queue<pair<char, int>> pq;\n for (int i = 0; i < 26; i++... | 8 | 0 | ['String', 'Python', 'C++', 'Java', 'JavaScript'] | 7 |
construct-string-with-repeat-limit | [Java] Using char counting array with O(1) space | java-using-char-counting-array-with-o1-s-ffcx | My easy-understanding solution, O(n) time complexity and O(26) -> O(1) space.\n\nclass Solution {\n public String repeatLimitedString(String s, int repeatLim | SiyuanXing | NORMAL | 2022-02-20T04:11:44.607792+00:00 | 2022-02-20T04:34:28.607677+00:00 | 893 | false | My easy-understanding solution, O(n) time complexity and O(26) -> O(1) space.\n```\nclass Solution {\n public String repeatLimitedString(String s, int repeatLimit) {\n int[] counter = new int[26];\n int max = 0;\n for (char ch : s.toCharArray()) {\n int curr = ch - \'a\';\n ... | 8 | 1 | ['Counting', 'Java'] | 4 |
construct-string-with-repeat-limit | Greedy Python Solution | 100% Runtime | greedy-python-solution-100-runtime-by-an-qenq | Upvote if you find this article helpful\u2B06\uFE0F.\n### STEPS:\n1. Basically we make a sorted table with freq of each character.\n2. Then we take the most lex | ancoderr | NORMAL | 2022-02-20T04:03:31.221909+00:00 | 2022-02-20T04:42:09.566575+00:00 | 1,104 | false | Upvote if you find this article helpful\u2B06\uFE0F.\n### STEPS:\n1. Basically we make a sorted table with freq of each character.\n2. Then we take the most lexicographically superior character (Call it A).\n3. If its freq is in limits, directly add it.\n4. If its freq is more than the limit. Add repeatLimit number of ... | 8 | 4 | ['Python', 'Python3'] | 2 |
construct-string-with-repeat-limit | EASY SOLUTION USING HEAP FOR BEGINNER'S!!! | easy-solution-using-heap-for-beginners-b-y5xg | IntuitionApproachComplexity
Time complexity:
Space complexity:
Code | aero_coder | NORMAL | 2024-12-17T01:08:05.768990+00:00 | 2024-12-17T21:39:02.321313+00:00 | 1,194 | 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)$$ --... | 7 | 0 | ['Heap (Priority Queue)', 'C++'] | 1 |
construct-string-with-repeat-limit | ✅Efficient🔥Beats 98,39%💡Beginner✅ C++ | Java | C# | Python🔥O(n) - O(1) | with-explanation-c-beats-944-on-o1-by-kh-h10u | Algorithm consists of 2 steps:Step 1) - Setup
Create 2 int[] arrays, symbolCount and indices with size of 26 (number of letters in the alphabet).
And fill them | khavv | NORMAL | 2024-12-17T22:36:48.285500+00:00 | 2024-12-19T10:02:20.970878+00:00 | 91 | false | 
### Algorithm consists of 2 steps:
***Step 1)*** - Setup
Create 2 `int[]` arrays, `symbolCount` and `indices` with size of 26 (number of letters in the alphabet).
And fill them in the following manner:
... | 5 | 0 | ['Python', 'C++', 'Java', 'C#'] | 0 |
construct-string-with-repeat-limit | Priority Queue (Max Heap) solution in easy way. (java, c++, python) | priority-queue-max-heap-solution-in-easy-lhqj | ApproachSteps to Solve the ProblemUnderstand the Problem:You are given a string s with lowercase letters.
You need to construct a new string where:
Any characte | Tanmay_bhure | NORMAL | 2024-12-17T08:48:48.005181+00:00 | 2024-12-17T08:48:48.005181+00:00 | 533 | false | # Approach\n# Steps to Solve the Problem\n\n**Understand the Problem:**\n\nYou are given a string s with lowercase letters.\nYou need to construct a new string where:\nAny character can appear consecutively at most limit times.\nThe resulting string should be lexicographically the largest possible.\nUse a greedy approa... | 5 | 0 | ['String', 'Heap (Priority Queue)', 'C++', 'Java', 'Python3'] | 0 |
construct-string-with-repeat-limit | ✅Simple Solution | Beginner Friendly | Easy Approach with Explanation✅ | simple-solution-beginner-friendly-easy-a-1o4f | IntuitionThe goal is to construct the largest lexicographical string such that no character appears more than repeatLimit times consecutively.To achieve this, w | Jithinp96 | NORMAL | 2024-12-17T08:01:12.720079+00:00 | 2024-12-17T08:01:12.720079+00:00 | 488 | false | # Intuition\nThe goal is to construct the largest lexicographical string such that no character appears more than repeatLimit times consecutively. \n\nTo achieve this, we can:\n1. Use a frequency count of each character in the string.\n2. Start from the largest character (lexicographically) and add it to the result as... | 5 | 0 | ['TypeScript', 'JavaScript'] | 0 |
construct-string-with-repeat-limit | Easy Solution✅| Tree Map✅| O(N*K)✅| Java Solution✅| | easy-solution-tree-map-onk-java-solution-2xv3 | Intuition
The goal is to construct the lexicographically largest string under the condition that no character can repeat more than repeatLimit times consecutive | PRASHANT_KUMAR03 | NORMAL | 2024-12-17T05:19:19.249524+00:00 | 2024-12-17T05:19:19.249524+00:00 | 171 | false | # Intuition\n- The goal is to construct the lexicographically largest string under the condition that no character can repeat more than repeatLimit times consecutively. To achieve this:\n\n1. **Prioritize the largest characters first to make the string lexicographically larger.**\n1. **Ensure that a character is append... | 5 | 0 | ['Java'] | 0 |
construct-string-with-repeat-limit | Python 3 | Priority queue | Easy understanding with comment | python-3-priority-queue-easy-understandi-46h9 | Approach\n Describe your approach to solving the problem. \nFirst we store all the characters in a heap.\nThen we pop them one by one to add them into the answe | PinkGlove | NORMAL | 2024-01-08T21:39:39.339790+00:00 | 2024-01-08T21:39:39.339824+00:00 | 276 | false | # Approach\n<!-- Describe your approach to solving the problem. -->\nFirst we store all the characters in a heap.\nThen we pop them one by one to add them into the answer.\nIf the count of the current character is smaller than limit, add it to the answer directly.\nIf the count of the current character is larger than l... | 5 | 0 | ['String', 'Heap (Priority Queue)', 'Counting', 'Python3'] | 2 |
construct-string-with-repeat-limit | Java solution-->(Classic implementation of Map+PriorityQueue) | java-solution-classic-implementation-of-uav91 | In the priority queue sorting part, Character.compare(b,a) can be replaced by return b-a; which reduces the runtime \n\n\nclass pair{\n Character ele;\n i | Ritabrata_1080 | NORMAL | 2022-09-20T09:22:38.662990+00:00 | 2022-09-20T09:36:08.647120+00:00 | 494 | false | # In the priority queue sorting part, Character.compare(b,a) can be replaced by return b-a; which reduces the runtime \n\n```\nclass pair{\n Character ele;\n int freq;\n pair(Character ele, int freq){\n this.ele = ele;\n this.freq = freq;\n }\n}\nclass Solution {\n public String repeatLimit... | 5 | 0 | ['Heap (Priority Queue)', 'Java'] | 0 |
construct-string-with-repeat-limit | 3 Liner Python Solution | 3-liner-python-solution-by-subin_nair-kpxe | \n def repeatLimitedString(self, s: str, k: int) -> str:\n leters , sol, stack, rep, prev , ll = (x for x in sorted(s, reverse=True)) , [], [], 0, Non | subin_nair | NORMAL | 2022-02-20T04:24:20.080858+00:00 | 2022-02-20T16:59:48.131520+00:00 | 190 | false | ```\n def repeatLimitedString(self, s: str, k: int) -> str:\n leters , sol, stack, rep, prev , ll = (x for x in sorted(s, reverse=True)) , [], [], 0, None ,len(s)\n while ele:=stack.pop() if rep <= k and stack else next(leters, 0):_,prev = sol.append(ele) if (rep:=1 if ele != prev else rep+1) <= k else... | 5 | 4 | ['Stack', 'Python'] | 1 |
construct-string-with-repeat-limit | Fastest Python3 Solution! 💯 | fastest-python3-solution-by-rinsane-3j5u | Complexity
Time complexity: O(n∗log(n))
Space complexity: O(1) (only need to store the characters that are atmost 26)
Code | rinsane | NORMAL | 2024-12-18T05:30:11.159976+00:00 | 2024-12-18T05:30:11.159976+00:00 | 69 | false | # Complexity\n- Time complexity: $$O(n*log(n))$$\n- Space complexity: $$O(1)$$ (only need to store the characters that are atmost 26)\n\n# Code\n```python3 []\nclass Solution:\n def repeatLimitedString(self, s: str, rl: int) -> str:\n heapq.heapify(chars:= list(map(lambda x: [-ord(x[0]), x[1]], Counter(s).ite... | 4 | 0 | ['Python3'] | 2 |
construct-string-with-repeat-limit | C++ Solution || 2 Approaches || Detailed Explanation | c-solution-2-approaches-detailed-explana-lsom | IntuitionThe key observation is that to make the string lexicographically largest, we need to use the highest possible characters first, as long as we respect t | Rohit_Raj01 | NORMAL | 2024-12-17T12:07:23.031732+00:00 | 2024-12-17T12:07:23.031732+00:00 | 159 | false | # Intuition\nThe key observation is that to make the string lexicographically largest, we need to use the highest possible characters first, as long as we respect the `repeatLimit` constraint. Once we hit the `repeatLimit`, we must insert the next lexicographically smaller character before continuing.\n\n# Approach 1: ... | 4 | 0 | ['Hash Table', 'String', 'Heap (Priority Queue)', 'Counting', 'C++'] | 0 |
construct-string-with-repeat-limit | Easyyy Pessyyy Solution ✅ | Beats 80% Runtime 🔥 | 100% Memory 🔥 | easyyy-pessyyy-solution-beats-80-runtime-x1ip | Intuition
The goal is to build the lexicographically largest string while ensuring no character appears more than repeatLimit times consecutively.
Count Charact | jaitaneja333 | NORMAL | 2024-12-17T05:34:31.115380+00:00 | 2024-12-17T05:34:31.115380+00:00 | 397 | false | \n# Intuition\n* The goal is to build the lexicographically largest string while ensuring no character appears more than repeatLimit times consecutively.\n\n### Count Character Frequency:\nUse a frequency array to count occurrences of each character.\n\n### Max-Heap for Character Priority:\nStore characters in a max-he... | 4 | 0 | ['Array', 'String', 'Greedy', 'Heap (Priority Queue)', 'Counting', 'C++', 'Java'] | 1 |
construct-string-with-repeat-limit | Easy Solution in C++. | easy-solution-in-c-by-dhanu07-lu9a | Intuitionthe problem of constructing the lexicographically largest string from the characters of s while adhering to the repeatLimit, we can use a greedy approa | Dhanu07 | NORMAL | 2024-12-17T05:05:05.086672+00:00 | 2024-12-17T05:12:32.261095+00:00 | 468 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nthe problem of constructing the lexicographically largest string from the characters of s while adhering to the repeatLimit, we can use a greedy approach combined with a max-heap (priority queue). The idea is to always try to add the larg... | 4 | 0 | ['Hash Table', 'String', 'Heap (Priority Queue)', 'Counting', 'C++'] | 4 |
construct-string-with-repeat-limit | faster than 80% of user's runtime | faster-than-80-of-users-runtime-by-dev_b-7cgw | Code | dev_bhatt202 | NORMAL | 2024-12-17T04:18:04.162255+00:00 | 2024-12-17T04:18:04.162255+00:00 | 117 | false | \n\n# Code\n```java []\nclass Solution {\n public String repeatLimitedString(String s, int repeatLimit) {\n int[] freq = new int[26];\n\n for (char c : s.toCharArray())\n freq[c - \'a\']++;\n\n int pendingLetterIndex = -1;\n StringBuilder sb = new StringBuilder();\n\n fo... | 4 | 0 | ['Hash Table', 'String', 'Greedy', 'Heap (Priority Queue)', 'Counting', 'Java'] | 1 |
construct-string-with-repeat-limit | max heap || easy to understand || must see | max-heap-easy-to-understand-must-see-by-fcc02 | 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 | akshat0610 | NORMAL | 2023-03-10T10:03:43.276628+00:00 | 2023-03-10T10:03:43.276668+00:00 | 368 | 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)$$ --... | 4 | 0 | ['String', 'Greedy', 'Heap (Priority Queue)', 'Counting', 'C++'] | 0 |
construct-string-with-repeat-limit | [C++] Simple C++ Code || No Priority Queue || 87% time || 85% space | c-simple-c-code-no-priority-queue-87-tim-vggb | If you like the implementation then Please help me by increasing my reputation. By clicking the up arrow on the left of my image.\n\nclass Solution {\npublic:\n | _pros_ | NORMAL | 2022-07-31T09:10:13.684238+00:00 | 2022-07-31T09:10:13.684277+00:00 | 580 | false | # **If you like the implementation then Please help me by increasing my reputation. By clicking the up arrow on the left of my image.**\n```\nclass Solution {\npublic:\n string repeatLimitedString(string s, int repeatLimit) {\n vector<int> alphabet(26,0);\n int f = 0, n = s.size();\n string ans ... | 4 | 0 | ['C', 'C++'] | 1 |
construct-string-with-repeat-limit | 💥💥✅ Simple Implementation using Priority Queue || T.C = O(N) || Clean Code | simple-implementation-using-priority-que-2n33 | \nclass Solution {\npublic:\n string repeatLimitedString(string s, int repeatLimit) {\n int freq[26] = {0};\n string ans = "";\n\t\t\n f | kirtanprajapati | NORMAL | 2022-02-23T05:29:28.333103+00:00 | 2022-02-23T05:29:28.333142+00:00 | 472 | false | ```\nclass Solution {\npublic:\n string repeatLimitedString(string s, int repeatLimit) {\n int freq[26] = {0};\n string ans = "";\n\t\t\n for(auto c : s){\n freq[c-\'a\']++;\n }\n \n priority_queue<pair<char,int>>pq;\n \n for(int i=0;i<26;i++){\n ... | 4 | 0 | ['C', 'Heap (Priority Queue)'] | 0 |
construct-string-with-repeat-limit | priorityqueue + Longest Happy String | priorityqueue-longest-happy-string-by-al-kako | Found it quite close to Longest Happy String: https://leetcode.com/problems/longest-happy-string/\n\nFor solution and explanation for it, please see this (Thank | alpha_gigachad | NORMAL | 2022-02-20T17:14:15.385183+00:00 | 2022-02-20T17:14:46.219378+00:00 | 395 | false | Found it quite close to `Longest Happy String`: https://leetcode.com/problems/longest-happy-string/\n\nFor solution and explanation for it, please see this (Thanks to @lechen999): https://leetcode.com/problems/longest-happy-string/discuss/564248/Python-HEAP-solution-with-explanation\n\nWe first count each char\'s appea... | 4 | 0 | ['Heap (Priority Queue)', 'Python'] | 0 |
construct-string-with-repeat-limit | Greedy + Priority Queue || Easy solution C++ | greedy-priority-queue-easy-solution-c-by-fzze | \nclass Solution {\npublic:\n string repeatLimitedString(string s, int k) \n {\n int n=s.size(),i;\n priority_queue <pair<char,int>> q;\n | Roar47 | NORMAL | 2022-02-20T04:03:30.415348+00:00 | 2022-02-20T04:03:30.415376+00:00 | 314 | false | ```\nclass Solution {\npublic:\n string repeatLimitedString(string s, int k) \n {\n int n=s.size(),i;\n priority_queue <pair<char,int>> q;\n vector <int> a(26);\n for(auto x : s)\n {\n a[x-\'a\']++;\n }\n for(i=0;i<26;i++)\n {\n if(a[i]... | 4 | 0 | ['Greedy', 'C', 'Heap (Priority Queue)'] | 0 |
construct-string-with-repeat-limit | 🔥BEATS 💯 % 🎯 |✨SUPER EASY BEGINNERS 👏 | beats-super-easy-beginners-by-codewithsp-ujsx | IntuitionThe goal is to construct a string using the characters of s such that:
No character appears consecutively more than repeatLimit times.
The resulting st | CodeWithSparsh | NORMAL | 2024-12-17T16:40:31.672200+00:00 | 2024-12-17T16:40:31.672200+00:00 | 214 | false | \n\n\n---\n\n\n\n\n# Intuition\nThe goal is to construct a string using the characters of `s` such that:\n1. No character appears consecutively more than `repeatLimit` times.\n2. The resulting string is lex... | 3 | 0 | ['String', 'Greedy', 'C', 'Counting', 'C++', 'Java', 'Go', 'Python3', 'JavaScript', 'Dart'] | 0 |
construct-string-with-repeat-limit | ✅✅easy and beginner friendly solution 🔥🔥 | easy-and-beginner-friendly-solution-by-i-eyuh | Intuition
The goal is to construct the lexicographically largest string such that no character appears more than repeatLimit times consecutively. To achieve thi | ishanbagra | NORMAL | 2024-12-17T10:10:39.110313+00:00 | 2024-12-17T10:10:39.110313+00:00 | 150 | false | Intuition\nThe goal is to construct the lexicographically largest string such that no character appears more than repeatLimit times consecutively. To achieve this, we prioritize the largest characters (lexicographically) and use a greedy approach to add them to the result. If a character reaches the repeatLimit, we tem... | 3 | 0 | ['Hash Table', 'String', 'Heap (Priority Queue)', 'C++'] | 2 |
construct-string-with-repeat-limit | Two pointer approach (barely passed TCs and surpassed no one :) *fire emojis* | two-pointer-approach-barely-passed-tcs-a-512r | Intuition and ApproachSorted string in lexicographically largest way to bring bigger chars ahead and together,
Two pointers i and j used, i is current index whi | om_kumar_saini | NORMAL | 2024-12-17T09:03:14.846904+00:00 | 2024-12-18T18:29:44.113711+00:00 | 82 | false | # Intuition and Approach\nSorted string in lexicographically largest way to bring bigger chars ahead and together,\nTwo pointers `i` and `j` used, `i` is current index while `j` is index of next different character that is ready to get swapped.\nIdea is to iterate till we cross `repeatLimit` then swap `s[i]` and `s[j]`... | 3 | 0 | ['Two Pointers', 'Sorting', 'C++'] | 1 |
construct-string-with-repeat-limit | ✅ 🔥 95% Faster with 90% less memory | Array | No inbuilt algorithm used ✅ 🔥 | 100-faster-with-less-memory-array-no-inb-pto5 | IntuitionWe can use a basic array to find the frequency of each character from 'a' to 'z'. Then, sort the result in descending order from 'z' to 'a' based on th | Surendaar | NORMAL | 2024-12-17T04:34:11.282728+00:00 | 2024-12-17T04:36:54.781044+00:00 | 294 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nWe can use a basic array to find the frequency of each character from \'a\' to \'z\'. Then, sort the result in descending order from \'z\' to \'a\' based on the given conditions of maximum repetition and lexicographical order.\n\n# Approa... | 3 | 0 | ['Array', 'Greedy', 'Java'] | 0 |
construct-string-with-repeat-limit | C++ Solution || without Heap, using Map | c-solution-without-heap-using-map-by-shi-mnrn | if it's help, please up ⬆ vote! ❤️Complexity
Overall Time Complexity: O(nlogk+26n) = O(nlogk)
Space complexity: O(k) (max 26 char)
Code | shishirRsiam | NORMAL | 2024-12-17T02:42:20.461481+00:00 | 2024-12-17T02:42:20.461481+00:00 | 328 | false | # if it\'s help, please up \u2B06 vote! \u2764\uFE0F\n\n\n\n\n## Complexity\n- Overall Time Complexity: **O(nlogk+26n) = O(nlogk)**\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity: O(k) (max 26 char)\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```cpp []\nclass Solutio... | 3 | 0 | ['Hash Table', 'String', 'Greedy', 'Counting', 'C++'] | 3 |
construct-string-with-repeat-limit | For beginners - Heap Basics/Python [15 lines code] | for-beginners-heappython-by-jayanth_y-ig6k | Code | jayanth_y | NORMAL | 2024-12-17T01:46:29.252718+00:00 | 2024-12-17T22:26:42.685415+00:00 | 330 | false | # Code\n```python3 []\nclass Solution:\n def repeatLimitedString(self, s: str, rl: int) -> str:\n h = [ (-ord(x), y) for x, y in Counter(s).items() ]\n heapq.heapify(h)\n \n res = ""\n while h:\n x, y = heapq.heappop(h)\n while y > rl and h: \n ... | 3 | 0 | ['Heap (Priority Queue)', 'Python3'] | 0 |
construct-string-with-repeat-limit | [Python3] Priority Queue + Greedy + Couting - Simple Solution | python3-priority-queue-greedy-couting-si-shnz | Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time | dolong2110 | NORMAL | 2023-07-11T18:42:43.698401+00:00 | 2023-07-11T18:42:43.698434+00:00 | 161 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity: $$O(NlogN)$$\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity: $$O(N)$$\n<!-- Add your space complexity ... | 3 | 1 | ['String', 'Greedy', 'Heap (Priority Queue)', 'Counting', 'Python3'] | 1 |
construct-string-with-repeat-limit | Java || Priority Queue || Hash Map Implementation | java-priority-queue-hash-map-implementat-cqif | \nclass Element {\n char letter;\n int frequency;\n Element(char letter, int frequency) {\n this.letter = letter;\n this.frequency = freq | devansh2805 | NORMAL | 2022-04-13T07:37:36.431672+00:00 | 2022-04-13T07:37:36.431711+00:00 | 292 | false | ```\nclass Element {\n char letter;\n int frequency;\n Element(char letter, int frequency) {\n this.letter = letter;\n this.frequency = frequency;\n }\n}\n\nclass ElementComparator implements Comparator<Element> {\n @Override\n public int compare(Element e1, Element e2) {\n return... | 3 | 0 | ['Heap (Priority Queue)', 'Java'] | 1 |
construct-string-with-repeat-limit | Simple Java Solution with O(n) time complexity. | simple-java-solution-with-on-time-comple-p8en | Bullets boints to make this question super easily\n\n Construct a string which contains all the alphabet from a to z and also a integer array of length 26.\n co | htksaurav | NORMAL | 2022-03-27T16:55:26.458055+00:00 | 2022-03-27T16:55:26.458098+00:00 | 378 | false | Bullets boints to make this question super easily\n\n* Construct a string which contains all the alphabet from a to z and also a integer array of length 26.\n* count repeatation of each character and stored in integer arrat.\n* Now loop over integer array in reverse order and apply following codition.\n* if repeated ch... | 3 | 0 | ['Java'] | 0 |
construct-string-with-repeat-limit | [C++] O(n) easy to understand. | c-on-easy-to-understand-by-lovebaonvwu-8gpu | \nclass Solution {\npublic:\n string repeatLimitedString(string s, int repeatLimit) {\n int cnt[26] = {0};\n \n for (auto c : s) {\n | lovebaonvwu | NORMAL | 2022-02-25T02:04:43.614328+00:00 | 2022-02-25T02:06:23.667267+00:00 | 259 | false | ```\nclass Solution {\npublic:\n string repeatLimitedString(string s, int repeatLimit) {\n int cnt[26] = {0};\n \n for (auto c : s) {\n ++cnt[c - \'a\'];\n }\n \n string ans;\n int prevsize = -1;\n \n int k = 0;\n while (ans.size() < s.... | 3 | 0 | [] | 0 |
construct-string-with-repeat-limit | Java Easy to Understand | PriorityQueue Solution | java-easy-to-understand-priorityqueue-so-bvew | Add all the characters to a PriorityQueue which sorts in reverse order.\n2. Remove all the elements in the queue and add them to a String if they are occuring l | sridhar2007 | NORMAL | 2022-02-20T11:52:29.944186+00:00 | 2022-02-20T18:05:30.387180+00:00 | 435 | false | 1. Add all the characters to a PriorityQueue which sorts in reverse order.\n2. Remove all the elements in the queue and add them to a String if they are occuring less then the limit\n3. Else push them to another PriorityQueue and add them if there is a new character to go before this.\n\nUpvote if you like the solutio... | 3 | 0 | ['Heap (Priority Queue)', 'Java'] | 1 |
construct-string-with-repeat-limit | [c++] Frequency Vector + Priority Queue | c-frequency-vector-priority-queue-by-gar-vgjq | \nclass Solution {\npublic:\n string repeatLimitedString(string s, int limit){\n vector<int> hash(26);\n for(int i=0;i<s.length();i++) hash[s[i | garvbareja | NORMAL | 2022-02-20T09:16:37.592248+00:00 | 2022-02-20T09:16:37.592273+00:00 | 210 | false | ```\nclass Solution {\npublic:\n string repeatLimitedString(string s, int limit){\n vector<int> hash(26);\n for(int i=0;i<s.length();i++) hash[s[i]-\'a\']++;\n priority_queue<pair<char,int>> pq; \n for(int i=0;i<26;i++) if(hash[i]) pq.push({\'a\'+i,hash[i]});\n string res="";\n ... | 3 | 1 | ['Greedy', 'C', 'Heap (Priority Queue)', 'C++'] | 1 |
construct-string-with-repeat-limit | C++ Easy Solution | c-easy-solution-by-simplekind-o1ev | \n#define deb(x) cout << #x << " = " << x << endl\nclass Solution {\npublic:\n string repeatLimitedString(string s, int rL) {\n map<char,int> m ;\n | simplekind | NORMAL | 2022-02-20T05:02:26.320151+00:00 | 2022-02-20T05:02:26.320178+00:00 | 440 | false | ```\n#define deb(x) cout << #x << " = " << x << endl\nclass Solution {\npublic:\n string repeatLimitedString(string s, int rL) {\n map<char,int> m ;\n for (int i =0;i<26;i++)m[i+\'a\']=0;\n for (int i = 0 ;i <s.length();i++)\n m[s[i]]++;\n string res = "";\n priority_que... | 3 | 0 | ['Greedy', 'C', 'Heap (Priority Queue)', 'C++'] | 0 |
construct-string-with-repeat-limit | C++ Easy Thinking and Approach | c-easy-thinking-and-approach-by-sanheen-ree9i | \n Count all characters in 26 buckets (\'a\' to \'z\').\n While we have characters left:\n\t \tFind the largest bucket i with characters.\n\t \tPick up to limit | sanheen-sethi | NORMAL | 2022-02-20T04:43:22.026528+00:00 | 2022-02-20T04:43:22.026566+00:00 | 473 | false | \n* Count all characters in 26 buckets (\'a\' to \'z\').\n* While we have characters left:\n\t* \tFind the largest bucket i with characters.\n\t* \tPick up to limit characters from that bucket.\n\t* \tFind the second largest bucket j with characters.\n\t* \tPick 1 character from the next largest bucket.\n\n\n```\n#defi... | 3 | 0 | ['Greedy', 'C', 'C++'] | 0 |
construct-string-with-repeat-limit | Simple JAVA Solution | simple-java-solution-by-mohit038-7lbr | \nclass Solution {\n public String repeatLimitedString(String s, int repeatLimit) {\n int i = 25, dp[] = new int[26];\n for (char c: s.toCharAr | mohit038 | NORMAL | 2022-02-20T04:02:19.366069+00:00 | 2022-02-20T04:03:42.341600+00:00 | 191 | false | ```\nclass Solution {\n public String repeatLimitedString(String s, int repeatLimit) {\n int i = 25, dp[] = new int[26];\n for (char c: s.toCharArray()) dp[c - \'a\']++;\n StringBuilder sb = new StringBuilder();\n while(i > -1) {\n while (dp[i] > 0) {\n int numbe... | 3 | 1 | [] | 1 |
construct-string-with-repeat-limit | C++ solution O(nlogn) | c-solution-onlogn-by-adarshxd-8y9p | \nclass Solution {\npublic:\n string repeatLimitedString(string s, int rl) {\n int limit = rl;\n int n= s.size();\n sort(s.begin(), s.en | AdarshxD | NORMAL | 2022-02-20T04:02:00.481397+00:00 | 2022-02-20T04:03:20.256551+00:00 | 365 | false | ```\nclass Solution {\npublic:\n string repeatLimitedString(string s, int rl) {\n int limit = rl;\n int n= s.size();\n sort(s.begin(), s.end());\n reverse(s.begin(), s.end());\n //construction of p\n string p; \n p.push_back(s[0]);\n int size = 1;\n unor... | 3 | 0 | ['C'] | 1 |
construct-string-with-repeat-limit | Simple Java Solution with Comments | simple-java-solution-with-comments-by-pa-7c4s | ```\nclass Solution {\n public String repeatLimitedString(String s, int repeatLimit) {\n TreeMap tm = new TreeMap<>(Collections.reverseOrder());\n | pavankumarchaitanya | NORMAL | 2022-02-20T04:00:48.617919+00:00 | 2022-02-20T04:00:48.617953+00:00 | 406 | false | ```\nclass Solution {\n public String repeatLimitedString(String s, int repeatLimit) {\n TreeMap<Character, Integer> tm = new TreeMap<>(Collections.reverseOrder());\n StringBuffer sb = new StringBuffer();\n boolean shouldGetLargest = true;\n for(char c: s.toCharArray())\n {\n ... | 3 | 1 | [] | 1 |
construct-string-with-repeat-limit | java easy solution || 8ms || construct string | java-easy-solution-8ms-construct-string-u0hkl | class Solution {\n public String repeatLimitedString(String s, int repeatLimit) {\n int[] freq = new int[26];\n for (char ch : s.toCharArray()) | Seema_Kumari1 | NORMAL | 2024-12-18T05:17:37.521901+00:00 | 2024-12-18T05:17:37.521945+00:00 | 4 | false | class Solution {\n public String repeatLimitedString(String s, int repeatLimit) {\n int[] freq = new int[26];\n for (char ch : s.toCharArray()) {\n freq[ch - \'a\']++;\n }\n\n StringBuilder result = new StringBuilder();\n int currentCharIndex = 25;\n\n while (curr... | 2 | 0 | ['Java'] | 0 |
construct-string-with-repeat-limit | ✅ Simple Java Solution | Map + PQ | simple-java-solution-map-pq-by-harsh__00-uiyo | CODE\nJava []\npublic String repeatLimitedString(String s, int repeatLimit) {\n\tMap<Character, Integer> map = new HashMap<>();\n\tfor(char ch : s.toCharArray() | Harsh__005 | NORMAL | 2024-12-17T18:00:39.387547+00:00 | 2024-12-17T18:00:39.387573+00:00 | 27 | false | ## **CODE**\n```Java []\npublic String repeatLimitedString(String s, int repeatLimit) {\n\tMap<Character, Integer> map = new HashMap<>();\n\tfor(char ch : s.toCharArray()){\n\t\tmap.put(ch, map.getOrDefault(ch, 0) + 1);\n\t}\n\n\tPriorityQueue<Character> pq = new PriorityQueue<>(Collections.reverseOrder());\n\tfor(char... | 2 | 0 | ['Java'] | 1 |
construct-string-with-repeat-limit | Easy Solution using map for beginners !!! | easy-solution-using-map-for-beginners-by-fuba | IntuitionWe want to build a string by repeating characters from the input string s, but no character should appear more than k times consecutively. The idea is | sachanharshit | NORMAL | 2024-12-17T22:53:58.957393+00:00 | 2024-12-17T22:53:58.957393+00:00 | 14 | false | # Intuition\nWe want to build a string by repeating characters from the input string s, but no character should appear more than k times consecutively. The idea is to always pick the largest available character and add it to the result, respecting the k limit. If a character appears more than k times, we handle it by i... | 2 | 0 | ['Hash Table', 'String', 'Greedy', 'Counting'] | 1 |
construct-string-with-repeat-limit | Swift | Greedy | swift-great-solution-if-i-may-say-so-by-9wenq | Code | pagafan7as | NORMAL | 2024-12-17T17:46:41.701427+00:00 | 2024-12-17T23:48:08.470877+00:00 | 22 | false | # Code\n```swift []\nclass Solution {\n func repeatLimitedString(_ s: String, _ repeatLimit: Int) -> String {\n var repeatCount = 0\n var overflow = [Character]()\n var res = ""\n\n // Greedily, go through each character by descending order.\n for c in s.sorted(by: >) {\n ... | 2 | 0 | ['Swift'] | 0 |
construct-string-with-repeat-limit | 🌟 Greedy Approach: Simple & Beginner-Friendly🚀 | 💥 Beats 95% 💪 | greedy-approach-simple-beginner-friendly-q6uv | Intuition 🤔:We aim to create the lexicographically largest string. Using a greedy approach:
Pick the largest character 🎯 (e.g., 'z') as much as allowed by repea | mansimittal935 | NORMAL | 2024-12-17T16:07:46.131722+00:00 | 2024-12-17T16:07:46.131722+00:00 | 87 | false | # Intuition \uD83E\uDD14:\nWe aim to create the lexicographically largest string. Using a greedy approach:\n\n1. Pick the largest character \uD83C\uDFAF (e.g., \'z\') as much as allowed by repeatLimit \u23F3.\n2. Once we hit the limit, move to the next largest character \uD83D\uDD3D.\n3. Repeat this until we\u2019ve us... | 2 | 0 | ['Hash Table', 'Greedy', 'Python', 'C++', 'Java', 'Go', 'Rust', 'Ruby', 'JavaScript', 'C#'] | 0 |
construct-string-with-repeat-limit | ✅ Easy to Understand | HashTable | Step by Step | Detailed Video Explanation🔥 | easy-to-understand-hashtable-step-by-ste-z2ka | IntuitionThe problem requires constructing a string such that no character repeats more than the given limit consecutively. My initial thought was to use a gree | sahilpcs | NORMAL | 2024-12-17T14:54:50.268865+00:00 | 2024-12-17T14:54:50.268865+00:00 | 60 | false | # Intuition\nThe problem requires constructing a string such that no character repeats more than the given limit consecutively. My initial thought was to use a greedy approach, always selecting the lexicographically largest available character that respects the constraints, and falling back to smaller characters when n... | 2 | 0 | ['Hash Table', 'String', 'Greedy', 'Java'] | 0 |
construct-string-with-repeat-limit | Simple Solution using PriorityQueue. | simple-solution-using-priorityqueue-by-s-6tfb | IntuitionThe problem involves creating a lexicographically largest string with a given constraint: no character should repeat more than repeatLimit times consec | surya04082004 | NORMAL | 2024-12-17T10:56:05.719785+00:00 | 2024-12-17T10:56:05.719785+00:00 | 8 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nThe problem involves creating a lexicographically largest string with a given constraint: no character should repeat more than `repeatLimit` times consecutively. This requires:\n\n- Tracking the frequency of each character.\n- Using a pri... | 2 | 0 | ['Java'] | 0 |
construct-string-with-repeat-limit | Kotlin | Rust | kotlin-rust-by-samoylenkodmitry-8frw | Join me on Telegramhttps://t.me/leetcode_daily_unstoppable/835Problem TLDRMax lexical ordered with repeat_limit string #medium #bucket_sortIntuitionAlways peek | SamoylenkoDmitry | NORMAL | 2024-12-17T08:58:23.791982+00:00 | 2024-12-17T08:58:34.082537+00:00 | 94 | false | \n\nhttps://youtu.be/tGaNBtqpNec\n\n#### Join me on Telegram\n\nhttps://t.me/leetcode_daily_unstoppable/835\n\n#### Problem TLDR\n\nMax lexical ordered with `repeat_limit` string #medium #bucket_sort\n\n#### ... | 2 | 0 | ['Bucket Sort', 'C++', 'Rust', 'Kotlin'] | 1 |
construct-string-with-repeat-limit | Very Easy Detailed Solution 🔥 || 🎯 Greedy Algorithm | very-easy-detailed-solution-greedy-algor-q1rc | IntuitionWe will make a frequency array to track the frequency of each character in the string, and will iterate from last in order the maintain the lexicograph | chaturvedialok44 | NORMAL | 2024-12-17T08:14:34.054936+00:00 | 2024-12-17T08:14:34.054936+00:00 | 81 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nWe will make a frequency array to track the frequency of each character in the string, and will iterate from last in order the maintain the lexicographical order. We will append the char untill reached the limit of its frequency gets 0. I... | 2 | 0 | ['Array', 'String', 'Greedy', 'Counting', 'Java'] | 2 |
construct-string-with-repeat-limit | Beginner Friendly || C++ || Priority Queue :) | beginner-friendly-c-priority-queue-by-ma-dy8r | Count Character Frequencies:
Use an unordered map mp to store the frequency of each character in the string s.
Max Heap (Priority Queue):
Push all characters an | mayankn31 | NORMAL | 2024-12-17T07:52:44.814648+00:00 | 2024-12-17T07:52:44.814648+00:00 | 95 | false | **Count Character Frequencies:**\n- Use an unordered map mp to store the frequency of each character in the string s.\n\n---\n\n\n**Max Heap (Priority Queue):**\n- Push all characters and their frequencies into a max heap (priority queue), sorting by character value. (lexographical)\n\n---\n\n\n**Construct Result:**\n\... | 2 | 0 | ['Hash Table', 'String', 'Heap (Priority Queue)', 'Counting', 'C++'] | 0 |
construct-string-with-repeat-limit | O(n) two pointer approach in Go | on-two-pointer-approach-in-go-by-zhengha-795v | IntuitionUse two pointers to track the current lex max rune and the next one, if the current one reaches repeatLimit, append the next one, in a loopApproachUse | zhenghaozhao | NORMAL | 2024-12-17T07:35:37.113026+00:00 | 2024-12-17T07:35:37.113026+00:00 | 140 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nUse two pointers to track the current lex max rune and the next one, if the current one reaches repeatLimit, append the next one, in a loop\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\nUse strings.Repeat for con... | 2 | 0 | ['Go'] | 0 |
construct-string-with-repeat-limit | Swift💯 | swift-by-upvotethispls-f4fs | Heap of Character Frequencies (accepted answer) | UpvoteThisPls | NORMAL | 2024-12-17T07:24:08.721196+00:00 | 2024-12-17T07:24:08.721196+00:00 | 52 | false | **Heap of Character Frequencies (accepted answer)**\n```\nimport Collections\nstruct HeapItem: Comparable {\n var ch: Character, count:Int\n init(_ a:Character, _ b:Int) {ch=a;count=b}\n static func <(l:Self, r:Self) -> Bool { l.ch < r.ch }\n}\nclass Solution {\n func repeatLimitedString(_ s: String, _ repe... | 2 | 0 | ['Swift'] | 0 |
construct-string-with-repeat-limit | C++ easy to understand solution | c-easy-to-understand-solution-by-dg15060-x2px | IntuitionWe need to build the lexicographically largest string with the repeatLimit number of occurences of a character at any time. The largest character shoul | dg150602 | NORMAL | 2024-12-17T07:19:51.753927+00:00 | 2024-12-17T07:19:51.753927+00:00 | 28 | false | # Intuition\nWe need to build the `lexicographically largest` string with the `repeatLimit` number of occurences of a character at any time. The largest character should come first. The trick is to keep the repetition of a particular character `rep` in `s` as `rep <= repeatLimit`\n\n# Approach\nWe store the occurences ... | 2 | 0 | ['Hash Table', 'String', 'Counting', 'Iterator', 'C++'] | 0 |
construct-string-with-repeat-limit | JAVA | Simple Solution | Frequency Table | java-simple-solution-frequency-table-by-lex3l | Code | SudhikshaGhanathe | NORMAL | 2024-12-17T07:12:17.183854+00:00 | 2024-12-17T07:12:17.183854+00:00 | 131 | false | \n# Code\n```java []\nclass Solution {\n public String repeatLimitedString(String s, int repeatLimit) {\n int [] vocab = new int[26]; //frequency table\n for (char ch : s.toCharArray()) vocab[ch - \'a\']++;\n StringBuilder sb = new StringBuilder();\n int idx = 25;\n while (idx >= ... | 2 | 0 | ['Hash Table', 'String', 'Counting', 'Java'] | 0 |
construct-string-with-repeat-limit | 2 solutions [array] + [maxHeap] | 2-solutions-array-maxheap-by-ajay_74-4ohc | Code | Ajay_74 | NORMAL | 2024-12-17T07:07:59.535810+00:00 | 2024-12-17T07:07:59.535810+00:00 | 109 | false | \n# Code\n```golang [Using Array]\nfunc repeatLimitedString(s string, repeatLimit int) string {\n\n\tfreq := make([]int, 26)\n\n\tfor i := 0; i < len(s); i++ {\n\t\tfreq[s[i]-\'a\']++\n\t}\n\n\tindex := 25\n\tresult := []byte{}\n\n\tfor index >= 0 {\n\t\tif freq[index] == 0 {\n\t\t\tindex--\n\t\t\tcontinue\n\t\t}\n\n\t... | 2 | 0 | ['Go'] | 1 |
construct-string-with-repeat-limit | Intuition, Approach and Solution using a Priority Queue in C++(Easy to Understand,Straightforward A) | intuition-approach-and-solution-using-a-cr9p8 | IntuitionAs it said lexicographically largest my main idea was to solve it using a max heap i.e. a priority queue.(As it will order the characters lexicographic | japitSaikap302 | NORMAL | 2024-12-17T06:16:59.045806+00:00 | 2024-12-17T06:16:59.045806+00:00 | 45 | false | # Intuition\nAs it said **lexicographically** largest my main idea was to solve it using a max heap i.e. a priority queue.(As it will order the characters lexicographical descending order and make it easy for me to get the correct order for the largest character)\n\n---\n\n\n# Approach\n1. Character Frequency Count\nUs... | 2 | 0 | ['Hash Table', 'String', 'Heap (Priority Queue)', 'C++'] | 0 |
construct-string-with-repeat-limit | Best solution so far without heap | best-solution-so-far-without-heap-by-rik-vn7m | Intuition: Use Freq array from 26 - 1, if freq > limit then simply grab one next char and come back to prev charComplexity
Time complexity: O(n * Min(freq, limi | rikam | NORMAL | 2024-12-17T05:32:11.676676+00:00 | 2024-12-17T05:32:11.676676+00:00 | 46 | false | # Intuition: Use Freq array from 26 - 1, if freq > limit then simply grab one next char and come back to prev char\n\n\n# Complexity\n- Time complexity: O(n * Min(freq, limit))\n- Space complexity: O(n)\n\n# Code\n```csharp []\npublic class Solution {\n public string RepeatLimitedString(string s, int repeatLimit) {\... | 2 | 0 | ['Iterator', 'C#'] | 0 |
construct-string-with-repeat-limit | Easy and Simple Greedy Solution with Linear Time | easy-and-simple-greedy-solution-with-lin-61c7 | SolutionApproach:
Frequency Count:
Use an array hash of size 26 to store the frequency of each letter ('a' to 'z').
Greedy Selection:
Select the largest a | pavansai11 | NORMAL | 2024-12-17T04:55:13.412015+00:00 | 2024-12-17T04:55:13.412015+00:00 | 104 | false | # Solution\n\n## Approach:\n1. **Frequency Count**:\n - Use an array `hash` of size 26 to store the frequency of each letter (\'a\' to \'z\').\n\n2. **Greedy Selection**:\n - Select the largest available character (from \'z\' to \'a\') that satisfies the repeat constraint.\n\n3. **Helper Function**:\n - `getLastC... | 2 | 0 | ['String', 'Greedy', 'Counting', 'C++'] | 0 |
construct-string-with-repeat-limit | Efficient Max-heap solution || Easy to understand | efficient-max-heap-solution-easy-to-unde-kzhx | IntuitionThe problem asks us to construct the lexicographically largest string from the characters in s such that no character repeats more than repeatLimit tim | _adarsh_01 | NORMAL | 2024-12-17T04:31:25.811576+00:00 | 2024-12-17T04:31:25.811576+00:00 | 17 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\nThe problem asks us to construct the lexicographically largest string from the characters in s such that no character repeats more than `repeatLimit` times consecutively. This involves:\n\n 1. Using a greedy approach to select the lexic... | 2 | 0 | ['Hash Table', 'String', 'Greedy', 'Heap (Priority Queue)', 'C++'] | 0 |
construct-string-with-repeat-limit | ✅ | Beats 95% | O(N * 26) time | O(26) Space | Greedy Strategy | beats-95-on-26-time-o26-space-greedy-str-imo4 | IntuitionFor Making a string Lexicographically largest, we will follow a Greedy Strategy
For making string lexicographically largest, always put largest alphabe | fake_name | NORMAL | 2024-12-17T04:12:50.894557+00:00 | 2024-12-17T05:42:33.733065+00:00 | 97 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nFor Making a string Lexicographically largest, we will follow a **Greedy Strategy**\n> For making string lexicographically largest, always put largest alphabet in the begining (if possible), otherwise, find the next largest alphabet. \n\n... | 2 | 0 | ['Hash Table', 'String', 'Greedy', 'Counting', 'C++'] | 0 |
construct-string-with-repeat-limit | Video | O(n) solution keep count char frequency array, with constant space | keep-count-char-frequency-array-on-solut-qdjj | YouTube solution Intuitionkeep it simple like below steps:-
create an array of 26 size and count the frequency of all characters.
int count[] = new int[26]
w | vinod_aka_veenu | NORMAL | 2024-12-17T02:09:31.448196+00:00 | 2024-12-17T03:34:19.794674+00:00 | 252 | false | # YouTube solution https://youtu.be/LGUrExMrW-c\n# Intuition\nkeep it simple like below steps:- \n\n1) create an array of 26 size and count the frequency of all characters.\nint count[] = new int[26]\n\n2) write a method that will give us next smaller char frequency in count array (remeber we are traversing from right ... | 2 | 0 | ['Java'] | 4 |
construct-string-with-repeat-limit | Construct String with Repeat Limit - Easy Explanation - Unique Solution👌🔜 | construct-string-with-repeat-limit-easy-ce3xl | IntuitionWe have to reorder the given string into lexicographically longer string so that the number of repeated element in a row should be atmost repeatLimit.H | RAJESWARI_P | NORMAL | 2024-12-17T01:18:01.282792+00:00 | 2024-12-17T01:18:01.282792+00:00 | 67 | false | # Intuition\nWe have to reorder the given string into lexicographically longer string so that the number of repeated element in a row should be atmost repeatLimit.\n\nHence this can be computed by noting frequency of all the characters in the string.\n<!-- Describe your first thoughts on how to solve this problem. -->\... | 2 | 0 | ['String', 'Counting', 'Java'] | 0 |
construct-string-with-repeat-limit | C++ O(1) space, O(N) time | explanation included | c-o1-space-on-time-explanation-included-zxxz0 | IntuitionWe need to construct a string that is lexicographically the largest and not have characters repeat more then a limit times in a row.ApproachWe will sta | LATUANKHANG | NORMAL | 2024-12-17T01:17:12.815692+00:00 | 2024-12-17T01:17:12.815692+00:00 | 397 | false | # Intuition\nWe need to construct a string that is lexicographically the largest and not have characters repeat more then a limit times in a row.\n\n# Approach\nWe will start with chracter \'z\' and make our way down the alphabet to ensure we arrive at the lexicographically largest string. \nWe first map out how many c... | 2 | 0 | ['C++'] | 2 |
construct-string-with-repeat-limit | Sorted Hash Map Python | Beats 99% 😎🐍 | sorted-hash-map-python-beats-99-by-rahil-u5gl | IntuitionThe problem requires constructing a string using characters from s such that:
No letter appears more than repeatLimit times consecutively.
The resultin | rahilshah01 | NORMAL | 2024-12-17T01:13:40.352576+00:00 | 2024-12-17T01:13:40.352576+00:00 | 174 | false | ## **Intuition**\nThe problem requires constructing a string using characters from `s` such that:\n1. No letter appears more than `repeatLimit` times consecutively.\n2. The resulting string is lexicographically the largest possible.\n\nThe lexicographical order constraint suggests that we should prioritize using larger... | 2 | 0 | ['Hash Table', 'Greedy', 'Counting', 'Counting Sort', 'Python3'] | 0 |
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