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k-diff-pairs-in-an-array | C++ solution | binary search | sorting | O(n) | c-solution-binary-search-sorting-on-by-r-wxo8 | \nint findPairs(vector<int>& nums, int k)\n{\n sort(nums.begin(), nums.end());\n int count = 0;\n for (int i = 0; i < nums.size(); i++)\n {\n | rajnishgeek | NORMAL | 2021-08-16T12:26:50.888713+00:00 | 2021-08-16T12:26:50.888778+00:00 | 241 | false | ```\nint findPairs(vector<int>& nums, int k)\n{\n sort(nums.begin(), nums.end());\n int count = 0;\n for (int i = 0; i < nums.size(); i++)\n {\n if (i > 0 && i < nums.size() && nums[i - 1] == nums[i])\n {\n while (i < nums.size() && nums[i - 1] == nums[i])\n i++;\n ... | 3 | 3 | ['Binary Search'] | 1 |
k-diff-pairs-in-an-array | C++ O(n) soln | c-on-soln-by-aditya0753-9bx3 | \nclass Solution {\npublic:\n int findPairs(vector<int>& nums, int k) {\n int n=nums.size();\n unordered_map<int,int> m;\n int c=0;\n | aditya0753 | NORMAL | 2021-08-07T03:28:27.498887+00:00 | 2021-08-07T03:28:27.498915+00:00 | 347 | false | ```\nclass Solution {\npublic:\n int findPairs(vector<int>& nums, int k) {\n int n=nums.size();\n unordered_map<int,int> m;\n int c=0;\n \n if(k<0) return 0;\n \n for(int i=0;i<n;i++){\n m[nums[i]]++;\n }\n \n if(k==0){\n for... | 3 | 0 | [] | 1 |
k-diff-pairs-in-an-array | [C++] Simple single loop approach | c-simple-single-loop-approach-by-ritik30-xyce | \nclass Solution {\npublic:\n int findPairs(vector<int>& nums, int k) {\n unordered_map<int,int> freq;\n for(int i=0;i<nums.size();i++){\n | ritik307 | NORMAL | 2021-07-14T18:51:44.581141+00:00 | 2021-07-14T18:51:44.581165+00:00 | 504 | false | ```\nclass Solution {\npublic:\n int findPairs(vector<int>& nums, int k) {\n unordered_map<int,int> freq;\n for(int i=0;i<nums.size();i++){\n freq[nums[i]]++;\n }\n int count=0;\n if(k!=0){\n for(auto itr:freq){\n if(freq.find(itr.first+k)!=freq... | 3 | 0 | ['Hash Table', 'C', 'C++'] | 0 |
k-diff-pairs-in-an-array | Cpp Solution beats 100% - O(nlogn) Two Pointer Approach [ proof attached ;) ] | cpp-solution-beats-100-onlogn-two-pointe-8c1f | Hello World !!. A two pointer based approach. Code is commented for better understanding. Hope it helps otherwise feel free to comment. \nThanks.\nHappy Coding | zeroAadi | NORMAL | 2021-06-26T08:51:31.623609+00:00 | 2021-07-11T05:21:39.245607+00:00 | 262 | false | Hello World !!. A two pointer based approach. Code is commented for better understanding. Hope it helps otherwise feel free to comment. \nThanks.\nHappy Coding.\n```\n int n=nums.size();\n\t \n\t // Sort the given vector\n sort(nums.begin(),nums.end());\n int i=0;\n int j=1;\n int c=0; ... | 3 | 1 | [] | 1 |
k-diff-pairs-in-an-array | Easy C++ Soln || nlogn || One pointer with Binary Search | easy-c-soln-nlogn-one-pointer-with-binar-x5um | \nclass Solution {\npublic:\n //one pointer with binary search nlogn\n int findPairs(vector<int>& nums, int k) {\n int ans=0;\n sort(nums.be | mitedyna | NORMAL | 2021-05-24T21:58:46.854304+00:00 | 2021-05-24T21:58:46.854334+00:00 | 292 | false | ```\nclass Solution {\npublic:\n //one pointer with binary search nlogn\n int findPairs(vector<int>& nums, int k) {\n int ans=0;\n sort(nums.begin(),nums.end());\n for(int i=0;i<nums.size()-1;i++){\n if(i>0 && nums[i]==nums[i-1])continue;\n int x=nums[i];\n in... | 3 | 0 | ['C', 'Binary Tree'] | 0 |
k-diff-pairs-in-an-array | C++ || Simple easy solution || Two pointer technique | c-simple-easy-solution-two-pointer-techn-m5tz | \nclass Solution {\npublic:\n int findPairs(vector<int>& nums, int k) {\n int n=nums.size();\n sort(nums.begin(),nums.end());\n int count | akanksha_06 | NORMAL | 2021-05-23T07:42:20.244377+00:00 | 2021-05-23T07:42:20.244425+00:00 | 214 | false | ```\nclass Solution {\npublic:\n int findPairs(vector<int>& nums, int k) {\n int n=nums.size();\n sort(nums.begin(),nums.end());\n int count=0;\n int i=0;\n int j=1;\n while(i<n && j<n)\n {\n if(i!=j && nums[j]-nums[i]==k){\n while(i+1 < n && ... | 3 | 1 | ['Two Pointers', 'C'] | 0 |
k-diff-pairs-in-an-array | C++ solution using set and hashmap. TC - O(nlogn) and SC - O(n) | c-solution-using-set-and-hashmap-tc-onlo-wdbw | \nclass Solution {\npublic:\n int findPairs(vector<int>& nums, int k) {\n unordered_map<int,int> mp;\n \n int s = nums.size();\n | sumit_code | NORMAL | 2021-04-17T04:38:21.912573+00:00 | 2021-04-17T04:38:21.912604+00:00 | 365 | false | ```\nclass Solution {\npublic:\n int findPairs(vector<int>& nums, int k) {\n unordered_map<int,int> mp;\n \n int s = nums.size();\n \n for(int i=0;i<s;i++)\n mp[nums[i]] = i;\n \n \n int res = 0;\n \n set<pair<int,int>> st;\n \n ... | 3 | 1 | ['C', 'Ordered Set'] | 1 |
minimum-deletions-to-make-string-k-special | [Java/C++/Python] Enumerate Minimum Frequency | javacpython-enumerate-minimum-frequency-038ab | Intuition\nFirst, we need to calculate character frequencies.\nwe delete characters for two reasons:\n1. Delete the small frequency to zero, so it doesn\'t coun | lee215 | NORMAL | 2024-03-17T04:09:32.411228+00:00 | 2024-03-17T07:38:17.504122+00:00 | 3,176 | false | # **Intuition**\nFirst, we need to calculate character frequencies.\nwe delete characters for two reasons:\n1. Delete the small frequency to zero, so it doesn\'t count for the rule.\n2. Delete the big frequency smaller, so it `big freq - small freq <= k`.\nAnd we won\'t reduce small freq to smaller positive.\n<br>\n\n#... | 57 | 1 | ['C', 'Python', 'Java'] | 18 |
minimum-deletions-to-make-string-k-special | Explained - using simple freq check | explained-using-simple-freq-check-by-kre-kh7f | Approach\n- Find the frequency of each of the characters in word\n- Sort the freq array\n- Consider each of the freq as the minFreq possible and with this consi | kreakEmp | NORMAL | 2024-03-17T04:02:18.060816+00:00 | 2024-03-17T05:09:08.157531+00:00 | 5,706 | false | # Approach\n- Find the frequency of each of the characters in word\n- Sort the freq array\n- Consider each of the freq as the minFreq possible and with this consideration we need to evaluate how many char from the higher frequency to be deleted to get the condition satisfied for all freq\n- Consider the next frequency ... | 34 | 0 | ['C++'] | 9 |
minimum-deletions-to-make-string-k-special | Simple Hashing || Consider All Possibilities | simple-hashing-consider-all-possibilitie-ws58 | # Intuition \n\n\n# Approach\n1. Initialize an unordered map mp to store the frequency of each character in the word.\n2. Iterate through the characters in th | AdityaRaj_cpp | NORMAL | 2024-03-17T04:02:01.092233+00:00 | 2024-03-17T04:02:01.092261+00:00 | 2,990 | false | <!-- # Intuition -->\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n1. Initialize an unordered map mp to store the frequency of each character in the word.\n2. Iterate through the characters in the word and update the frequency in the map.\n3. Create a vector v to store the frequenc... | 29 | 1 | ['Hash Table', 'Sorting', 'C++'] | 7 |
minimum-deletions-to-make-string-k-special | C++||With explanation | cwith-explanation-by-baibhavkr143-w5sc | 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 | baibhavkr143 | NORMAL | 2024-03-17T04:00:46.049068+00:00 | 2024-03-17T04:07:15.792122+00:00 | 1,532 | 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. $$O(n)$$ --... | 15 | 1 | ['C++'] | 4 |
minimum-deletions-to-make-string-k-special | Simple java solution | simple-java-solution-by-siddhant_1602-nps2 | Complexity\n- Time complexity: O(n)\n\n- Space complexity: O(1)\n\n# Code\n\nclass Solution {\n public int minimumDeletions(String word, int k) {\n in | Siddhant_1602 | NORMAL | 2024-03-17T04:02:39.059413+00:00 | 2024-03-17T04:02:39.059446+00:00 | 816 | false | # Complexity\n- Time complexity: $$O(n)$$\n\n- Space complexity: $$O(1)$$\n\n# Code\n```\nclass Solution {\n public int minimumDeletions(String word, int k) {\n int a[]=new int[26];\n for(char c:word.toCharArray())\n {\n a[c-\'a\']++;\n }\n Arrays.sort(a);\n List<... | 14 | 0 | ['Java'] | 1 |
minimum-deletions-to-make-string-k-special | Python 3 || 7 lines, Counter || T/S: 98% / 29% | python-3-7-lines-counter-ts-98-29-by-spa-s5dn | Here\'s the plan:\n\n\n1. We use Counter to determine the counts of the characters as the list vals.\n\n1. We construct a function minimumDeletions that, given | Spaulding_ | NORMAL | 2024-03-17T20:07:49.067850+00:00 | 2024-05-24T20:15:43.657009+00:00 | 313 | false | Here\'s the plan:\n\n\n1. We use `Counter` to determine the counts of the characters as the list `vals`.\n\n1. We construct a function `minimumDeletions` that, given a range [*num... num+k*], determines the minimum count of deletions to ensure that character counts remaining after these deletions are in that interval.... | 9 | 0 | ['Python3'] | 2 |
minimum-deletions-to-make-string-k-special | Easy C++ Solution using Dynamic Programming and Sorting | easy-c-solution-using-dynamic-programmin-gavh | Intuition\nThe problem requires finding the minimum number of characters needed to delete from a string to make it k-special. A string is considered k-special i | vanshwari | NORMAL | 2024-03-17T04:03:17.198623+00:00 | 2024-04-03T05:51:53.683977+00:00 | 549 | false | # Intuition\nThe problem requires finding the minimum number of characters needed to delete from a string to make it k-special. A string is considered k-special if the absolute difference in frequencies of any two characters in the string is less than or equal to k.\n\n# Approach\n1. Count the frequencies of each chara... | 9 | 0 | ['C++'] | 1 |
minimum-deletions-to-make-string-k-special | C++ 100% Faster | Easy Step By Step Explanation | c-100-faster-easy-step-by-step-explanati-brnt | Intuition\nThe key to solve this problem lies in identifying the discrepancies between character frequencies. A string is k-special if the absolute difference b | VYOM_GOYAL | NORMAL | 2024-03-17T04:00:46.362211+00:00 | 2024-03-18T01:39:33.640927+00:00 | 860 | false | # Intuition\nThe key to solve this problem lies in identifying the discrepancies between character frequencies. A string is k-special if the absolute difference between the frequencies of any two characters (|freq(i) - freq(j)|) is less than or equal to k. Therefore, we need to find the minimum number of deletions requ... | 8 | 1 | ['Array', 'Math', 'C++'] | 2 |
minimum-deletions-to-make-string-k-special | Python Easy Solution | python-easy-solution-by-shree_govind_jee-y5l6 | Code\nJava\n\nclass Solution {\n public int minimumDeletions(String word, int k) {\n int[] freq = new int[26];\n for(char ch:word.toCharArray() | Shree_Govind_Jee | NORMAL | 2024-03-17T04:04:20.333307+00:00 | 2024-03-17T04:04:20.333335+00:00 | 781 | false | # Code\n**Java**\n```\nclass Solution {\n public int minimumDeletions(String word, int k) {\n int[] freq = new int[26];\n for(char ch:word.toCharArray()){\n freq[ch-\'a\']++;\n }\n \n// Sorted in Increasing order\n Arrays.sort(freq);\n int min_del = Intege... | 7 | 0 | ['Array', 'Python', 'Python3'] | 3 |
minimum-deletions-to-make-string-k-special | Dynamic Programming Easy to Understand Solution | dynamic-programming-easy-to-understand-s-tmj7 | Intuition\nThe function f is recursively called to find the minimum number of deletions required. It considers the leftmost and rightmost characters \nIf the di | asitvts | NORMAL | 2024-03-17T04:30:14.007159+00:00 | 2024-03-17T04:30:14.007177+00:00 | 390 | false | # Intuition\nThe function f is recursively called to find the minimum number of deletions required. It considers the leftmost and rightmost characters \nIf the difference between their frequencies is less equal to k, no need to check further.\nElse perform deletion\nDeletion can be done in two different ways\n- Either ... | 6 | 0 | ['Dynamic Programming', 'Sorting', 'C++'] | 0 |
minimum-deletions-to-make-string-k-special | Easy to Understand O(N) TC and O(1) SC Beats 100% | easy-to-understand-on-tc-and-o1-sc-beats-yogf | https://leetcode.com/problems/minimum-deletions-to-make-string-k-special/submissions/1206029598/\n# Intuition\n Describe your first thoughts on how to solve thi | sainadth | NORMAL | 2024-03-17T06:13:53.920826+00:00 | 2024-03-17T10:51:27.294504+00:00 | 283 | false | [https://leetcode.com/problems/minimum-deletions-to-make-string-k-special/submissions/1206029598/]()\n# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nIntution is to select a non zero frequency and make all the other frequencies to be in the range of `[frequency, frequency + k]`\n# Appr... | 5 | 0 | ['C++'] | 1 |
minimum-deletions-to-make-string-k-special | [Python3] DFS + Backtracking | python3-dfs-backtracking-by-dolong2110-xtre | 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 | 2024-03-17T04:05:05.493830+00:00 | 2024-03-17T14:57:55.505353+00:00 | 311 | 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. $$O(n)$$ --... | 5 | 0 | ['Python3'] | 0 |
minimum-deletions-to-make-string-k-special | Simple C++ solution, very intuitive | simple-c-solution-very-intuitive-by-fell-hfpr | Intuition\n Describe your first thoughts on how to solve this problem. \nAt the beginning the problem seems a little bit scary, infact it tells us to consider a | Felle33 | NORMAL | 2024-03-17T20:24:25.951043+00:00 | 2024-03-17T20:25:39.717850+00:00 | 316 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nAt the beginning the problem seems a little bit scary, infact it tells us to consider all the pairwise indeces and confront them.\nIn reality we don\'t have to confront all the possible indeces, but just the letters.\n\n**The main intuiti... | 4 | 0 | ['String', 'Greedy', 'C++'] | 1 |
minimum-deletions-to-make-string-k-special | Short explanation + Comments || 87ms Solution || Using 2-Dimensional DP || O(N) time and O(1) space | short-explanation-comments-87ms-solution-mw95 | Approach\n Describe your approach to solving the problem. \nFirst, calculate frequencies of all the characters.\n\nObservation - For each pair of indices in the | neil_paul | NORMAL | 2024-03-17T06:45:19.936798+00:00 | 2024-03-17T06:47:00.336053+00:00 | 146 | false | # Approach\n<!-- Describe your approach to solving the problem. -->\nFirst, calculate frequencies of all the characters.\n\nObservation - For each pair of indices in the frequency array, we can either delete all characters of the lower frequency, or reduce just enough characters of the higher frequency as to make their... | 4 | 0 | ['Python3'] | 2 |
minimum-deletions-to-make-string-k-special | Binary Search || Prefix Sum || nlog(n) solution || Iterate over all possible values as the minimum | binary-search-prefix-sum-nlogn-solution-gm8u8 | Intuition\n Describe your first thoughts on how to solve this problem. \n\nAfter computing the frequency of characters in the string, we can convert the problem | hliu4 | NORMAL | 2024-03-17T04:24:43.097857+00:00 | 2024-03-17T23:18:50.275409+00:00 | 434 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\nAfter computing the frequency of characters in the string, we can convert the problem to the following:\n\nGiven an array nums and an integer k, find the minimum number of operations to make |nums[i] - nums[j]| <= k for any pair of i an... | 4 | 0 | ['Binary Search', 'Prefix Sum', 'Java'] | 0 |
minimum-deletions-to-make-string-k-special | Java Clean Solution || Weekly Contest | java-clean-solution-weekly-contest-by-sh-flki | Complexity\n- Time complexity:O(n+Soting + 26*26)\n Add your time complexity here, e.g. O(n) \n\n- Space complexity:O(26)\n Add your space complexity here, e.g. | Shree_Govind_Jee | NORMAL | 2024-03-17T04:03:03.470518+00:00 | 2024-03-17T04:03:03.470561+00:00 | 454 | false | # Complexity\n- Time complexity:$$O(n+Soting + 26*26)$$\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:$$O(26)$$\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\nclass Solution {\n public int minimumDeletions(String word, int k) {\n int[] freq = new int[26]... | 4 | 0 | ['Array', 'Python', 'Java', 'Python3'] | 2 |
minimum-deletions-to-make-string-k-special | Beats 💯% of all other | Beginner Friendly✅ | beats-of-all-other-beginner-friendly-by-0gexr | Intuition\n\n\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n- Create a freq map for every charecter\n- Sort the value of each cha | viresh_dev | NORMAL | 2024-03-17T05:22:50.420348+00:00 | 2024-03-17T05:22:50.420382+00:00 | 180 | false | # Intuition\n\n\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n- Create a freq map for every charecter\n- Sort the value of each charecter\n- chec... | 3 | 0 | ['Array', 'Hash Table', 'Greedy', 'Sorting', 'Python3'] | 2 |
minimum-deletions-to-make-string-k-special | Easy Java Solution | easy-java-solution-by-mayur0106-oyas | \n# Code\n\n\nclass Solution {\n public int minimumDeletions(String word, int k) {\n \n // store the frequency of the character\n Intege | mayur0106 | NORMAL | 2024-03-17T04:14:43.765684+00:00 | 2024-03-17T04:14:43.765712+00:00 | 249 | false | \n# Code\n```\n\nclass Solution {\n public int minimumDeletions(String word, int k) {\n \n // store the frequency of the character\n Integer []arr=new Integer[26];\n for(int i=0;i<arr.length;i++) {\n arr[i]=0;\n }\n for(int i=0;i<word.length();i++)\n {\n ... | 3 | 0 | ['Java'] | 0 |
minimum-deletions-to-make-string-k-special | 🔥 [CPP] | Freq Array (Max Freq) | cpp-freq-array-max-freq-by-rushi_mungse-cwgx | \n\n\nclass Solution {\npublic:\n int minimumDeletions(string word, int k) {\n int max_freq = 0;\n vector<int> freq(26);\n for(auto &w : | rushi_mungse | NORMAL | 2024-03-17T04:01:12.098681+00:00 | 2024-03-17T04:01:12.098730+00:00 | 369 | false | \n\n```\nclass Solution {\npublic:\n int minimumDeletions(string word, int k) {\n int max_freq = 0;\n vector<int> freq(26);\n for(auto &w : word) max_freq = max(max_freq, ++freq[w - \'a\']);\n \n int ans = INT_MAX;\n for(int make = 0; make <= max_freq; make++) {\n ... | 3 | 0 | ['C++'] | 1 |
minimum-deletions-to-make-string-k-special | Very easy DP Code || Intuitive || Explained | very-easy-dp-code-intuitive-explained-by-bvjb | Complexity\n- Time complexity: O(n + 2626) \n\n- Space complexity: O(2626) \n\n# Code\n\nclass Solution {\n int solve(int i, int j, int k, vector<int>& arr, | coder_rastogi_21 | NORMAL | 2024-03-20T15:03:39.051381+00:00 | 2024-03-20T15:03:39.051414+00:00 | 18 | false | # Complexity\n- Time complexity: $$O(n + 26*26)$$ \n\n- Space complexity: $$O(26*26)$$ \n\n# Code\n```\nclass Solution {\n int solve(int i, int j, int k, vector<int>& arr, vector<vector<int>>& dp)\n {\n if(i == j || arr[j] - arr[i] <= k) //base case\n return 0;\n \n if(dp[i][j] != ... | 2 | 0 | ['Dynamic Programming', 'C++'] | 0 |
minimum-deletions-to-make-string-k-special | Easy HashMap Java Solution | easy-hashmap-java-solution-by-deep_ptl-5xqo | Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n\n1) Initialize a HashMap freqMap to store the frequency of each characte | deep_ptl | NORMAL | 2024-03-17T22:14:47.101314+00:00 | 2024-03-17T22:14:47.101344+00:00 | 43 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n\n1) Initialize a HashMap freqMap to store the frequency of each character in the word.\n2) Iterate through each character in the input word and update the frequency count in the freqMap.\n3) For each character c in the word... | 2 | 0 | ['Array', 'Hash Table', 'Java'] | 0 |
minimum-deletions-to-make-string-k-special | DP | Alternative Approach | C++ | dp-alternative-approach-c-by-sparker_724-qji2 | It can be solved with simple brute force iteration. Here is an alternative approach whic uses basic dynamic programming\n# Intuition\nFor any two numbers of the | Sparker_7242 | NORMAL | 2024-03-17T08:08:31.923455+00:00 | 2024-03-17T08:08:31.923474+00:00 | 29 | false | It can be solved with simple brute force iteration. Here is an alternative approach whic uses basic dynamic programming\n# Intuition\nFor any two numbers of the frequency array, there are two possibilities - either remove the smaller one or decrease the larger one.\n\n\n# Approach\nWrite a recursive code for the two by... | 2 | 0 | ['C++'] | 0 |
minimum-deletions-to-make-string-k-special | Video Explanation (Along with a HARDER VERSION) | video-explanation-along-with-a-harder-ve-qwzx | Explanation \n\nClick here for the video\n\n# Code\n\nclass Solution {\npublic:\n int minimumDeletions(string word, int k) {\n vector<int> frq(26);\n | codingmohan | NORMAL | 2024-03-17T05:47:04.168850+00:00 | 2024-03-17T05:47:04.168883+00:00 | 93 | false | # Explanation \n\n[Click here for the video](https://youtu.be/NAWNz8caFK8)\n\n# Code\n```\nclass Solution {\npublic:\n int minimumDeletions(string word, int k) {\n vector<int> frq(26);\n for (auto c : word) frq[c-\'a\'] ++; \n \n int ans = 1e5;\n \n for (int l = 1; l <= ... | 2 | 0 | ['C++'] | 0 |
minimum-deletions-to-make-string-k-special | Intuitive recursion-based, whether to delete the least frequency one or keep it. | intuitive-recursion-based-whether-to-del-8leb | Intuition\nAfter sorting by the frequencies in descrending order, recursively consider keeping the letter with the least frequency or delete it.\n\n# Approach\n | practice_harder | NORMAL | 2024-03-17T04:17:29.684662+00:00 | 2024-03-17T06:43:06.507679+00:00 | 300 | false | # Intuition\nAfter sorting by the frequencies in descrending order, recursively consider keeping the letter with the least frequency or delete it.\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n1. Count the frequencies for 26 letters\n2. Sort the frequencies in descrending order\n3. Two ways to... | 2 | 0 | ['C++'] | 0 |
minimum-deletions-to-make-string-k-special | Only Using Map Explained ✅✔️ | only-using-map-explained-by-nandunk-mgnw | 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 | nandunk | NORMAL | 2024-03-17T04:08:38.364062+00:00 | 2024-03-17T04:08:38.364113+00:00 | 477 | 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-deletions-to-make-string-k-special | [Java Solution] Dynamic Programming With Clear Explanation | java-solution-dynamic-programming-with-c-a0lb | Intuition\nThe challenge involves balancing the frequencies of characters in a string to make it "k-special," where the frequency difference between any two cha | trtdytr | NORMAL | 2024-03-17T04:07:40.509011+00:00 | 2024-03-17T04:37:48.725879+00:00 | 111 | false | # Intuition\nThe challenge involves balancing the frequencies of characters in a string to make it "k-special," where the frequency difference between any two characters does not exceed `k`. The initial step is to understand the frequency distribution of characters, guiding the deletions needed to achieve this balance.... | 2 | 0 | ['Dynamic Programming', 'Java'] | 1 |
minimum-deletions-to-make-string-k-special | O(n) Solution | Rust | on-solution-rust-by-prog_jacob-kuch | Complexity\n- Time complexity: O(n)\n Add your time complexity here, e.g. O(n) \n\n- Space complexity: O(1)\n Add your space complexity here, e.g. O(n) \n\n# Co | Prog_Jacob | NORMAL | 2024-03-17T04:05:04.440808+00:00 | 2024-03-17T16:30:42.275672+00:00 | 15 | false | # Complexity\n- Time complexity: $$O(n)$$\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity: $$O(1)$$\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\nimpl Solution {\n pub fn minimum_deletions(word: String, k: i32) -> i32 {\n let mut map = [0; 26];\n l... | 2 | 0 | ['Rust'] | 0 |
minimum-deletions-to-make-string-k-special | check all ranges from 0 to k to (maxfreq - k) to (maxfreq) | check-all-ranges-from-0-to-k-to-maxfreq-qizje | 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 | batman654 | NORMAL | 2024-03-17T04:04:04.377691+00:00 | 2024-03-17T04:35:35.081055+00:00 | 130 | 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-deletions-to-make-string-k-special | Hasp Map Solution | hasp-map-solution-by-tlecodes-xbxg | 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 | tlecodes | NORMAL | 2024-10-08T11:46:18.983050+00:00 | 2024-10-08T11:46:18.983067+00:00 | 8 | 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-deletions-to-make-string-k-special | [Python] count freq | python-count-freq-by-pbelskiy-9h5b | \nclass Solution:\n def minimumDeletions(self, word: str, k: int) -> int:\n\n @cache\n def dfs(left, right):\n d = a[right] - a[left | pbelskiy | NORMAL | 2024-03-23T14:43:14.540076+00:00 | 2024-03-23T14:43:14.540093+00:00 | 6 | false | ```\nclass Solution:\n def minimumDeletions(self, word: str, k: int) -> int:\n\n @cache\n def dfs(left, right):\n d = a[right] - a[left]\n if d <= k:\n return 0\n\n return min(\n dfs(left + 1, right) + a[left], # remove\n df... | 1 | 0 | ['Python'] | 0 |
minimum-deletions-to-make-string-k-special | Leelavardhan's Java Solution 5ms | leelavardhans-java-solution-5ms-by-klu_2-79h1 | Intuition\n Describe your first thoughts on how to solve this problem. \n Calculate the frequency of each Character and store them into a array and do sort and | klu_2200031318 | NORMAL | 2024-03-23T11:02:29.342672+00:00 | 2024-03-23T11:02:29.342702+00:00 | 51 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n Calculate the frequency of each Character and store them into a array and do sort and follow the approach.\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\nThe question is about to delete minimum characters so that... | 1 | 0 | ['Java'] | 0 |
minimum-deletions-to-make-string-k-special | [C++ / Go] Sliding Window - O(n) time / O(26) space solution | c-go-sliding-window-on-time-o26-space-so-c03i | Approach\n- Count the number of occurrences of each character and then sort them in ascending order.\n- Iterate l through the counting array cnt:\n - Maintai | mikazuki4712 | NORMAL | 2024-03-19T17:05:12.931445+00:00 | 2024-03-19T17:05:12.931482+00:00 | 39 | false | # Approach\n- Count the number of occurrences of each character and then sort them in ascending order.\n- Iterate `l` through the counting array `cnt`:\n - Maintain a sliding window with 2 ends `l` and `r`, where `cnt[r] - cnt[l] <= k` and `r` is maximum.\n - Here we perform deletions to ensure that all occurrenc... | 1 | 0 | ['Sliding Window', 'C++', 'Go'] | 0 |
minimum-deletions-to-make-string-k-special | 7 Line Dynamic Programming Solution using Counter, O(n) runtime. | 7-line-dynamic-programming-solution-usin-psnk | I and J are restricted by O(26) for each since there is 26 char in the alphabet. So our runtime is O(n) from len(word) = n and spacetime is O(26 * 26) ~ O(1). \ | robert961 | NORMAL | 2024-03-18T19:38:37.544522+00:00 | 2024-03-18T19:38:37.544552+00:00 | 46 | false | I and J are restricted by O(26) for each since there is 26 char in the alphabet. So our runtime is O(n) from len(word) = n and spacetime is O(26 * 26) ~ O(1). \n# Code\n```\nclass Solution:\n def minimumDeletions(self, word: str, k: int) -> int:\n cnt = Counter(word)\n c = deque(sorted(list(cnt.values(... | 1 | 0 | ['Dynamic Programming', 'Python3'] | 0 |
minimum-deletions-to-make-string-k-special | Recursive C++ Solution | recursive-c-solution-by-anupsingh556-q3hb | Intuition\n- sort frequency array in descending order\n- we can either reduce minimum freq to 0 or make max freq to min(freq) + k\n- use recursion to find minim | anupsingh556 | NORMAL | 2024-03-18T09:07:36.673589+00:00 | 2024-03-18T09:07:36.673615+00:00 | 8 | false | # Intuition\n- sort frequency array in descending order\n- we can either reduce minimum freq to 0 or make max freq to min(freq) + k\n- use recursion to find minimum from both operations\n\n# Code\n```\nclass Solution {\npublic:\n int minimumDeletions(string word, int k) {\n vector<int> freq(26, 0);\n f... | 1 | 0 | ['C++'] | 0 |
minimum-deletions-to-make-string-k-special | Easy C++ solution | Beats 100% solutions | easy-c-solution-beats-100-solutions-by-a-j2a2 | Intuition\nCalculate the frequencies of all characters in the string. Since we can only delete a character (that is reduce the frequency of character), we can l | ankitjangir001 | NORMAL | 2024-03-18T04:21:17.919380+00:00 | 2024-03-18T04:21:17.919425+00:00 | 48 | false | # Intuition\nCalculate the frequencies of all characters in the string. Since we can only delete a character (that is reduce the frequency of character), we can loop for all frequencies of chars keeping an `target` value and adjusting all frequencies suitable to that `target`. If for a `target`, the freq of char is les... | 1 | 0 | ['C++'] | 0 |
minimum-deletions-to-make-string-k-special | Simple python3 solution | 85 ms - faster than 100.00% solutions | simple-python3-solution-85-ms-faster-tha-xyzq | Complexity\n- Time complexity: O(n + m) = O(n)\n Add your time complexity here, e.g. O(n) \n\n- Space complexity: O(m) = O(1)\n Add your space complexity here, | tigprog | NORMAL | 2024-03-17T23:17:25.778171+00:00 | 2024-03-17T23:17:25.778207+00:00 | 161 | false | # Complexity\n- Time complexity: $$O(n + m) = O(n)$$\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity: $$O(m) = O(1)$$\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\nwhere `m = 26` - number of lowercase English letters\n\n# Code\n``` python3 []\nclass Solution:\n def minimumDe... | 1 | 0 | ['Two Pointers', 'Sliding Window', 'Sorting', 'Counting', 'Python3'] | 0 |
minimum-deletions-to-make-string-k-special | ⚡Greedy || 🌟Sorting || 🔥HashMap | greedy-sorting-hashmap-by-adish_21-3ka5 | \n\n# Complexity\n\n- Time complexity:\nO(26 * 26)\n\n- Space complexity:\nO(26)\n\n\n# Code\n## Please Upvote if it helps\uD83E\uDD17\n\nclass Solution {\npubl | aDish_21 | NORMAL | 2024-03-17T07:36:23.041447+00:00 | 2024-03-17T08:44:49.870294+00:00 | 111 | false | \n\n# Complexity\n```\n- Time complexity:\nO(26 * 26)\n\n- Space complexity:\nO(26)\n```\n\n# Code\n## Please Upvote if it helps\uD83E\uDD17\n```\nclass Solution {\npublic:\n int minimumDeletions(string word, int k) {\n int n = word.size(), mini = INT_MAX;\n unordered_map<char, int> mp;\n for(au... | 1 | 0 | ['Greedy', 'Ordered Map', 'Sorting', 'C++'] | 0 |
minimum-deletions-to-make-string-k-special | Minimum Deletions to Make String K-Special | minimum-deletions-to-make-string-k-speci-mus9 | Intuition\nvery simple\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 | darthVader26 | NORMAL | 2024-03-17T05:43:33.619235+00:00 | 2024-03-17T05:43:33.619282+00:00 | 159 | false | # Intuition\nvery simple\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. $$O(n)$$ -->\n\n# Code\n```\nclass Solution {\n public int solve(i... | 1 | 0 | ['Dynamic Programming', 'Memoization', 'Java'] | 1 |
minimum-deletions-to-make-string-k-special | DP Memoization | dp-memoization-by-kushal1605-4fvo | 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 | Kushal1605 | NORMAL | 2024-03-17T05:18:28.200224+00:00 | 2024-03-17T05:18:28.200249+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 | ['Java'] | 0 |
minimum-deletions-to-make-string-k-special | Minimum Deletions to make String K-Special: Efficient HashMap Algo | minimum-deletions-to-make-string-k-speci-ywsa | Intuition\nThe intuition behind this code is to find the minimum number of deletions required to make the frequency of each character in the string word at most | madhusudanrathi99 | NORMAL | 2024-03-17T05:05:28.938889+00:00 | 2024-03-17T05:05:28.938907+00:00 | 58 | false | # Intuition\nThe intuition behind this code is to find the minimum number of deletions required to make the frequency of each character in the string `word` at most `k`.\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n1. **Initialize Variables**: \nThe code initializes a dictionary `... | 1 | 0 | ['Hash Table', 'String', 'Sorting', 'C#'] | 0 |
minimum-deletions-to-make-string-k-special | Explained-Solution with Intuition || AC ✅ | explained-solution-with-intuition-ac-by-g1sh1 | Intuition\n * We want to make freq[i] - freq[j] <=k for all i and j. \n * find frequency of all characters and sort them. \n * At max we have 26 characters s | satyam_9911 | NORMAL | 2024-03-17T04:28:10.007787+00:00 | 2024-03-17T04:33:29.564010+00:00 | 98 | false | # Intuition\n * We want to make ```freq[i] - freq[j] <=k ``` for all `i` and `j`. \n * find frequency of all characters and sort them. \n * At max we have 26 characters so , size of the `freq` can be of `26`.\n * Now , as we can only decrease the freq of characaters , so first we choose `freq[0]` as `minimum`.\n \... | 1 | 0 | ['Hash Table', 'Sorting', 'Prefix Sum', 'C++'] | 0 |
minimum-deletions-to-make-string-k-special | [With Explanation] Frequency + Sort | with-explanation-frequency-sort-by-manis-wrcj | Code Explanation\n\nFill the Frequency Array: Iterate through each character of the input word and increment the count in corresponding position.\n\nSort the Fr | manisreekar | NORMAL | 2024-03-17T04:22:45.705035+00:00 | 2024-03-17T04:56:44.159117+00:00 | 102 | false | # Code Explanation\n\n**Fill the Frequency Array:** Iterate through each character of the input word and increment the count in corresponding position.\n\n**Sort the Frequency Array:** The count array is sorted so that it simplifies the process of calculating the minimum deletions required.\n\n**Calculate Minimum Delet... | 1 | 0 | ['Java'] | 0 |
minimum-deletions-to-make-string-k-special | Python: frequencies of characters, linear solution | python-frequencies-of-characters-linear-kugvt | Intuition\n\nIt is clear that the string does not matter and you need to calculate the frequencies of each characters and sort them.\n\nLets assume that your ar | salvadordali | NORMAL | 2024-03-17T04:14:42.223532+00:00 | 2024-03-17T04:14:42.223554+00:00 | 33 | false | # Intuition\n\nIt is clear that the string does not matter and you need to calculate the frequencies of each characters and sort them.\n\nLets assume that your array is `1, 3, 5, 5, 7, 8, 9, 11, 14` and `k = 4`.\n\nIf you start at position `2`, then you need to remove all the frequencies below this position (cost will ... | 1 | 0 | ['Python'] | 1 |
minimum-deletions-to-make-string-k-special | Hashmap + Sort + Binary Search + Recursion JS | hashmap-sort-binary-search-recursion-js-6hw8a | \n# Code\n\n/**\n * @param {string} word\n * @param {number} k\n * @return {number}\n */\nvar minimumDeletions = function(word, k) {\n const map = new Map()\ | geeni_10 | NORMAL | 2024-03-17T04:04:27.365576+00:00 | 2024-03-17T04:04:27.365604+00:00 | 111 | false | \n# Code\n```\n/**\n * @param {string} word\n * @param {number} k\n * @return {number}\n */\nvar minimumDeletions = function(word, k) {\n const map = new Map()\n \n for (const char of word) {\n if (map.has(char)) {\n map.set(char, map.get(char) + 1)\n } else {\n map.set(char... | 1 | 0 | ['JavaScript'] | 2 |
minimum-deletions-to-make-string-k-special | C++ || EASY || Hashing || beginner friendly | c-easy-hashing-beginner-friendly-by-been-gd53 | \n\n# Code\n\nclass Solution {\npublic:\n int minimumDeletions(string word, int k) {\n // Count the frequencies of characters in the word\n vec | SprihaAnand | NORMAL | 2024-03-17T04:01:39.423026+00:00 | 2024-03-17T04:01:39.423059+00:00 | 73 | false | \n\n# Code\n```\nclass Solution {\npublic:\n int minimumDeletions(string word, int k) {\n // Count the frequencies of characters in the word\n vector<int> freq(26, 0);\n for (char c : word) {\n freq[c - \'a\']++;\n }\n \n // Extract the frequencies and sort them i... | 1 | 0 | ['Hash Table', 'Hash Function', 'C++'] | 0 |
minimum-deletions-to-make-string-k-special | Python3 solution || Beginner friendly | python3-solution-beginner-friendly-by-sy-o33k | \nclass Solution(object):\n def minimumDeletions(self, word, k):\n a=list(Counter(word).values())\n a.sort(reverse=True)\n ans=float("in | Syamkrishnareddypulagam | NORMAL | 2024-03-17T04:00:56.999743+00:00 | 2024-03-17T04:04:25.309300+00:00 | 221 | false | ```\nclass Solution(object):\n def minimumDeletions(self, word, k):\n a=list(Counter(word).values())\n a.sort(reverse=True)\n ans=float("inf")\n for i in range(len(a)):\n temp=0\n for j in a:\n """\n j can be in 3 cases. \n ... | 1 | 1 | ['Array', 'Hash Table', 'Math', 'Sorting', 'Python3'] | 1 |
minimum-deletions-to-make-string-k-special | Interesting efficient O(n) O(1) solution | interesting-efficient-on-o1-solution-by-mttbw | IntuitionApproachComplexity
Time complexity: O(n)
Space complexity: O(1)
Code | shmirrakhimov | NORMAL | 2025-03-27T18:34:48.567035+00:00 | 2025-03-27T18:34:48.567035+00:00 | 2 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
# Approach
<!-- Describe your approach to solving the problem. -->
# Complexity
- Time complexity: O(n)
<!-- Add your time complexity here, e.g. $$O(n)$$ -->
- Space complexity: O(1)
<!-- Add your space complexity here, e.g. $$O(n)$$ -->... | 0 | 0 | ['Hash Table', 'Sorting', 'Counting', 'JavaScript'] | 0 |
minimum-deletions-to-make-string-k-special | 🚀 Mastering Character Deletions: The Ultimate Strategy to Minimize Changes! 🔥Easy Explanation! | mastering-character-deletions-the-ultima-lco1 | IntuitionThe goal of this problem is to minimize the number of character deletions required to make the frequency difference between the most and least occurrin | Saurabhabd_360-45 | NORMAL | 2025-02-13T14:01:53.961030+00:00 | 2025-02-13T14:01:53.961030+00:00 | 4 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
The goal of this problem is to minimize the number of character deletions required to make the frequency difference between the most and least occurring characters at most k. This is done by removing characters with either the lowest or hig... | 0 | 0 | ['String', 'Dynamic Programming', 'Sorting', 'Counting', 'Java'] | 0 |
minimum-deletions-to-make-string-k-special | O(n) with sorting beating 100% | on-with-sorting-beating-100-by-lilongxue-vz92 | IntuitionCount the freq of each letter, and handle the freqs.Approach
Count the freq of each letter.
Sort the freqs.
For each index in freqs, we must try to dro | lilongxue | NORMAL | 2025-02-02T18:20:52.146052+00:00 | 2025-02-02T18:20:52.146052+00:00 | 4 | false | # Intuition
Count the freq of each letter, and handle the freqs.
# Approach
1. Count the freq of each letter.
2. Sort the freqs.
3. For each index in freqs, we must try to drop off the freqs to its left, and meanwhile, we must do some cutting from the right of the remaining array.
4. For each index, the above two opera... | 0 | 0 | ['JavaScript'] | 0 |
minimum-deletions-to-make-string-k-special | Memoization DP JAVA | memoization-dp-java-by-prachikumari-wdlf | IntuitionTarget : To get minDeletion to get the character freq in range k
In order to minimise :
we can either delete the character with least freq , hence del | PrachiKumari | NORMAL | 2025-01-09T07:35:25.376605+00:00 | 2025-01-09T07:35:25.376605+00:00 | 2 | false | # Intuition
Target : To get minDeletion to get the character freq in range k
In order to minimise :
- we can either `delete the character with least freq` , hence del of character with least freq
-` Delete the character with highest freq `
# Approach
- Get all character freq in Array
- Then Sort the freqArray in as... | 0 | 0 | ['Java'] | 0 |
minimum-deletions-to-make-string-k-special | Check all ranges | check-all-ranges-by-theabbie-zkxm | \nclass Solution:\n def minimumDeletions(self, word: str, k: int) -> int:\n n = len(word)\n ctr = Counter(word)\n res = float(\'inf\')\n | theabbie | NORMAL | 2024-12-24T06:17:07.944740+00:00 | 2024-12-24T06:17:07.944774+00:00 | 1 | false | ```\nclass Solution:\n def minimumDeletions(self, word: str, k: int) -> int:\n n = len(word)\n ctr = Counter(word)\n res = float(\'inf\')\n for minf in range(1, n + 1):\n maxf = minf + k\n curr = 0\n for c in ctr:\n if ctr[c] < minf:\n ... | 0 | 0 | ['Python'] | 0 |
minimum-deletions-to-make-string-k-special | Constant time-space complexity Solution ( sigma = size Lang = 26 [a-z] lowercase ) leveraging maps | constant-time-space-complexity-solution-6b7he | Intuition and ApproachSee problem descriptionComplexity
Time complexity:
O(1)
Space complexity:
O(1) ( Explicit )
O(1) ( Implicit )
Code | 2018hsridhar | NORMAL | 2024-12-21T20:49:25.473765+00:00 | 2024-12-21T20:49:25.473765+00:00 | 5 | false | # Intuition and Approach
See problem description
# Complexity
- Time complexity:
$$O(1)$$
- Space complexity:
$$O(1)$$ ( Explicit )
$$O(1)$$ ( Implicit )
# Code
```python3 []
'''
URL := https://leetcode.com/problems/minimum-deletions-to-make-string-k-special/description/
3085. Minimum Deletions to Make String K-Spec... | 0 | 0 | ['Python3'] | 0 |
minimum-deletions-to-make-string-k-special | Simple Java Solution | simple-java-solution-by-sakshikishore-nu6o | Code | sakshikishore | NORMAL | 2024-12-14T19:12:01.222292+00:00 | 2024-12-14T19:12:01.222292+00:00 | 3 | false | # Code\n```java []\nclass Solution {\n int result=Integer.MAX_VALUE;\n public void Solve(int arr[],int i, int j, int score,int k)\n {\n \n if(i>=j)\n {\n result=Math.min(result,score);\n return;\n }\n if(arr[j]-arr[i]<=k)\n {\n \n ... | 0 | 0 | ['Java'] | 0 |
minimum-deletions-to-make-string-k-special | Using hashmap Step wise commented code | using-hashmap-brute-force-by-sapilol-w3gx | null | LeadingTheAbyss | NORMAL | 2024-12-11T09:15:38.096598+00:00 | 2024-12-11T09:17:00.002265+00:00 | 4 | false | # Code\n```cpp []\nclass Solution {\npublic:\n int minimumDeletions(string word, int k) {\n // Step 1: Count the frequency of each character in the word\n unordered_map<char, int> mpp;\n for (char c : word) mpp[c]++; // Iterate through the word and increment the count of each character\n\n ... | 0 | 0 | ['C++'] | 0 |
minimum-deletions-to-make-string-k-special | C++ | c-by-tinachien-l2xk | \nclass Solution {\npublic:\n int minimumDeletions(string word, int k) {\n unordered_map<char, int>Map;\n for(auto ch : word){\n Map | TinaChien | NORMAL | 2024-12-04T10:32:14.450551+00:00 | 2024-12-04T10:32:14.450584+00:00 | 1 | false | ```\nclass Solution {\npublic:\n int minimumDeletions(string word, int k) {\n unordered_map<char, int>Map;\n for(auto ch : word){\n Map[ch]++;\n }\n vector<int>Count;\n for(auto [_, t] : Map){\n Count.push_back(t);\n }\n sort(Count.begin(), Count... | 0 | 0 | [] | 0 |
minimum-deletions-to-make-string-k-special | Python 1-liner | python-1-liner-by-dsapelnikov-h6kt | Code\npython3 []\nclass Solution:\n def minimumDeletions(self, word: str, k: int) -> int:\n return (freq := Counter(list(word)).values()) and min(sum( | dsapelnikov | NORMAL | 2024-11-21T22:46:09.625143+00:00 | 2024-11-21T22:46:09.625173+00:00 | 2 | false | # Code\n```python3 []\nclass Solution:\n def minimumDeletions(self, word: str, k: int) -> int:\n return (freq := Counter(list(word)).values()) and min(sum(f if f < ref else max(f - ref - k, 0) for f in freq) for ref in freq)\n``` | 0 | 0 | ['Python3'] | 0 |
minimum-deletions-to-make-string-k-special | brute force | brute-force-by-czxoxo-trd7 | 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 | czxoxo | NORMAL | 2024-11-18T22:07:24.425516+00:00 | 2024-11-18T22:07:24.425549+00:00 | 0 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ --... | 0 | 0 | ['C++'] | 0 |
minimum-deletions-to-make-string-k-special | Freq count - Java O(26*26)|O(26) | freq-count-java-o2626o26-by-wangcai20-5k7k | Intuition\n Describe your first thoughts on how to solve this problem. \n Build freq array. \n Loop over each freq, calulate count by adding all lower freq and | wangcai20 | NORMAL | 2024-10-25T16:19:56.058780+00:00 | 2024-10-25T16:21:26.832855+00:00 | 3 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n* Build freq array. \n* Loop over each freq, calulate `count` by adding all lower freq and higher freq beyond current freq + k.\n* Find the minimum `count`.\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Compl... | 0 | 0 | ['Java'] | 0 |
minimum-deletions-to-make-string-k-special | [Java] ✅ 100% ✅ SLIDING WINDOW ✅ FASTEST ✅ BEST ✅ CLEAN CODE | java-100-sliding-window-fastest-best-cle-r4ia | Approach\n1. Get the frequency of the characters: charFreq[].\n2. Sort and trim (do not include 0 freq) in the charFreq. Now charFreq is like [1,1,3,4,7,8,9]\n3 | StefanelStan | NORMAL | 2024-10-25T01:43:17.857953+00:00 | 2024-10-25T01:43:17.857978+00:00 | 2 | false | # Approach\n1. Get the frequency of the characters: charFreq[].\n2. Sort and trim (do not include 0 freq) in the charFreq. Now charFreq is like [1,1,3,4,7,8,9]\n3. Looking at K (max diff), we see this is actually a window. (left, right): [1,1,3,4,7,8,9]\n - at freq of 1 (left) , you can have a max freq of 1 + k (rig... | 0 | 0 | ['Two Pointers', 'Sliding Window', 'Java'] | 0 |
minimum-deletions-to-make-string-k-special | Beats 100% -----> Simple Approach | beats-100-simple-approach-by-vatan999-hh8v | Intuition\nThe problem involves reducing character frequencies to a manageable level by performing deletions. The goal is to minimize the number of deletions re | vatan999 | NORMAL | 2024-10-22T14:27:49.364482+00:00 | 2024-10-22T14:27:49.364518+00:00 | 0 | false | # Intuition\nThe problem involves reducing character frequencies to a manageable level by performing deletions. The goal is to minimize the number of deletions required to ensure that the difference between the frequencies of any two characters does not exceed a given threshold `k`.\n\n# Approach\n1. **Count Frequencie... | 0 | 0 | ['C++'] | 0 |
minimum-deletions-to-make-string-k-special | sexy + hot code | sexy-hot-code-by-rajan_singh5639-7gdk | 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 | rajan_singh5639 | NORMAL | 2024-09-17T13:50:15.617869+00:00 | 2024-09-17T13:50:15.617903+00:00 | 0 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ --... | 0 | 0 | ['C++'] | 0 |
minimum-deletions-to-make-string-k-special | HOT + SEXY CODE | hot-sexy-code-by-rajan_singh5639-eydk | 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 | rajan_singh5639 | NORMAL | 2024-09-17T11:50:06.074739+00:00 | 2024-09-17T11:50:06.074781+00:00 | 0 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ --... | 0 | 0 | ['C++'] | 0 |
minimum-deletions-to-make-string-k-special | 5 ms Beats 97.91% | 5-ms-beats-9791-by-pribyte1-2vqg | Code\njava []\nclass Solution {\n public int minimumDeletions(String word, int k) {\n int freq[]=new int[26];\n for(int i=0;i<word.length();i++ | pribyte1 | NORMAL | 2024-09-03T04:32:26.152557+00:00 | 2024-09-03T04:32:26.152598+00:00 | 1 | false | # Code\n```java []\nclass Solution {\n public int minimumDeletions(String word, int k) {\n int freq[]=new int[26];\n for(int i=0;i<word.length();i++){\n freq[word.charAt(i)-\'a\']++;\n }\n Arrays.sort(freq);\n int res=Integer.MAX_VALUE;\n for(int i=0;i<26;i++){\n ... | 0 | 0 | ['Java'] | 0 |
minimum-deletions-to-make-string-k-special | Beats 100% time | 86% space || easy solution using frequency array | beats-100-time-86-space-easy-solution-us-x1h6 | Complexity\n- Time complexity: 0(n)\n- Beats complexity: 0(everybody)\n- Space complexity: 0(n)\n\n# Code\ncpp []\nclass Solution {\npublic:\n int minimumDel | Nazar_Zakrevski | NORMAL | 2024-08-29T09:04:33.546853+00:00 | 2024-08-29T09:04:33.546879+00:00 | 4 | false | # Complexity\n- Time complexity: 0(n)\n- Beats complexity: 0(everybody)\n- Space complexity: 0(n)\n\n# Code\n```cpp []\nclass Solution {\npublic:\n int minimumDeletions(string word, int k) {\n ios::sync_with_stdio(false);\n cin.tie(nullptr);\n cout.tie(nullptr);\n vector<int> arr(26);\n ... | 0 | 0 | ['C++'] | 0 |
minimum-deletions-to-make-string-k-special | C++ || simple solution | c-simple-solution-by-rohityadav2002-6tty | 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 | RohitYadav2002 | NORMAL | 2024-08-20T15:09:52.597273+00:00 | 2024-08-20T15:09:52.597308+00:00 | 2 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ --... | 0 | 0 | ['C++'] | 0 |
minimum-deletions-to-make-string-k-special | Easy Solution C++ Greedy✅✅✅ | easy-solution-c-greedy-by-jayesh_06-aoy4 | \n\n# Code\n\nclass Solution {\npublic:\n\n int minimumDeletions(string word, int k) {\n int n = word.size();\n map<char, int> mp;\n for | Jayesh_06 | NORMAL | 2024-08-05T11:48:54.491459+00:00 | 2024-08-05T11:48:54.491496+00:00 | 0 | false | \n\n# Code\n```\nclass Solution {\npublic:\n\n int minimumDeletions(string word, int k) {\n int n = word.size();\n map<char, int> mp;\n for (int i = 0; i < n; i++) {\n mp[word[i]]++;\n }\n int i = 1, mini = n;\n int j = min(i + k, n);\n while (j <= n) {\n ... | 0 | 0 | ['Hash Table', 'String', 'Greedy', 'Sorting', 'Counting', 'C++'] | 0 |
minimum-deletions-to-make-string-k-special | O(n) Soln | on-soln-by-ozzyozbourne-gq7r | Complexity\n- Time complexity:O(n)\n Add your time complexity here, e.g. O(n) \n\n- Space complexity:O(1)\n Add your space complexity here, e.g. O(n) \n\n# Code | ozzyozbourne | NORMAL | 2024-08-01T00:08:48.506164+00:00 | 2024-08-01T00:10:15.295598+00:00 | 11 | false | # Complexity\n- Time complexity:$$O(n)$$\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:$$O(1)$$\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n``` python []\nclass Solution:\n def minimumDeletions(self, word: str, k: int) -> int:\n map, ans = [0] * 26, float(... | 0 | 0 | ['Java', 'Python3', 'Rust'] | 0 |
minimum-deletions-to-make-string-k-special | Sort char frequency and check window with k | sort-char-frequency-and-check-window-wit-6ian | 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 | linda2024 | NORMAL | 2024-07-16T17:26:33.465064+00:00 | 2024-07-16T17:26:33.465112+00:00 | 3 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ --... | 0 | 0 | ['C#'] | 0 |
minimum-deletions-to-make-string-k-special | Python || Simple and Easy Solution | python-simple-and-easy-solution-by-vilap-q3ix | 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 | vilaparthibhaskar | NORMAL | 2024-07-02T09:44:09.332254+00:00 | 2024-07-02T09:44:09.332296+00:00 | 8 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ --... | 0 | 0 | ['Hash Table', 'Greedy', 'Sorting', 'Python3'] | 0 |
minimum-deletions-to-make-string-k-special | More Intuitive Dp approach | more-intuitive-dp-approach-by-divyam_1-cqhu | 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 | Divyam_1 | NORMAL | 2024-06-23T02:41:34.694949+00:00 | 2024-06-23T02:41:34.694979+00:00 | 4 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ --... | 0 | 0 | ['C++'] | 0 |
minimum-deletions-to-make-string-k-special | Solution explained with diagrams. | solution-explained-with-diagrams-by-nan-y6h97 | Approach\n\n\n# Code\n\nclass Solution:\n def minimumDeletions(self, word: str, k: int) -> int:\n counts = Counter(word)\n values = list(sorted | nan-do | NORMAL | 2024-06-21T04:04:12.567502+00:00 | 2024-06-21T04:04:12.567530+00:00 | 8 | false | # Approach\n\n\n# Code\n```\nclass Solution:\n def minimumDeletions(self, word: str, k: int) -> int:\n counts = Counter(word)\n values = list(sorted(counts.values(), reverse=True))\n\n ... | 0 | 0 | ['Python3'] | 0 |
minimum-deletions-to-make-string-k-special | 3 solutions | Greedy, DFS/Rec, Memoization (DP) | 3-solutions-greedy-dfsrec-memoization-dp-nmhq | Solution-1: Greedy (WA -- 688 / 732 TCs passed -- 93.98%)\n+ IDEA: At each stage, we\'re greedily picking the smallest required change. i.e either mn (min eleme | shahsb | NORMAL | 2024-06-17T05:37:16.729432+00:00 | 2024-07-02T03:35:51.635719+00:00 | 7 | false | # Solution-1: Greedy (WA -- 688 / 732 TCs passed -- 93.98%)\n+ **IDEA:** At each stage, we\'re greedily picking the smallest required change. i.e either `mn` (min element) or `mx-mn-k` (removing diff from max element).\n+ **NOTE:** Kindly note that this solution only passes 93.98% of TCs. Intention behind sharing this ... | 0 | 0 | ['Dynamic Programming', 'Greedy', 'Depth-First Search', 'Recursion', 'Memoization', 'C'] | 0 |
minimum-deletions-to-make-string-k-special | TOO EASY DP | too-easy-dp-by-mohdibrahim12123-e5ki | Intuition\ndp\n\n# Approach\ndp\n\n# Complexity\n- Time complexity:\no(2626)\n\n- Space complexity:\no(2626)\n\n# Code\n\nclass Solution {\npublic:\n int hel | mohdibrahim12123 | NORMAL | 2024-06-16T17:41:26.270998+00:00 | 2024-06-16T17:41:26.271065+00:00 | 3 | false | # Intuition\ndp\n\n# Approach\ndp\n\n# Complexity\n- Time complexity:\no(26*26)\n\n- Space complexity:\no(26*26)\n\n# Code\n```\nclass Solution {\npublic:\n int helper(int i,int j,vector<int> &freq,int k,vector<vector<int>> &dp)\n {\n if(i==j || freq[j]-freq[i]<=k)\n return 0;\n if(dp[i][j]!=... | 0 | 0 | ['C++'] | 0 |
minimum-deletions-to-make-string-k-special | [Rust] Beats 100% -- it's not O(n^2) if you only have 26 letters | rust-beats-100-its-not-on2-if-you-only-h-tz7w | Intuition\nFirst thing, we need to count the number of frequency of each letter in word, and sort them so we can easily identify low frequencies and high freque | tiedyedvortex | NORMAL | 2024-06-15T14:10:31.713108+00:00 | 2024-06-15T14:10:31.713143+00:00 | 1 | false | # Intuition\nFirst thing, we need to count the number of frequency of each letter in word, and sort them so we can easily identify low frequencies and high frequencies.\n\nThen, when faced with a frequency distribution, we have two approaches to reduce it down to some range.\n\nOne option would be to keep the lowest-fr... | 0 | 0 | ['Rust'] | 0 |
minimum-deletions-to-make-string-k-special | 3085. Minimum Deletions to Make String K-Special | 3085-minimum-deletions-to-make-string-k-4l5od | 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 | Priyankaaaaa | NORMAL | 2024-06-10T14:47:17.942656+00:00 | 2024-06-10T14:47:17.942686+00:00 | 5 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ --... | 0 | 0 | ['Java'] | 0 |
minimum-deletions-to-make-string-k-special | SImple solution | simple-solution-by-risabhuchiha-16h0 | \n\nclass Solution {\n Integer[][]dp=new Integer[26][26];\n public int minimumDeletions(String word, int k) {\n int[]a=new int[26];\n for(ch | risabhuchiha | NORMAL | 2024-06-09T01:54:39.259569+00:00 | 2024-06-09T01:54:39.259601+00:00 | 4 | false | \n```\nclass Solution {\n Integer[][]dp=new Integer[26][26];\n public int minimumDeletions(String word, int k) {\n int[]a=new int[26];\n for(char ch:word.toCharArray()){\n a[ch-\'a\']++;\n}\n//if(word.length()==1)return 0;\nint ans=0;\nArrays.sort(a);\nint j=0;\nwhile(a[j]==0)j++;\n //ret... | 0 | 0 | ['Java'] | 0 |
minimum-deletions-to-make-string-k-special | simple java solution | simple-java-solution-by-lavlesh_pandey-jtbh | 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 | lavlesh_pandey | NORMAL | 2024-06-08T08:16:52.981763+00:00 | 2024-06-08T08:16:52.981784+00:00 | 2 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ --... | 0 | 0 | ['Array', 'String', 'Java'] | 0 |
minimum-deletions-to-make-string-k-special | ✅counting->DP || cPP | counting-dp-cpp-by-darkenigma-8con | \n# Code\n\nclass Solution {\npublic:\n\n int solve(vector<int>& arr,vector<vector<int>>& dp,int i,int j,int k){\n if(i==j || arr[j]-arr[i]<=k)return | darkenigma | NORMAL | 2024-06-04T18:13:16.628082+00:00 | 2024-06-04T18:13:16.628119+00:00 | 0 | false | \n# Code\n```\nclass Solution {\npublic:\n\n int solve(vector<int>& arr,vector<vector<int>>& dp,int i,int j,int k){\n if(i==j || arr[j]-arr[i]<=k)return 0;\n if(dp[i][j]!=-1)return dp[i][j];\n return dp[i][j]=min(arr[i]+solve(arr,dp,i+1,j,k),arr[j]-arr[i]-k+solve(arr,dp,i,j-1,k));\n }\n in... | 0 | 0 | ['Hash Table', 'String', 'Sorting', 'Counting', 'C++'] | 0 |
minimum-deletions-to-make-string-k-special | O(n)...bahu saras solutioon | onbahu-saras-solutioon-by-gungun16-rtlu | 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 | Gungun16 | NORMAL | 2024-05-16T04:41:08.991524+00:00 | 2024-05-16T04:41:08.991564+00:00 | 0 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ --... | 0 | 0 | ['Java'] | 0 |
minimum-deletions-to-make-string-k-special | gujju_ben_ni_gamzzat # Bahu saras O(n) ma solution... | gujju_ben_ni_gamzzat-bahu-saras-on-ma-so-vc3a | Intuition\n Describe your first thoughts on how to solve this problem. \nDarek character ni frequency par jau 6u and then check karu 6u ke ae character ni frequ | Gungun16 | NORMAL | 2024-05-16T03:52:30.242011+00:00 | 2024-05-16T03:52:30.242042+00:00 | 1 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nDarek character ni frequency par jau 6u and then check karu 6u ke ae character ni frequency jetli bija badha j characters ni frequency ne karvi hoi to su karvanu.\n\nSau pratham to tamare badha j characters ni frequecny ne calculate kari ... | 0 | 0 | ['Java'] | 0 |
minimum-deletions-to-make-string-k-special | Minimum prefix removal | minimum-prefix-removal-by-vokasik-u8ip | The long description asks us to find minimum number of chars to be removed so that the diff between any 2 frequencies in word <= k.\n2. Find frequencies with Co | vokasik | NORMAL | 2024-05-10T23:04:02.058579+00:00 | 2024-05-10T23:04:02.058598+00:00 | 5 | false | 1. The long description asks us to find **minimum** number of chars to be removed so that the diff between any 2 frequencies in `word` <= k.\n2. Find frequencies with Counter/whatever\n3. Now we need to remove minimum chars and satisfy the condition: `|any_freq - any_other_freq| <= k`\n\nLet\'s look at a few examples t... | 0 | 0 | ['Prefix Sum', 'Python', 'Python3'] | 0 |
minimum-deletions-to-make-string-k-special | Shortest, Easiest, Clean & Clear | To the Point & Beginners Friendly Approach (❤️ ω ❤️) | shortest-easiest-clean-clear-to-the-poin-657a | Code\n\nclass Solution {\npublic:\n int minimumDeletions(string word, int k) {\n vector<int>v(26,0);\n for(auto i:word)v[i-\'a\']++;\n i | Nitansh_Koshta | NORMAL | 2024-04-28T08:12:53.268313+00:00 | 2024-04-28T08:12:53.268339+00:00 | 2 | false | # Code\n```\nclass Solution {\npublic:\n int minimumDeletions(string word, int k) {\n vector<int>v(26,0);\n for(auto i:word)v[i-\'a\']++;\n int y=INT_MAX;\n for(int i=0;i<v.size();i++){\n int x=0;\n for(int j=0;j<v.size();j++){\n if(i!=j&&v[j]<v[i])x+=... | 0 | 0 | ['C++'] | 0 |
minimum-deletions-to-make-string-k-special | images | notes | c++ | greedy | images-notes-c-greedy-by-rahullodhi12-ehl5 | Notes\n\n\n\n# Code\n\nclass Solution {\npublic:\n int minimumDeletions(string word, int k) {\n int ans = word.length();\n vector<int> freq(26, | rahullodhi12 | NORMAL | 2024-04-22T12:39:43.270443+00:00 | 2024-04-22T12:39:43.270475+00:00 | 8 | false | # Notes\n\n\n\n# Code\n```\nclass Solution {\npublic:\n int... | 0 | 0 | ['Hash Table', 'String', 'Greedy', 'Sorting', 'Counting', 'C++'] | 0 |
minimum-deletions-to-make-string-k-special | O(N) Easiest java solution. | on-easiest-java-solution-by-marshadow54-z2o5 | 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 | marshadow54 | NORMAL | 2024-04-20T12:37:43.654390+00:00 | 2024-04-20T12:37:43.654435+00:00 | 3 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ --... | 0 | 0 | ['Java'] | 0 |
minimum-deletions-to-make-string-k-special | Move window start over frequencies, 94% speed | move-window-start-over-frequencies-94-sp-x7uh | image.png\n\n\n# Code\n\nclass Solution:\n def minimumDeletions(self, word: str, k: int) -> int:\n freq = list(Counter(word).values())\n ans = | evgenysh | NORMAL | 2024-04-17T15:20:10.753257+00:00 | 2024-04-17T15:20:10.753294+00:00 | 9 | false | image.png\n\n\n# Code\n```\nclass Solution:\n def minimumDeletions(self, word: str, k: int) -> int:\n freq = list(Counter(word).values())\n ans = len(word)\n for start in set(freq):\... | 0 | 0 | ['Python3'] | 0 |
minimum-deletions-to-make-string-k-special | Proof of correctness | proof-of-correctness-by-ram_raghav-byn8 | As suggested by the hint, the following result can be used to solve the problem.\n\n*\nLet $new\_word$ be a k-simple word formed after perfoming the minimum num | Ram_Raghav | NORMAL | 2024-04-13T23:55:59.339579+00:00 | 2024-04-14T00:03:28.318445+00:00 | 11 | false | As suggested by the hint, the following result can be used to solve the problem.\n\n****\nLet $new\\_word$ be a k-simple word formed after perfoming the minimum number of deletions from word. Then there exists a character $c$ in $new\\_word$ that is never deleted. i.e., its frequency in $new\\_word$ is equal to its fre... | 0 | 0 | ['Python3'] | 0 |
minimum-deletions-to-make-string-k-special | sliding window python | sliding-window-python-by-russellleung-6w00 | class Solution:\n def minimumDeletions(self, word: str, k: int) -> int:\n \n counter=Counter(word)\n values=sorted(counter.values())\n | russellleung | NORMAL | 2024-04-10T03:44:45.261896+00:00 | 2024-04-10T03:44:45.261921+00:00 | 0 | false | class Solution:\n def minimumDeletions(self, word: str, k: int) -> int:\n \n counter=Counter(word)\n values=sorted(counter.values())\n i=0\n total=0\n ans=float("inf")\n for j in range(len(values)):\n while i<len(values) and values[i]-values[j]<=k:\n ... | 0 | 0 | [] | 0 |
minimum-deletions-to-make-string-k-special | 👌Runtime 14 ms Beats 77.08% of users with Go | runtime-14-ms-beats-7708-of-users-with-g-kgdr | Code\n\nfunc minimumDeletions(word string, k int) int {\n\tfreq := make([]int, 26)\n\tfor _, ch := range word {\n\t\tfreq[ch-\'a\']++\n\t}\n\n\tsort.Ints(freq)\ | pvt2024 | NORMAL | 2024-04-09T01:20:26.343986+00:00 | 2024-04-09T01:20:26.344015+00:00 | 5 | false | # Code\n```\nfunc minimumDeletions(word string, k int) int {\n\tfreq := make([]int, 26)\n\tfor _, ch := range word {\n\t\tfreq[ch-\'a\']++\n\t}\n\n\tsort.Ints(freq)\n\tminDel := int(^uint(0) >> 1) // MaxInt\n\n\tfor i := len(freq) - 1; i >= 0; i-- {\n\t\ttrgt := freq[i]\n\t\tdels := 0\n\n\t\tfor _, f := range freq {\n\... | 0 | 0 | ['Go'] | 0 |
minimum-deletions-to-make-string-k-special | C++ | Greedy | c-greedy-by-pikachuu-79nr | Code\n\nclass Solution {\npublic:\n int minimumDeletions(string word, int k) {\n int freq[26] = {0};\n for(int i = 0; i < word.size(); i++) {\n | pikachuu | NORMAL | 2024-04-05T16:43:15.847611+00:00 | 2024-04-05T16:43:15.847643+00:00 | 7 | false | # Code\n```\nclass Solution {\npublic:\n int minimumDeletions(string word, int k) {\n int freq[26] = {0};\n for(int i = 0; i < word.size(); i++) {\n freq[word[i] - \'a\']++;\n }\n int ans = INT_MAX;\n for(int i = 0; i < 26; i++) {\n int del = 0;\n f... | 0 | 0 | ['String', 'Greedy', 'Counting', 'C++'] | 0 |
minimum-deletions-to-make-string-k-special | [C++][Prefix sum] | cprefix-sum-by-mumrocks-t70n | \nclass Solution {\npublic:\n int minimumDeletions(string word, int k) {\n /*\n a->1 b->2 c->3 d->5\n */\n vector<int> cnt(26);\n | MumRocks | NORMAL | 2024-04-03T14:09:15.762038+00:00 | 2024-04-03T14:09:15.762066+00:00 | 2 | false | ```\nclass Solution {\npublic:\n int minimumDeletions(string word, int k) {\n /*\n a->1 b->2 c->3 d->5\n */\n vector<int> cnt(26);\n for(const auto w: word) cnt[w-\'a\']++;\n \n vector<int> dp;\n for (int i=0;i<26;i++) if (cnt[i]>0) dp.push_back(cnt[i]);\n ... | 0 | 0 | [] | 0 |
minimum-deletions-to-make-string-k-special | Easy C++ Prefix sum + Greedy | easy-c-prefix-sum-greedy-by-parteek_code-w707 | \n\n# Code\n\nclass Solution {\npublic:\n int minimumDeletions(string word, int k) {\n int n = word.size();\n if(n<=1) return 0;\n vecto | parteek_coder | NORMAL | 2024-04-03T04:39:01.506368+00:00 | 2024-04-03T04:39:01.506397+00:00 | 6 | false | \n\n# Code\n```\nclass Solution {\npublic:\n int minimumDeletions(string word, int k) {\n int n = word.size();\n if(n<=1) return 0;\n vector<int>alpha(26,0);\n vector<int>prefix(26,0);\n for(auto ch: word){\n alpha[ch-\'a\']++;\n }\n\n sort(alpha.begin(),al... | 0 | 0 | ['C++'] | 0 |
minimum-deletions-to-make-string-k-special | Python Medium | python-medium-by-lucasschnee-w2ac | \nclass Solution:\n def minimumDeletions(self, word: str, k: int) -> int:\n c = Counter(word)\n arr = list(c.values())\n arr.sort()\n | lucasschnee | NORMAL | 2024-04-03T02:03:36.413365+00:00 | 2024-04-03T02:03:36.413416+00:00 | 9 | false | ```\nclass Solution:\n def minimumDeletions(self, word: str, k: int) -> int:\n c = Counter(word)\n arr = list(c.values())\n arr.sort()\n N = len(arr)\n \'\'\'\n [1, 2]\n \n can also make chars 0\n \'\'\'\n # median = arr[N//2]\n # if N % 2 ... | 0 | 0 | ['Python3'] | 0 |
minimum-deletions-to-make-string-k-special | scala solution | scala-solution-by-vititov-62bg | \nobject Solution {\n def minimumDeletions(word: String, k: Int): Int = {\n lazy val freqs = word.groupMapReduce(identity)(_ => 1)(_ + _).values.to(List)\n | vititov | NORMAL | 2024-03-30T20:59:23.984248+00:00 | 2024-03-30T20:59:23.984282+00:00 | 2 | false | ```\nobject Solution {\n def minimumDeletions(word: String, k: Int): Int = {\n lazy val freqs = word.groupMapReduce(identity)(_ => 1)(_ + _).values.to(List)\n freqs.distinct.map{f => freqs.collect{\n case g if g<f => g\n case g if g-f > k => g-f-k \n }.sum\n }.min\n }\n}\n``` | 0 | 0 | ['Scala'] | 0 |
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