Easy C++ Solution using Dynamic Programming and Sorting
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
# 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 character in the string `word`.\n2. Sort the frequencies in ascending order.\n3. Use dynamic programming to find the minimum number of characters needed to delete to make the substring k-special.\n4. If the absolute difference between frequencies of characters at indices `i` and `j` is less than or equal to `k`, no characters need to be deleted.\n5. Otherwise, calculate the number of characters needed to delete to make the substring k-special.\n6. Return the minimum number of characters needed to delete.\n\n# Complexity\n- Time complexity: O(N) as we are traversing the string.\n- Space complexity: O(26^2) = O(1) since the DP table has a fixed size and additional space is used for storing intermediate results, which doesn\'t scale with the input size.\n\n# Code\n```cpp\nclass Solution {\npublic:\n    // Helper function to recursively solve the problem using dynamic programming\n    int minDeletions(int i, int j, int A[], int k, vector<vector<int>>& dp) {\n        // Base case: If the range [i, j] is empty or has only one element,\n        // no characters need to be deleted, so return 0.\n        if (i >= j) return 0;\n        \n        // If the frequency of the character at index i is zero, skip it.\n        if (A[i] == 0) return minDeletions(i + 1, j, A, k, dp);\n        \n        // If the absolute difference between frequencies of characters at indices i and j\n        // is less than or equal to k, no characters need to be deleted, so return 0.\n        if (A[j] - A[i] <= k) return 0;\n        \n        // If the result for the current range [i, j] is already calculated, return it.\n        if (dp[i][j] != -1) return dp[i][j];\n        \n        // Calculate the number of characters needed to delete to make the substring k-special.\n        // Two options:\n        // 1. Delete characters from the left side of the substring.\n        // 2. Delete characters from the right side of the substring.\n        int takeLeft = A[i] + minDeletions(i + 1, j, A, k, dp);\n        int takeRight = A[j] - A[i] - k + minDeletions(i, j - 1, A, k, dp);\n        \n        // Store the minimum number of characters needed to delete for the current range [i, j].\n        dp[i][j] = min(takeLeft, takeRight);\n        \n        // Return the minimum number of characters needed to delete for the current range [i, j].\n        return dp[i][j];\n    }\n    \n    // Main function to find the minimum number of characters to delete\n    int minimumDeletions(string word, int k) {\n        // Array to store the frequencies of characters\n        int A[26] = {0};\n        // Count the frequencies of characters in the word\n        for (char ch : word) {\n            A[ch - \'a\']++;\n        }\n        // Sort the frequencies in ascending order\n        sort(A, A + 26);\n        \n        // Initialize the indices i and j\n        int i = 0, j = 25;\n        \n        // Initialize the DP table with -1 (indicating that the result is not calculated yet)\n        vector<vector<int>> dp(26, vector<int>(26, -1));\n        \n        // Call the helper function to solve the problem\n        return minDeletions(i, j, A, k, dp);\n    }\n};\n```
['C++']