Minimum Deletions to Make String K-Special

Problem Description

You are given a string word and an integer k. We consider word to be k-special if |freq(word[i]) - freq(word[j])| <= k for all indices i and j in the string. Here, freq(x) denotes the frequency of the character x in word. Return the minimum number of characters you need to delete to make word k-special.

Key Insights

• The problem involves character frequency counts and requires ensuring that the difference in frequencies of any two characters does not exceed k.
• A character frequency distribution can be used to determine how many characters need to be deleted to achieve the k-special condition.
• A greedy approach can be used to minimize deletions by focusing on the most frequent characters.

Space and Time Complexity

Time Complexity: O(n + m log m), where n is the length of the string and m is the number of unique characters (due to sorting). Space Complexity: O(m), where m is the number of unique characters stored in the frequency map.

Solution

To solve the problem, we can follow these steps:

1. Count the frequency of each character using a hash table (dictionary).
2. Store the frequencies in a list and sort them.
3. Use a sliding window approach to find the longest contiguous subarray of frequencies that meets the k-special condition.
4. Calculate the minimum deletions needed to achieve the k-special condition by determining how many characters are not included in this subarray.

We utilize a hash table (for frequency counting) and sorting (for arranging frequencies) to implement this solution effectively.

Code Solutions