3085. Minimum Deletions to Make String K-Special

Problem Link

Description

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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, and |y| denotes the absolute value of y.

Return the minimum number of characters you need to delete to make word k-special.

 

Example 1:

Input: word = "aabcaba", k = 0

Output: 3

Explanation: We can make word 0-special by deleting 2 occurrences of "a" and 1 occurrence of "c". Therefore, word becomes equal to "baba" where freq('a') == freq('b') == 2.

Example 2:

Input: word = "dabdcbdcdcd", k = 2

Output: 2

Explanation: We can make word 2-special by deleting 1 occurrence of "a" and 1 occurrence of "d". Therefore, word becomes equal to "bdcbdcdcd" where freq('b') == 2, freq('c') == 3, and freq('d') == 4.

Example 3:

Input: word = "aaabaaa", k = 2

Output: 1

Explanation: We can make word 2-special by deleting 1 occurrence of "b". Therefore, word becomes equal to "aaaaaa" where each letter's frequency is now uniformly 6.

 

Constraints:

• 1 <= word.length <= 105
• 0 <= k <= 105
• word consists only of lowercase English letters.

Solution

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Python3

class Solution:
    def minimumDeletions(self, word: str, k: int) -> int:
        N = len(word)
        freq = [0] * 26
        res = N
 
        for x in word:
            freq[ord(x) - ord('a')] += 1
        
        for lo in range(N + 1):
            upper = lo + k
            count = 0
 
            for v in freq:
                if v < lo:
                    count += v
                elif v > upper:
                    count += v - upper
            
            res = min(res, count)
        
        return res