3360. Minimum Deletions To Make String K Special¶

Difficulty: Medium

3360. Minimum Deletions to Make String K-Special

Medium

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¶

class Solution {
public:
    int minimumDeletions(string word, int k) {
        vector<int> freq(26, 0);
        for (char c : word) freq[c - 'a']++;
        int res = word.size();
        for (int j = 0; j < 26; j++) {
            int mn = freq[j], cnt = 0;
            for (int i : freq) {
                if (i > mn + k) 
                    cnt += i - (mn + k);
                else if (i < mn) 
                    cnt += i;
            }
            res = min(res, cnt);
        }
        return res;
    }
};

Complexity Analysis¶

Time Complexity: O(?)
Space Complexity: O(?)

Approach¶

Detailed explanation of the approach will be added here