LeetCode 3389: Minimum Operations to Make Character Frequencies Equal Solution

Master LeetCode problem 3389 (Minimum Operations to Make Character Frequencies Equal), a hard challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

3389. Minimum Operations to Make Character Frequencies Equal

Hard

Hash Table String Dynamic Programming Counting Enumeration

Problem Explanation

Explanation:

To solve this problem, we can iterate through the input string s and count the frequency of each character. Then, we find the most frequent character and calculate the total number of operations required to make all characters occur the same number of times as the most frequent character. The operations can be either deletion, insertion, or changing a character to its next letter in the alphabet.

Count the frequency of each character in the input string s.
Find the most frequent character.
Calculate the total number of operations required to make all characters occur the same number of times as the most frequent character.

Time Complexity: O(n) where n is the length of the input string s. Space Complexity: O(1) since we are using a fixed-size array to count the frequencies of characters.

Solution Code

class Solution {
    public int minOperations(String s) {
        int[] freq = new int[26];
        for (char c : s.toCharArray()) {
            freq[c - 'a']++;
        }
        
        int maxFreq = 0;
        for (int f : freq) {
            maxFreq = Math.max(maxFreq, f);
        }
        
        int operations = 0;
        for (int f : freq) {
            if (f > 0 && f < maxFreq) {
                operations += f;
            } else if (f > maxFreq) {
                operations += f - maxFreq;
            }
        }
        
        return operations;
    }
}

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