LeetCode 1737: Change Minimum Characters to Satisfy One of Three Conditions Solution

Master LeetCode problem 1737 (Change Minimum Characters to Satisfy One of Three Conditions), a medium 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.

1737. Change Minimum Characters to Satisfy One of Three Conditions

Medium

Hash Table String Counting Prefix Sum

Problem Explanation

Explanation:

To solve this problem, we can iterate over all possible letters and find the best way to achieve one of the three conditions. We need to consider changing letters in both strings a and b to satisfy the conditions. We can calculate the required operations for each condition separately and return the minimum number of operations needed.

Calculate the number of operations needed to make every letter in a strictly less than every letter in b.
Calculate the number of operations needed to make every letter in b strictly less than every letter in a.
Calculate the number of operations needed to make both a and b consist of only one distinct letter.
Return the minimum number of operations among the three conditions.

Solution Code

class Solution {
    public int minCharacters(String a, String b) {
        int[] freqA = new int[26];
        int[] freqB = new int[26];
        
        for (char c : a.toCharArray()) {
            freqA[c - 'a']++;
        }
        
        for (char c : b.toCharArray()) {
            freqB[c - 'a']++;
        }
        
        int minOps = Integer.MAX_VALUE;
        
        for (int i = 0; i < 26; i++) {
            minOps = Math.min(minOps, a.length() + b.length() - freqA[i] - freqB[i]);
            if (i > 0) {
                freqA[i] += freqA[i - 1];
                freqB[i] += freqB[i - 1];
            }
            if (i < 25) {
                minOps = Math.min(minOps, a.length() - freqA[i] + freqB[i]);
                minOps = Math.min(minOps, b.length() - freqB[i] + freqA[i]);
            }
        }
        
        return minOps;
    }
}

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