LeetCode-in-Java

1737. Change Minimum Characters to Satisfy One of Three Conditions

Medium

You are given two strings a and b that consist of lowercase letters. In one operation, you can change any character in a or b to any lowercase letter.

Your goal is to satisfy one of the following three conditions:

• Every letter in a is strictly less than every letter in b in the alphabet.
• Every letter in b is strictly less than every letter in a in the alphabet.
• Both a and b consist of only one distinct letter.

Return the minimum number of operations needed to achieve your goal.

Example 1:

Input: a = “aba”, b = “caa”

Output: 2

Explanation: Consider the best way to make each condition true:

1) Change b to “ccc” in 2 operations, then every letter in a is less than every letter in b.

2) Change a to “bbb” and b to “aaa” in 3 operations, then every letter in b is less than every letter in a.

3) Change a to “aaa” and b to “aaa” in 2 operations, then a and b consist of one distinct letter. The best way was done in 2 operations (either condition 1 or condition 3).

Example 2:

Input: a = “dabadd”, b = “cda”

Output: 3

Explanation: The best way is to make condition 1 true by changing b to “eee”.

Constraints:

• 1 <= a.length, b.length <= 105
• a and b consist only of lowercase letters.

Solution

public class Solution {
    public int minCharacters(String s1, String s2) {
        int[] a = new int[26];
        int[] b = new int[26];
        int l1 = s1.length();
        int l2 = s2.length();
        for (char i : s1.toCharArray()) {
            a[i - 'a']++;
        }
        for (char i : s2.toCharArray()) {
            b[i - 'a']++;
        }
        int min = Integer.MAX_VALUE;
        int t1 = 0;
        int t2 = 0;
        int max = -1;
        for (int i = 0; i < 25; i++) {
            t1 += a[i];
            t2 += b[i];
            min = Math.min(min, Math.min(t1 + l2 - t2, t2 + l1 - t1));
            max = Math.max(max, a[i] + b[i]);
        }
        max = Math.max(max, a[25] + b[25]);
        return Math.min(min, l1 + l2 - max);
    }
}

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