LeetCode 3088: Make String Anti-palindrome Solution

Master LeetCode problem 3088 (Make String Anti-palindrome), 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.

3088. Make String Anti-palindrome

Hard

String Greedy Sorting Counting Sort

Problem Explanation

Explanation

To solve this problem, we need to find the lexicographically smallest anti-palindrome string that can be formed from the given string. An anti-palindrome string is a string that is not a palindrome. We can achieve this by iterating through the input string and changing characters if needed to make it an anti-palindrome.

We iterate from left to right and also from right to left in parallel to check if the characters at the corresponding positions are the same.
If a character matches at both ends, we replace one of them with a character that is lexicographically smaller but not equal to the character on the opposite end.
If we cannot find such a character, we replace the character at the right end with 'a', as it is the smallest lexicographically.

The time complexity of this solution is O(n) where n is the length of the input string. The space complexity is O(n) as we are storing the modified string.

Solution Code

class Solution {
    public String makeAntiPalindrome(String s) {
        char[] chars = s.toCharArray();
        int n = chars.length;

        for (int i = 0, j = n - 1; i < j; i++, j--) {
            if (chars[i] == chars[j]) {
                if (chars[i] == 'a') {
                    chars[j] = 'b';
                } else {
                    chars[j] = 'a';
                }
            }
        }

        return String.valueOf(chars);
    }
}

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