Problem

Given a palindromic string of lowercase English letters palindrome, replace exactly one character with any lowercase English letter so that the resulting string is not a palindrome and that it is the lexicographically smallest one possible.

Return the resulting string. If there is no way to replace a character to make it not a palindrome, return anempty string.

A string a is lexicographically smaller than a string b (of the same length) if in the first position where a and b differ, a has a character strictly smaller than the corresponding character in b. For example, "abcc" is lexicographically smaller than "abcd" because the first position they differ is at the fourth character, and 'c' is smaller than 'd'.

Examples

Example 1

1
2
3
4

Input: palindrome = "abccba"
Output: "aaccba"
Explanation: There are many ways to make "abccba" not a palindrome, such as "_z_ bccba", "a _a_ ccba", and "ab _a_ cba".
Of all the ways, "aaccba" is the lexicographically smallest.

Example 2

1
2
3

Input: palindrome = "a"
Output: ""
Explanation: There is no way to replace a single character to make "a" not a palindrome, so return an empty string.

Constraints

  • 1 <= palindrome.length <= 1000
  • palindrome consists of only lowercase English letters.

Solution

Method 1 – Greedy Replacement

Intuition

To make the palindrome not a palindrome and lexicographically smallest, change the first non-‘a’ character in the first half to ‘a’. If all are ‘a’, change the last character to ‘b’.

Approach

  1. If the string length is 1, return “”.
  2. For each character in the first half, if it is not ‘a’, replace it with ‘a’ and return.
  3. If all are ‘a’, change the last character to ‘b’.

Code

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12

class Solution:
    def breakPalindrome(self, palindrome: str) -> str:
        n = len(palindrome)
        if n == 1:
            return ""
        arr = list(palindrome)
        for i in range(n // 2):
            if arr[i] != 'a':
                arr[i] = 'a'
                return ''.join(arr)
        arr[-1] = 'b'
        return ''.join(arr)
 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15

class Solution {
    public String breakPalindrome(String palindrome) {
        int n = palindrome.length();
        if (n == 1) return "";
        char[] arr = palindrome.toCharArray();
        for (int i = 0; i < n / 2; i++) {
            if (arr[i] != 'a') {
                arr[i] = 'a';
                return new String(arr);
            }
        }
        arr[n - 1] = 'b';
        return new String(arr);
    }
}

Complexity

  • ⏰ Time complexity: O(n)
  • 🧺 Space complexity: O(n)