1328. Break a Palindrome

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Description

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 an empty 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'.

 

Example 1:

Input: palindrome = "abccba"
Output: "aaccba"
Explanation: There are many ways to make "abccba" not a palindrome, such as "zbccba", "aaccba", and "abacba".
Of all the ways, "aaccba" is the lexicographically smallest.

Example 2:

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.

Solutions

Solution 1: Greedy

First, we check if the length of the string is $1$. If it is, we directly return an empty string.

Otherwise, we traverse the first half of the string from left to right, find the first character that is not 'a', and change it to 'a'. If no such character exists, we change the last character to 'b'.

The time complexity is $O(n)$, and the space complexity is $O(n)$. Here, $n$ is the length of the string.

Python3

class Solution:
    def breakPalindrome(self, palindrome: str) -> str:
        n = len(palindrome)
        if n == 1:
            return ""
        s = list(palindrome)
        i = 0
        while i < n // 2 and s[i] == "a":
            i += 1
        if i == n // 2:
            s[-1] = "b"
        else:
            s[i] = "a"
        return "".join(s)

Java

class Solution {
    public String breakPalindrome(String palindrome) {
        int n = palindrome.length();
        if (n == 1) {
            return "";
        }
        char[] s = palindrome.toCharArray();
        int i = 0;
        while (i < n / 2 && s[i] == 'a') {
            ++i;
        }
        if (i == n / 2) {
            s[n - 1] = 'b';
        } else {
            s[i] = 'a';
        }
        return String.valueOf(s);
    }
}

C++

class Solution {
public:
    string breakPalindrome(string palindrome) {
        int n = palindrome.size();
        if (n == 1) {
            return "";
        }
        int i = 0;
        while (i < n / 2 && palindrome[i] == 'a') {
            ++i;
        }
        if (i == n / 2) {
            palindrome[n - 1] = 'b';
        } else {
            palindrome[i] = 'a';
        }
        return palindrome;
    }
};

Go

func breakPalindrome(palindrome string) string {
	n := len(palindrome)
	if n == 1 {
		return ""
	}
	i := 0
	s := []byte(palindrome)
	for i < n/2 && s[i] == 'a' {
		i++
	}
	if i == n/2 {
		s[n-1] = 'b'
	} else {
		s[i] = 'a'
	}
	return string(s)
}

TypeScript

function breakPalindrome(palindrome: string): string {
    const n = palindrome.length;
    if (n === 1) {
        return '';
    }
    const s = palindrome.split('');
    let i = 0;
    while (i < n >> 1 && s[i] === 'a') {
        i++;
    }
    if (i == n >> 1) {
        s[n - 1] = 'b';
    } else {
        s[i] = 'a';
    }
    return s.join('');
}

Rust

impl Solution {
    pub fn break_palindrome(palindrome: String) -> String {
        let n = palindrome.len();
        if n == 1 {
            return "".to_string();
        }
        let mut s: Vec<char> = palindrome.chars().collect();
        let mut i = 0;

        while i < n / 2 && s[i] == 'a' {
            i += 1;
        }

        if i == n / 2 {
            s[n - 1] = 'b';
        } else {
            s[i] = 'a';
        }

        s.into_iter().collect()
    }
}