Introduction

In this blog post, we will explore how to solve the LeetCode problem "1328" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

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.

Explanation

Here's the solution:

• Find the first non-'a' character from the beginning: Iterate through the first half of the palindrome. The first character that's not 'a' is the optimal place to make a change to achieve the smallest lexicographical order. Replace it with 'a'.
• Handle the all 'a' case: If the string consists only of 'a's, change the last character to 'b'. This ensures it's not a palindrome and is lexicographically smallest.

• Edge case for single-character palindromes: If the length is 1, return an empty string, as no change can make it non-palindromic.

• Time Complexity: O(n), where n is the length of the palindrome.

• Space Complexity: O(n), converting the string to list, or O(1) if modifying the string in place in languages that support this.

Code

    def breakPalindrome(palindrome: str) -> str:
    """
    Breaks a palindromic string to the lexicographically smallest non-palindromic string.
    """
    n = len(palindrome)
    if n <= 1:
        return ""

    chars = list(palindrome)  # Convert to list for easy modification

    for i in range(n // 2):
        if chars[i] != 'a':
            chars[i] = 'a'
            return "".join(chars)

    chars[n - 1] = 'b'  # If all chars are 'a', change the last one to 'b'
    return "".join(chars)