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In this approach, we iterate through the first half of the string and try to replace the first non-'a' character with 'a'. This ensures that we're making the lexicographically smallest change as early as possible in the string. If we can't find a character that is not 'a' in the first half, we replace the last character with 'b' to ensure it's not a palindrome.
Time Complexity: O(n)
Space Complexity: O(1)
1function breakPalindrome(palindrome) {
2 if (palindrome.length === 1) return "";
3
4 let arr = palindrome.split('');
5 for (let i = 0; i < Math.floor(palindrome.length / 2); i++) {
6 if (arr[i] !== 'a') {
7 arr[i] = 'a';
8 return arr.join('');
9 }
10 }
11
12 arr[arr.length - 1] = 'b';
13 return arr.join('');
14}In JavaScript, we split the string into an array for manipulation and join it back into a string after making the necessary change.
This approach focuses on changing the center character if possible, because altering the center in an odd-length palindrome often results in a non-palindrome quickly. For even lengths, alterations must be controlled to preserve lexicographical order.
Time Complexity: O(n)
Space Complexity: O(1)
1class
Similar to other implementations, this Java code modifies the center character for palindromes of odd length.