Palindrome by Front Insertion

Given a string s consisting of only lowercase English letters, find the minimum number of characters that need to be added to the front of s to make it a palindrome.
Note: A palindrome is a string that reads the same forward and backward.

Examples:

Input: s = "abc"
Output: 2
Explanation: We can make above string palindrome as "cbabc", by adding 'b' and 'c' at front.

Input: s = "aacecaaaa"
Output: 2
Explanation: We can make above string palindrome as "aaaacecaaaa" by adding two a's at front of string.

[Naive Approach] Checking all prefixes - O(n^2) Time and O(1) Space

The idea is based on the observation that we need to find the longest prefix from given string which is also a palindrome. Then minimum front characters to be added to make given string palindrome will be the remaining characters.

2

[Expected Approach 1] Using lps array of KMP Algorithm - O(n) Time and O(n) Space

The key observation is that the longest palindromic prefix of a string becomes the longest palindromic suffix of its reverse.
Given a string s = "aacecaaaa", its reverse revS = "aaaacecaa". The longest palindromic prefix of s is "aacecaa".

To find this efficiently, we use the LPS array from the KMP algorithm. We concatenate the original string with a special character and its reverse: s + '$' + revS.
The LPS array for this combined string helps identify the longest prefix of s that matches a suffix of revS, which also represents the palindromic prefix of s.

The last value of the LPS array tells us how many characters already form a palindrome at the beginning. Thus, the minimum number of characters to add to make s a palindrome is s.length() - lps.back().

2

[Expected Approach 2] Using Manacher's Algorithm

The idea is to use Manacher’s algorithm to efficiently find all palindromic substrings in linear time.
We transform the string by inserting special characters (#) to handle both even and odd length palindromes uniformly.
After preprocessing, we scan from the end of the original string and use the palindrome radius array to check if the prefix s[0...i] is a palindrome. The first such index i gives us the longest palindromic prefix, and we return n - (i + 1) as the minimum characters to add.

2

Time Complexity: O(n), manacher's algorithm runs in linear time by expanding palindromes at each center without revisiting characters, and the prefix check loop performs O(1) operations per character over n characters.
Auxiliary Space: O(n), used for the modified string and the palindrome length array p[], both of which grow linearly with the input size.