[NeetCode 150] Longest Palindromic Substring

Longest Palindromic Substring

Given a string s, return the longest substring of s that is a palindrome.

A palindrome is a string that reads the same forward and backward.

If there are multiple palindromic substrings that have the same length, return any one of them.

Example 1:

Input: s = "ababd"

Output: "bab"

Explanation: Both “aba” and “bab” are valid answers.

Example 2:

Input: s = "abbc"

Output: "bb"

Constraints:

1 <= s.length <= 1000
s contains only digits and English letters.

Solution

A classic algorithm for palindromic is manacher algorithm, which can compute the longest radius of palindrome taking each character as the center in O(n)O(n)O(n) time complexity.

Because we assume that all palindromes have its center, the length of all palindromes should be odd. To achieve this property, we can insert special characters, like # between each character in the original string, e.g. abba->#a#b#b#a#.

Now, assuming d[i]d[i]d[i] denotes the longest radius of palindrome whose center locates at iii, how can we use the information acquired before? Thinking about the property of palindrome, if iii is in a palindrome centered on j (j+d[j]>i)j\ (j+d[j]>i)j (j+d[j]>i), then iii’s symmetric character of jjj, j+j−ij+j-ij+ji would be helpful. The overlapping part of palindromes centered on j+j−ij+j-ij+ji and jjj would be the same for palindrome centered on iii. The next we need to do is to see if we can expand d[i]d[i]d[i] to exceed j+d[j]j+d[j]j+d[j] and update jjj, which denotes the center of palindrome whose right endpoint is the largest. Because each time we update d[i]d[i]d[i], iii or j+d[j]j+d[j]j+d[j] will be at least be plused by 1. So, the whole process can be done in O(n)O(n)O(n) time complexity.

Code

class Solution:
    def longestPalindrome(self, s: str) -> str:
        seq = ['#']
        for c in s:
            seq.append(c)
            seq.append('#')
        d = [0]*len(seq)
        center = 0
        max_center = 0
        for i in range(1, len(seq)):
            if center + d[center] > i:
                d[i] = min(center+d[center]-i, d[center+center-i])
            while i-d[i]-1 >= 0 and i+d[i]+1 < len(seq) and seq[i-d[i]-1] == seq[i+d[i]+1]:
                d[i] += 1
            if center + d[center] < i + d[i]:
                center = i
            if d[i] >= d[max_center]:
                max_center = i
        print(seq)
        print(d)
        
        ans = []
        for i in range(max_center-d[max_center], max_center+d[max_center]+1):
            if seq[i] != '#':
                ans.append(seq[i])
        
        return ''.join(ans)

        
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