This approach is based on the observation that a palindrome mirrors around its center. Therefore, if we choose a center, we can expand outward to check for the longest possible palindrome. We can have centers between each two characters as well as on each character to cater for even and odd length palindromes.
Time Complexity: O(n^2) because we potentially expand around n centers.
Space Complexity: O(1), aside from the space required for the output string.
1class Solution {
2 public String longestPalindrome(String s) {
3 if (s == null || s.length() < 1) return "";
4 int start = 0, end = 0;
5 for (int i = 0; i < s.length(); i++) {
6 int len1 = expandAroundCenter(s, i, i);
7 int len2 = expandAroundCenter(s, i, i + 1);
8 int len = Math.max(len1, len2);
9 if (len > end - start) {
10 start = i - (len - 1) / 2;
11 end = i + len / 2;
12 }
13 }
14 return s.substring(start, end + 1);
15 }
16
17 private int expandAroundCenter(String s, int left, int right) {
18 while (left >= 0 && right < s.length() && s.charAt(left) == s.charAt(right)) {
19 left--;
20 right++;
21 }
22 return right - left - 1;
23 }
24}This Java solution adopts the expandaround center method by iterating over potential centers and logging the maximal palindromic span found for each. It then computes the longest such span and derives the substring.
In this approach, a 2D DP table is constructed where dp[i][j] is true if the string s[i...j] is a palindrome. Each entry is dependent on smaller substring checks. This method leverages overlapping subproblems.
Time Complexity: O(n^2) due to the complete 2D table scan.
Space Complexity: O(n^2) as the DP table fully holds substring truths.
1class Solution:
2 def longestPalindrome(self, s: str) -> str:
3 n = len(s)
4 if n == 0: return ""
5 dp = [[False] * n for _ in range(n)]
6 start, max_length = 0, 1
7
8 for i in range(n):
9 dp[i][i] = True
10
11 for begin in range(n - 1, -1, -1):
12 for end in range(begin + 1, n):
13 if s[begin] == s[end]:
14 if end - begin == 1 or dp[begin + 1][end - 1]:
15 dp[begin][end] = True
16 if end - begin + 1 > max_length:
17 start = begin
18 max_length = end - begin + 1
19
20 return s[start:start + max_length]Using Python, the DP strategy uses a list of lists capturing boolean statuses for substring palindromes, evaluating each with smaller pointers and checking towards determining max solutions.