Given a string s, return the longest palindromic substring in s.

Test Cases

Example 1:

Input: s = "babad"
Output: "bab"
Explanation: "aba" is also a valid answer.

Example 2:

Input: s = "cbbd"
Output: "bb"

Solution

class Solution {
public:
    string longestPalindrome(string s) {
        int n = s.length(), start = 0, end=0;
        bool dp[n][n];
        memset(dp, false, sizeof(dp));
        for(int i=n-1; i>=0; i--) {
            for(int j=i; j<n; j++) {
                if (s[i] == s[j] && (j-i <= 2 || dp[i+1][j-1])) {
                    dp[i][j] = true;
                }
                if (dp[i][j] && j-i > end-start) {
                    end = j;
                    start = i;
                }
            }
        }
        return s.substr(start, end-start+1);
    }
};
class Solution {
    public String longestPalindrome(String s) {
        int best = 0, start = 0, end = 0;
        for(int i=0; i<s.length(); i++) {
            int left = i-1;
            while(i < s.length() - 1 && s.charAt(i) == s.charAt(i+1)) {
                i++;
            }

            int right = i+1;
            while(left >= 0 && right < s.length() && s.charAt(left) == s.charAt(right)) {
                left--;
                right++;
            }

            if (right-left > best) {
                best = right - left;
                start = left+1;
                end = right;
            }
        }
        return s.substring(start, end);
    }
}

// Alternate solution using dp
// class Solution {
//     public String longestPalindrome(String s) {
//         int n = s.length(), start = 0, end=0;
//         boolean[][] dp = new boolean[n][n];
//         for(int i=n-1; i>=0; i--) {
//             for(int j=i; j<n; j++) {
//                 if (s.charAt(i) == s.charAt(j) && (j-i <= 2 || dp[i+1][j-1])) {
//                     dp[i][j] = true;
//                 }
//                 if (dp[i][j] && j-i > end-start) {
//                     end = j;
//                     start = i;
//                 }
//             }
//         }
//         return s.substring(start, end+1);
//     }
// }
class Solution:
    def longestPalindrome(self, s: str) -> str:
        best, start, end = 0, 0, 0
        for i in range(len(s)):
            left = i-1
            while i < len(s) - 1 and s[i] == s[i+1]:
                i += 1
            
            right = i+1
            while left >= 0 and right < len(s) and s[left] == s[right]:
                left -= 1
                right += 1
            
            if right - left > best:
                best, start, end = right-left, left+1, right
        return s[start:end]
Time Complexity: O(n2)
Space Complexity: O(n2)