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# Longest Palindromic Substring LeetCode Solution

## Problem – Longest Palindromic Substring

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

Example 1:

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

Example 2:

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

Constraints:

• `1 <= s.length <= 1000`
• `s` consist of only digits and English letters.

### Longest Palindromic Substring LeetCode Solution in Java

``````public class Solution {
private int lo, maxLen;

public String longestPalindrome(String s) {
int len = s.length();
if (len < 2)
return s;

for (int i = 0; i < len-1; i++) {
extendPalindrome(s, i, i);  //assume odd length, try to extend Palindrome as possible
extendPalindrome(s, i, i+1); //assume even length.
}
return s.substring(lo, lo + maxLen);
}

private void extendPalindrome(String s, int j, int k) {
while (j >= 0 && k < s.length() && s.charAt(j) == s.charAt(k)) {
j--;
k++;
}
if (maxLen < k - j - 1) {
lo = j + 1;
maxLen = k - j - 1;
}
}}``````

### Longest Palindromic Substring LeetCode Solution in Python

``````def longestPalindrome(self, s):
res = ""
for i in xrange(len(s)):
# odd case, like "aba"
tmp = self.helper(s, i, i)
if len(tmp) > len(res):
res = tmp
# even case, like "abba"
tmp = self.helper(s, i, i+1)
if len(tmp) > len(res):
res = tmp
return res

# get the longest palindrome, l, r are the middle indexes
# from inner to outer
def helper(self, s, l, r):
while l >= 0 and r < len(s) and s[l] == s[r]:
l -= 1; r += 1
return s[l+1:r]
``````

### Longest Palindromic Substring LeetCode Solution in C++

``````string longestPalindrome(string s) {
if (s.empty()) return "";
if (s.size() == 1) return s;
int min_start = 0, max_len = 1;
for (int i = 0; i < s.size();) {
if (s.size() - i <= max_len / 2) break;
int j = i, k = i;
while (k < s.size()-1 && s[k+1] == s[k]) ++k; // Skip duplicate characters.
i = k+1;
while (k < s.size()-1 && j > 0 && s[k + 1] == s[j - 1]) { ++k; --j; } // Expand.
int new_len = k - j + 1;
if (new_len > max_len) { min_start = j; max_len = new_len; }
}
return s.substr(min_start, max_len);
}``````
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