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# Longest Palindromic Substring

Given a string s, find the longest palindromic substring in s. You may assume that the maximum length of s is 1000.

Example 1:

```Input: "babad"
Output: "bab"```

Note: "aba" is also a valid answer.

Example 2:

```Input: "cbbd"
Output: "bb"```

Solution 1 in C++

```#include <iostream>
#include <algorithm>

using namespace std;

class Solution {
public:
string longestPalindrome(string s) {
int str_len = s.length();
string myPalindrome;
string ret;
int currentPalindromeLen;

int str_remaining=str_len;
for(int i=0; i<str_len; i++)
{
myPalindrome="";
currentPalindromeLen=0;
for(int j=i; j<str_len; j++)
{
myPalindrome += s[j];
currentPalindromeLen++;
int idx1=0, idx2=currentPalindromeLen-1;
while(idx1<idx2)
{
if (myPalindrome[idx1]!=myPalindrome[idx2]) break;
idx1++;
idx2--;
}
if (idx1>=idx2)
{
if (myPalindrome.length()>=ret.length()) ret = myPalindrome;
}
}
str_remaining--;
if (str_remaining<ret.length()) break;
}
return ret;
}
};

int main(void)
{
Solution s;

string testcase1 = "babad"; // the answer should be "bab" or "aba"
cout << testcase1 << " -> " << s.longestPalindrome(testcase1) << endl;

string testcase2 = "cbbd"; // the answer should be "bb"
cout << testcase2 << " -> " << s.longestPalindrome(testcase2) << endl;
}```

Solution 2 in Java - Time Complexity = O(n^2)

```public String longestPalindrome(String s) {
if (s == null || s.length() < 1) return "";
int start = 0, end = 0;
for (int i = 0; i < s.length(); i++) {
int len1 = expandAroundCenter(s, i, i);
int len2 = expandAroundCenter(s, i, i + 1);
int len = Math.max(len1, len2);
if (len > end - start) {
start = i - (len - 1) / 2;
end = i + len / 2;
}
}
return s.substring(start, end + 1);
}

private int expandAroundCenter(String s, int left, int right) {
int L = left, R = right;
while (L >= 0 && R < s.length() && s.charAt(L) == s.charAt(R)) {
L--;
R++;
}
return R - L - 1;
}```