## Problem:

Given a string , find the longest palindromic substring in that string.

For example , for the input “babad”

the longest palindromic substring is “bab” or “aba”.

For the input “cbbd” the longest palindromic substring is “bb”

## Assumptions:

- The string has at least one character and at the most 1000 characters
- The string consists of digits and English letters

Try out the solution here:

https://leetcode.com/problems/longest-palindromic-substring/

## Solution:

## Hint:

For every character , find if the left and right characters match and keep expanding the palindromic window for odd palindromes (length of the palindrome is odd)

For every two characters , find if the left and right characters match and keep expanding the palindromic window for even palindromes

## ALGORITHM:

**STEP 1: **If the string consists of a single character return it

**STEP 2**: For ever character check for odd and even palindrome substrings

**STEP 2a:** For odd palindromes , check if the left and right characters to the current character match , if so keep expanding the window and track the window starting index and its length if it is greater than already found out windows

**STEP 2b:** For even palindromes , check if the left and right characters to the current character and its next character match , if so keep expanding the window and track the window length like the previous step.

**STEP 3:** Return the maximum window using the window start index and its length.

## Code:

```
class Solution {
int maxLeft = 0;
int maxLen = 0;
public String longestPalindrome(String s) {
if(s.length() ==1 ) return s;
char[] array = s.toCharArray();
for(int i=0;i<s.length();i++){
check(array,i,i); //odd palindromes
check(array,i,i+1); //even palindromes
}
return s.substring(maxLeft, maxLeft+ maxLen);
}
void check(char[] s,int i,int j){
while(i>=0 && j < s.length && s[i] == s[j]){
i--;
j++;
}
if(maxLen < j-i-1){
maxLeft = i+1;
maxLen = j-i-1;
}
}
}
```

In the above code while calculating the starting window index we do

```
maxLeft = i + 1
```

because while expanding the window in the previous step we land at one character behind because of the incrementing.

Similarly while calculation the max length , we do

```
maxLen = j - i +1
```

because we land at one character behind for the start and one character more for the end while incrementing and decrementing in the previous step.

That’s it!