Length of longest palindromic sub-string : Recursion

Last Updated : 27 Feb, 2026

Given a string S, the task is to find the length of the longest substring which is a palindrome.

Examples: 

Input: S = "aaaabbaa" 
Output: 6 
Explanation: Substring "aabbaa" is the longest palindromic sub-string.

Input: S = "banana" 
Output: 5 
Explanation: Substring "anana" is the longest palindromic sub-string. 

[Approach] Using Recursion - O(2^n) Time and O(n) Space

The idea is to use recursion to break the problem into smaller sub-problems. To break the problem into two smaller sub-problems, compare the start and end characters of the string until the starting index becomes greater than the ending index. 

  • If the starting and ending characters are equal, then recursively call for the substring by excluding the starting and ending characters.
  • If starting and ending characters are not equal, then recursively call for the substring by excluding the starting and ending characters one at a time. 
C++ 
//Driver Code Starts
#include <iostream>
#include <string>
#include <algorithm>
using namespace std;

//Driver Code Ends

int longestPalindromic(string str, int i, int j, int count)
{
    // Base case: invalid substring
    if (i > j)
        return count;

    // Base case: single character is palindrome of length 1
    if (i == j)
        return count + 1;

    // If characters match, extend palindrome
    if (str[i] == str[j]) {

        count = longestPalindromic(str, i + 1, j - 1, count + 2);

        return max(count,
                   max(longestPalindromic(str, i + 1, j, 0),
                       longestPalindromic(str, i, j - 1, 0)));
    }

    // If characters don't match, explore both possibilities
    return max(longestPalindromic(str, i + 1, j, 0),
               longestPalindromic(str, i, j - 1, 0));
}

//Driver Code Starts

int main()
{
    string str = "aaaabbaa";
    int n = str.length();

    cout << longestPalindromic(str, 0, n - 1, 0);

    return 0;
}

//Driver Code Ends
Java 
//Driver Code Starts
import java.util.*;

//Driver Code Ends

public class GfG {

    static int longestPalindromic(String str, int i, int j, int count)
    {
        // Base case: invalid substring
        if (i > j)
            return count;

        // Base case: single character is palindrome of length 1
        if (i == j)
            return count + 1;

        // If characters match, extend palindrome
        if (str.charAt(i) == str.charAt(j)) {

            count = longestPalindromic(str, i + 1, j - 1, count + 2);

            return Math.max(count,
                   Math.max(longestPalindromic(str, i + 1, j, 0),
                            longestPalindromic(str, i, j - 1, 0)));
        }

        // If characters don't match, explore both possibilities
        return Math.max(longestPalindromic(str, i + 1, j, 0),
                        longestPalindromic(str, i, j - 1, 0));
}

//Driver Code Starts

    public static void main(String[] args)
    {
        String str = "aaaabbaa";
        int n = str.length();

        System.out.println(longestPalindromic(str, 0, n - 1, 0));
    }
}

//Driver Code Ends
Python 
def longestPalindromic(str, i, j, count):
    # Base case: invalid substring
    if i > j:
        return count

    # Base case: single character is palindrome of length 1
    if i == j:
        return count + 1

    # If characters match, extend palindrome
    if str[i] == str[j]:

        count = longestPalindromic(str, i + 1, j - 1, count + 2)

        return max(count,
                   max(longestPalindromic(str, i + 1, j, 0),
                       longestPalindromic(str, i, j - 1, 0)))

    # If characters don't match, explore both possibilities
    return max(longestPalindromic(str, i + 1, j, 0),
               longestPalindromic(str, i, j - 1, 0))

if __name__ == "__main__":
    str = "aaaabbaa"
    n = len(str)
    
    print(longestPalindromic(str, 0, n - 1, 0))
C# 
//Driver Code Starts
using System;

//Driver Code Ends

class GfG
{
    static int longestPalindromic(string str, int i, int j, int count)
    {
        // Base case: invalid substring
        if (i > j)
            return count;

        // Base case: single character is palindrome of length 1
        if (i == j)
            return count + 1;

        // If characters match, extend palindrome
        if (str[i] == str[j])
        {
            count = longestPalindromic(str, i + 1, j - 1, count + 2);

            return Math.Max(count,
                   Math.Max(longestPalindromic(str, i + 1, j, 0),
                            longestPalindromic(str, i, j - 1, 0)));
        }

        // If characters don't match, explore both possibilities
        return Math.Max(longestPalindromic(str, i + 1, j, 0),
                        longestPalindromic(str, i, j - 1, 0));
}

//Driver Code Starts

    static void Main()
    {
        string str = "aaaabbaa";
        int n = str.Length;

        Console.WriteLine(longestPalindromic(str, 0, n - 1, 0));
    }
}
//Driver Code Ends
JavaScript 
function longestPalindromic(str, i, j, count)
{
    // Base case: invalid substring
    if (i > j)
        return count;

    // Base case: single character is palindrome of length 1
    if (i == j)
        return count + 1;

    // If characters match, extend palindrome
    if (str[i] === str[j]) {

        count = longestPalindromic(str, i + 1, j - 1, count + 2);

        return Math.max(count,
                   Math.max(longestPalindromic(str, i + 1, j, 0),
                            longestPalindromic(str, i, j - 1, 0)));
    }

    // If characters don't match, explore both possibilities
    return Math.max(longestPalindromic(str, i + 1, j, 0),
                    longestPalindromic(str, i, j - 1, 0));
}

let str = "aaaabbaa";
let n = str.length;

console.log(longestPalindromic(str, 0, n - 1, 0));

Output
6

Please note that the above solution can be optimized. Please refer Longest Palindromic Substring for more details.

Comment

Explore