Longest Palindromic Substring | Set 1

Given a string, find the longest substring which is palindrome.
For example,

Input: Given string :"forgeeksskeegfor", 
Output: "geeksskeeg"

Input: Given string :"Geeks", 
Output: "ee"

Method 1: Brute Force.
Approach: The simple approach is to check each substring whether the substring is a palindrome or not. To do this first, run three nested loops, the outer two loops pick all substrings one by one by fixing the corner characters, the inner loop checks whether the picked substring is palindrome or not.

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// A C++ solution for longest palindrome
#include <bits/stdc++.h>
using namespace std;
  
// Function to print a substring str[low..high]
void printSubStr(string str, int low, int high)
{
    for (int i = low; i <= high; ++i)
        cout << str[i];
}
  
// This function prints the
// longest palindrome substring
// It also returns the length
// of the longest palindrome
int longestPalSubstr(string str)
{
    // get length of input string
    int n = str.size();
  
    // All substrings of length 1
    // are palindromes
    int maxLength = 1, start = 0;
  
    // Nested loop to mark start and end index
    for (int i = 0; i < str.length(); i++) {
        for (int j = i; j < str.length(); j++) {
            int flag = 1;
  
            // Check palindrome
            for (int k = 0; k < (j - i + 1) / 2; k++)
                if (str[i + k] != str[j - k])
                    flag = 0;
  
            // Palindrome
            if (flag && (j - i + 1) > maxLength) {
                start = i;
                maxLength = j - i + 1;
            }
        }
    }
  
    cout << "Longest palindrome substring is: ";
    printSubStr(str, start, start + maxLength - 1);
  
    // return length of LPS
    return maxLength;
}
  
// Driver Code
int main()
{
    string str = "forgeeksskeegfor";
    cout << "\nLength is: "
         << longestPalSubstr(str);
    return 0;
}

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Complexity Analysis:

  • Time complexity: O(n^3).
    Three nested loops are needed to find the longest palindromic substring in this approach, so the time complexity is O(n^3).
  • Auxiliary complexity: O(1).
    As no extra space is needed.


Method 2: Dynamic Programming.
Approach: The time complexity can be reduced by storing results of sub-problems. The idea is similar to this post.



  1. Maintain a boolean table[n][n] that is filled in bottom up manner.
  2. The value of table[i][j] is true, if the substring is palindrome, otherwise false.
  3. To calculate table[i][j], check the value of table[i+1][j-1], if the value is true and str[i] is same as str[j], then we make table[i][j] true.
  4. Otherwise, the value of table[i][j] is made false.
  5. We have to fill table previously for substring of length = 1 and length =2 because
    as we are finding , if table[i+1][j-1] is true or false , so in case of
    (i) length == 1 , lets say i=2 , j=2 and i+1,j-1 doesn’t lies between [i , j]
    (ii) length == 2 ,lets say i=2 , j=3 and i+1,j-1 again doesn’t lies between [i , j].

Below is the implementation of the above approach:

C/C++

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// A C++ dynamic programming
// solution for longest palindrome
  
#include <bits/stdc++.h>
using namespace std;
  
// Function to print a substring
// str[low..high]
void printSubStr(
    string str, int low, int high)
{
    for (int i = low; i <= high; ++i)
        cout << str[i];
}
  
// This function prints the
// longest palindrome substring
// It also returns the length of
// the longest palindrome
int longestPalSubstr(string str)
{
    // get length of input string
    int n = str.size();
  
    // table[i][j] will be false if substring
    // str[i..j] is not palindrome.
    // Else table[i][j] will be true
    bool table[n][n];
  
    memset(table, 0, sizeof(table));
  
    // All substrings of length 1
    // are palindromes
    int maxLength = 1;
  
    for (int i = 0; i < n; ++i)
        table[i][i] = true;
  
    // check for sub-string of length 2.
    int start = 0;
    for (int i = 0; i < n - 1; ++i) {
        if (str[i] == str[i + 1]) {
            table[i][i + 1] = true;
            start = i;
            maxLength = 2;
        }
    }
  
    // Check for lengths greater than 2.
    // k is length of substring
    for (int k = 3; k <= n; ++k) {
        // Fix the starting index
        for (int i = 0; i < n - k + 1; ++i) {
            // Get the ending index of substring from
            // starting index i and length k
            int j = i + k - 1;
  
            // checking for sub-string from ith index to
            // jth index iff str[i+1] to str[j-1] is a
            // palindrome
            if (table[i + 1][j - 1] && str[i] == str[j]) {
                table[i][j] = true;
  
                if (k > maxLength) {
                    start = i;
                    maxLength = k;
                }
            }
        }
    }
  
    cout << "Longest palindrome substring is: ";
    printSubStr(str, start, start + maxLength - 1);
  
    // return length of LPS
    return maxLength;
}
  
// Driver Code
int main()
{
    string str = "forgeeksskeegfor";
    cout << "\nLength is: "
         << longestPalSubstr(str);
    return 0;
}

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Java

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// Java Solution
public class LongestPalinSubstring {
    // A utility function to print
    // a substring str[low..high]
    static void printSubStr(
        String str, int low, int high)
    {
        System.out.println(
            str.substring(
                low, high + 1));
    }
  
    // This function prints the longest
    // palindrome substring of str[].
    // It also returns the length of the
    // longest palindrome
    static int longestPalSubstr(String str)
    {
        // get length of input string
        int n = str.length();
  
        // table[i][j] will be false if
        // substring str[i..j] is not palindrome.
        // Else table[i][j] will be true
        boolean table[][] = new boolean[n][n];
  
        // All substrings of length 1 are palindromes
        int maxLength = 1;
        for (int i = 0; i < n; ++i)
            table[i][i] = true;
  
        // check for sub-string of length 2.
        int start = 0;
        for (int i = 0; i < n - 1; ++i) {
            if (str.charAt(i) == str.charAt(i + 1)) {
                table[i][i + 1] = true;
                start = i;
                maxLength = 2;
            }
        }
  
        // Check for lengths greater than 2.
        // k is length of substring
        for (int k = 3; k <= n; ++k) {
  
            // Fix the starting index
            for (int i = 0; i < n - k + 1; ++i) {
                // Get the ending index of substring from
                // starting index i and length k
                int j = i + k - 1;
  
                // checking for sub-string from ith index to
                // jth index iff str.charAt(i+1) to
                // str.charAt(j-1) is a palindrome
                if (table[i + 1][j - 1]
                    && str.charAt(i) == str.charAt(j)) {
                    table[i][j] = true;
  
                    if (k > maxLength) {
                        start = i;
                        maxLength = k;
                    }
                }
            }
        }
        System.out.print("Longest palindrome substring is; ");
        printSubStr(str, start,
                    start + maxLength - 1);
  
        // return length of LPS
        return maxLength;
    }
  
    // Driver program to test above functions
    public static void main(String[] args)
    {
  
        String str = "forgeeksskeegfor";
        System.out.println("Length is: " + longestPalSubstr(str));
    }
}
  
// This code is contributed by Sumit Ghosh

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Python

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# Python program
  
import sys
  
# A utility function to print a
# substring str[low..high]
def printSubStr(st, low, high) :
    sys.stdout.write(st[low : high + 1])
    sys.stdout.flush()
    return ''
  
# This function prints the longest palindrome
# substring of st[]. It also returns the length
# of the longest palindrome
def longestPalSubstr(st) :
    n = len(st) # get length of input string
  
    # table[i][j] will be false if substring 
    # str[i..j] is not palindrome. Else 
    # table[i][j] will be true
    table = [[0 for x in range(n)] for y
                          in range(n)] 
      
    # All substrings of length 1 are
    # palindromes
    maxLength = 1
    i = 0
    while (i < n) :
        table[i][i] = True
        i = i + 1
      
    # check for sub-string of length 2.
    start = 0
    i = 0
    while i < n - 1 :
        if (st[i] == st[i + 1]) :
            table[i][i + 1] = True
            start = i
            maxLength = 2
        i = i + 1
      
    # Check for lengths greater than 2. 
    # k is length of substring
    k = 3
    while k <= n :
        # Fix the starting index
        i = 0
        while i < (n - k + 1) :
              
            # Get the ending index of 
            # substring from starting 
            # index i and length k
            j = i + k - 1
      
            # checking for sub-string from
            # ith index to jth index iff 
            # st[i + 1] to st[(j-1)] is a 
            # palindrome
            if (table[i + 1][j - 1] and 
                      st[i] == st[j]) :
                table[i][j] = True
      
                if (k > maxLength) :
                    start = i
                    maxLength = k
            i = i + 1
        k = k + 1
    print "Longest palindrome substring is: ", printSubStr(st, start,
                                               start + maxLength - 1)
  
    return maxLength # return length of LPS
  
  
# Driver program to test above functions
st = "forgeeksskeegfor"
l = longestPalSubstr(st)
print "Length is:", l
  
# This code is contributed by Nikita Tiwari.

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C#

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// C# Solution
using System;
  
class GFG {
  
    // A utility function to print a
    // substring str[low...( high - (low+1))]
    static void printSubStr(string str, int low,
                            int high)
    {
        Console.WriteLine(str.Substring(low,
                                        high - low + 1));
    }
  
    // This function prints the longest
    // palindrome substring of str[].
    // It also returns the length of the
    // longest palindrome
    static int longestPalSubstr(string str)
    {
  
        // Get length of input string
        int n = str.Length;
  
        // Table[i, j] will be false if substring
        // str[i..j] is not palindrome. Else
        // table[i, j] will be true
        bool[, ] table = new bool[n, n];
  
        // All substrings of length 1 are palindromes
        int maxLength = 1;
        for (int i = 0; i < n; ++i)
            table[i, i] = true;
  
        // Check for sub-string of length 2.
        int start = 0;
  
        for (int i = 0; i < n - 1; ++i) {
            if (str[i] == str[i + 1]) {
                table[i, i + 1] = true;
                start = i;
                maxLength = 2;
            }
        }
  
        // Check for lengths greater than 2.
        // k is length of substring
        for (int k = 3; k <= n; ++k) {
  
            // Fix the starting index
            for (int i = 0; i < n - k + 1; ++i) {
  
                // Get the ending index of substring from
                // starting index i and length k
                int j = i + k - 1;
  
                // Checking for sub-string from ith index
                // to jth index iff str.charAt(i+1) to
                // str.charAt(j-1) is a palindrome
                if (table[i + 1, j - 1] && str[i] == str[j]) {
                    table[i, j] = true;
                    if (k > maxLength) {
                        start = i;
                        maxLength = k;
                    }
                }
            }
        }
        Console.Write("Longest palindrome substring is: ");
        printSubStr(str, start, start + maxLength - 1);
  
        // Return length of LPS
        return maxLength;
    }
  
    // Driver code
    public static void Main(string[] args)
    {
        string str = "forgeeksskeegfor";
  
        Console.WriteLine("Length is: " + longestPalSubstr(str));
    }
}
  
// This code is contributed by SoumikMondal

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Output:

Longest palindrome substring is: geeksskeeg
Length is: 10

Complexity Analysis:

  • Time complexity: O(n^2).
    Two nested traversals are needed.
  • Auxiliary Space: O(n^2).
    Matrix of size n*n is needed to store the dp array.

A better space complexity approach can be found in Set-2.

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.

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