Longest Palindromic Substring using Dynamic Programming

Difficulty Level : Hard
Last Updated : 27 Jan, 2023

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

Below is the implementation of the above approach:

C++

// 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;
}

Java

// A Java solution for longest palindrome
import java.util.*;
 
class GFG{
 
// Function to print a subString str[low..high]
static void printSubStr(String str, int low, int high)
{
    for (int i = low; i <= high; ++i)
        System.out.print(str.charAt(i));
}
 
// This function prints the
// longest palindrome subString
// It also returns the length
// of the longest palindrome
static int longestPalSubstr(String str)
{
    // get length of input String
    int n = str.length();
 
    // 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.charAt(i + k) != str.charAt(j - k))
                    flag = 0;
 
            // Palindrome
            if (flag!=0 && (j - i + 1) > maxLength) {
                start = i;
                maxLength = j - i + 1;
            }
        }
    }
 
    System.out.print("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";
    System.out.print("\nLength is: "
         + longestPalSubstr(str));
}
}
 
// This code is contributed by shikhasingrajput

Python3

# A Python3 solution for longest palindrome
 
# Function to print a subString str[low..high]
def printSubStr(str, low, high):
     
    for i in range(low, high + 1):
        print(str[i], end = "")
 
# This function prints the
# longest palindrome subString
# It also returns the length
# of the longest palindrome
def longestPalSubstr(str):
     
    # Get length of input String
    n = len(str)
     
    # All subStrings of length 1
    # are palindromes
    maxLength = 1
    start = 0
     
    # Nested loop to mark start
    # and end index
    for i in range(n):
        for j in range(i, n):
            flag = 1
             
            # Check palindrome
            for k in range(0, ((j - i) // 2) + 1):
                if (str[i + k] != str[j - k]):
                    flag = 0
 
            # Palindrome
            if (flag != 0 and (j - i + 1) > maxLength):
                start = i
                maxLength = j - i + 1
                 
    print("Longest palindrome subString is: ", end = "")
    printSubStr(str, start, start + maxLength - 1)
 
    # Return length of LPS
    return maxLength
 
# Driver Code
if __name__ == '__main__':
 
    str = "forgeeksskeegfor"
     
    print("\nLength is: ", longestPalSubstr(str))
 
# This code is contributed by 29AjayKumar

C#

// A C# solution for longest palindrome
using System;
 
class GFG{
 
// Function to print a subString str[low..high]
static void printSubStr(String str, int low, int high)
{
    for (int i = low; i <= high; ++i)
        Console.Write(str[i]);
}
 
// This function prints the
// longest palindrome subString
// It also returns the length
// of the longest palindrome
static int longestPalSubstr(String str)
{
    // get length of input String
    int n = str.Length;
 
    // 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!=0 && (j - i + 1) > maxLength) {
                start = i;
                maxLength = j - i + 1;
            }
        }
    }
 
    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.Write("\nLength is: "
         + longestPalSubstr(str));
}
}
 
// This code is contributed by shikhasingrajput

Javascript

<script>
// A Javascript solution for longest palindrome
 
// Function to print a subString str[low..high]
function printSubStr(str,low,high)
{
    for (let i = low; i <= high; ++i)
        document.write(str[i]);
}
 
// This function prints the
// longest palindrome subString
// It also returns the length
// of the longest palindrome
function longestPalSubstr(str)
{
    // get length of input String
    let n = str.length;
  
    // All subStrings of length 1
    // are palindromes
    let maxLength = 1, start = 0;
  
    // Nested loop to mark start and end index
    for (let i = 0; i < str.length; i++) {
        for (let j = i; j < str.length; j++) {
            let flag = 1;
  
            // Check palindrome
            for (let k = 0; k < (j - i + 1) / 2; k++)
                if (str[i + k] != str[j - k])
                    flag = 0;
  
            // Palindrome
            if (flag!=0 && (j - i + 1) > maxLength) {
                start = i;
                maxLength = j - i + 1;
            }
        }
    }
  
    document.write("Longest palindrome subString is: ");
    printSubStr(str, start, start + maxLength - 1);
  
    // return length of LPS
    return maxLength;
}
 
// Driver Code
let str = "forgeeksskeegfor";
document.write("<br>Length is: "
         + longestPalSubstr(str));
 
// This code is contributed by rag2127
</script>

Output

Longest palindrome substring is: geeksskeeg
Length is: 10

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.

Maintain a boolean table[n][n] that is filled in bottom up manner.
The value of table[i][j] is true, if the substring is palindrome, otherwise false.
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.
Otherwise, the value of table[i][j] is made false.
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++

// 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 if 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;
}

Java

// 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 if 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

Python

# 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 if
            # 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.

C#

// 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 if 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

Javascript

<script>
// Javascript Solution
     
    // A utility function to print
    // a substring str[low..high]
   function printSubStr(str,low,high)
   {
           document.write(
            str.substring(
                low, high + 1)+"<br>");
   }
    
   // This function prints the longest
    // palindrome substring of str[].
    // It also returns the length of the
    // longest palindrome
   function longestPalSubstr(str)
   {
    
           // get length of input string
        let n = str.length;
  
        // table[i][j] will be false if
        // substring str[i..j] is not palindrome.
        // Else table[i][j] will be true
        let table = new Array(n);
        for(let i = 0; i < n; i++)
        {
            table[i] = new Array(n);
        }
  
        // All substrings of length 1 are palindromes
        let maxLength = 1;
        for (let i = 0; i < n; ++i)
            table[i][i] = true;
  
        // check for sub-string of length 2.
        let start = 0;
        for (let 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 (let k = 3; k <= n; ++k) {
  
            // Fix the starting index
            for (let i = 0; i < n - k + 1; ++i)
            {
             
                // Get the ending index of substring from
                // starting index i and length k
                let j = i + k - 1;
  
                // checking for sub-string from ith index to
                // jth index if 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;
                    }
                }
            }
        }
        document.write("Longest palindrome substring is; ");
        printSubStr(str, start,
                    start + maxLength - 1);
  
        // return length of LPS
        return maxLength;
   }
    
   // Driver program to test above functions
   let str = "forgeeksskeegfor";
   document.write("Length is: " + longestPalSubstr(str));
 
// This code is contributed by avanitrachhadiya2155
</script>

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.

METHOD 3: Using loops

APPROACH: First we will run a loop for iterating every character. Then we will run another loop inside it to check that is there any other character similar to the current character. If it is, then it is possible that they both are first and last character of longest substring. We will store that substring and check whether that substring is longest or not. If yes then we will store that substring and keep iterating.

C++

// A C++ solution for longest palindrome
#include <bits/stdc++.h>
using namespace std;
// This function prints the
// longest palindrome substring
// It also returns the length
// of the longest palindrome
int longestPalSubstr(string str)
{
    //Stores Longest Pallidrome Substring
    string longest = "";
    int n = str.length();
    int j;
    //To store substring which we think can be a pallindrome
    string subs = "";
    //To strore reverse of substring we think can be pallidrome
    string subsrev = "";
    for(int i = 0; i < n; i++){
        j = n-1;
        while(i < j){
            //Checking whether the character at i and j are same. If they are same then that substring can be LPS
            if((str[i] == str[j]) && (longest.length() < (j-i+1))){
               subs = str.substr(i,(j-i+1));
               //cout<<subs<<" ";
               subsrev = subs;
               reverse(subsrev.begin(), subsrev.end());
               if(subs == subsrev){
                  longest = subs;
               }
            }
            j--;
        }
    }
   //If no longest substring then we will return first character(In Leetcode it was a testcase so...)
    if(longest.length() == 0){
        longest = str[0];
    }
    cout << "Longest palindrome substring is: " << longest;
 
    // return length of LPS
    return longest.length();
}
 
// Driver Code
int main()
{
    string str = "forgeeksskeegfor";
    cout << "\nLength is: "
        << longestPalSubstr(str);
    return 0;
}

Python3

# Python code for the above approach
def longestPalSubstr(s):
    n = len(s)
     
    # Stores Longest Pallidrome Substring
    longest = ""
    j = 0
    # To store substring which we think can be a pallindrome
    subs = ""
    # To strore reverse of substring we think can be pallidrome
    subsrev = ""
    for i in range(n):
        j = n-1
        while i < j:
            # Checking whether the character
            # at i and j are same. If they are
            # same then that substring can be LPS
            if(s[i] == s[j] and len(longest) < (j-i+1)):
               subs = s[i:(j+1)]
               subsrev = subs[::-1]
               if(subs == subsrev):
                  longest = subs
            j -= 1
             
    # If no longest substring then we will
    # return first character(In Leetcode it was a testcase so...)
    if(len(longest) == 0):
        longest = s[0]
    print("Longest palindrome substring is: " + longest)
     
    # return length of LPS
    return len(longest)
 
# Driver Code
stri = "forgeeksskeegfor"
print("Length is: " + str(longestPalSubstr(stri)))
 
# This code is contributed by lokeshpotta20.

Javascript

// A javascript solution for longest palindrome
 
// This function prints the
// longest palindrome substring
// It also returns the length
// of the longest palindrome
function longestPalSubstr(str)
{
    //Stores Longest Pallidrome Substring
    let longest = "";
    let n = str.length;
    let j;
    //To store substring which we think can be a pallindrome
    let subs = "";
    //To strore reverse of substring we think can be pallidrome
    let subsrev = "";
    for(let i = 0; i < n; i++){
        j = n-1;
        while(i < j){
            //Checking whether the character at i and j are same. If they are same then that substring can be LPS
            if((str[i] == str[j]) && (longest.length < (j-i+1))){
               subs = str.substring(i,(j+1));
               subsrev = subs;
               subsrev=subsrev.split('').reverse().join('');
               if(subs == subsrev){
                  longest = subs;
               }
            }
            j--;
        }
    }
   //If no longest substring then we will return
   // first character(In Leetcode it was a testcase so...)
    if(longest.length == 0){
        longest = str[0];
    }
    console.log("Longest palindrome substring is: " + longest);
 
    // return length of LPS
    return longest.length;
}
 
// Driver Code
let str = "forgeeksskeegfor";
console.log("Length is: "+ longestPalSubstr(str));

Output

Longest palindrome substring is: geeksskeeg
Length is: 10

Time and Space Complexity Analysis:

Time Complexity – Worst case -> O(n^3) because used 2 nested loops and a reverse function in 2nd nested loop.

But better than first solution because we were checking palindrome for all substrings but here we are checking for some cases only.

Auxiliary Space – O(3*(length of longest palindrome)) – To store answer, substring which we think can be palindrome and reverse of string which we think is a palindrome.

My Personal Notes