Home Data Structures and Algorithms Recursive solution to find the Longest Palindromic Subsequence (LPS)

Recursive solution to find the Longest Palindromic Subsequence (LPS)

Using the Memoization Technique of Dynamic Programming:

The naive solution for this problem is to generate all subsequences of the given sequence and find the longest palindromic subsequence. This solution is exponential in terms of time complexity. Let us see how this problem possesses both important properties of a Dynamic Programming (DP) Problem and can efficiently be solved using Dynamic Programming.

Following is a general recursive solution with all cases handled. 

  • Case1: Every single character is a palindrome of length 1
    • L(i, i) = 1 (for all indexes i in given sequence)
  • Case2: If first and last characters are not same
    • If (X[i] != X[j])  L(i, j) = max{L(i + 1, j), L(i, j – 1)} 
  • Case3: If there are only 2 characters and both are same
    • Else if (j == i + 1) L(i, j) = 2  
  • Case4: If there are more than two characters, and first and last characters are same
    • Else L(i, j) =  L(i + 1, j – 1) + 2 

Below is the implementation for the above approach:

C++ 
// C++ program of above approach
#include <bits/stdc++.h>
using namespace std;

// A utility function to get max
// of two integers
int max(int x, int y) { return (x > y) ? x : y; }

// Returns the length of the longest
// palindromic subsequence in seq
int lps(char* seq, int i, int j)
{

    // Base Case 1: If there is
    // only 1 character
    if (i == j)
        return 1;

    // Base Case 2: If there are only 2
    // characters and both are same
    if (seq[i] == seq[j] && i + 1 == j)
        return 2;

    // If the first and last characters match
    if (seq[i] == seq[j])
        return lps(seq, i + 1, j - 1) + 2;

    // If the first and last characters
    // do not match
    return max(lps(seq, i, j - 1), lps(seq, i + 1, j));
}

// Driver program to test above functions
int main()
{
    char seq[] = "w3wiki";
    int n = strlen(seq);
    cout << "The length of the LPS is "
         << lps(seq, 0, n - 1);
    return 0;
}
C 
// C program of above approach
#include <stdio.h>
#include <string.h>

// A utility function to get max of two integers
int max(int x, int y) { return (x > y) ? x : y; }

// Returns the length of the longest palindromic subsequence
// in seq
int lps(char* seq, int i, int j)
{
    // Base Case 1: If there is only 1 character
    if (i == j)
        return 1;

    // Base Case 2: If there are only 2 characters and both
    // are same
    if (seq[i] == seq[j] && i + 1 == j)
        return 2;

    // If the first and last characters match
    if (seq[i] == seq[j])
        return lps(seq, i + 1, j - 1) + 2;

    // If the first and last characters do not match
    return max(lps(seq, i, j - 1), lps(seq, i + 1, j));
}

/* Driver program to test above functions */
int main()
{
    char seq[] = "w3wiki";
    int n = strlen(seq);
    printf("The length of the LPS is %d",
           lps(seq, 0, n - 1));
    getchar();
    return 0;
}
Java 
// Java program of above approach
import java.io.*;
import java.util.*;

class GFG {

    // A utility function to get max of two integers
    static int max(int x, int y) { return (x > y) ? x : y; }
    // Returns the length of the longest palindromic
    // subsequence in seq

    static int lps(char seq[], int i, int j)
    {
        // Base Case 1: If there is only 1 character
        if (i == j) {
            return 1;
        }

        // Base Case 2: If there are only 2 characters and
        // both are same
        if (seq[i] == seq[j] && i + 1 == j) {
            return 2;
        }

        // If the first and last characters match
        if (seq[i] == seq[j]) {
            return lps(seq, i + 1, j - 1) + 2;
        }

        // If the first and last characters do not match
        return max(lps(seq, i, j - 1), lps(seq, i + 1, j));
    }

    /* Driver program to test above function */
    public static void main(String[] args)
    {
        String seq = "w3wiki";
        int n = seq.length();
        System.out.printf("The length of the LPS is %d",
                          lps(seq.toCharArray(), 0, n - 1));
    }
}
Python3 
# Python 3 program of above approach

# A utility function to get max
# of two integers


def max(x, y):
    if(x > y):
        return x
    return y

# Returns the length of the longest
# palindromic subsequence in seq


def lps(seq, i, j):

    # Base Case 1: If there is
    # only 1 character
    if (i == j):
        return 1

    # Base Case 2: If there are only 2
    # characters and both are same
    if (seq[i] == seq[j] and i + 1 == j):
        return 2

    # If the first and last characters match
    if (seq[i] == seq[j]):
        return lps(seq, i + 1, j - 1) + 2

    # If the first and last characters
    # do not match
    return max(lps(seq, i, j - 1),
               lps(seq, i + 1, j))


# Driver Code
if __name__ == '__main__':
    seq = "w3wiki"
    n = len(seq)
    print("The length of the LPS is",
          lps(seq, 0, n - 1))

# This code contributed by Rajput-Ji
C# 
// C# program of the above approach
using System;

public class GFG {

    // A utility function to get max of two integers
    static int max(int x, int y) { return (x > y) ? x : y; }
    // Returns the length of the longest palindromic
    // subsequence in seq

    static int lps(char[] seq, int i, int j)
    {
        // Base Case 1: If there is only 1 character
        if (i == j) {
            return 1;
        }

        // Base Case 2: If there are only 2 characters and
        // both are same
        if (seq[i] == seq[j] && i + 1 == j) {
            return 2;
        }

        // If the first and last characters match
        if (seq[i] == seq[j]) {
            return lps(seq, i + 1, j - 1) + 2;
        }

        // If the first and last characters do not match
        return max(lps(seq, i, j - 1), lps(seq, i + 1, j));
    }

    /* Driver program to test above function */
    public static void Main()
    {
        String seq = "w3wiki";
        int n = seq.Length;
        Console.Write("The length of the LPS is "
                      + lps(seq.ToCharArray(), 0, n - 1));
    }
}

// This code is contributed by Rajput-Ji
Javascript 
// A utility function to get max of two integers  
    function max(x, y)
    {
        return (x > y) ? x : y; 
    }
    
    // Returns the length of the longest palindromic subsequence in seq     
    function lps(seq, i, j)
    {
        // Base Case 1: If there is only 1 character 
        if (i == j)
        { 
            return 1; 
        } 
  
        // Base Case 2: If there are only 2 characters and both are same  
            if (seq[i] == seq[j] && i + 1 == j)
            { 
                return 2; 
            } 
      
        // If the first and last characters match  
            if (seq[i] == seq[j]) 
            { 
                return lps(seq, i + 1, j - 1) + 2; 
            } 
      
        // If the first and last characters do not match  
            return max(lps(seq, i, j - 1), lps(seq, i + 1, j)); 
    }
    
    /* Driver program to test above function */
    let seq = "w3wiki"; 
    let n = seq.length;
    console.log("The length of the LPS is ", lps(seq.split(""), 0, n - 1));
    
    // This code is contributed by avanitrachhadiya2155

Output
The length of the LPS is 5

Time complexity: O(2n), where ‘n’ is the length of the input sequence.
Auxiliary Space: O(n2) as we are using a 2D array to store the solutions of the subproblems.

Longest Palindromic Subsequence (LPS)

Given a string ‘S’, find the length of the Longest Palindromic Subsequence in it.

The Longest Palindromic Subsequence (LPS) is the problem of finding a maximum-length subsequence of a given string that is also a Palindrome.

Longest Palindromic Subsequence

Examples:

Input: S = “w3wiki”
Output: 5
Explanation: The longest palindromic subsequence we can get is of length 5. There are more than 1 palindromic subsequences of length 5, for example: EEKEE, EESEE, EEFEE, …etc.

Input: S = “BBABCBCAB”
Output: 7
Explanation: As “BABCBAB” is the longest palindromic subsequence in it. “BBBBB” and “BBCBB” are also palindromic subsequences of the given sequence, but not the longest ones.

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Using the Memoization Technique of Dynamic Programming:

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