Longest Palindromic Subsequence | DP-12

Given a sequence, find the length of the longest palindromic subsequence in it.

As another example, if the given sequence is “BBABCBCAB”, then the output should be 7 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.
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.
1) Optimal Substructure: 
Let X[0..n-1] be the input sequence of length n and L(0, n-1) be the length of the longest palindromic subsequence of X[0..n-1]. 
If last and first characters of X are same, then L(0, n-1) = L(1, n-2) + 2. 
Else L(0, n-1) = MAX (L(1, n-1), L(0, n-2)). 

Following is a general recursive solution with all cases handled. 

// Every single character is a palindrome of length 1
L(i, i) = 1 for all indexes i in given sequence

// 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)} 

// If there are only 2 characters and both are same
Else if (j == i + 1) L(i, j) = 2  

// If there are more than two characters, and first and last 
// characters are same
Else L(i, j) =  L(i + 1, j - 1) + 2 

2) Overlapping Subproblems: Following is a simple recursive implementation of the LPS problem. The implementation simply follows the recursive structure mentioned above. 

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[] = "GEEKSFORGEEKS";
    int n = strlen(seq);
    cout << "The length of the LPS is "
         << lps(seq, 0, n-1);
    return 0;
}

// This code is contributed
// by Akanksha Rai

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[] = "GEEKSFORGEEKS";
    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 = "GEEKSFORGEEKS";
        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 = "GEEKSFORGEEKS"
    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 = "GEEKSFORGEEKS";
        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

PHP

<?php
// PHP program of above approach

// 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 Code
$seq = "GEEKSFORGEEKS";
$n = strlen($seq);
echo "The length of the LPS is ".
            lps($seq, 0, $n - 1);

// This code is contributed by ita_c
?>

Javascript

<script>

    // 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 = "GEEKSFORGEEKS";
    let n = seq.length;
    document.write("The length of the LPS is ", lps(seq.split(""), 0, n - 1));

    // This code is contributed by avanitrachhadiya2155
</script>

Output

The length of the LPS is 5

Time complexity: The time complexity of the above algorithm is exponential O(2^n), where ‘n’ is the length of the input sequence.

Space complexity: The space complexity of the above algorithm is O(n^2) as we are using a 2D array to store the subproblems solutions.

Considering the above implementation, the following is a partial recursion tree for a sequence of length 6 with all different characters. 

               L(0, 5)
             /        \ 
            /          \  
        L(1,5)          L(0,4)
       /    \            /    \
      /      \          /      \
  L(2,5)    L(1,4)  L(1,4)  L(0,3)

In the above partial recursion tree, L(1, 4) is being solved twice. If we draw the complete recursion tree, then we can see that there are many subproblems that are solved again and again. Since the same subproblems are called again, this problem has the Overlapping Subproblems property. So LPS problem has both properties (see this and this) of a dynamic programming problem. Like other typical Dynamic Programming(DP) problems, recomputations of the same subproblems can be avoided by constructing a temporary array L[][] in a bottom-up manner.
3) Dynamic Programming Solution:

C++

// A Dynamic Programming based C++ program for LPS problem
// Returns the length of the longest palindromic subsequence in seq
#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 *str)
{
   int n = strlen(str);
   int i, j, cl;
   int L[n][n];  // Create a table to store results of subproblems


   // Strings of length 1 are palindrome of length 1
   for (i = 0; i < n; i++)
      L[i][i] = 1;

    // Build the table. Note that the lower diagonal values of table are
    // useless and not filled in the process. The values are filled in a
    // manner similar to Matrix Chain Multiplication DP solution (See
    // https://www.geeksforgeeks.org/matrix-chain-multiplication-dp-8/). cl is length of
    // substring
    for (cl=2; cl<=n; cl++)
    {
        for (i=0; i<n-cl+1; i++)
        {
            j = i+cl-1;
            if (str[i] == str[j] && cl == 2)
               L[i][j] = 2;
            else if (str[i] == str[j])
               L[i][j] = L[i+1][j-1] + 2;
            else
               L[i][j] = max(L[i][j-1], L[i+1][j]);
        }
    }

    return L[0][n-1];
}

/* Driver program to test above functions */
int main()
{
    char seq[] = "GEEKSFORGEEKS";
    int n = strlen(seq);
    printf ("The length of the LPS is %d", lps(seq));
    getchar();
    return 0;
}

Java

// A Dynamic Programming based Java
// Program for the Egg Dropping Puzzle
import java.io.*;
import java.util.*;

class LPS {

    // 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(String seq)
    {
        int n = seq.length();
        int i, j, cl;
        // Create a table to store results of subproblems
        int L[][] = new int[n][n];

        // Strings of length 1 are palindrome of length 1
        for (i = 0; i < n; i++)
            L[i][i] = 1;

        // Build the table. Note that the lower
        // diagonal values of table are
        // useless and not filled in the process.
        // The values are filled in a manner similar
        //  to Matrix Chain Multiplication DP solution (See
        // https://www.geeksforgeeks.org/matrix-chain-multiplication-dp-8/).
        // cl is length of substring
        for (cl = 2; cl <= n; cl++) {
            for (i = 0; i < n - cl + 1; i++) {
                j = i + cl - 1;
                if (seq.charAt(i) == seq.charAt(j)
                    && cl == 2)
                    L[i][j] = 2;
                else if (seq.charAt(i) == seq.charAt(j))
                    L[i][j] = L[i + 1][j - 1] + 2;
                else
                    L[i][j] = max(L[i][j - 1], L[i + 1][j]);
            }
        }

        return L[0][n - 1];
    }

    /* Driver program to test above functions */
    public static void main(String args[])
    {
        String seq = "GEEKSFORGEEKS";
        int n = seq.length();
        System.out.println("The length of the LPS is "
                           + lps(seq));
    }
}
/* This code is contributed by Rajat Mishra */

Python3

# // A Dynamic Programming based Python program for LPS problem
#  Returns the length of the longest palindromic subsequence in seq


#  Returns the length of the longest palindromic subsequence in seq
def lps(str):
    n = len(str)

    L = [[0 for i in range(n)]for j in range(n)]

    # Strings of length 1 are palindrome of length 1
    for i in range(n):
        L[i][i] = 1  # Create a table to store results of subproblems

    #  Build the table. Note that the lower diagonal values of table are
    #  useless and not filled in the process. The values are filled in a
    #  manner similar to Matrix Chain Multiplication DP solution (See
    #  https://www.geeksforgeeks.org/matrix-chain-multiplication-dp-8/). cl is length of
    #  substring
    for cl in range(2, n+1):
        for i in range(n-cl+1):
            j = i+cl-1
            if str[i] == str[j] and cl == 2:
                L[i][j] = 2
            elif str[i] == str[j]:
                L[i][j] = L[i+1][j-1]+2
            else:
                L[i][j] = max(L[i][j-1], L[i+1][j])

    return L[0][n-1]


#  Driver program to test above functions
if __name__ == "__main__":
    seq = "GEEKSFORGEEKS"
    n = len(seq)
    print("The length of the LPS is {}".format(lps(seq)))

# This Code is Contributed By Vivek Maddeshiya

C#

// A Dynamic Programming based C# Program
// for the Egg Dropping Puzzle
using System;

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(string seq)
    {
    int n = seq.Length;
    int i, j, cl;

    // Create a table to store results
    // of subproblems
    int [,]L = new int[n,n];

    // Strings of length 1 are
    // palindrome of length 1
    for (i = 0; i < n; i++)
        L[i,i] = 1;

        // Build the table. Note that the
        // lower diagonal values of table
        // are useless and not filled in
        // the process. The values are
        // filled in a manner similar to
        // Matrix Chain Multiplication DP
        // solution (See
        // https://www.geeksforgeeks.org/matrix-chain-multiplication-dp-8/
        // cl is length of substring
        for (cl = 2; cl <= n; cl++)
        {
            for (i = 0; i < n-cl+1; i++)
            {
                j = i + cl - 1;

                if (seq[i] == seq[j] &&
                                  cl == 2)
                    L[i,j] = 2;
                else if (seq[i] == seq[j])
                    L[i,j] = L[i+1,j-1] + 2;
                else
                    L[i,j] =
                     max(L[i,j-1], L[i+1,j]);
            }
        }

        return L[0,n-1];
    }

    /* Driver program to test above
    functions */
    public static void Main()
    {
        string seq = "GEEKSFORGEEKS";
        int n = seq.Length;
        Console.Write("The length of the "
                  + "LPS is "+ lps(seq));
    }
}

// This code is contributed by nitin mittal.

PHP

<?php
// A Dynamic Programming based
// PHP program for LPS problem
// Returns the length of the
// longest palindromic
// subsequence in seq

// 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($str)
{
$n = strlen($str);
$i; $j; $cl;

// Create a table to store
// results of subproblems
$L[][] = array(array());


// Strings of length 1 are
// palindrome of length 1
for ($i = 0; $i < $n; $i++)
    $L[$i][$i] = 1;

    // Build the table. Note that
    // the lower diagonal values
    // of table are useless and
    // not filled in the process.
    // The values are filled in a
    // manner similar to Matrix
    // Chain Multiplication DP
    // solution (See
    // https://www.geeksforgeeks.org/matrix-chain-multiplication-dp-8/).
    // cl is length of substring
    for ($cl = 2; $cl <= $n; $cl++)
    {
        for ($i = 0; $i < $n - $cl + 1; $i++)
        {
            $j = $i + $cl - 1;
            if ($str[$i] == $str[$j] &&
                            $cl == 2)
            $L[$i][$j] = 2;
            else if ($str[$i] == $str[$j])
            $L[$i][$j] = $L[$i + 1][$j - 1] + 2;
            else
            $L[$i][$j] = max($L[$i][$j - 1],
                             $L[$i + 1][$j]);
        }
    }

    return $L[0][$n - 1];
}

// Driver Code
$seq = 'GEEKSFORGEEKS';
$n = strlen($seq);
echo "The length of the " .
      "LPS is ", lps($seq);

// This code is contributed
// by shiv_bhakt.
?>

Javascript

<script>
// A Dynamic Programming based Javascript
// Program for the Egg Dropping Puzzle

// 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)
{
    let n = seq.length;
    let i, j, cl;
    // Create a table to store results of subproblems
    let L = new Array(n);
    for(let x=0;x<n;x++)
    {
        L[x] = new Array(n);
        for(let y = 0; y < n; y++)
            L[x][y] = 0;
    }

    // Strings of length 1 are palindrome of length 1
    for (i = 0; i < n; i++)
        L[i][i] = 1;

        // Build the table. Note that the lower
        // diagonal values of table are
        // useless and not filled in the process.
        // The values are filled in a manner similar
        //  to Matrix Chain Multiplication DP solution (See
        // https://www.geeksforgeeks.org/matrix-chain-multiplication-dp-8/).
        // cl is length of substring
        for (cl = 2; cl <= n; cl++)
        {
            for (i = 0; i < n -cl + 1; i++)
            {
                j = i + cl - 1;
                if (seq[i] == seq[j] && cl == 2)
                    L[i][j] = 2;
                else if (seq[i] == seq[j])
                    L[i][j] = L[i + 1][j - 1] + 2;
                else
                    L[i][j] = max(L[i][j - 1], L[i + 1][j]);
            }
        }

        return L[0][n - 1];
}

 /* Driver program to test above functions */
let seq = "GEEKSFORGEEKS";
let n = seq.length;
document.write("The length of the lps is "+ lps(seq));

// This code is contributed by rag2127
</script>

Output

The length of the LPS is 5

Time Complexity: O(n^2), which is much better than the worst-case time complexity of Naive Recursive implementation.
Auxiliary Space: O(n^2),  Creating a table

Using Memoization Technique of Dynamic programming: The idea used here is to reverse the given input string and check the length of the longest common subsequence. That would be the answer for the longest palindromic subsequence.

C++

// A Dynamic Programming based C++ program for LPS problem
// Returns the length of the longest palindromic subsequence
// in seq
#include <bits/stdc++.h>
using namespace std;

int dp[1001][1001];

// Returns the length of the longest palindromic subsequence
// in seq
int lps(string& s1, string& s2, int n1, int n2)
{
    if (n1 == 0 || n2 == 0) {
        return 0;
    }
    if (dp[n1][n2] != -1) {
        return dp[n1][n2];
    }
    if (s1[n1 - 1] == s2[n2 - 1]) {
        return dp[n1][n2] = 1 + lps(s1, s2, n1 - 1, n2 - 1);
    }
    else {
        return dp[n1][n2] = max(lps(s1, s2, n1 - 1, n2),
                                lps(s1, s2, n1, n2 - 1));
    }
}

/* Driver program to test above functions */
int main()
{
    string seq = "GEEKSFORGEEKS";
    int n = seq.size();
    dp[n][n];
    memset(dp, -1, sizeof(dp));
    string s2 = seq;
    reverse(s2.begin(), s2.end());
    cout << "The length of the LPS is "
         << lps(s2, seq, n, n) << endl;
    return 0;
}

// This code is contributed by Arun Bang

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, int dp[][])
    {

        // Base Case 1: If there is only 1 character
        if (i == j) {
            return dp[i][j] = 1;
        }

        // Base Case 2: If there are only 2 characters and
        // both are same
        if (seq[i] == seq[j] && i + 1 == j) {
            return dp[i][j] = 2;
        }
        // Avoid extra call for already calculated subproblems, Just return the saved ans from cache
          if(dp[i][j] != -1){
          return dp[i][j];
        }
        // If the first and last characters match
        if (seq[i] == seq[j]) {
            return dp[i][j] = lps(seq, i + 1, j - 1, dp) + 2;
        }

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

    /* Driver program to test above function */
    public static void main(String[] args)
    {
        String seq = "GEEKSFORGEEKS";
        int n = seq.length();
          int dp[][] = new int[n][n];
          for(int[] arr: dp)
          Arrays.fill(arr, -1);
        System.out.printf("The length of the LPS is %d",
                          lps(seq.toCharArray(), 0, n - 1, dp));
    }
}

// This code is contributed by gauravrajput1

Python3

# A Dynamic Programming based Python program for LPS problem
# Returns the length of the longest palindromic subsequence
# in seq

dp = [[-1 for i in range(1001)]for j in range(1001)]

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


def lps(s1, s2, n1, n2):

    if (n1 == 0 or n2 == 0):
        return 0

    if (dp[n1][n2] != -1):
        return dp[n1][n2]

    if (s1[n1 - 1] == s2[n2 - 1]):
        dp[n1][n2] = 1 + lps(s1, s2, n1 - 1, n2 - 1)
        return dp[n1][n2]
    else:
        dp[n1][n2] = max(lps(s1, s2, n1 - 1, n2), lps(s1, s2, n1, n2 - 1))
        return dp[n1][n2]

# Driver program to test above functions


seq = "GEEKSFORGEEKS"
n = len(seq)

s2 = seq
s2 = s2[::-1]
print(f"The length of the LPS is {lps(s2, seq, n, n)}")

# This code is contributed by shinjanpatra

C#

// C# code to implement the approach
using System;
using System.Numerics;
using System.Collections.Generic;

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 Code
    public static void Main(string[] args)
    {
        string seq = "GEEKSFORGEEKS";
        int n = seq.Length;
        Console.Write("The length of the LPS is "
                      + lps(seq.ToCharArray(), 0, n - 1));
    }
}

// This code is contributed by sanjoy_62.

Javascript

<script>

// A Dynamic Programming based JavaScript program for LPS problem
// Returns the length of the longest palindromic subsequence
// in seq
let dp;

// Returns the length of the longest palindromic subsequence
// in seq
function lps(s1, s2, n1, n2)
{
    if (n1 == 0 || n2 == 0) {
        return 0;
    }
    if (dp[n1][n2] != -1) {
        return dp[n1][n2];
    }
    if (s1[n1 - 1] == s2[n2 - 1]) {
        return dp[n1][n2] = 1 + lps(s1, s2, n1 - 1, n2 - 1);
    }
    else {
        return dp[n1][n2] = Math.max(lps(s1, s2, n1 - 1, n2),
                                lps(s1, s2, n1, n2 - 1));
    }
}

/* Driver program to test above functions */

let seq = "GEEKSFORGEEKS";
let n = seq.length;
dp = new Array(1001);
for(let i=0;i<1001;i++){
    dp[i] = new Array(1001).fill(-1);
}
let s2 = seq;
s2 = s2.split('').reverse().join('');
document.write("The length of the LPS is " + lps(s2, seq, n, n),"</br>");

// This code is contributed by shinjanpatra

</script>

Output

The length of the LPS is 5

Time Complexity: O(n²)
Auxiliary Space: O(n²)

Without reversing of an input string: in the above solution, we first reverse the input string and then pass it to the lps function but if we use string character indexing properly then we have no need to reverse the input string. so this way we did not need the reverse of the string, which means we extra saving O(n) space for which need to store the reverse of the input string.

reverseOfS[n2-1] = s[n-n2]

C++

// A Dynamic Programming based C++ program for LPS problem Returns the length of the longest palindromic subsequence in seq
#include <bits/stdc++.h>
using namespace std;

int dp[1001][1001];
int n;

// Returns the length of the longest palindromic subsequence in seq
int lps(string& s, int n1, int n2)
{
    if (n1 == 0 || n2 == 0) {
        return 0;
    }
    if (dp[n1][n2] != -1) {
        return dp[n1][n2];
    }
    // here instead of using reverse of s as s2
    // we use s[n-1-n2] which is similar to revOfS[n2-1]
    if (s[n1 - 1] == s[n - n2]) {
        return dp[n1][n2] = 1 + lps(s, n1 - 1, n2 - 1);
    }
    else {
        return dp[n1][n2] = max(lps(s, n1 - 1, n2),
                                lps(s, n1, n2 - 1));
    }
}

/* Driver program to test above functions */
int main()
{
    string s = "GEEKSFORGEEKS";
    n = s.size();
    dp[n][n];
    memset(dp, -1, sizeof(dp));
    cout << "The length of the LPS is " << lps(s, n, n) << endl;
    return 0;
}

Java

// A Dynamic Programming based Java program for LPS problem Returns the length of the longest palindromic subsequence in seq
public class LPS {
    static int dp[][] = new int[1001][1001];

    // Returns the length of the longest palindromic subsequence in seq
    static int lps(String s, int n1, int n2) {
        if (n1 == 0 || n2 == 0) {
            return 0;
        }
        if (dp[n1][n2] != -1) {
            return dp[n1][n2];
        }
        // here instead of using reverse of s as s2
        // we use s[n-1-n2] which is similar to revOfS[n2-1]
        if (s.charAt(n1 - 1) == s.charAt(s.length() - n2)) {
            return dp[n1][n2] = 1 + lps(s, n1 - 1, n2 - 1);
        }
        else {
            return dp[n1][n2] = Math.max(lps(s, n1 - 1, n2), lps(s, n1, n2 - 1));
        }
    }

    /* Driver program to test above functions */
    public static void main(String[] args) {
        String s = "GEEKSFORGEEKS";
        int n = s.length();
        // Initialize dp array with -1
        for (int i = 0; i < dp.length; i++) {
            for (int j = 0; j < dp[0].length; j++) {
                dp[i][j] = -1;
            }
        }
        System.out.println("The length of the LPS is " + lps(s, n, n));
    }
}
// This code contributed by SRJ

Python3

def lps(s, n1, n2):

    # Base Case 1: If there is only 1 character
    if (n1 == 0 or n2 == 0):
        return 0

    # Check if the values are same
    # if same, then its a palindrome
    if (s[n1 - 1] == s[n - n2]):
        return 1 + lps(s, n1 - 1, n2 - 1)

    else:
        return max(lps(s, n1 - 1, n2), lps(s, n1, n2 - 1))

# Driver program to test above functions
s = "GEEKSFORGEEKS"
n = len(s)

dp = [[-1 for j in range(n + 1)] for i in range(n + 1)]

print("The length of the LPS is " + str(lps(s, n, n)))

# This code is contributed by factworx412

Javascript

function lps(s, n1, n2)
{
    // Base Case 1: If there is only 1 character
    if (n1 == 0 || n2 == 0)
        return 0;

    // Check if the values are same
    // if same, then its a palindrome
    if (s[n1 - 1] == s[n - n2])
        return 1 + lps(s, n1 - 1, n2 - 1) ;

    else
        return Math.max(lps(s, n1 - 1, n2), lps(s, n1, n2 - 1)) ;
}

// Driver program to test above functions
let str = "GEEKSFORGEEKS";
let n = str.length;

document.write("The length of the LPS is " + lps(str, n, n).toString());

C#

// A Dynamic Programming based C# program for LPS problem Returns the length of the longest palindromic subsequence in seq
using System;
public class LPSProgram {
    static int[,] dp;
    static int n;

    // Returns the length of the longest palindromic subsequence in seq
    static int lps(string s, int n1, int n2)
    {
        if (n1 == 0 || n2 == 0) {
            return 0;
        }
        if (dp[n1,n2] != -1) {
            return dp[n1,n2];
        }
        // here instead of using reverse of s as s2
        // we use s[n-1-n2] which is similar to revOfS[n2-1]
        if (s[n1 - 1] == s[n - n2]) {
            return dp[n1,n2] = 1 + lps(s, n1 - 1, n2 - 1);
        }
        else {
            return dp[n1,n2] = Math.Max(lps(s, n1 - 1, n2),
                                         lps(s, n1, n2 - 1));
        }
    }

    /* Driver program to test above functions */
    static void Main()
    {
        string s = "GEEKSFORGEEKS";
        n = s.Length;
        dp = new int[n + 1, n + 1];
        for (int i = 0; i <= n; i++) {
            for (int j = 0; j <= n; j++) {
                dp[i,j] = -1;
            }
        }
        Console.WriteLine("The length of the LPS is " + lps(s, n, n));
    }
}

Output

The length of the LPS is 5

Time Complexity : O(n^2)
Auxiliary Space : O(n^2) 

ANOTHER APPROACH USING TABULATION:

Intuition:

1. We make a the reverse of the string and store in string R.
2. Then we declare a 2-D array of size n^2.
3. we fill the array such that the ans of Longest Palindromic Subsequence is stored at arr[n][n].
4. Following is the logic behind the approach:

•         if(s.charAt(i-1)==r.charAt(j-1)) then we fill the psoition of dp[i][j] as 1+dp[i-1][j-1] as both the characters are same and could contribute to the LPS.
• else we fill  dp[i][j] = Math.max(dp[i][j-1],dp[i-1][j] to fill whats the LPS encountered so far as the characters are not matching.

Implementation:

Java

// A Dynamic Programming based Java program for LPS problem
// Returns the length of the longest palindromic subsequence
import java.io.*;
import java.util.*;

class GFG {
    public static int longestPalinSubseq(String S)
    {
        // code here
        String R
            = new StringBuilder(S).reverse().toString();

        int dp[][]
            = new int[S.length() + 1][R.length() + 1];

        for (int i = 1; i <= S.length(); i++) {
            for (int j = 1; j <= R.length(); j++) {
                if (S.charAt(i - 1) == R.charAt(j - 1))
                    dp[i][j] = 1 + dp[i - 1][j - 1];
                else
                    dp[i][j] = Math.max(dp[i][j - 1],
                                        dp[i - 1][j]);
            }
        }
        return dp[S.length()][R.length()];
    }
    public static void main(String[] args)
    {
        String s = "GEEKSFORGEEKS";
        System.out.println("The length of the LPS is "
                           + longestPalinSubseq(s));
    }
}
// This code is contributed by Raunak Singh

Output

The length of the LPS is 5

Time Complexity : O(n^2)
Space Complexity : O(n^2) ,since we are using a 2-D array.

Print Longest Palindromic Subsequence
Longest palindrome subsequence with O(n) space
Please write comments if you find anything incorrect, or if you want to share more information about the topic discussed above.
References:
http://users.eecs.northwestern.edu/~dda902/336/hw6-sol.pdf