# 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 = “GEEKSFORGEEKS”
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.

## Recursive solution to find the Longest Palindromic Subsequence (LPS):

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 ` `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;` `}`

## C

 `// C program of above approach` `#include ` `#include `   `// 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`

## 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 = ``"GEEKSFORGEEKS"``; ` `    ``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 ‘nâ€™ 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.

## Using the 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.

Below is the implementation for the above approach:

## C++

 `// A Dynamic Programming based C++ program` `// for LPS problem returns the length of` `// the longest palindromic subsequence` `// in seq` `#include ` `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;` `}`

## 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

 `// 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(``''``);` `console.log(``"The length of the LPS is "` `+ lps(s2, seq, n, n),``"
"``);` ` `  `// This code is contributed by shinjanpatra`

Output

```The length of the LPS is 5

```

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

## Using the Tabulation technique of Dynamic programming to find LPS:

In the earlier sections, we discussed recursive and dynamic programming approaches with memoization for solving the Longest Palindromic Subsequence (LPS) problem. Now, we will shift our focus to the Bottom-up dynamic programming method.

Below is the implementation for the above approach:

## C++

 `// A Dynamic Programming based C++ program for LPS problem` `// Returns the length of the longest palindromic subsequence` `#include ` `#include // for memset` `#include ` `#include `   `using` `namespace` `std;`   `int` `longestPalinSubseq(string S)` `{` `    ``string R = S;` `    ``reverse(R.begin(), R.end());`   `    ``// dp[i][j] will store the length of the longest` `    ``// palindromic subsequence for the substring` `    ``// starting at index i and ending at index j` `    ``int` `dp[S.length() + 1][R.length() + 1];`   `    ``// Initialize dp array with zeros` `    ``memset``(dp, 0, ``sizeof``(dp));`   `    ``// Filling up DP table based on conditions discussed` `    ``// in the above approach` `    ``for` `(``int` `i = 1; i <= S.length(); i++) {` `        ``for` `(``int` `j = 1; j <= R.length(); j++) {` `            ``if` `(S[i - 1] == R[j - 1])` `                ``dp[i][j] = 1 + dp[i - 1][j - 1];` `            ``else` `                ``dp[i][j] = max(dp[i][j - 1], dp[i - 1][j]);` `        ``}` `    ``}`   `    ``// At the end, DP table will contain the LPS` `    ``// So just return the length of LPS` `    ``return` `dp[S.length()][R.length()];` `}`   `// Driver code` `int` `main()` `{` `    ``string s = ``"GEEKSFORGEEKS"``;` `    ``cout << ``"The length of the LPS is "` `         ``<< longestPalinSubseq(s) << endl;`   `    ``return` `0;` `}`   `// This code is contributed by akshitaguprzj3`

## 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)` `    ``{` `        ``String R` `            ``= ``new` `StringBuilder(S).reverse().toString();`   `        ``// dp[i][j] will store the length of the longest` `        ``// palindromic subsequence for the substring` `        ``// starting at index i and ending at index j` `        ``int` `dp[][]` `            ``= ``new` `int``[S.length() + ``1``][R.length() + ``1``];`   `        ``// Filling up DP table based on conditions discussed` `        ``// in above approach` `        ``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]);` `            ``}` `        ``}`   `        ``// At the end DP table will contain the LPS` `        ``// So just return the length of LPS` `        ``return` `dp[S.length()][R.length()];` `    ``}` `  `  `    ``// Driver code` `    ``public` `static` `void` `main(String[] args)` `    ``{` `        ``String s = ``"GEEKSFORGEEKS"``;` `        ``System.out.println(``"The length of the LPS is "` `                           ``+ longestPalinSubseq(s));` `    ``}` `}`

## Python3

 `def` `longestPalinSubseq(S):` `    ``R ``=` `S[::``-``1``]`   `    ``# dp[i][j] will store the length of the longest` `    ``# palindromic subsequence for the substring` `    ``# starting at index i and ending at index j` `    ``dp ``=` `[[``0``] ``*` `(``len``(R) ``+` `1``) ``for` `_ ``in` `range``(``len``(S) ``+` `1``)]`   `    ``# Filling up DP table based on conditions discussed` `    ``# in the above approach` `    ``for` `i ``in` `range``(``1``, ``len``(S) ``+` `1``):` `        ``for` `j ``in` `range``(``1``, ``len``(R) ``+` `1``):` `            ``if` `S[i ``-` `1``] ``=``=` `R[j ``-` `1``]:` `                ``dp[i][j] ``=` `1` `+` `dp[i ``-` `1``][j ``-` `1``]` `            ``else``:` `                ``dp[i][j] ``=` `max``(dp[i][j ``-` `1``], dp[i ``-` `1``][j])`   `    ``# At the end, DP table will contain the LPS` `    ``# So just return the length of LPS` `    ``return` `dp[``len``(S)][``len``(R)]`     `# Driver code` `s ``=` `"GEEKSFORGEEKS"` `print``(``"The length of the LPS is"``, longestPalinSubseq(s))`   `# This code is contributed by shivamgupta310570`

## C#

 `using` `System;`   `public` `class` `GFG {`   `    ``// Function to find the length of the longest` `    ``// palindromic subsequence` `    ``static` `int` `LongestPalinSubseq(``string` `S)` `    ``{` `        ``char``[] charArray = S.ToCharArray();` `        ``Array.Reverse(charArray);` `        ``string` `R = ``new` `string``(charArray);`   `        ``// dp[i][j] will store the length of the longest` `        ``// palindromic subsequence for the substring` `        ``// starting at index i and ending at index j` `        ``int``[, ] dp = ``new` `int``[S.Length + 1, R.Length + 1];`   `        ``// Initialize dp array with zeros` `        ``for` `(``int` `i = 0; i <= S.Length; i++) {` `            ``for` `(``int` `j = 0; j <= R.Length; j++) {` `                ``dp[i, j] = 0;` `            ``}` `        ``}`   `        ``// Filling up DP table based on conditions discussed` `        ``// in the above approach` `        ``for` `(``int` `i = 1; i <= S.Length; i++) {` `            ``for` `(``int` `j = 1; j <= R.Length; j++) {` `                ``if` `(S[i - 1] == R[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]);` `            ``}` `        ``}`   `        ``// At the end, DP table will contain the LPS` `        ``// So just return the length of LPS` `        ``return` `dp[S.Length, R.Length];` `    ``}`   `    ``// Driver code` `    ``public` `static` `void` `Main(``string``[] args)` `    ``{` `        ``string` `s = ``"GEEKSFORGEEKS"``;` `        ``Console.WriteLine(``"The length of the LPS is "` `                          ``+ LongestPalinSubseq(s));` `    ``}` `}`   `// This code is contributed by shivamgupta310570`

## Javascript

 `// A Dynamic Programming based C++ program for LPS problem` `// Returns the length of the longest palindromic subsequence`   `function` `longestPalinSubseq(S)` `{` `    ``let R = S.split(``''``).reverse().join(``''``);`   `    ``// dp[i][j] will store the length of the longest` `    ``// palindromic subsequence for the substring` `    ``// starting at index i and ending at index j` `    `  `    ``// Initialize dp array with zeros` `    ``let dp = ``new` `Array(S.length + 1).fill(0).map(() => ``new` `Array(R.length + 1).fill(0));`   `    ``// Filling up DP table based on conditions discussed` `    ``// in the above approach` `    ``for` `(let i = 1; i <= S.length; i++) {` `        ``for` `(let j = 1; j <= R.length; j++) {` `            ``if` `(S[i - 1] == R[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]);` `        ``}` `    ``}`   `    ``// At the end, DP table will contain the LPS` `    ``// So just return the length of LPS` `    ``return` `dp[S.length][R.length];` `}`   `// Driver code` `let s = ``"GEEKSFORGEEKS"``;` `console.log(``"The length of the LPS is "` `+ longestPalinSubseq(s));`

Output

```The length of the LPS is 5

```

Time Complexity : O(n2)
Auxiliary Space: O(n2), since we use a 2-D array.

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