Given a sequence, find the length of the longest palindromic subsequence in it. 
Examples:
Input : abbaab Output : 4 Input : neveropen Output : 5
We have discussed a Dynamic Programming solution for Longest Palindromic Subsequence which is based on below recursive formula.
// 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
The solution discussed above takes O(n2) extra space. In this post a space optimized solution is discussed that requires O(n) extra space. The idea is to create a one dimensional array a[] of same size as given string. We make sure that a[i] stores length of longest palindromic subsequence of prefix ending with i (or substring s[0..i]).Â
C++
// A Space optimized Dynamic Programming based C++ // program for LPS problem #include <bits/stdc++.h> using namespace std;   // Returns the length of the longest palindromic // subsequence in str int lps(string &s) {     int n = s.length();       // a[i] is going to store length of longest     // palindromic subsequence of substring s[0..i]     int a[n];       // Pick starting point     for (int i = n - 1; i >= 0; i--) {           int back_up = 0;                     // Pick ending points and see if s[i]         // increases length of longest common         // subsequence ending with s[j].         for (int j = i; j < n; j++) {               // similar to 2D array L[i][j] == 1             // i.e., handling substrings of length             // one.             if (j == i)                 a[j] = 1;               // Similar to 2D array L[i][j] = L[i+1][j-1]+2             // i.e., handling case when corner characters             // are same.             else if (s[i] == s[j])             {                                   // value a[j] is depend upon previous                 // unupdated value of a[j-1] but in                 // previous loop value of a[j-1] is                 // changed. To store the unupdated                 // value of a[j-1] back_up variable                 // is used.                 int temp = a[j];                 a[j] = back_up + 2;                 back_up = temp;             }               // similar to 2D array L[i][j] = max(L[i][j-1],             // a[i+1][j])             else            {                 back_up = a[j];                 a[j] = max(a[j - 1], a[j]);             }         }     }           return a[n - 1]; }   /* Driver program to test above functions */int main() {     string str = "GEEKSFORGEEKS";     cout << lps(str);     return 0; } |
Java
// A Space optimized Dynamic Programming // based Java program for LPS problem   class GFG {       // Returns the length of the longest     // palindromic subsequence in str     static int lps(String s)     {         int n = s.length();       // a[i] is going to store length     // of longest palindromic subsequence     // of substring s[0..i]         int a[] = new int[n];           // Pick starting point         for (int i = n - 1; i >= 0; i--)             {             int back_up = 0;       // Pick ending points and see if s[i]     // increases length of longest common     // subsequence ending with s[j].     for (int j = i; j < n; j++) {       // similar to 2D array L[i][j] == 1     // i.e., handling substrings of length     // one.         if (j == i)         a[j] = 1;       // Similar to 2D array L[i][j] = L[i+1][j-1]+2     // i.e., handling case when corner characters     // are same.     else if (s.charAt(i) == s.charAt(j))         {         int temp = a[j];         a[j] = back_up + 2;         back_up = temp;         }       // similar to 2D array L[i][j] = max(L[i][j-1],     // a[i+1][j])       else               {                 back_up = a[j];                 a[j] = Math.max(a[j - 1], a[j]);             }           }     }     return a[n - 1];     }   /* Driver program to test above functions */    public static void main(String[] args)     {         String str = "GEEKSFORGEEKS";         System.out.println(lps(str));     } }   //This article is contributed by prerna saini. |
Python3
# A Space optimized Dynamic Programming # based Python3 program for LPS problem   # Returns the length of the longest # palindromic subsequence in str def lps(s):           n = len(s)       # a[i] is going to store length     # of longest palindromic subsequence     # of substring s[0..i]     a = [0] * n       # Pick starting point     for i in range(n-1, -1, -1):           back_up = 0      # Pick ending points and see if s[i]     # increases length of longest common     # subsequence ending with s[j].         for j in range(i, n):       # similar to 2D array L[i][j] == 1     # i.e., handling substrings of length     # one.             if j == i:                 a[j] = 1       # Similar to 2D array L[i][j] = L[i+1][j-1]+2     # i.e., handling case when corner characters     # are same.             elif s[i] == s[j]:                 temp = a[j]                 a[j] = back_up + 2                back_up = temp       # similar to 2D array L[i][j] = max(L[i][j-1],     # a[i+1][j])             else:                 back_up = a[j]                 a[j] = max(a[j - 1], a[j])       return a[n - 1]     # Driver Code string = "GEEKSFORGEEKS"print(lps(string))     # This code is contributed by Ansu Kumari. |
C#
// A Space optimized Dynamic Programming // based C# program for LPS problem using System;   class GFG {       // Returns the length of the longest     // palindromic subsequence in str     static int lps(string s)     {         int n = s.Length;       // a[i] is going to store length     // of longest palindromic subsequence     // of substring s[0..i]         int []a = new int[n];           // Pick starting point         for (int i = n - 1; i >= 0; i--)             {             int back_up = 0;       // Pick ending points and see if s[i]     // increases length of longest common     // subsequence ending with s[j].     for (int j = i; j < n; j++) {       // similar to 2D array L[i][j] == 1     // i.e., handling substrings of length     // one.         if (j == i)         a[j] = 1;       // Similar to 2D array L[i][j] = L[i+1][j-1]+2     // i.e., handling case when corner characters     // are same.     else if (s[i] == s[j])         {         int temp = a[j];         a[j] = back_up + 2;         back_up = temp;         }       // similar to 2D array L[i][j] = max(L[i][j-1],     // a[i+1][j])     else            {                 back_up = a[j];                 a[j] = Math.Max(a[j - 1], a[j]);             }         }     }     return a[n - 1];     }       // Driver program     public static void Main()     {         string str = "GEEKSFORGEEKS";         Console.WriteLine(lps(str));     } }   // This code is contributed by vt_m. |
PHP
<?php // A Space optimized Dynamic // Programming based PHP // program for LPS problem   // Returns the length of // the longest palindromic . // subsequence in str function lps($s) {     $n = strlen($s);       // Pick starting point     for ($i = $n - 1;          $i >= 0; $i--)     {         $back_up = 0;                   // Pick ending points and         // see if s[i] increases         // length of longest common         // subsequence ending with s[j].         for ($j = $i; $j < $n; $j++)         {               // similar to 2D array             // L[i][j] == 1 i.e.,             // handling substrings             // of length one.             if ($j == $i)                 $a[$j] = 1;               // Similar to 2D array             // L[i][j] = L[i+1][j-1]+2             // i.e., handling case when             // corner characters are same.             else if ($s[$i] == $s[$j])             {                                   // value a[j] is depend                 // upon previous unupdated                 // value of a[j-1] but in                 // previous loop value of                 // a[j-1] is changed. To                 // store the unupdated value                 // of a[j-1] back_up variable                 // is used.                 $temp = $a[$j];                 $a[$j] = $back_up + 2;                 $back_up = $temp;             }               // similar to 2D array             // L[i][j] = max(L[i][j-1],             // a[i+1][j])             else            {                 $back_up = $a[$j];                 $a[$j] = max($a[$j - 1],                              $a[$j]);             }         }     }           return $a[$n - 1]; }   // Driver Code $str = "GEEKSFORGEEKS"; echo lps($str);   // This code is contributed // by shiv_bhakt. ?> |
Javascript
<script>   // A Space optimized Dynamic Programming // based Python3 program for LPS problem   // Returns the length of the longest // palindromic subsequence in str function lps(s){           let n = s.length       // a[i] is going to store length     // of longest palindromic subsequence     // of substring s[0..i]     let a = new Array(n).fill(0);       // Pick starting point     for(let i = n - 1; i >= 0; i--){           let back_up = 0       // Pick ending points and see if s[i]     // increases length of longest common     // subsequence ending with s[j].         for (let j = i; j < n; j++){       // similar to 2D array L[i][j] == 1     // i.e., handling substrings of length     // one.             if(j == i)                 a[j] = 1       // Similar to 2D array L[i][j] = L[i+1][j-1]+2     // i.e., handling case when corner characters     // are same.             else if(s[i] == s[j]){                 let temp = a[j]                 a[j] = back_up + 2                 back_up = temp             }       // similar to 2D array L[i][j] = max(L[i][j-1],     // a[i+1][j])             else{                 back_up = a[j]                 a[j] = Math.max(a[j - 1], a[j])             }         }     }       return a[n - 1] }   // Driver Code let string = "GEEKSFORGEEKS"document.write(lps(string),"</br>")   // This code is contributed by shinjanpatra   </script> |
5
Time Complexity : O(n*n) Auxiliary Space : O(n)
Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
