Longest palindrome subsequence with O(n) space

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

Input : abbaab
Output : 4

Input : geeksforgeeks
Output : 5

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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]).

C++

 // A Space optimized Dynamic Programming based C++ // program for LPS problem #include 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 =  * 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

 = 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. ?>

5

Time Complexity : O(n*n)
Auxiliary Space : O(n)

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