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 subseuqnce 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 term of time complexity. Let us see how this problem possesses both important properties of a Dynamic Programming (DP) Problem and can efficiently 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)).

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

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 simple recursive implementation of the LPS problem. The implementation simply follows the recursive structure mentioned above.

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; }    // This code is contributed // by Akanksha Rai

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    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 egers  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 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



Output:

The length of the LPS is 5

Considering the above implementation, 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 which are solved again and again. Since same suproblems are called again, this problem has Overlapping Subprolems 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 same subproblems can be avoided by constructing a temporary array L[][] in bottom up manner.

Dynamic Programming Solution

C++

 // A Dynamic Programming based C++ program for LPS problem // Returns the length of the longest palindromic subsequence in seq #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 *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 lentgh 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://tutorialspoint.dev/slugresolver/matrix-chain-multiplication-dp-8/). cl is length of     // substring     for (cl=2; cl<=n; cl++)     {         for (i=0; i

Java

 // A Dynamic Programming based Java  // Program for the Egg Dropping Puzzle 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 lentgh 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         // cl is length of substring         for (cl=2; cl<=n; cl++)         {             for (i=0; i

Python

 # A Dynamic Programming based Python  # program for LPS problem Returns the length #  of the longest palindromic subsequence in seq def lps(str):     n = len(str)        # Create a table to store results of subproblems     L = [[0 for x in range(n)] for x in range(n)]        # Strings of length 1 are palindrome of length 1     for i in range(n):         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     # 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[n-1]    # Driver program to test above functions seq = "GEEKS FOR GEEKS" n = len(seq) print("The length of the LPS is " + str(lps(seq)))    # This code is contributed by Bhavya Jain

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 lentgh 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         // 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 = "GEEKS FOR GEEKS";         int n = seq.Length;         Console.Write("The lnegth of the "                   + "lps is "+ lps(seq));     } }    // This code is contributed by nitin mittal.

PHP

 \$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 lentgh 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      // 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[\$n - 1]; }    // Driver Code \$seq = 'GEEKS FOR GEEKS'; \$n = strlen(\$seq); echo "The lnegth of the " .        "LPS is ", lps(\$seq);    // This code is contributed // by shiv_bhakt. ?>

Output:

The lnegth of the LPS is 7

Time Complexity of the above implementation is O(n^2) which is much better than the worst case time complexity of Naive Recursive implementation.

Print Longest Palindromic Subsequence

Longest palindrome subsequence with O(n) space