Java Program for 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.
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.

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// 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));
    }
}
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Output:
The length of the LPS is 5

Dynamic Programming Solution

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// 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
        // 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 lnegth of the lps is " + lps(seq));
    }
}
/* This code is contributed by Rajat Mishra */
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Output:
The lnegth of the lps is 5

Please refer complete article on Longest Palindromic Subsequence | DP-12 for more details!





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