LongestPalindromicSubsequence.java

package com.thealgorithms.dynamicprogramming;

/**
 * Algorithm explanation
 * https://www.educative.io/edpresso/longest-palindromic-subsequence-algorithm
 */
public final class LongestPalindromicSubsequence {
    private LongestPalindromicSubsequence() {
    }

    public static void main(String[] args) {
        String a = "BBABCBCAB";
        String b = "BABCBAB";

        String aLPS = lps(a);
        String bLPS = lps(b);

        System.out.println(a + " => " + aLPS);
        System.out.println(b + " => " + bLPS);
    }

    public static String lps(String original) throws IllegalArgumentException {
        StringBuilder reverse = new StringBuilder(original);
        reverse = reverse.reverse();
        return recursiveLPS(original, reverse.toString());
    }

    private static String recursiveLPS(String original, String reverse) {
        String bestResult = "";

        // no more chars, then return empty
        if (original.length() == 0 || reverse.length() == 0) {
            bestResult = "";
        } else {
            // if the last chars match, then remove it from both strings and recur
            if (original.charAt(original.length() - 1) == reverse.charAt(reverse.length() - 1)) {
                String bestSubResult = recursiveLPS(original.substring(0, original.length() - 1), reverse.substring(0, reverse.length() - 1));

                bestResult = reverse.charAt(reverse.length() - 1) + bestSubResult;
            } else {
                // otherwise (1) ignore the last character of reverse, and recur on original and
                // updated reverse again (2) ignore the last character of original and recur on the
                // updated original and reverse again then select the best result from these two
                // subproblems.

                String bestSubResult1 = recursiveLPS(original, reverse.substring(0, reverse.length() - 1));
                String bestSubResult2 = recursiveLPS(original.substring(0, original.length() - 1), reverse);
                if (bestSubResult1.length() > bestSubResult2.length()) {
                    bestResult = bestSubResult1;
                } else {
                    bestResult = bestSubResult2;
                }
            }
        }

        return bestResult;
    }
}