JavaAlgorithms/DynamicProgramming/LongestPalindromicSubsequence.java

package DynamicProgramming;

/**
 * Algorithm explanation https://www.educative.io/edpresso/longest-palindromic-subsequence-algorithm
 */
public class 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;
  }
}