Problem:
Score : 100 points

Problem Statement

You are given strings s and t.
Find one longest string that is a subsequence of both s and t.

Notes

A subsequence of a string x is the string obtained by removing zero or more characters from x and concatenating the remaining characters without changing the order.

Constraints

s and t are strings consisting of lowercase English letters.

1 \leq |s|, |t| \leq 3000

Input

Input is given from Standard Input in the following format:

s
t

Output

Print one longest string that is a subsequence of both s and t.
If there are multiple such strings, any of them will be accepted.

Sample Input 1

axyb
abyxb

Sample Output 1

axb

The answer is axb or ayb; either will be accepted.

Sample Input 2

aa
xayaz

Sample Output 2

aa

Sample Input 3

a
z

Sample Output 3

The answer is (an empty string).

Sample Input 4

abracadabra
avadakedavra

Sample Output 4

aaadara

Solution in Java8 (OpenJDK 1.8.0) (judge verdict: Time Limit Exceeded)

import java.util.Scanner;

public class Main {

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        String s = sc.next();
        String t = sc.next();
        String[][] dp = new String[s.length() + 1][t.length() + 1];
        for (int i = 0 ; i < s.length() + 1 ; i++) {
            for (int j = 0; j < t.length() + 1; j++) {
                dp[i][j] = "";
            }
        }
        for (int i = 0 ; i < s.length(); i++) {
            for (int j = 0 ; j < t.length(); j++) {
                if (s.substring(i, i + 1).equals(t.substring(j, j+1))) {
                    dp[i +1][j + 1] = dp[i][j] + s.substring(i, i + 1);
                } else {
                    if (dp[i][j + 1].length() > dp[i + 1][j].length()) {
                        dp[i +1][j + 1] = dp[i][j + 1];
                    } else {
                        dp[i +1][j + 1] = dp[i + 1][j];
                    }
                }
            }
        }
        System.out.println(dp[s.length()][t.length()]);
    }

}