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 Java (OpenJDK 11.0.6) (judge verdict: Wrong Answer)

import java.util.*;

public class Main
{
	public static void main(String[] args)
	{
		Scanner sc = new Scanner(System.in);
		String s1 = sc.next();
		String s2 = sc.next();

		int[][] dp = new int[s1.length()+1][s2.length()+1];

		for(int i=1;i<=s1.length();i++)
		{
			for(int j=1;j<=s2.length();j++)
			{
				dp[i][j] = Math.max(dp[i-1][j],dp[i][j-1]);

					if(s1.charAt(i-1)==s2.charAt(j-1))
					{
						dp[i][j] = 1 + dp[i-1][j-1];
					}
					
			}
		}
       
  

       String ans = "";
       int i =dp.length-1;
       int j =dp[0].length-1;

       while(i>=1||j>=1)
       {
       	  if(i>=1&&j>=1&&s1.charAt(i-1)==s2.charAt(j-1)&&dp[i][j]==dp[i-1][j-1]+1)
       	  {
       	  	  ans = ans + s1.charAt(i-1);
       	  }

       	  if(i>=0&&j>=1&&dp[i][j]==dp[i][j-1])
       	  {
              j=j-1;
       	  }
       	  else
       	  {
             i=i-1;
       	  }
       	 
       }
       StringBuilder str = new StringBuilder();
       str.append(ans);

       System.out.println(str.reverse());
 	}
}