Problem:
MathJax.Hub.Config({ tex2jax: { inlineMath: [["$","$"], ["\\(","\\)"]], processEscapes: true }});

Longest Common Subsequence

For given two sequences $X$ and $Y$, a sequence $Z$ is a common subsequence of $X$ and $Y$ if $Z$ is a subsequence of both $X$ and $Y$. For example, if $X = \{a,b,c,b,d,a,b\}$ and $Y = \{b,d,c,a,b,a\}$, the sequence $\{b,c,a\}$ is a common subsequence of both $X$ and $Y$. On the other hand, the sequence $\{b,c,a\}$ is not a longest common subsequence (LCS) of $X$ and $Y$, since it has length 3 and the sequence $\{b,c,b,a\}$, which is also common to both $X$ and $Y$, has length 4. The sequence $\{b,c,b,a\}$ is an LCS of $X$ and $Y$, since there is no common subsequence of length 5 or greater.

Write a program which finds the length of LCS of given two sequences $X$ and $Y$. The sequence consists of alphabetical characters.

Input

The input consists of multiple datasets. In the first line, an integer $q$ which is the number of datasets is given. In the following $2 \times q$ lines, each dataset which consists of the two sequences $X$ and $Y$ are given.

Output

For each dataset, print the length of LCS of $X$ and $Y$ in a line.

Constraints

$1 \leq q \leq 150$

$1 \leq$ length of $X$ and $Y$ $\leq 1,000$

$q \leq 20$ if the dataset includes a sequence whose length is more than 100

Sample Input 1

3
abcbdab
bdcaba
abc
abc
abc
bc

Sample Output 1

4
3
2

Reference

Introduction to Algorithms, Thomas H. Cormen, Charles E. Leiserson, Ronald L. Rivest, and Clifford Stein. The MIT Press.

Solution in JAVA (judge verdict: Accepted)



import java.util.*;

public class Main{

	public static void main(String[] args) {
		Scanner input = new Scanner(System.in);
		int q = input.nextInt();
		String a;
		String b;
		for(int i=0; i<q; i++) {
			 a = input.next();
			 b = input.next();
		     System.out.println(getLcs(a,b));
		}
	}
		public static int getLcs(String a, String b) {
		int l1 = a.length();
		int l2 = b.length();
		int[][] lcs = new int [a.length()+1][b.length()+1];
		for(int i=1; i<=l1; i++) {
			for(int j=1; j<=l2; j++) {
				if(a.charAt(i-1) == b.charAt(j-1)) {
 				 	lcs[i][j] = 1 + lcs[i-1][j-1];} 
				else {
					lcs[i][j] = Math.max(lcs[i][j-1], lcs[i-1][j]);
				}
			}
		}
		return lcs[l1][l2];
	}	
}