# Question: In this project, you will implement the dynamic programming-based solution to find the longest co……

In this project, you will implement the dynamic programming-based solution to find the longest common subsequence (LCS) of two sequences. Your inputs will be the two sequences (as Strings) and the outputs are the longest common subsequence (printed as a String) and the final matrix (printed as a two-dimensional array) depicting the length of the longest common subsequences (as shown in the slides) for all possible subsequences of the two input sequences The two input sequences to be used by each student are shown below. The LCS expected for the two sequences is also shown.

package regex;

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import java.util.Scanner;
import java.lang.*;
import static java.lang.Integer.max;

/**
*
* @author prathmesh
*/
public class chegg {

/**
* @param args the command line arguments
*/
String x,y,ans=””,ans2=””;
chegg(String a,String b){
this.x = a;
this.y = b;
}
int recur(String x,String y,int i,int j){
if(i==0 || j==0){
return 0;
}
if(x.charAt(i-1) == y.charAt(j-1)){
dp[i][j] = 1 + recur(x,y,i-1,j-1);
}
else{
dp[i][j] = max(recur(x,y,i-1,j),recur(x,y,i,j-1));
}
return dp[i][j];

}
void lcs(){
for(int i=0;i<100;i++){
for(int j=0;j<100;j++){
dp[i][j] = 0;
}
}
recur(x,y,x.length(),y.length());
}
void printdp(){

for(int i=0;i<=x.length();i++){
for(int j=0;j<=y.length();j++){
System.out.print( dp[i][j]+ ” “);

}
System.out.println(“”);
}
recur(x,y,x.length(),y.length());
}
String printlcs(){
int i=x.length();
int j=y.length();

while(i>0 && j>0){
if(x.charAt(i-1)==y.charAt(j-1)){
ans += x.charAt(i-1);
i–;j–;
}
else if(dp[i-1][j] > dp[i][j-1]){
i–;
}
else{
j–;
}
}
// As the LCS string is in reverse;
for(i=ans.length()-1;i>=0;i–){
ans2 += ans.charAt(i);
}
return ans2;
}
int dp[][] = new int[100][100];
public static void main(String[] args) {
Scanner input = new Scanner(System.in);
String x = input.next();
String y = input.next();
chegg c = new chegg(x,y);
c.lcs();
System.out.println( “Longest subsequence : ” + c.printlcs());
c.printdp();

}

}

As you can see, there are mainly three functions”-

1)lcs

It calculates and builds the dynamic programming table, also initilizes all the values to 0.

2) printlcs

It traverses the table and creates the answer by the entries in the DP table.

We start from the bottom and traverse up by going towards the direction which has a higher value.

3)printdp

To print the dp table.

Note: Final alignment can be worked out simply by looking at the strings manually

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