Question

Write an algorithm that takes two strings X and Y and a positive integer k as input, and determines whether Y can be turned into X by at most k insertions or deletions. For example: let X = “abacad", Y = “cebacad" and k = 3 then your algorithm should give a yes answer because Y can be turned into X in 3 steps by a left-to-right processing as follows: delete leftmost c delete “e" insert “a" Give an tight asymptotic bound on the running time of your algorithm. JAVA CODE

VERY URGENT THANKS

Answer 1

import java.io.*;

import java.util.Scanner;

  

class LeastCommonSubsequence {

      

    // Returns length of length common  

    // subsequence for a[0..m-1],  

    // b[0..n-1]

    static int lcs(String X, String Y)

    {

        int m = X.length();

        int n = Y.length();

        int MAT[][] = new int[m+1][n+1];

        int i, j;

  

        // Following steps build MAT[m+1][n+1] in

        // bottom up fashion. Note that MAT[i][j]

        // contains length of LCS of a[0..i-1]

        // and b[0..j-1]  

        for (i = 0; i <= m; i++)

        {

            for (j = 0; j <= n; j++)

            {

                if (i == 0 || j == 0)

                    MAT[i][j] = 0;

                else if (X.charAt(i-1) == Y.charAt(j-1))

                    MAT[i][j] = MAT[i-1][j-1] + 1;

                else

                    MAT[i][j] = Math.max(MAT[i-1][j], MAT[i][j-1]);

            }

        }

        // MAT[m][n] contains length of LCS  

        // for X[0..n-1] and Y[0..m-1]  

            return m+n- 2*MAT[m][n]; //least number of insertion or deletion

        }   

    // Driver program to test above

    public static void main(String[] args)

    {

        String X, Y;

        int k;

        Scanner sc = new Scanner(System.in);

        System.out.print("Enter 1st string X: ");        

        X = sc.next();

        System.out.print("Enter 2nd string Y: ");        

        Y = sc.next();

        System.out.print("Enter value of k: ");        

        k = sc.nextInt();

        int l = lcs(X,Y);

        if(k<l){

            System.out.println("No");

        }

        else

        {

            System.out.println("Yes");     

        }

    }  

}