Question: Code Skeleton: import java.util.Arrays; public class kopy { /* merges b[] into a[], assumes a[] has enough space (at its end) to hold elements of

 Code Skeleton: import java.util.Arrays; public class kopy { /* merges b[]

Code Skeleton:

import java.util.Arrays;

public class kopy {

/* merges b[] into a[], assumes a[] has enough space (at its end) to hold elements of b[]. m is the number of sorted elements in a[](from index 0 to m-1) and n is the size of b[] */ public static void merge(int a[], int m, int b[], int n) { //write here }

static void merge (int al, int m, int bll, int n) merges a sorted array bl into another sorted array all. Assume all has enough space (at its end) to hold elements of b[]. m is the number of sorted elements in all (from index 0 to m-1) and n is the size of bl. For instance, given b [ 11, 22, 23] and a-[1, 12, 15, 17, 0, 0, 0] where n-3 and m-4, the output should be a [1, 11, 12,15, 17, 22, 23]. This is basically the merge method we had for mergesort, but instead of creating a new array, it uses the space available at the end of all. (50 points)

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