Finish and test the following two functions append and merge in the skeleton file: (1) function int* append(int*,int, int*,int); which accepts two dynamic arrays and return a new array by appending the second array to the first array.
(2) function int* merge(int*,int, int*,int); which accepts two sorted arrays and returns a new merged sorted array.
#include
using namespace std;
int* append(int*,int, int*,int);
int* merge(int*,int, int*,int);
void print(int*,int);
int main()
{ int a[] = {11,33,55,77,99};
int b[] = {22,44,66,88};
print(a,5);
print(b,4);
int* c = append(a,5,b,4); // c points to the appended array=
print(c,9);
int* d = merge(a,5,b,4);
print(d,9);
}
void print(int* a, int n)
{ cout << "{" << a[0];
for (int i=1; i cout << "," << a[i];
cout << "}\n"; }
int* append(int* a, int m, int* b, int n)
{
// wru=ite your codes in the text fields
}
int* merge(int* a, int m, int* b, int n)
{
// wru=ite your codes in the text fields
}

Explanation:

#include <iostream>

using namespace std;

int* append(int*,int,int*,int);

int* merge(int*,int,int*,int);

void print(int*,int);

int main()

{ int a[] = {11,33,55,77,99};

int b[] = {22,44,66,88};

print(a,5);

print(b,4);

int* c = append(a,5,b,4); // c points to the appended array=

print(c,9);

int* d = merge(a,5,b,4);

print(d,9);

}

void print(int* a, int n)

{ cout << "{" << a[0];

for (int i=1; i<n; i++)

cout << "," << a[i];

cout << "}\n";

}

int* append(int* a, int m, int* b, int n)

{

int * p= (int *)malloc(sizeof(int)*(m+n));

int i,index=0;

for(i=0;i<m;i++)

p[index++]=a[i];

for(i=0;i<n;i++)

p[index++]=b[i];

return p;

}

int* merge(int* a, int m, int* b, int n)

{

int i, j, k;

j = k = 0;

int *mergeRes = (int *)malloc(sizeof(int)*(m+n));

for (i = 0; i < m + n;) {

if (j < m && k < n) {

if (a[j] < b[k]) {

mergeRes[i] = a[j];

j++;

}

else {

mergeRes[i] = b[k];

k++;

}

i++;

}

// copying remaining elements from the b

else if (j == m) {

for (; i < m + n;) {

mergeRes[i] = b[k];

k++;

i++;

}

}

// copying remaining elements from the a

else {

for (; i < m + n;) {

mergeRes[i] = a[j];

j++;

i++;

}

}

}

return mergeRes;

}

the formula for work is

f*d

therefore work=2.0n*3.0=6n*m