Question

CAN YOU PLEASE WRITE THIS CODE IN A DIFFERENT WAY 'EASIER AND BETTER'

QUESTION

Using C++

11. Write a function that will merge the contents of two sorted (ascending order) arrays of type double values, storing the result in an array out- put parameter (still in ascending order). The function shouldn’t assume that both its input parameter arrays are the same length but can assume

First array

04

Second array

Result array

that one array doesn’t contain two copies of the same value. The result array should also contain no duplicate values.

(Hint: When one of the input arrays has been exhausted, don’t forget to copy the remaining data in the other array into the result array.) Test your function with cases in which (1) the first array is exhausted first, (2) the second array is exhausted first, and (3) the two arrays are exhausted at the same time (that is, they end with the same value). Remember that the arrays input to this function must already be sorted.

CODE THAT NEED TO BE WRITTEN IN A DIFFERENT WAY

#include <iostream>

using namespace std;

//Function to merge 2 sorted arrays
int * merge(int *a1, int n1, int *a2, int n2){
//Initialize array to store merged array
static int ans[100];
int i1=0,i2=0,ind=0;
//Loop through both arrays
while(i1<n1&&i2<n2){
//Add minimum element to the array
if(a1[i1]<a2[i2]){
ans[ind++] = a1[i1++];
}
else{
ans[ind++] = a2[i2++];
}
}
//Add remaining elements to the array
while(i1<n1){
ans[ind++] = a1[i1++];
}
while(i2<n2){
ans[ind++] = a2[i2++];
}
return ans;
}

int main()
{
int a1[] = {1,3,5,8};
int a2[] = {2,4,6,7,9,10};
int *ans = merge(a1,4,a2,6);
for(int i=0;i<10;i++){
cout<<ans[i]<<" ";
}
cout<<endl;
return 0;
}

Answer 1

• There is a function called std::sort() that can do the sorting job. We have to put a1 and a2 one after other in a new array and sort the new array using this function.

Modified program:

#include <iostream>
#include <bits/stdc++.h>
using namespace std;

int * merge(int *a1, int n1, int *a2, int n2){
int* a3 = new int[n1+n2];
a3 = a1; // assign starting address to a1
a2 = a3+n1; // assign address of n1th element to staring address to a2
sort(a3, a3+n1+n2);
return a3;
}

int main()
{
int a1[] = {1,3,5,8};
int a2[] = {2,4,6,7,9,10};
int *ans = merge(a1,4,a2,6);
for(int i=0;i<10;i++){
cout<<ans[i]<<" ";
}
cout<<endl;
return 0;
}

output: