Question

• We are given an array of data1. We need to sort the only even positioned elements in the ascending order and the odd positioned elements same as the original order. You can apply insertion sorting.

Test Cases:

Test Case	Expected output
data1=new int [] {-1, 4, 5, 0}	-1, 0, 5,4 (20 points)
data1=new int [] {-1, 2, 3, -1, 5}	-1, -1, 3, 2, 5 (20 points)
If you code the for loop in insertionSort() which goes through only even positioned numbers, you get 20 points.	20 points

Answer 1

Q. We are given an array of data1. We need to sort the only even positioned elements in the ascending order and the odd positioned elements same as the original order. You can apply insertion sorting.

Test Case	Expected output
data1=new int [] {-1, 4, 5, 0}	-1, 0, 5,4 (20 points)
data1=new int [] {-1, 2, 3, -1, 5}	-1, -1, 3, 2, 5 (20 points)
If you code the for loop in insertionSort() which goes through only even positioned numbers, you get 20 points.	20 points

Ans- The below C++ code will give you the desired result and it considers only even positioned elements in the ascending order and the odd positioned elements same as the original order.

CODE:

#include <bits/stdc++.h>
using namespace std;
  
/* Function to sort an array using insertion sort*/
void insertionSort(int arr[], int n)
{
int i, key, j;
for (i = 1; i < n; i=i+2)
{
key = arr[i];
j = i - 1;
  
/* Move elements of arr[0..i-1], that are
greater than key, to one position ahead
of their current position */
while (j >= 0 && arr[j] > key)
{
arr[j + 1] = arr[j];
j = j - 1;
}
arr[j + 1] = key;
}
}
  
// A function to print an array of size n
void printArray(int arr[], int n)
{
int i;
for (i = 0; i < n; i++)
cout << arr[i] << " ";
cout << endl;
}
  
/* main code */
int main()
{
int arr[] = {-1,2,3,-1,5};
int n = sizeof(arr) / sizeof(arr[0]);
  
insertionSort(arr, n);
printArray(arr, n);
  
return 0;
}

OUTPUTS: