Question

a. Sort the list A[ ]={ 20, 13,4, 34, 5, 15, 90, 100, 75, 102, 112, 1} using Insertion Sort and determine the total number of comparisons made (do not count swaps)

b. Sort the list stated in 5a) but using Merge Sort

Answer 1

Below is the c++ solution with output screenshot

Code (a) insertion sort :

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

// 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;
}

/* Function to sort an array using insertion sort*/
void insertionSort(int arr[], int n)
{
        int i, key, j, comparision=0;
        for (i = 1; i < n; i++)
        {
                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)
                {
                    comparision++;
                        arr[j + 1] = arr[j];
                        j = j - 1;
                }
                arr[j + 1] = key;
        }
    cout<<"\nArray after insertion sort is:\n";
    printArray(arr, n);
    cout<<"total number of comparisons made = "<<comparision;
}

// main method
int main()
{
        int arr[] = { 20, 13, 4, 34, 5, 15, 90, 100, 75, 102, 112, 1};
        int n = sizeof(arr) / sizeof(arr[0]);
    cout<<"Original array:\n";
    printArray(arr,n);
        insertionSort(arr, n);
        return 0;
}

Output : insertion sort

Total number of comparisons made = 21

Code (b) Merge sort :

// C++ program for Merge Sort
#include<iostream>
using namespace std;

// Merges two subarrays of arr[].
// First subarray is arr[l..m]
// Second subarray is arr[m+1..r]
void merge(int arr[], int l, int m, int r)
{
    int n1 = m - l + 1;
    int n2 = r - m;

    // Create temp arrays
    int L[n1], R[n2];

    // Copy data to temp arrays L[] and R[]
    for(int i = 0; i < n1; i++)
        L[i] = arr[l + i];
    for(int j = 0; j < n2; j++)
        R[j] = arr[m + 1 + j];

    // Merge the temp arrays back into arr[l..r]

    // Initial index of first subarray
    int i = 0;

    // Initial index of second subarray
    int j = 0;

    // Initial index of merged subarray
    int k = l;

    while (i < n1 && j < n2)
    {
        if (L[i] <= R[j])
        {
            arr[k] = L[i];
            i++;
        }
        else
        {
            arr[k] = R[j];
            j++;
        }
        k++;
    }

    // Copy the remaining elements of
    // L[], if there are any
    while (i < n1)
    {
        arr[k] = L[i];
        i++;
        k++;
    }

    // Copy the remaining elements of
    // R[], if there are any
    while (j < n2)
    {
        arr[k] = R[j];
        j++;
        k++;
    }
}

// l is for left index and r is
// right index of the sub-array
// of arr to be sorted */
void mergeSort(int arr[], int l, int r)
{
    if (l < r)
    {

        // Same as (l+r)/2, but avoids
        // overflow for large l and h
        int m = l + (r - l) / 2;

        // Sort first and second halves
        mergeSort(arr, l, m);
        mergeSort(arr, m + 1, r);

        merge(arr, l, m, r);
    }
}

// UTILITY FUNCTIONS
// Function to print an array
void printArray(int A[], int size)
{
    for(int i = 0; i < size; i++)
        cout << A[i] << " ";
}

int main()
{
    int arr[] = { 20, 13, 4, 34, 5, 15, 90, 100, 75, 102, 112, 1 };
    int arr_size = sizeof(arr) / sizeof(arr[0]);

    cout << "Given array is :\n";
    printArray(arr, arr_size);

    mergeSort(arr, 0, arr_size - 1);

    cout << "\n\nArray after merge sort is :\n";
    printArray(arr, arr_size);
    return 0;
}

Output (merge sort) :