Skip to content
Perform Set Operations on Arrays

Perform Set Operations on Arrays

In this problem, we’ll be performing some basic set operations on the given arrays like Union, Intersection, and Difference. I found this problem when I was writing a practice test on ADT (Array as a Data Structure).

Examples Set Operations

Array1: 2 9 21 28 35
Array2: 2 3 9 18 28

Union(A1∪A2)
2 3 9 18 21 28 35

Intersection(A1∩A2)
2 9 28

Difference(A1-A2)
21 35

Union ∪

A union of two sets will contain the result of elements present in both the sets, but by their nature sets do not contain duplicates hence no repeated values.

If, A = {1,2,3} B={1,3,5} then A ∪ B = {1,2,3,5}

Intersection ∩

As the name says, the intersection of two sets will give us the values/elements present in both the sets.

If, C = {6,8,9} C={8,9,11} then C ∩ D = {8,9}

Difference –

The difference operation is nothing but subtraction, this usually results in the elements of first set excluding similar elements present in the other set.

If, E = {4,8,11} F={8,9,12} then E – F = {4,11}

Program for Set Operations

I’ve written this code in C however the logic remains the same in C++, python, or any other language

#include <stdio.h>
#include <stdlib.h>

//set operations
struct Array
{
    int A[10];
    int size;
    int length;
};

void Display(struct Array arr)
{
    int i;
    printf("Elements are\n");
    for (i = 0; i < arr.length; i++)
        printf("%d ", arr.A[i]);
}

struct Array *Union(struct Array *arr1, struct Array *arr2)
{
    int i, j, k;
    i = j = k = 0;
    //join both arrays
    struct Array *arr3 = (struct Array *)malloc(sizeof(struct Array));
    while (i < arr1->length && j < arr2->length)
    {

        if (arr1->A[i] < arr2->A[j])
            arr3->A[k++] = arr1->A[i++];
        else if (arr2->A[j] < arr1->A[i])
            arr3->A[k++] = arr2->A[j++];
        else
        {
            arr3->A[k++] = arr1->A[i++];
            j++;
        }
    }
    for (; i < arr1->length; i++)
        arr3->A[k++] = arr1->A[i];
    for (; j < arr2->length; j++)
        arr3->A[k++] = arr2->A[j];
    arr3->length = k;
    arr3->size = 10;
    return arr3;
}

struct Array *Intersection(struct Array *arr1, struct Array *arr2)
{
    int i, j, k;
    i = j = k = 0;
    //add if  both are same, present in both
    struct Array *arr3 = (struct Array *)malloc(sizeof(struct Array));
    while (i < arr1->length && j < arr2->length)
    {
        if (arr1->A[i] < arr2->A[j])
            i++;
        else if (arr2->A[j] < arr1->A[i])
            j++;
        else if (arr1->A[i] == arr2->A[j])
        {
            arr3->A[k++] = arr1->A[i++];
            j++;
        }
    }
    arr3->length = k;
    arr3->size = 10;
    return arr3;
}

struct Array *Difference(struct Array *arr1, struct Array *arr2)
{
    int i, j, k;
    i = j = k = 0;
    //a-b=c
    struct Array *arr3 = (struct Array *)malloc(sizeof(struct Array));
    while (i < arr1->length && j < arr2->length)
    {
        //if a<b then a is already crossed or not present hence add it
        if (arr1->A[i] < arr2->A[j])
            arr3->A[k++] = arr1->A[i++];
        else if (arr2->A[j] < arr1->A[i])
            j++;
        else
        {
            i++;
            j++;
        }
    }
    for (; i < arr1->length; i++)
        arr3->A[k++] = arr1->A[i];
    arr3->length = k;
    arr3->size = 10;
    return arr3;
}

int main()
{
    struct Array arr1 = {{2, 9, 21, 28, 35}, 10, 5};
    struct Array arr2 = {{2, 3, 9, 18, 28}, 10, 5};
    Display(arr1);
    Display(arr2);
    struct Array *arr3;
    arr3 = Union(&arr1, &arr2);
    Display(*arr3);
    arr3 = Intersection(&arr1, &arr2);
    Display(*arr3);
    arr3 = Difference(&arr1, &arr2);
    Display(*arr3);
    return 0;
}

Conclusion

This post [Performing Set Operations on Arrays] is a part of my #30DaysChallenge to write a blog post every day on what I learn daily, All the Best~ Abhiram Reddy.

Tags: