C Program To Find Union Of Two Sets | C Programming

The collection of well-defined distinct objects is known as a set. The word well-defined refers to a specific property that makes it easy to identify whether the given object belongs to the set or not. The word ‘distinct’ means that the objects of a set must be all different. 

Union Of Sets:-

The union of two given sets is the smallest set which contains all the elements of both the sets.

To find the union of two given sets A and B is a set that consists of all the elements of A and all the elements of B such that no element is repeated.

The symbol for denoting the union of sets is ‘∪’. 

For example;

Let set A = {2, 4, 5, 6}
and set B = {4, 6, 7, 8}

Taking every element of both the sets A and B, without repeating any element, we get a new set = {2, 4, 5, 6, 7, 8}

This new set contains all the elements of set A and all the elements of set B with no repetition of elements and is named as the union of set A and B.

Source Code

#include <stdio.h>
#include <conio.h>
int main()
{
    int a[10], b[10], m, n, i, j;
    int c[20], k = 0, flag = 0;
    int ch;
    printf("Enter the number of elements in first set:\n");
    scanf("%d", &m);
    printf("Enter the elements:\n");
    for (i = 0; i < m; i++)
    {
        scanf("%d", &a[i]);
    }
    printf("\nElement of First set:\n");
    for (i = 0; i < m; i++)
    {
        printf("%d\t", a[i]);
    }
    printf("\nEnter the number of elements in second set:\n");
    scanf("%d", &n);
    printf("Enter the elements:\n");
    for (i = 0; i < n; i++)
    {
        scanf("%d", &b[i]);
    }
    printf("\nElement of Second set:\n");
    for (i = 0; i < n; i++)
    {
        printf("%d\t", b[i]);
    }
    for (i = 0; i < m; i++)
    {
        c[k] = a[i];
        k++;
    }
    for (i = 0; i < n; i++)
    {
        flag = 0;
        for (j = 0; j < m; j++)
        {
            if (b[i] == c[j])
            {
                flag = 1;
                break;
            }
        }
        if (flag == 0)
        {
            c[k] = b[i];
            k++;
        }
    }
    printf("\nElement of resultant set\n");
    for (i = 0; i < k; i++)
    {
        printf("%d\t", c[i]);
    }
}

OUTPUT

Enter the number of elements in the first set:

5

Enter the elements:

1

2

3

4

5

Element of the First set:

1       2       3       4       5

Enter the number of elements in the second set:

5

Enter the elements:

6

7

8

5

4

Element of the Second set:

6       7       8       5       4

Element of the resultant set

1       2       3       4       5       6       7       8