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
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Arogya Thapa Magar

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