Project 1 - Creation of user defined data type to implement the user interfaces for working with ‘Set’ (Mathematical Set theory)

An Interface is a collection of loosely bound items that have a common functionality or attributes. Interfaces contain method signatures, properties, events etc. Interfaces are used so that one class or struct can implement multiple behaviors. C# doesn’t support the concept of Multiple Inheritance because of the ambiguity it causes. But a lot of real-life objects inherit properties of more than just one type, therefore interfaces are used instead of extending classes. 

An Interface consists only of the signatures and not its implementation, therefore, any class or struct that implements it has to provide the implementation by overriding.

Giving two list of data as input to get the output as union & intersection for given list of data.

// C program to find union
// and intersection of two unsorted
// linked lists
#include
#include
#include
/* Link list node */
struct Node {
int data;
struct Node* next;
};

/* A utility function to insert a
node at the beginning ofa linked list*/
void push(struct Node** head_ref, int new_data);

/* A utility function to check if
given data is present in a list */
bool isPresent(struct Node* head, int data);

/* Function to get union of two
linked lists head1 and head2 */
struct Node* getUnion(
struct Node* head1,
struct Node* head2)
{
struct Node* result = NULL;
struct Node *t1 = head1, *t2 = head2;

// Insert all elements of
// list1 to the result list
while (t1 != NULL) {
push(&result, t1->data);
t1 = t1->next;
}

// Insert those elements of list2
// which are not present in result list
while (t2 != NULL) {
if (!isPresent(result, t2->data))
push(&result, t2->data);
t2 = t2->next;
}

return result;
}

/* Function to get intersection of
two linked lists head1 and head2 */
struct Node* getIntersection(struct Node* head1,
struct Node* head2)
{
struct Node* result = NULL;
struct Node* t1 = head1;

// Traverse list1 and search each element of it in
// list2. If the element is present in list 2, then
// insert the element to result
while (t1 != NULL) {
if (isPresent(head2, t1->data))
push(&result, t1->data);
t1 = t1->next;
}

return result;
}

/* A utility function to insert a
node at the beginning of a linked list*/
void push(struct Node** head_ref, int new_data)
{
/* allocate node */
struct Node* new_node
= (struct Node*)malloc(
sizeof(struct Node));

/* put in the data */
new_node->data = new_data;

/* link the old list off the new node */
new_node->next = (*head_ref);

/* move the head to point to the new node */
(*head_ref) = new_node;
}

/* A utility function to print a linked list*/
void printList(struct Node* node)
{
while (node != NULL) {
printf("%d ", node->data);
node = node->next;
}
}

/* A utility function that returns true if data is
present in linked list else return false */
bool isPresent(struct Node* head, int data)
{
struct Node* t = head;
while (t != NULL) {
if (t->data == data)
return 1;
t = t->next;
}
return 0;
}

/* Driver program to test above function*/
int main()
{
/* Start with the empty list */
struct Node* head1 = NULL;
struct Node* head2 = NULL;
struct Node* intersecn = NULL;
struct Node* unin = NULL;

/*create a linked lists 10->15->5->20 */
push(&head1, 20);
push(&head1, 4);
push(&head1, 15);
push(&head1, 10);

/*create a linked lists 8->4->2->10 */
push(&head2, 10);
push(&head2, 2);
push(&head2, 4);
push(&head2, 8);

intersecn = getIntersection(head1, head2);
unin = getUnion(head1, head2);

printf("\n First list is \n");
printList(head1);

printf("\n Second list is \n");
printList(head2);

printf("\n Intersection list is \n");
printList(intersecn);

printf("\n Union list is \n");
printList(unin);

return 0;
}