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

Aim:

• Implementation of interface for mathematical Set Theory using C programming.
• Giving two list of data as input to get the output as union & intersection for given list of data.

Key Highlights of Project:

• Creating two list of data using linked list method.
• Insert the node to 1st and 2nd list accordingly.
• Intersection of data of two lists.
• Getting Union of two list.

Theory:

The union of two sets A and B is the set of all those elements which are either in A or in B, i.e. A ∪ B,

whereas the intersection of two sets A and B is the set of all elements which are common. A ∩ B

Program:

// C program to find union and intersection of two unsorted linked lists
#include<stdio.h>
#include<stdlib.h>
#include<stdbool.h>
// Link list node
struct Node
{
int data;
struct Node* next;
};
void push(struct Node** head_ref, int new_data);//function declaration
int isPresent(struct Node* head, int data);
struct Node* getUnion(struct Node* head1, struct Node* head2)// Function to get union of two linked lists head1 and 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;
}
struct Node* getIntersection(struct Node* head1, struct Node* head2)// Function to get intersection of two linked lists head1 and 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;
}
void push(struct Node** head_ref, int new_data)// A utility function to insert a node at the beginning of a linked list
{
struct Node* new_node = (struct Node*)malloc(sizeof(struct Node));// allocate node
new_node->data = new_data; // put in the data
new_node->next = (*head_ref);// link the old list off the new node
(*head_ref) = new_node;// move the head to point to the new node
}
void printList(struct Node* node)// function to print a linked list
{
while (node != NULL)
{
printf("%d ", node->data);
node = node->next;
}
}
int isPresent(struct Node* head, int data)// function that returns true if data is present in linked list else return false
{
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()
{
int n,m,temp;
printf("enter the value of number of value in link list(both):-");
fflush(stdout);
scanf("%d %d", &n, &m);
// Start structure pointer with the empty list
struct Node* head1 = NULL;
struct Node* head2 = NULL;
struct Node* intersecn = NULL;
struct Node* unin = NULL;
for(int i=0;i<n;i++)//for loop for link list1
{
printf("Enter the value in link list1 %d:-",i);
fflush(stdout);
scanf("%d", &temp);
push(&head1, temp);
}
for(int i=0;i<m;i++)//for loop for link list2
{
printf("Enter the value in link list2 %d:-",i);
fflush(stdout);
scanf("%d", &temp);
push(&head2, temp);

}
intersecn = getIntersection(head1, head2);//call the function of intersection
unin = getUnion(head1, head2);//call the function of union
printf("\n First list is ");
printList(head1); //function calling with structure pointer as argument
printf("\n Second list is ");
printList(head2);
printf("\n Intersection list is ");
printList(intersecn);
printf("\n Union list is ");
printList(unin);
return 0;
}

Output:

Result:The following opeartiopns has done successfully

• Intersection of data of two list.
• Union of data of two list