Challenge

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

Code:

// C program to find union

// and intersection of two unsorted

// linked lists

#include <stdbool.h>

#include<stdio.h>

#include<stdlib.h>

/* Link list node */

struct Node {

int data;

struct Node* next;

};

/* A utility function to insert a

node at the beginning of a linked list*/

/* 2. put in the data */

//new_node->data = new_data//

/* 3. Make next of new node as head */

//new_node->next = (*file_ref)//

/* 4. move the head to point to the new node */

//(*file_ref) = new_node//

void push(struct Node** file_ref, int new_data);

/* A utility function to check if

given data is present in a list */

bool isPresent(struct Node* file, int data);

/* Function to get union of two

linked lists file1 and file2 */

struct Node* getUnion(struct Node* file1,

struct Node* file2)

{

struct Node* result = NULL;

struct Node *t1 = file1, *t2 = file2;

// 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 file1 and file2 */

struct Node* getIntersection(struct Node* file1,

struct Node* file2)

{

struct Node* result = NULL;

struct Node* t1 = file1;

struct Node* t2=file2;

// 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(result, t1->data))

push(&result, t1->data);

t1 = t1->next;

}

while (t2 != NULL) {

if (!isPresent(result, t2->data))

push(&result, t2->data);

t2 = t2->next;

}

return result;

}

/* A utility function to insert a

node at the beginning of a linked list*/

void push(struct Node** file_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 = (*file_ref);

/* move the head to point to the new node */

(*file_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* file, int data)

{

struct Node* t = file;

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* file1 = NULL;

struct Node* file2 = NULL;

struct Node* intersecn = NULL;

struct Node* unin = NULL;

/*create a linked lists 15310245->15310246->15310247->15310248 */

push(&file1, 15310245);

push(&file1, 15310246);

push(&file1, 15310247);

push(&file1, 15310248);

/*create a linked lists 15310245->15310248->15310249->15310250 */

push(&file2, 15310245);

push(&file2, 15310248);

push(&file2, 15310249);

push(&file2, 15310250);

intersecn = getIntersection(file1, file2);

unin = getUnion(file1,file2);

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

printList(file1);

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

printList(file2);

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

printList(intersecn);

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

printList(unin);

return 0;

}

comment:

  • Time Complexity: O(m*n).
    O(m+n) it means that the algorithm will take O(m) time when m>n and O(n) when m<n.
  • For union: For every element in list-2 we check if that element is already present in the resultant list made using list-1.
    For intersection: For every element in list-1 we check if that element is also present in list-2.

Output:

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