Union and Intersection of two Linked Lists

Given two Linked Lists, create union and intersection lists that contain union and intersection of the elements present in the given lists. Order of elements in output lists doesn’t matter.

Example:

Input:
   List1: 10->15->4->20
   lsit2:  8->4->2->10
Output:
   Intersection List: 4->10
   Union List: 2->8->20->4->15->10

Method 1 (Simple)
The following are simple algorithms to get union and intersection lists respectively.

Intersection (list1, list2)
Initialize the result list as NULL. Traverse list1 and look for its every element in list2, if the element is present in list2, then add the element to result.

Union (list1, list2):
Initialize the result list as NULL. Traverse list1 and add all of its elements to the result.
Traverse list2. If an element of list2 is already present in result then do not insert it to result, otherwise insert.



This method assumes that there are no duplicates in the given lists.

Thanks to Shekhu for suggesting this method. Following are C and Java implementations of this method.

C/C++

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// C/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 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;
}
  
/* Drier 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 lits 10->15->5->20 */
    push(&head1, 20);
    push(&head1, 4);
    push(&head1, 15);
    push(&head1, 10);
  
    /*create a linked lits 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;
}

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Java

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// Java program to find union and
// intersection of two unsorted
// linked lists
class LinkedList {
    Node head; // head of list
  
    /* Linked list Node*/
    class Node {
        int data;
        Node next;
        Node(int d)
        {
            data = d;
            next = null;
        }
    }
  
    /* Function to get Union of 2 Linked Lists */
    void getUnion(Node head1, Node head2)
    {
        Node t1 = head1, t2 = head2;
  
        // insert all elements of list1 in the result
        while (t1 != null) {
            push(t1.data);
            t1 = t1.next;
        }
  
        // insert those elements of list2
        // that are not present
        while (t2 != null) {
            if (!isPresent(head, t2.data))
                push(t2.data);
            t2 = t2.next;
        }
    }
  
    void getIntersection(Node head1, Node head2)
    {
        Node result = null;
        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(t1.data);
            t1 = t1.next;
        }
    }
  
    /* Utility function to print list */
    void printList()
    {
        Node temp = head;
        while (temp != null) {
            System.out.print(temp.data + " ");
            temp = temp.next;
        }
        System.out.println();
    }
  
    /*  Inserts a node at start of linked list */
    void push(int new_data)
    {
        /* 1 & 2: Allocate the Node &
                  Put in the data*/
        Node new_node = new Node(new_data);
  
        /* 3. Make next of new Node as head */
        new_node.next = head;
  
        /* 4. Move the head to point to new Node */
        head = new_node;
    }
  
    /* A utilty function that returns true 
       if data is present in linked list 
       else return false */
    boolean isPresent(Node head, int data)
    {
        Node t = head;
        while (t != null) {
            if (t.data == data)
                return true;
            t = t.next;
        }
        return false;
    }
  
    /* Drier program to test above functions */
    public static void main(String args[])
    {
        LinkedList llist1 = new LinkedList();
        LinkedList llist2 = new LinkedList();
        LinkedList unin = new LinkedList();
        LinkedList intersecn = new LinkedList();
  
        /*create a linked lits 10->15->5->20 */
        llist1.push(20);
        llist1.push(4);
        llist1.push(15);
        llist1.push(10);
  
        /*create a linked lits 8->4->2->10 */
        llist2.push(10);
        llist2.push(2);
        llist2.push(4);
        llist2.push(8);
  
        intersecn.getIntersection(llist1.head, llist2.head);
        unin.getUnion(llist1.head, llist2.head);
  
        System.out.println("First List is");
        llist1.printList();
  
        System.out.println("Second List is");
        llist2.printList();
  
        System.out.println("Intersection List is");
        intersecn.printList();
  
        System.out.println("Union List is");
        unin.printList();
    }
} /* This code is contributed by Rajat Mishra */

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Output:

 First list is
10 15 4 20
 Second list is
8 4 2 10
 Intersection list is
4 10
 Union list is
2 8 20 4 15 10

Complexity Analysis:

  • Time Complexity: O(m*n).
    Here ‘m’ and ‘n’ are number of elements present in first and second lists respectively.
    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.
  • Auxiliary Space: O(1).
    No use of any data structure for storing values.

Method 2 (Use Merge Sort)
In this method, algorithms for Union and Intersection are very similar. First, we sort the given lists, then we traverse the sorted lists to get union and intersection.
The following are the steps to be followed to get union and intersection lists.

  1. Sort the first Linked List using merge sort. This step takes O(mLogm) time. Refer this post for details of this step.
  2. Sort the second Linked List using merge sort. This step takes O(nLogn) time. Refer this post for details of this step.
  3. Linearly scan both sorted lists to get the union and intersection. This step takes O(m + n) time. This step can be implemented using the same algorithm as sorted arrays algorithm discussed here.

Time complexity of this method is O(mLogm + nLogn) which is better than method 1’s time complexity.

Method 3 (Use Hashing)
Union (list1, list2)
Initialize the result list as NULL and create an empty hash table. Traverse both lists one by one, for each element being visited, look the element in the hash table. If the element is not present, then insert the element to the result list. If the element is present, then ignore it.

Intersection (list1, list2)
Initialize the result list as NULL and create an empty hash table. Traverse list1. For each element being visited in list1, insert the element in the hash table. Traverse list2, for each element being visited in list2, look the element in the hash table. If the element is present, then insert the element to the result list. If the element is not present, then ignore it.

Both of the above methods assume that there are no duplicates.

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// Java code for Union and Intersection of two
// Linked Lists
import java.util.HashMap;
import java.util.HashSet;
  
class LinkedList {
    Node head; // head of list
  
    /* Linked list Node*/
    class Node {
        int data;
        Node next;
        Node(int d)
        {
            data = d;
            next = null;
        }
    }
  
    /* Utility function to print list */
    void printList()
    {
        Node temp = head;
        while (temp != null) {
            System.out.print(temp.data + " ");
            temp = temp.next;
        }
        System.out.println();
    }
  
    /* Inserts a node at start of linked list */
    void push(int new_data)
    {
        /* 1 & 2: Allocate the Node &
        Put in the data*/
        Node new_node = new Node(new_data);
  
        /* 3. Make next of new Node as head */
        new_node.next = head;
  
        /* 4. Move the head to point to new Node */
        head = new_node;
    }
  
    public void append(int new_data)
    {
        if (this.head == null) {
            Node n = new Node(new_data);
            this.head = n;
            return;
        }
        Node n1 = this.head;
        Node n2 = new Node(new_data);
        while (n1.next != null) {
            n1 = n1.next;
        }
  
        n1.next = n2;
        n2.next = null;
    }
  
    /* A utilty function that returns true if data is
    present in linked list else return false */
    boolean isPresent(Node head, int data)
    {
        Node t = head;
        while (t != null) {
            if (t.data == data)
                return true;
            t = t.next;
        }
        return false;
    }
  
    LinkedList getIntersection(Node head1, Node head2)
    {
        HashSet<Integer> hset = new HashSet<>();
        Node n1 = head1;
        Node n2 = head2;
        LinkedList result = new LinkedList();
  
        // loop stores all the elements of list1 in hset
        while (n1 != null) {
            if (hset.contains(n1.data)) {
                hset.add(n1.data);
            }
            else {
                hset.add(n1.data);
            }
            n1 = n1.next;
        }
  
        // For every element of list2 present in hset
        // loop inserts the element into the result
        while (n2 != null) {
            if (hset.contains(n2.data)) {
                result.push(n2.data);
            }
            n2 = n2.next;
        }
        return result;
    }
  
    LinkedList getUnion(Node head1, Node head2)
    {
        // HashMap that will store the
        // elements of the lists with their counts
        HashMap<Integer, Integer> hmap = new HashMap<>();
        Node n1 = head1;
        Node n2 = head2;
        LinkedList result = new LinkedList();
  
        // loop inserts the elements and the count of
        // that element of list1 into the hmap
        while (n1 != null) {
            if (hmap.containsKey(n1.data)) {
                int val = hmap.get(n1.data);
                hmap.put(n1.data, val + 1);
            }
            else {
                hmap.put(n1.data, 1);
            }
            n1 = n1.next;
        }
  
        // loop further adds the elements of list2 with
        // their counts into the hmap
        while (n2 != null) {
            if (hmap.containsKey(n2.data)) {
                int val = hmap.get(n2.data);
                hmap.put(n2.data, val + 1);
            }
            else {
                hmap.put(n2.data, 1);
            }
            n2 = n2.next;
        }
  
        // Eventually add all the elements
        // into the result that are present in the hmap
        for (int a : hmap.keySet()) {
            result.append(a);
        }
        return result;
    }
  
    /* Driver program to test above functions */
    public static void main(String args[])
    {
        LinkedList llist1 = new LinkedList();
        LinkedList llist2 = new LinkedList();
        LinkedList union = new LinkedList();
        LinkedList intersection = new LinkedList();
  
        /*create a linked list 10->15->4->20 */
        llist1.push(20);
        llist1.push(4);
        llist1.push(15);
        llist1.push(10);
  
        /*create a linked list 8->4->2->10 */
        llist2.push(10);
        llist2.push(2);
        llist2.push(4);
        llist2.push(8);
  
        intersection
            = intersection.getIntersection(llist1.head,
                                           llist2.head);
        union = union.getUnion(llist1.head, llist2.head);
  
        System.out.println("First List is");
        llist1.printList();
  
        System.out.println("Second List is");
        llist2.printList();
  
        System.out.println("Intersection List is");
        intersection.printList();
  
        System.out.println("Union List is");
        union.printList();
    }
}
// This code is contributed by Kamal Rawal

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Output:

First List is
10 15 4 20 
Second List is
8 4 2 10 
Intersection List is
10 4 
Union List is
2 4 20 8 10 15 

Complexity Analysis:

  • Time Complexity: O(m+n).
    Here ‘m’ and ‘n’ are number of elements present in first and second lists respectively.
    For union: Traverse both the lists, store the elements in Hash-map and update the respective count.
    For intersection: First traverse list-1, store its elements in Hash-map and then for every element in list-2 check if it is already present in the map. This takes O(1) time.
  • Auxiliary Space:O(m+n).
    Use of Hash-map data structure for storing values.

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.

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Improved By : bidibaaz123, Akanksha_Rai