Union and Intersection of two Linked Lists

11 Jul, 2022
Given two Linked Lists, create union and intersection lists that contain union and intersection of the elements present in the given lists. The order of elements in output lists doesn’t matter.Example: 
Input:
   List1: 10->15->4->20
   List2:  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 every element in list2, if the element is present in list2, then add the element to the result.Union (list1, list2):Initialize a new list ans and store first and second list data to set to remove duplicate data and then store it into our new list ans and return its head.
C++
C
Java
C#
Javascript
// C++ program to find union// and intersection of two unsorted// linked lists#include "bits/stdc++.h"using namespace std;  /* Link list node */struct Node {    int data;    struct Node* next;    Node(int x)    {        data = x;        next = NULL;    }};  /* A utility function to insert anode at the beginning ofa linked list*/void push(struct Node** head_ref, int new_data);  /* A utility function to check ifgiven data is present in a list */bool isPresent(struct Node* head, int data);  /* Function to get union of twolinked lists head1 and head2 */struct Node* getUnion(struct Node* head1,                      struct Node* head2){    struct Node* ans = new Node(-1);    struct Node* head = ans;    set<int> st;    while (head1 != NULL) {        st.insert(head1->data);        head1 = head1->next;    }    while (head2 != NULL) {        st.insert(head2->data);        head2 = head2->next;    }    for (auto it : st) {        struct Node* t = new Node(it);        ans->next = t;        ans = ans->next;    }    head = head->next;    return head;}  /* Function to get intersection oftwo 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 anode 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) {        cout << " " << node->data;        node = node->next;    }}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);    cout << "\n First list is " << endl;    printList(head1);    cout << "\n Second list is " << endl;    printList(head2);    cout << "\n Intersection list is " << endl;    printList(intersecn);    cout << "\n Union list is " << endl;    printList(unin);    return 0;}  // This code is contributed by zishanahmad786
// 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;}  /* 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;}
// Java program to find union and// intersection of two unsorted// linked listsclass 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 utility 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;    }      /* Driver 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 lists 10->15->4->20 */        llist1.push(20);        llist1.push(4);        llist1.push(15);        llist1.push(10);          /*create a linked lists 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 */
// C# program to find union and// intersection of two unsorted// linked listsusing System;class LinkedList {      public Node head; // head of list      /* Linked list Node*/    public class Node {        public int data;        public Node next;        public 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 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) {            Console.Write(temp.data + " ");            temp = temp.next;        }        Console.WriteLine();    }      /*  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 utility function that returns true       if data is present in linked list       else return false */    bool isPresent(Node head, int data)    {        Node t = head;        while (t != null) {            if (t.data == data)                return true;            t = t.next;        }        return false;    }      /* Driver code*/    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 lists 10->15->5->20 */        llist1.push(20);        llist1.push(4);        llist1.push(15);        llist1.push(10);          /*create a linked lists 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);          Console.WriteLine("First List is");        llist1.printList();          Console.WriteLine("Second List is");        llist2.printList();          Console.WriteLine("Intersection List is");        intersecn.printList();          Console.WriteLine("Union List is");        unin.printList();    }}  // This code is contributed by rutvik_56.
<script>    //Javascript program to find union and    // intersection of two unsorted    // linked lists          /* Linked list Node*/    class Node {        constructor(d)        {            this.data = d;            this.next = null;        }    }          class LinkedList{        constructor(){            this.head=null;        }                  /*  Inserts a node at start of linked list */        push(new_data)        {            /* 1 & 2: Allocate the Node &                      Put in the data*/            var new_node = new Node(new_data);                   /* 3. Make next of new Node as head */            new_node.next = this.head;                   /* 4. Move the head to point to new Node */            this.head = new_node;        }               /* A utility function that returns true           if data is present in linked list           else return false */        isPresent(head, data)        {            var t = head;            while (t != null) {                if (t.data == data)                    return true;                t = t.next;            }            return false;        }                  /* Function to get Union of 2 Linked Lists */        getUnion(head1, head2)        {            var t1 = head1, t2 = head2;                   // insert all elements of list1 in the result            while (t1 != null) {                this.push(t1.data);                t1 = t1.next;            }                   // insert those elements of list2            // that are not present            while (t2 != null) {                if (!this.isPresent(this.head, t2.data))                    this.push(t2.data);                t2 = t2.next;            }        }                  getIntersection(head1, head2)        {            var result = null;            var 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 (this.isPresent(head2, t1.data))                    this.push(t1.data);                t1 = t1.next;            }        }                  /* Utility function to print list */        printList()        {            var temp = this.head;            while (temp != null) {                document.write(temp.data + " ");                temp = temp.next;            }        }    }          const llist1 = new LinkedList();    const llist2 = new LinkedList();    const unin = new LinkedList();    const intersecn = new LinkedList();          /*create a linked lists 10->15->4->20 */    llist1.push(20);    llist1.push(4);    llist1.push(15);    llist1.push(10);          /*create a linked lists 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);          document.write("First List is ");    llist1.printList();          document.write("Second List is ");    llist2.printList();          document.write("Intersection List is ");    intersecn.printList();          document.write("Union List is ");    unin.printList();          //This code is contributed by shruti456rawal</script>
 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 the 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.
Sort the first Linked List using merge sort. This step takes O(mLogm) time. Refer this post for details of this step.Sort the second Linked List using merge sort. This step takes O(nLogn) time. Refer this post for details of this step.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.
Sort the first Linked List using merge sort. This step takes O(mLogm) time. Refer this post for details of this step.
Sort the second Linked List using merge sort. This step takes O(nLogn) time. Refer this post for details of this step.
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.
The 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 at the element in the hash table. If the element is not present, then insert the element into 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. 
Java
C#
// Java code for Union and Intersection of two// Linked Listsimport 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 utility 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
// C# code for Union and Intersection of two// Linked Listsusing System;using System.Collections;using System.Collections.Generic;  class LinkedList{    public Node head; // head of list      /* Linked list Node*/    public class Node     {        public int data;        public Node next;        public Node(int d)        {            data = d;            next = null;        }    }      /* Utility function to print list */    void printList()    {        Node temp = head;        while (temp != null)         {            Console.Write(temp.data + " ");            temp = temp.next;        }        Console.WriteLine();    }      /* 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 utility function that returns true if data is    present in linked list else return false */    bool 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<int> hset = new HashSet<int>();        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        SortedDictionary<int, int> hmap = new SortedDictionary<int, int>();        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)) {                hmap[n1.data]++;            }            else {                hmap[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)) {                hmap[n2.data]++;            }            else {                hmap[n2.data]= 1;            }            n2 = n2.next;        }          // Eventually add all the elements        // into the result that are present in the hmap        foreach(int a in hmap.Keys) {            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);        Console.WriteLine("First List is");        llist1.printList();        Console.WriteLine("Second List is");        llist2.printList();        Console.WriteLine("Intersection List is");        intersection.printList();        Console.WriteLine("Union List is");        union.printList();    }}  //This code is contributed by pratham76
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 the 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.
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shruti456rawal
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24*7 Innovation Labs
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VMWare
Flipkart
Accolite
Amazon
Microsoft
24*7 Innovation Labs
Walmart
Komli Media
Taxi4Sure
Linked List
Hash
Sorting
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