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.

**Examples:**

Input:List1: 10 -> 15 -> 4 -> 20 List2: 8 -> 4 -> 2 -> 10Output:Intersection List: 4 -> 10 Union List: 2 -> 8 -> 20 -> 4 -> 15 -> 10Explanation:In this two lists 4 and 10 nodes are common. The union lists contains all the nodes of both the lists.Input:List1: 1 -> 2 -> 3 -> 4 List2: 3 -> 4 -> 8 -> 10Output:Intersection List: 3 -> 4 Union List: 1 -> 2 -> 3 -> 4 -> 8 -> 10Explanation:In this two lists 4 and 3 nodes are common. The union lists contains all the nodes of both the lists.

We have already discussed Method-1 and Method-2 of this question.

In this post, its Method-3 (Using Hashing) is discussed with a Time Complexity of O(m+n) i.e. better than both methods discussed earlier.

Implementation:1- Start traversing both the lists. a) Store the current element of both lists with its occurrence in the map. 2- For Union: Store all the elements of the map in the resultant list. 3- For Intersection: Store all the elements only with an occurrence of 2 as 2 denotes that they are present in both the lists.

Below is the C++ implementation of the above steps.

`// C++ program to find union and intersection of` `// two unsorted linked lists in O(m+n) time.` `#include <bits/stdc++.h>` `using` `namespace` `std;` ` ` `/* Link list node */` `struct` `Node {` ` ` `int` `data;` ` ` `struct` `Node* next;` `};` ` ` `/* 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;` `}` ` ` `/* Utility function to store the ` ` ` `elements of both list */` `void` `storeEle(` `struct` `Node* head1, ` `struct` `Node* head2,` ` ` `unordered_map<` `int` `, ` `int` `>& eleOcc)` `{` ` ` `struct` `Node* ptr1 = head1;` ` ` `struct` `Node* ptr2 = head2;` ` ` ` ` `// Traverse both lists` ` ` `while` `(ptr1 != NULL || ptr2 != NULL) {` ` ` `// store element in the map` ` ` `if` `(ptr1 != NULL) {` ` ` `eleOcc[ptr1->data]++;` ` ` `ptr1 = ptr1->next;` ` ` `}` ` ` ` ` `// store element in the map` ` ` `if` `(ptr2 != NULL) {` ` ` `eleOcc[ptr2->data]++;` ` ` `ptr2 = ptr2->next;` ` ` `}` ` ` `}` `}` ` ` `/* Function to get the union of two ` ` ` `linked lists head1 and head2 */` `struct` `Node* getUnion(` ` ` `unordered_map<` `int` `, ` `int` `> eleOcc)` `{` ` ` `struct` `Node* result = NULL;` ` ` ` ` `// Push all the elements into` ` ` `// the resultant list` ` ` `for` `(` `auto` `it = eleOcc.begin(); it != eleOcc.end(); it++)` ` ` `push(&result, it->first);` ` ` ` ` `return` `result;` `}` ` ` `/* Function to get the intersection of ` ` ` `two linked lists head1 and head2 */` `struct` `Node* getIntersection(` ` ` `unordered_map<` `int` `, ` `int` `> eleOcc)` `{` ` ` `struct` `Node* result = NULL;` ` ` ` ` `// Push a node with an element` ` ` `// having occurrence of 2 as that` ` ` `// means the current element is` ` ` `// present in both the lists` ` ` `for` `(` `auto` `it = eleOcc.begin();` ` ` `it != eleOcc.end(); it++)` ` ` `if` `(it->second == 2)` ` ` `push(&result, it->first);` ` ` ` ` `// return resultant list` ` ` `return` `result;` `}` ` ` `/* A utility function to print a linked list*/` `void` `printList(` `struct` `Node* node)` `{` ` ` `while` `(node != NULL) {` ` ` `printf` `(` `"%d "` `, node->data);` ` ` `node = node->next;` ` ` `}` `}` ` ` `// Prints union and intersection of` `// lists with head1 and head2.` `void` `printUnionIntersection(Node* head1,` ` ` `Node* head2)` `{` ` ` `// Store all the elements of` ` ` `// both lists in the map` ` ` `unordered_map<` `int` `, ` `int` `> eleOcc;` ` ` `storeEle(head1, head2, eleOcc);` ` ` ` ` `Node* intersection_list = getIntersection(eleOcc);` ` ` `Node* union_list = getUnion(eleOcc);` ` ` ` ` `printf` `(` `"\nIntersection list is \n"` `);` ` ` `printList(intersection_list);` ` ` ` ` `printf` `(` `"\nUnion list is \n"` `);` ` ` `printList(union_list);` `}` ` ` `/* Driver program to test above function*/` `int` `main()` `{` ` ` `/* Start with the empty list */` ` ` `struct` `Node* head1 = NULL;` ` ` `struct` `Node* head2 = NULL;` ` ` ` ` `/* create a linked list 11->10->15->4->20 */` ` ` `push(&head1, 1);` ` ` `push(&head1, 2);` ` ` `push(&head1, 3);` ` ` `push(&head1, 4);` ` ` `push(&head1, 5);` ` ` ` ` `/* create a linked list 8->4->2->10 */` ` ` `push(&head2, 1);` ` ` `push(&head2, 3);` ` ` `push(&head2, 5);` ` ` `push(&head2, 6);` ` ` ` ` `printf` `(` `"First list is \n"` `);` ` ` `printList(head1);` ` ` ` ` `printf` `(` `"\nSecond list is \n"` `);` ` ` `printList(head2);` ` ` ` ` `printUnionIntersection(head1, head2);` ` ` ` ` `return` `0;` `}` |

**Output:**

First list is 5 4 3 2 1 Second list is 6 5 3 1 Intersection list is 3 5 1 Union list is 3 4 6 5 2 1

We can also handle the case of duplicates by maintaining separate Hash for both the lists.

**Complexity Analysis:**

**Time Complexity:**O(m+n).

Here ‘m’ and ‘n’ are number of elements present in first and second lists respectively.**Reason:**

For Union: Traverse both the lists, store the elements in Hash-map and update the respective count.

For Intersection: Check if count of an element in hash-map is ‘2’.**Auxiliary Space:**O(m+n).

Use of Hash-map data structure for storing values.

This article is contributed by **Sahil Chhabra**. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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