# Union and Intersection of two Linked List using Hashing

• Difficulty Level : Basic
• Last Updated : 27 Mar, 2023

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 doesnâ€™t matter. Examples:

Input:
List1: 10 -> 15 -> 4 -> 20
List2: 8 -> 4 -> 2 -> 10
Output:
Intersection List: 4 -> 10
Union List: 2 -> 8 -> 20 -> 4 -> 15 -> 10
Explanation: 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 -> 10
Output:
Intersection List: 3 -> 4
Union List: 1 -> 2 -> 3 -> 4 -> 8 -> 10
Explanation: 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.

## Javascript

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.