Union and Intersection of two linked lists | Set-3 (Hashing)
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.
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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.
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.
- 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: 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.
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