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.
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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