Last Updated: 5 Jan, 2021
Remove Duplicates From an Unsorted Linked List
Moderate
Asked in companies
You are given a linked list of N nodes. Your task is to remove the duplicate nodes from the linked list such that every element in the linked list occurs only once i.e. in case an element occurs more than once, only keep its first occurrence in the list.
For example :
Assuming the linked list is 3 -> 2 -> 3 -> 4 -> 2 -> 3 -> NULL.
Number ‘2’ and ‘3’ occurs more than once. Hence we remove the duplicates and keep only their first occurrence. So, our list becomes : 3 -> 2 -> 4 -> NULL.
Input Format :
The first line of input contains an integer ‘T’ representing the number of test cases.
The first and the only line of every test case contains the elements of the linked list separated by a single space and terminated by -1. Hence, -1 would never be a list element.
Output Format :
For each test case, the resulting linked list is printed.
Note :
When multiple nodes have the same element, the node which appeared first is kept, all other duplicates are removed i.e. the order of the nodes should be preserved.
You don’t need to print anything. It has already been taken care of. Just implement the given function.
Constraints :
1 <= T <= 100
1 <= N <= 10 ^ 4
1 <= data <= 10 ^ 5
Where ‘N’ is the number of nodes in the list and 'data' is the value of list nodes.
Time Limit: 1sec
Approaches
- A brute force approach could be to use two nested loops to check whether an element occurs multiple times or not.
- The outer loop picks the elements one by one from the linked list.
- And the inner loop iterates over the rest of the linked list to compare the picked element with the rest of the elements.
- During comparison if the same element is found (means it’s a duplicate), we remove that element from the list.
- Return the head of the resulting linked list.
- Instead of iterating over the complete list to determine whether the element is duplicate or not, a better approach is to use the concept of hashing.
- The idea is to maintain a hashtable to keep track of the elements that have already been encountered.
- We traverse through the linked list.
- For every element, we check if it is present in the hashtable or not.
- If yes, it means that we have encountered this element before. Hence, this is a duplicate. So, remove it from the list.
- Otherwise, this is the first time we encounter the element. So, add it to the hash table.
- In, this way we can remove all the duplicate elements.
- Return the head of the resulting linked list.