How do you remove duplicates from an unsorted singly linked list without extra storage?

• A. Use a hash set to track seen values.
• B. Sort the list to bring duplicates together.
• C. Use two nested loops to compare each node with the rest; O(n^2) time, O(1) space.
• D. Slide a window pointer and delete duplicates.

Answer: (C)

To remove duplicates without extra storage on an unsorted singly linked list, the in-place approach compares each node with all nodes that follow it and removes any duplicates as it goes. Start with the first node as the “current.” For each current node, scan the remainder of the list with a secondary pointer that trails behind a candidate duplicate. If you find a node with the same value as current, bypass it by adjusting the next pointer (and free the removed node, if you’re managing memory). Move the trailing pointer forward when there’s no match. Then advance current to the next node and repeat until you reach the end.

This method uses only a fixed set of pointers, so the extra space is O(1). However, because you may traverse the remaining list for every node, the time complexity is O(n^2) in the worst case. Using a hash set would need extra storage, which isn’t allowed here, and while sorting the list first could group duplicates together, the straightforward no-storage solution typically taught is the in-place, two-nested-loop approach with quadratic time.