How do you remove duplicates from a sorted singly linked list?

• A. Traverse with a pointer; if current value equals next value, bypass the duplicate by linking current.next to current.next.next; O(n).
• B. Use a hash set to track seen values.
• C. Sort the list, then remove duplicates.
• D. Always delete the current node after checking next.

Answer: (A)

In a sorted singly linked list, duplicates appear next to each other, so you can remove them in one pass by adjusting next pointers as you go. Use a pointer to the current node and check the next node: if the current value equals the next value, skip the next node by setting current.next to current.next.next. Do not advance when you remove a node; after skipping, check again since another duplicate might follow. If they’re different, simply move to the next node. This approach keeps the first occurrence of each value and removes all duplicates in a single traversal.

It runs in O(n) time because each node is visited a constant number of times, and uses O(1) extra space since only a couple of pointers are needed. Using a hash set would add extra space without benefit here, sorting again is unnecessary because the list is already sorted, and deleting the current node every time would discard non-duplicate nodes and complicate maintenance of the list structure.