Which statement about removing duplicates from a sorted linked list is accurate?

• A. We traverse once, skipping immediate duplicates; overall O(n) time and O(1) space.
• B. Requires O(n^2) time.
• C. Requires extra space proportional to duplicates.
• D. Cannot be done in O(n) time.

Answer: (A)

In a sorted linked list, duplicates show up next to each other, so you can remove them with a single pass by rewiring the pointers. Start at the head and walk through the list; if the current node’s value equals its next node’s value, bypass the duplicate by setting current.next to current.next.next. If they’re different, simply move forward. This approach keeps the list intact and removes duplicates in place.

Because you only inspect each node once (and maybe skip over some next nodes in the same step), every node is touched a constant number of times, giving a linear time complexity, O(n). You don’t allocate new nodes or extra data structures, so the extra space used is constant, O(1).

The other statements would imply extra work or storage that isn’t necessary here—nested passes would slow things down to O(n^2), keeping duplicates would require storing them, and claiming it can’t be done in O(n) ignores the simple in-place single-pass solution.