Use a two-pointer technique with a dummy node to handle edge cases like removing the head. Advance the fast pointer n+1 steps ahead, then move both pointers until fast reaches the end; the slow pointer will be just before the node to remove.
Pro tip: Always create a dummy node pointing to the head to simplify removal of the head node and avoid separate edge-case handling. Also, clarify with the interviewer whether n is guaranteed to be valid and whether the list is singly linked.
Ask if n is always valid (1 ≤ n ≤ length) and if the list can be empty. Discuss edge cases like removing the head or the only node.
Explain that a one-pass two-pointer approach is optimal, achieving O(L) time and O(1) space, versus a two-pass approach that requires knowing the length first.
Create a dummy node pointing to head. Initialize fast and slow pointers at dummy. Move fast n+1 steps ahead, then move both until fast is null. Remove slow.next by updating slow.next = slow.next.next.
Walk through examples: remove middle node, remove head (n = length), remove tail (n = 1), and single-node list. Verify pointers and return dummy.next.
State time complexity O(L) and space O(1). Mention that a two-pass approach is simpler but less efficient; the two-pointer method is preferred for interviews.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.