Classic two-pointer setup but I fumbled the off-by-one on the first pass.
Use the two-pointer technique: advance a fast pointer k nodes ahead, then move both pointers until the fast pointer reaches the end. The slow pointer will then be at the node to remove. Handle edge cases like removing the head or when k exceeds the list length.
Pro tip: Clarify with the interviewer whether k is 1-indexed from the end (e.g., k=1 means last node) and whether the list is singly or doubly linked. Also, discuss time and space complexity upfront to show awareness.
Confirm the definition of k (1-indexed from end) and list type. Identify edge cases: empty list, k=1 (remove last), k=length (remove head), k>length (invalid).
Select the two-pointer (fast/slow) technique for O(n) time and O(1) space. Alternatively, mention the two-pass approach but note its inefficiency.
Initialize both pointers at a dummy head. Move fast k+1 steps ahead. Then move both until fast is null. Remove the next node of slow by updating pointers.
Use a dummy node to simplify head removal. After removal, return dummy.next as the new head. Verify with examples.
State time O(n) and space O(1). Walk through test cases: k=1, k=length, k>length, and normal case.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.