Seems straightforward until you realize the head itself might need to be removed.
Use a dummy node pointing to the head to simplify edge cases, then traverse the list with a current pointer, skipping nodes whose value equals the target. Return dummy.next as the new head.
Pro tip: Explicitly discuss edge cases like removing the head node or all nodes, and mention that the dummy node technique avoids separate handling for the head. Also, note that the solution runs in O(n) time and O(1) space.
Ask clarifying questions: Is the list singly linked? Can the head be null? Should we free memory? Confirm the target value and that in-place modification is required.
Explain that you'll create a dummy node pointing to the head to handle cases where the head itself needs removal. This simplifies pointer manipulation.
Iterate through the list with a current pointer starting at the dummy. While current.next exists, if current.next.val equals target, skip it by setting current.next = current.next.next; otherwise, advance current.
After traversal, return dummy.next as the new head of the modified list. Discuss time and space complexity: O(n) time, O(1) space.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.