I got the happy path pretty fast, two pointers walking both lists simultaneously, swap and advance.
Clarify the problem constraints and edge cases, then walk through a two-pointer iterative solution that interleaves nodes in-place without extra memory. Emphasize O(n) time and O(1) space, and discuss how to handle unequal lengths by appending the remainder.
Pro tip: Mention that you can avoid a separate 'append remainder' step by using a dummy head and a tail pointer, which simplifies the code and handles all cases uniformly. Also, explicitly state that you're modifying the lists in-place, which is often preferred for memory efficiency.
Ask if the lists can be empty, if they can have different lengths, and if in-place modification is acceptable. Confirm that interleaving should start with the first list.
Decide between iterative and recursive solutions. For interviews, an iterative two-pointer approach with O(1) extra space is usually optimal.
Use a dummy head and a tail pointer to build the merged list. Iterate while both lists have nodes, alternately appending nodes from each list, then append any remaining nodes.
Walk through examples with equal lengths, list1 longer, list2 longer, and empty lists to verify correctness and edge case handling.
State that time complexity is O(n+m) where n and m are the lengths of the lists, and space complexity is O(1) since we only rearrange pointers.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.