The length-difference trick clicked for me pretty fast: get both lengths, advance the pointer on the longer list by the difference, then walk both in sync until you hit the same node reference.
First, clarify the problem constraints and edge cases, then propose the two-pointer length-difference approach: compute lengths, advance the longer list's pointer by the difference, then move both pointers until they meet. Explain correctness, complexity, and handle edge cases like empty lists and no intersection.
Pro tip: Mention that the two-pointer approach works because after aligning the starting positions, the remaining lengths are equal, so they must meet at the intersection if one exists. Also, note that this is optimal for O(1) space and O(m+n) time.
Restate the problem to ensure understanding: two singly linked lists may intersect at some node, and we need the first common node by reference. Confirm that lists can be empty, have no intersection, or have very different lengths.
Describe the two-pointer length-difference method: compute lengths of both lists, advance the pointer of the longer list by the length difference, then traverse both simultaneously until pointers are equal or null.
Explain why this works: after aligning the starting points, the remaining segments have equal length, so if an intersection exists, the pointers will meet at the first common node; if not, both will reach null.
State time complexity O(m+n) because each list is traversed at most twice, and space complexity O(1) since only pointers and length variables are used.
Discuss edge cases: empty lists (return null), no intersection (return null), intersection at head (handled naturally), and very different lengths (length difference adjustment handles it).
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.