Apple·Software Engineer·Technical Phone Screen·Intermediate

Intermediate

May 2026

Summary

Apple coding screen, one question, merge two sorted linked lists. Pretty standard stuff but worth writing up since the details matter more than people think.

Questions Asked (1)

Given two sorted linked lists, merge them into a single sorted linked list by splicing together the nodes from both lists.

Algorithms & Data Structures

Author's notes

Classic problem, I knew it, but I fumbled a bit on the in-place splicing part.

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AI HintsAI Generated

Suggested Approach

Use a two-pointer technique with a dummy head to iteratively compare the current nodes of both lists, splicing the smaller node into the merged list. Handle edge cases like empty lists and ensure the remaining nodes are appended efficiently.

Pro tip: Emphasize that this solution runs in O(n+m) time with O(1) extra space by reusing nodes, and mention that it's the core of merge sort's merge step, showing you understand its broader applications.

1. Clarify and Confirm

Ask clarifying questions about input (e.g., sorted ascending, possible null lists) and output (e.g., return head of merged list). Confirm that splicing means reusing nodes, not creating new ones.

2. Plan the Approach

Explain the two-pointer strategy with a dummy head to simplify edge cases. Outline that you'll compare nodes and advance pointers until one list is exhausted.

3. Walk Through an Example

Trace through a simple example (e.g., 1->3->5 and 2->4->6) to demonstrate the step-by-step merging and pointer updates.

4. Code the Solution

Write clean code with a dummy node, a current pointer, and a while loop. After the loop, attach the remaining non-null list.

5. Analyze Complexity and Edge Cases

State time O(n+m) and space O(1). Discuss edge cases: one or both lists empty, lists of different lengths, and duplicate values.

Key Points to Mention

Use of a dummy head node to simplify insertion at the beginning and avoid null checks.
Two-pointer technique to traverse both lists simultaneously.
In-place splicing by adjusting next pointers, achieving O(1) extra space.
Time complexity O(n+m) where n and m are the lengths of the input lists.
Handling of edge cases: empty lists, one list exhausted early, and duplicates.
The algorithm is stable and preserves the relative order of equal elements.

AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.