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Apple·Software Engineer·Technical Phone Screen·Intermediate

IntermediatePrefer not to say
Jun 2026

Summary

Apple SWE interview that leaned heavily on sorting fundamentals. The main question was straightforward enough but the follow-up about merging pre-sorted lists is where things got interesting and where I probably left some points on the table.

Questions Asked (2)

Q1

Given two unsorted lists of strings, return a single list sorted by ascending string length. You need to define your tie-breaking rule, analyze time and space complexity, and explain whether your sort is stable.

Algorithms & Data StructuresTechnical Trade-offs
Author's notes

I went with a standard sort using string length as the key and said I'd preserve relative order for ties, which makes it stable.

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AI HintsAI Generated

Suggested Approach

Start by clarifying the problem: merge the two lists, then sort by string length. Explicitly state your tie-breaking rule (e.g., lexicographical order for equal lengths) and justify it. Then analyze time and space complexity, and discuss stability, noting that Python's sort is stable but your tie-breaking rule may affect stability.

Pro tip: Mention that if the input lists are already sorted by length, you can merge them in O(n) time using a two-pointer approach, which is more efficient than sorting the combined list. This shows you consider optimal solutions based on input properties.

1. Clarify requirements and assumptions

Ask if the lists can be modified, if additional space is allowed, and confirm the definition of 'string length' (e.g., number of characters). Also clarify if the output should be a new list or if in-place is acceptable.

2. Define tie-breaking rule

Choose a rule for strings of equal length, such as lexicographical order. Explain why you chose it (e.g., deterministic output, user expectation) and note that it affects stability.

3. Describe algorithm and implementation

Outline the steps: concatenate the two lists, then sort using a key function that returns (length, string) if tie-breaking by lexicographical order. Mention that you can use a stable sort like Python's Timsort.

4. Analyze time and space complexity

State that concatenation takes O(n+m) time and space, sorting takes O((n+m) log(n+m)) time, and the overall space is O(n+m) for the new list. If using in-place sort, space is O(1) extra (excluding output).

5. Discuss stability

Explain that Python's sort is stable, but if you use a tie-breaking rule that includes the string itself, the relative order of equal-length strings from the original lists may change. If stability is required, you can sort by length only, preserving original order for ties.

Key Points to Mention

  • Time complexity: O((n+m) log(n+m)) due to sorting, where n and m are the lengths of the input lists.
  • Space complexity: O(n+m) for the merged list; if sorting in-place, extra space is O(1) or O(log(n+m)) depending on the algorithm.
  • Stability: Python's sort is stable, but tie-breaking by string value can disrupt original order; sorting by length only preserves stability.
  • Tie-breaking rule: lexicographical order is common, but you can also preserve input order or use another criterion.
  • Alternative approach: if inputs are already sorted by length, merge in O(n+m) time using two pointers.
  • Edge cases: empty lists, strings of same length, very long strings, and Unicode characters affecting length.

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

Q2

If each of the two input lists is already individually sorted by string length using the same tie-breaking rule, how would you efficiently merge them into a globally sorted list? Compare in-place versus extra-memory approaches and their complexities.

Algorithms & Data StructuresTechnical Trade-offs
Author's notes

Classic merge from merge sort, I knew that.

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AI HintsAI Generated

Suggested Approach

Start by recognizing this as a merge of two sorted lists, where the comparison key is (string length, tie-breaker). Explain that the optimal approach depends on constraints: if extra memory is allowed, use a standard two-pointer merge into a new list; if in-place is required, discuss techniques like merging from the end or using rotation-based methods. Compare time and space complexities, and justify your recommendation based on typical engineering trade-offs.

Pro tip: Mention that in-place merging of two sorted lists is non-trivial and often not worth the complexity unless memory is extremely constrained; in most practical scenarios, the extra-memory approach is preferred for its simplicity and O(n) time. Also, clarify the tie-breaking rule upfront to avoid ambiguity.

1. Clarify the problem and constraints

Confirm that both lists are sorted by the same key (string length, then tie-breaker) and ask about memory constraints, list sizes, and whether the lists are arrays or linked lists.

2. Describe the extra-memory merge

Explain the two-pointer technique: initialize pointers at the start of each list, compare elements by the key, and append the smaller to a new list. This runs in O(n+m) time and O(n+m) space.

3. Discuss in-place merging approaches

For arrays, mention that merging from the end works if one array has extra capacity; otherwise, algorithms like rotate-and-merge or shell sort-based methods can achieve O(1) extra space but with higher time complexity (e.g., O(nm) or O(n log n) with more complex algorithms).

4. Compare complexities and trade-offs

Contrast time and space: extra-memory is O(n+m) time and O(n+m) space; in-place is O(1) space but often O(nm) time for naive approaches, or O(n log n) with advanced algorithms. Discuss stability and practical considerations.

5. Recommend an approach

Conclude that unless memory is severely constrained, the extra-memory merge is preferable due to simplicity and optimal time. If in-place is required, suggest using a temporary buffer if allowed, or a rotation-based method for arrays.

Key Points to Mention

  • Two-pointer merge technique for sorted lists
  • Time complexity O(n+m) for extra-memory approach
  • Space complexity O(n+m) vs O(1) for in-place
  • In-place merging challenges: shifting elements, rotation, and higher time complexity
  • Stability of the merge (preserving original order for equal keys)
  • Applicability to arrays vs linked lists (linked lists allow O(1) space merge easily)

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