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Apple·Software Engineer·Technical Phone Screen·Intermediate

Intermediate
May 2026

Summary

Apple SWE interview with a string manipulation problem that looks deceptively simple but has a few edge cases that'll bite you if you're not careful. The core challenge is managing conflicts across abbreviations iteratively, which is more involved than it first appears.

Questions Asked (1)

Q1

Given an array of distinct strings, generate the shortest possible unique abbreviation for each word. An abbreviation is formed by taking a prefix of length k, appending the count of skipped characters, then the last character. If two words share the same abbreviation, increase the prefix length for those conflicting words until all abbreviations are unique. If an abbreviation isn't shorter than the original word, just return the original word.

Algorithms & Data StructuresTechnical Trade-offs
Author's notes

My first instinct was to precompute all abbreviations at k=1 and then resolve conflicts in a second pass, but I kept second-guessing whether I needed to re-check global uniqueness after each round of conflict resolution.

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AI HintsAI Generated

Suggested Approach

Start by clarifying the abbreviation rules and edge cases, then propose a solution using a hash map to group words by their initial abbreviations. For conflicts, iteratively increase the prefix length for those words until all abbreviations are unique, and finally compare each abbreviation's length to the original word, returning the shorter one.

Pro tip: Demonstrate awareness of trade-offs: for example, mention that while the iterative approach is simple, it could be optimized by sorting words or using a trie to reduce redundant comparisons, especially for large datasets.

1. Clarify requirements and edge cases

Ask questions to confirm the abbreviation format, handling of short words (e.g., length ≤ 2), and whether the output should preserve the original order.

2. Design the algorithm

Outline a plan: compute initial abbreviations for all words, group words by abbreviation, and for each group with size > 1, increment the prefix length for those words until unique.

3. Implement efficiently

Use a hash map to track abbreviations and a list to store words per abbreviation. For conflicts, update prefix lengths and recompute abbreviations, ensuring no infinite loops.

4. Handle final output

For each word, compare the length of its unique abbreviation with the original word; if the abbreviation is not shorter, return the original word.

5. Analyze complexity and test

Discuss time and space complexity, and walk through test cases like words with common prefixes, single-character words, and all words conflicting.

Key Points to Mention

  • Use of hash map to group words by abbreviation for efficient conflict detection.
  • Iterative approach to increase prefix length only for conflicting words, avoiding unnecessary work.
  • Edge cases: words of length 1 or 2, where abbreviation may not be shorter.
  • Time complexity: worst-case O(n * L) where n is number of words and L is max word length, but often much faster.
  • Space complexity: O(n * L) for storing abbreviations and groups.
  • Potential optimizations: sorting words by length or using a trie to find minimal unique prefixes.

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