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Google·Software Engineer·Technical Phone Screen·Intermediate

IntermediatePrefer not to say
Apr 2026Remote

Summary

Google SWE technical phone screen, one algorithmic question that looks deceptively simple until you realize the search space is exponential. Spent most of the time just trying to figure out what the question was actually asking.

Questions Asked (1)

Q1

Given a string s, determine whether every subsequence of length 3 or more is a valid English word, using an oracle function isWord(). Return true if all such subsequences are valid, false otherwise.

Algorithms & Data StructuresTechnical Trade-offs
Author's notes

The question sounds almost philosophical at first.

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AI HintsAI Generated

Suggested Approach

First, clarify the problem constraints and edge cases, such as the definition of a subsequence and the behavior of the oracle. Then, propose an efficient algorithm that minimizes calls to isWord(), likely by checking only minimal-length subsequences and using early termination. Finally, analyze the time complexity and discuss potential optimizations or trade-offs.

Pro tip: Emphasize that you would first check subsequences of length 3, as any longer subsequence contains a length-3 subsequence; if all length-3 subsequences are valid words, then all longer ones are automatically valid. This reduces the problem to checking O(n^3) subsequences, but you can further optimize by noting that if any length-3 subsequence is invalid, you can return false immediately.

1. Clarify the problem

Ask questions to confirm the definition of a subsequence (not necessarily contiguous), the oracle's behavior (e.g., time complexity, reliability), and edge cases like strings shorter than 3.

2. Identify the key insight

Realize that if all subsequences of length 3 are valid words, then any longer subsequence is also valid because it contains a length-3 subsequence. Thus, it suffices to check only length-3 subsequences.

3. Design the algorithm

Iterate over all triples of indices i < j < k, form the subsequence, and call isWord(). If any call returns false, return false immediately; otherwise, return true after all checks.

4. Analyze complexity and optimize

The naive approach is O(n^3) calls to isWord(). Discuss potential optimizations, such as early termination, caching results, or pruning based on character frequencies, and consider the trade-offs.

5. Handle edge cases and conclude

If the string length is less than 3, return true vacuously. Also consider duplicate characters and ensure the algorithm correctly handles them. Summarize the solution and its complexity.

Key Points to Mention

  • Definition of subsequence: characters in order but not necessarily contiguous.
  • The key insight: checking all length-3 subsequences is sufficient because any longer subsequence contains a length-3 subsequence.
  • Time complexity: O(n^3) calls to isWord() in the worst case, but early termination can improve average performance.
  • Space complexity: O(1) extra space if generating subsequences on the fly, or O(n^3) if storing them.
  • Potential optimizations: caching isWord() results, pruning based on character set, or using a trie if the dictionary is known.
  • Edge cases: strings of length < 3, repeated characters, and the oracle's reliability/performance.

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