The question sounds almost philosophical at first.
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.
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.
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.
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.
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.
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.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.