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Anthropic·Software Engineer·Technical Phone Screen·Senior

Senior
May 2026

Summary

Coding screen at Anthropic for a software engineering role, centered entirely on building a tokenizer from scratch. Three progressively trickier parts, each a follow-up to the last. Not the hardest interview I've had but the edge cases definitely kept me honest.

Questions Asked (3)

Q1

Given an input string and a fixed vocabulary, implement a greedy longest-match tokenizer. At each position, find the longest vocabulary token that starts there; if nothing matches, emit -1 for that character and move forward by one.

Algorithms & Data StructuresTechnical Trade-offs
Author's notes

Two nested loops, nothing fancy.

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AI HintsAI Generated

Suggested Approach

Clarify the problem constraints (vocabulary size, string length, character set) and discuss the naive O(n * L * V) approach, then optimize using a trie for O(n * L) time. Walk through the algorithm step-by-step, handle edge cases, and analyze time/space complexity.

Pro tip: Mention that a trie can be built once and reused for multiple strings, and that the greedy approach may not yield the globally optimal tokenization—this shows awareness of trade-offs beyond the immediate problem.

1. Clarify requirements and constraints

Ask about vocabulary size, maximum token length, input string length, and character set. Confirm that the greedy algorithm is required (not optimal tokenization) and that -1 is emitted for unmatched characters.

2. Design the algorithm

Propose a trie (prefix tree) built from the vocabulary. At each position, traverse the trie to find the longest matching token; if none, emit -1 and advance by one.

3. Implement and test

Write clean code with helper functions for trie construction and matching. Test with edge cases: empty string, no matches, overlapping tokens, and tokens that are prefixes of others.

4. Analyze complexity and trade-offs

State time complexity O(n * L) where L is max token length, and space O(V * L) for the trie. Discuss alternatives like sorting vocabulary by length or using a hash set for each length.

Key Points to Mention

  • Trie data structure for efficient longest-prefix matching
  • Time complexity: O(n * L) with trie vs O(n * L * V) naive
  • Space complexity: O(V * L) for trie storage
  • Handling of unmatched characters by emitting -1 and advancing one position
  • Edge cases: empty string, empty vocabulary, tokens that are prefixes of others
  • Trade-offs: greedy vs optimal tokenization, and potential for dynamic programming if optimality is needed

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

Q2

Modify the tokenizer so that consecutive unknown-character outputs (the -1s) get merged into a single -1 rather than one per character.

Algorithms & Data Structures
Author's notes

Easier than it sounds.

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AI HintsAI Generated

Suggested Approach

Clarify the tokenizer's input/output contract and the meaning of -1 (unknown token). Then propose a single-pass algorithm that tracks whether the previous output was -1, appending -1 only when transitioning from a known token to an unknown one, and handle edge cases like leading/trailing unknowns.

Pro tip: Mention that merging unknowns can be done in-place or with a new list, and discuss whether the tokenizer should preserve the count of unknowns for debugging or metrics. This shows awareness of production trade-offs beyond the basic algorithm.

1. Clarify requirements and edge cases

Ask whether -1 represents a single unknown token or a placeholder for any unknown character, and confirm expected behavior for consecutive unknowns at the start, middle, or end of input.

2. Design the algorithm

Use a single pass with a boolean flag (e.g., prev_was_unknown) to decide whether to append -1. Alternatively, use a stack or list and merge after tokenization.

3. Implement and test

Write clean code with clear variable names, and test with cases like all unknowns, alternating known/unknown, and empty input.

4. Analyze complexity and trade-offs

State O(n) time and O(1) extra space (if in-place) or O(n) space (if new list). Discuss whether merging should happen during tokenization or as a post-processing step.

Key Points to Mention

  • Single-pass O(n) algorithm with a flag to track previous token
  • Edge cases: leading, trailing, and multiple consecutive unknowns
  • In-place modification vs. creating a new list
  • Preserving original token count for debugging or metrics
  • Potential impact on downstream tasks that expect one -1 per unknown character
  • Testing strategy including unit tests for various input patterns

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

Q3

When the vocabulary is very small compared to the input length, how would you optimize the inner loop? What's the key insight that lets you cap the search early?

Algorithms & Data StructuresTechnical Trade-offs
Author's notes

This is where it got interesting.

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AI HintsAI Generated

Suggested Approach

First, clarify the problem context—likely string matching or dynamic programming where the inner loop iterates over a small vocabulary. Then, explain how to optimize by precomputing or indexing vocabulary entries, and highlight the key insight: the maximum useful match length is bounded by the longest vocabulary word, so you can cap the inner loop at that length.

Pro tip: Mention that this optimization is crucial in production systems like tokenizers or spell checkers, and that you'd validate the trade-off between precomputation memory and runtime speed. Also, note that if the vocabulary is truly tiny, a simple linear scan with early termination may suffice, avoiding over-engineering.

1. Clarify the problem and constraints

Ask or state the context: what is the input, what is the vocabulary, and what is the inner loop doing? Confirm that vocabulary size is small relative to input length.

2. Identify the bottleneck

Explain that the naive inner loop checks every vocabulary word at each position, leading to O(n * V) time. Since V is small but n is large, the constant factor matters.

3. Apply the key insight: cap by max word length

The longest vocabulary word has length L. Any match starting at position i cannot extend beyond i+L-1. So the inner loop only needs to consider substrings up to length L, reducing work to O(n * L).

4. Optimize further with data structures

Use a trie or hash set of vocabulary words to check matches in O(L) time per position, or precompute a set of all substrings up to length L. This avoids iterating over the entire vocabulary.

5. Discuss trade-offs and alternatives

Mention that if L is also small, the optimization is straightforward. If not, consider more advanced structures like Aho-Corasick. Also, note that early termination (breaking when no prefix matches) can further speed up.

Key Points to Mention

  • The maximum match length is bounded by the longest word in the vocabulary (L).
  • The inner loop can be capped at L, reducing complexity from O(n*V) to O(n*L).
  • Using a trie or hash set allows O(L) lookup per position instead of scanning all V words.
  • Early termination: if no vocabulary word starts with the current prefix, break immediately.
  • Trade-offs: precomputation memory vs. runtime speed; when V is tiny, simple linear scan may be fine.
  • Real-world applications: tokenization, spell checking, and DNA sequence matching.

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