Google·Machine Learning Engineer·Technical Phone Screen·Senior

Senior

Apr 2026

Summary

Google MLE interview with a chunking algorithm problem, a follow-up variant of a previous round's question. The twist this time was a header constraint that changes how you think about the minimum chunk calculation.

Questions Asked (1)

You have a data header and a data payload. The header cannot be split across chunks. Given a max chunk size, find the minimum number of chunks needed to fit everything. When can the header share a chunk with the beginning of the data, and how does that affect your calculation?

Algorithms & Data StructuresTechnical Trade-offs

Author's notes

This is basically a part 2 of the classic chunk-splitting problem, but the header constraint is where it gets interesting.

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AI HintsAI Generated

Suggested Approach

First, clarify the problem constraints: header size H, payload size P, and max chunk size C. Then derive the minimum number of chunks by considering whether the header and payload can share a chunk. The key is to check if H + P <= C (then 1 chunk) or if H + (C - H) <= C (always true, so header can share with some payload if H < C). Actually, the header can share a chunk with the beginning of the data if the header size is less than C, because then there is remaining space in the first chunk for some payload. The minimum number of chunks is then 1 + ceil((P - (C - H)) / C) if H < C, else ceil(H/C) + ceil(P/C) but header cannot be split, so if H > C, it's impossible? Wait, the problem says 'header cannot be split across chunks', so if H > C, it's impossible to fit the header in any chunk. So we assume H <= C. Then the minimum chunks is 1 + ceil((P - (C - H)) / C) if P > C - H, else 1. But if H = C, then no room for payload in first chunk, so chunks = 1 + ceil(P/C). So the answer depends on whether H < C and if there is remaining space. The candidate should explain the formula and edge cases.

Pro tip: Demonstrate awareness of real-world constraints: in networking, headers often have fixed sizes and chunk boundaries may be dictated by MTU; mentioning that you'd validate assumptions (e.g., H <= C) and handle edge cases (H = C, P = 0) shows maturity.

1. Clarify constraints and variables

Define H (header size), P (payload size), C (max chunk size). State assumptions: H <= C (otherwise impossible), and chunks are contiguous.

2. Determine if header can share a chunk

The header can share a chunk with the beginning of the data if H < C, because there is remaining space (C - H) in the first chunk for payload. If H = C, no sharing is possible.

3. Compute minimum chunks

If H < C: first chunk holds header and min(P, C-H) payload. Remaining payload = max(0, P - (C-H)). Additional chunks = ceil(remaining / C). Total = 1 + additional. If H = C: total = 1 + ceil(P/C).

4. Handle edge cases

Consider P = 0 (only header, 1 chunk), H = C (no sharing), and H > C (impossible). Also consider if P fits entirely in first chunk (total = 1).

5. Summarize and verify

State the formula clearly and test with examples (e.g., H=10, P=100, C=20 -> first chunk 10+10=20, remaining 90 -> 5 chunks, total 6).

Key Points to Mention

Header cannot be split, so H must be <= C for feasibility.
Sharing is possible only if H < C, leaving space C - H for payload in the first chunk.
Minimum chunks = 1 + ceil(max(0, P - (C - H)) / C) when H < C; else 1 + ceil(P/C) when H = C.
Edge cases: P = 0, H = C, H > C (impossible), and P <= C - H (then 1 chunk).
Trade-off: sharing reduces chunk count but may increase complexity in reassembly; consider if header must be aligned.
Real-world analogy: TCP/IP headers and MTU, or file storage with metadata blocks.

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