Google·Machine Learning Engineer·Technical Phone Screen·Senior

Senior

Jun 2026

Summary

Google ML Engineer interview with a probability/dynamic programming question that felt more like a math puzzle than anything ML-specific. Clean problem statement but the edge cases will get you if you're not careful.

Questions Asked (1)

You start at position 0 on an infinite 1D number line. Each turn you roll a fair K-sided die and move forward by the result. You win if you land anywhere in the range [mn, mx], and lose if you overshoot past mx without ever hitting that range. Given K, mn, and mx, compute the probability of winning.

Algorithms & Data StructuresTechnical Trade-offs

Author's notes

Took me a minute to see this as a DP problem.

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AI HintsAI Generated

Suggested Approach

Model the process as a Markov chain or dynamic programming problem where the state is the current position. Define f(i) as the probability of winning from position i, with base cases f(i)=1 for i in [mn, mx] and f(i)=0 for i > mx. Then solve the recurrence f(i) = (1/K) * sum_{d=1}^K f(i+d) for i < mn, working backwards from mx down to 0.

Pro tip: Mention that the recurrence can be solved in O(mx) time using a sliding window sum, and note that for large mx, you can use matrix exponentiation or find a closed-form solution by analyzing the characteristic equation. This shows you think about scalability and mathematical optimization.

1. Define the state and recurrence

Let f(i) be the probability of winning starting from position i. Clearly state the base cases: f(i)=1 for mn ≤ i ≤ mx, and f(i)=0 for i > mx. For i < mn, f(i) = (1/K) * sum_{d=1}^K f(i+d).

2. Identify dependencies and computation order

Observe that f(i) depends only on f(i+1) through f(i+K). Thus, we can compute f(i) backwards from i = mx down to 0. This avoids infinite recursion and ensures each state is computed once.

3. Optimize the computation

Naively summing K terms for each i gives O(K * mx) time. Use a sliding window sum to compute each f(i) in O(1) amortized time, achieving O(mx) overall. For very large mx, consider matrix exponentiation or solving the linear recurrence.

4. Handle edge cases and validate

Check cases where mn=0 (immediate win), mx < 0 (impossible), or K=1 (deterministic). Validate with small examples by brute force or simulation to ensure correctness.

5. Discuss complexity and trade-offs

State time and space complexity: O(mx) time and O(mx) space for DP, which can be reduced to O(K) space with sliding window. Discuss alternative approaches like matrix exponentiation (O(K^3 log mx)) for very large mx, and when each is preferable.

Key Points to Mention

Dynamic programming with state as current position and recurrence f(i) = (1/K) * sum_{d=1}^K f(i+d).
Base cases: f(i)=1 for i in [mn, mx], f(i)=0 for i > mx.
Sliding window optimization to achieve O(mx) time instead of O(K * mx).
Space optimization: only need to keep track of last K values, reducing space to O(K).
Edge cases: mn=0, mx < 0, K=1, and large mx requiring matrix exponentiation or closed-form solution.
Validation through simulation or exact computation for small inputs to ensure correctness.

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