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Weride·Software Engineer·Technical Phone Screen·Intermediate

IntermediatePrefer not to say
May 2026

Summary

Interviewed at Weride for a software engineering role and got hit with a classic algorithms question dressed up with a twist I wasn't fully ready for.

Questions Asked (1)

Q1

Compute the n-th Fibonacci number in O(log n) time.

Algorithms & Data Structures
Author's notes

I knew the recursive and iterative solutions cold, but O(log n) is a different beast.

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AI HintsAI Generated

Suggested Approach

Start by clarifying the problem constraints (e.g., n can be large, so O(n) is too slow). Then explain that matrix exponentiation or fast doubling can compute F(n) in O(log n) by using the identities F(2k) = F(k)*(2*F(k+1) - F(k)) and F(2k+1) = F(k+1)^2 + F(k)^2. Finally, discuss implementation details like handling large numbers and recursion depth.

Pro tip: Mention that fast doubling is generally more efficient than matrix exponentiation because it uses fewer multiplications and avoids 2x2 matrix operations, and be prepared to discuss how to handle very large n (e.g., using memoization or iterative implementation to avoid stack overflow).

1. Clarify requirements and constraints

Ask about the expected range of n, whether the result should be modulo something, and if recursion is acceptable. This shows you think about edge cases and practical limits.

2. Explain the O(log n) approach

Describe either matrix exponentiation or fast doubling, highlighting the logarithmic time complexity. Derive the key identities or matrix power to demonstrate understanding.

3. Walk through the algorithm

Outline the steps of the chosen method, e.g., for fast doubling: recursively compute F(k) and F(k+1) for k = n/2, then use formulas to get F(n). Mention base cases.

4. Analyze complexity and trade-offs

State that time complexity is O(log n) and space complexity is O(log n) for recursion or O(1) for iterative. Compare with naive O(n) and discuss when each method is preferable.

5. Discuss implementation details

Mention handling large numbers (e.g., using arbitrary precision or modulo), avoiding recursion depth issues, and potential optimizations like iterative fast doubling.

Key Points to Mention

  • Matrix exponentiation: [[1,1],[1,0]]^n gives F(n+1) and F(n).
  • Fast doubling identities: F(2k) = F(k)*(2*F(k+1) - F(k)), F(2k+1) = F(k+1)^2 + F(k)^2.
  • Time complexity O(log n) due to halving n at each step.
  • Space complexity: O(log n) for recursive, O(1) for iterative.
  • Handling large n: use modulo arithmetic or big integers as needed.
  • Edge cases: n=0 returns 0, n=1 returns 1; ensure base cases are correct.

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