Weride·Software Engineer·Technical Phone Screen·Intermediate

Intermediate

Jun 2026

Summary

Weride coding interview, one algorithmic question on fast exponentiation. Pretty standard stuff for a systems-adjacent role but the negative exponent edge case is where people slip up.

Questions Asked (1)

Implement fast power: given a base x and integer exponent n, compute x^n in O(log n) time. Make sure to handle negative exponents correctly.

Algorithms & Data StructuresTechnical Trade-offs

Author's notes

The squaring part came naturally enough, split even/odd exponents, recurse down, done.

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Suggested Approach

Explain the binary exponentiation (exponentiation by squaring) algorithm, which reduces the time complexity to O(log n) by repeatedly squaring the base and halving the exponent. Then, address negative exponents by computing the power for the absolute value of n and taking the reciprocal, ensuring to handle edge cases like n=0 and potential integer overflow.

Pro tip: Mention that using a recursive approach can be elegant but iterative avoids stack overflow; also, discuss how to handle large exponents with modular arithmetic if needed, showing awareness of practical constraints.

1. Clarify requirements and edge cases

Ask about constraints: type of x and n (integer, float), range of n, possibility of overflow, and whether modular arithmetic is required. Confirm handling of n=0, x=0, and negative exponents.

2. Explain the core algorithm

Describe binary exponentiation: if n is even, x^n = (x^(n/2))^2; if n is odd, x^n = x * x^(n-1). This reduces the number of multiplications to O(log n).

3. Handle negative exponents

For n < 0, compute x^|n| and return 1 / result. Note that this requires x != 0 and may introduce floating-point results if x is integer.

4. Implement iteratively or recursively

Show code for either approach. Iterative: use a loop with bit manipulation of n. Recursive: use divide-and-conquer. Discuss trade-offs (stack depth vs. clarity).

5. Analyze complexity and test

State time O(log n) and space O(1) iterative or O(log n) recursive. Walk through examples like x=2, n=10 and x=2, n=-3 to verify correctness.

Key Points to Mention

Binary exponentiation (exponentiation by squaring) reduces multiplications from O(n) to O(log n).
Negative exponents require computing the reciprocal of the positive exponent result.
Edge cases: n=0 returns 1 (except 0^0 undefined), x=0 with negative n is undefined.
Potential integer overflow when x and n are large; consider using modular exponentiation or big integers.
Iterative implementation avoids recursion stack overhead and is often preferred in production code.
Time complexity O(log n) and space complexity O(1) for iterative version.

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