Meta·Software Engineer·Technical Phone Screen·Intermediate

IntermediatePrefer not to say

May 2026Remote

Summary

Meta coding screen, pretty much just this one problem the whole time. Fast exponentiation sounds straightforward until you're actually on the clock trying to remember the edge cases.

Questions Asked (1)

Implement a power function that computes x raised to the n, handling negative exponents and the full 32-bit integer range for n, with better than linear time complexity.

Algorithms & Data StructuresTechnical Trade-offs

Author's notes

I knew the recursive halving approach going in but fumbled the negative exponent case for longer than I'd like to admit.

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AI HintsAI Generated

Suggested Approach

Start by clarifying requirements: handle negative exponents, full 32-bit integer range for n (including -2^31), and achieve better than linear time. Then implement binary exponentiation (exponentiation by squaring) iteratively to achieve O(log n) time, carefully handling edge cases like n = -2^31 and x = 0.

Pro tip: Mention that converting n to its absolute value can overflow for n = -2^31, so use a long or handle the negation carefully. Also, discuss the trade-off between iterative and recursive implementations, noting that recursion uses O(log n) stack space while iteration is O(1) space.

1. Clarify requirements and edge cases

Ask about input types (float or integer?), constraints (e.g., x = 0, n = 0, n = -2^31), and expected precision. Confirm that better than linear time means O(log n) is desired.

2. Choose the algorithm

Select binary exponentiation (exponentiation by squaring) to achieve O(log n) time. Decide between iterative and recursive approaches based on space complexity and simplicity.

3. Handle negative exponents and overflow

For negative n, compute the positive power and take the reciprocal. Be cautious with n = -2^31: use a 64-bit integer to avoid overflow when negating.

4. Implement and test

Write clean code with clear variable names. Test with edge cases: x=0, n=0, n negative, n = -2^31, and large n. Verify time complexity.

5. Analyze complexity and trade-offs

State that time complexity is O(log n) and space is O(1) for iterative or O(log n) for recursive. Discuss potential floating-point precision issues if x is a float.

Key Points to Mention

Binary exponentiation (exponentiation by squaring) reduces time complexity from O(n) to O(log n).
Handling negative exponents by computing the positive power and then taking the reciprocal.
The overflow issue when n = -2^31: negating it overflows a 32-bit signed integer, so use a 64-bit integer or handle separately.
Edge cases: x = 0, n = 0, n negative, and large n.
Iterative vs recursive implementation: iterative uses O(1) space, recursive uses O(log n) stack space.
Floating-point precision considerations if x is a float, and potential use of built-in functions like pow() in production.

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