Meta·Software Engineer·Technical Phone Screen·Intermediate

IntermediatePrefer not to say

May 2026

Summary

Meta SWE coding round, just the one question about implementing power function. Pretty standard fast exponentiation stuff but the edge cases are where they actually care.

Questions Asked (1)

Implement a function that calculates x raised to the power n, handling negative exponents and running in O(log n) time.

Algorithms & Data Structures

Author's notes

The O(log n) part is the whole point.

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Suggested Approach

Use binary exponentiation (exponentiation by squaring) to achieve O(log n) time. Handle negative exponents by taking the reciprocal of the base and negating the exponent. Discuss edge cases like n=0, x=0, and integer overflow.

Pro tip: Mention that for negative exponents, you can compute the positive power first and then take the reciprocal, but be careful with integer division if using integer types. Also, consider using long long for the exponent to avoid overflow when negating INT_MIN.

1. Clarify requirements and edge cases

Ask about input types (integer/float), constraints, and expected behavior for edge cases like 0^0, negative base, and large exponents.

2. Explain the binary exponentiation algorithm

Describe how to recursively or iteratively compute x^n by halving the exponent and squaring the base, reducing time complexity to O(log n).

3. Handle negative exponents

If n is negative, compute the power for |n| and then return 1/result. Ensure the exponent is stored in a type that can handle negation safely.

4. Implement the solution

Write clean code, using recursion or iteration, and include checks for base cases (n=0 returns 1, n=1 returns x).

5. Test and analyze

Walk through test cases (positive, negative, zero exponent, large n) and confirm O(log n) time and O(1) space (or O(log n) for recursion).

Key Points to Mention

Binary exponentiation (exponentiation by squaring) reduces time complexity from O(n) to O(log n).
Negative exponents require computing the reciprocal: x^(-n) = 1 / x^n.
Edge cases: n=0 (returns 1), x=0 with n>0 (returns 0), x=0 with n<0 (undefined/infinity).
Integer overflow: use long long for exponent and consider using double for result if needed.
Recursive vs iterative implementation: recursion uses O(log n) stack space, iteration uses O(1) space.
Modular exponentiation variant if the problem involves modulo arithmetic.

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