Meta·Software Engineer·Technical Phone Screen·Senior

SeniorPrefer not to say

Apr 2026

Summary

Meta software engineering interview with two algorithm problems back to back. The questions weren't unreasonable but the depth they wanted on edge cases and complexity analysis was pretty serious. Felt more like a design conversation than a pure coding screen.

Questions Asked (2)

Implement a fast exponentiation function pow(x, n) that runs in O(log |n|) time using exponentiation by squaring. Handle edge cases like negative exponents, x = 0, n = 0, and very large exponents. Also discuss iterative vs. recursive approaches and numerical precision concerns.

Algorithms & Data StructuresTechnical Trade-offs

Author's notes

I jumped straight to the recursive version and it went fine, but then they asked me to do it iteratively and I fumbled for a bit.

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AI HintsAI Generated

Suggested Approach

Start by clarifying the problem constraints and edge cases, then explain the exponentiation by squaring algorithm, emphasizing how it reduces the exponent by half each step to achieve O(log |n|) time. Discuss both iterative and recursive implementations, highlighting trade-offs in space and readability, and address numerical precision and overflow concerns.

Pro tip: Mention that for negative exponents, you can compute pow(x, -n) and then take the reciprocal, but be careful with integer overflow when n is the minimum integer value. Also, note that for floating-point x, repeated squaring can accumulate rounding errors, so consider using a compensated summation or higher precision if needed.

1. Clarify requirements and edge cases

Ask about input types (integer vs. floating-point), expected output precision, and constraints on n (e.g., 32-bit integer). Explicitly list edge cases: x=0, n=0, negative n, and very large |n|.

2. Explain the core algorithm

Describe exponentiation by squaring: if n is even, pow(x, n) = pow(x*x, n/2); if n is odd, pow(x, n) = x * pow(x*x, (n-1)/2). Emphasize that each step halves the exponent, giving O(log |n|) time.

3. Handle negative exponents and edge cases

For negative n, compute pow(x, -n) and return 1/result. Handle n=0 (return 1), x=0 with n>0 (return 0), and x=0 with n<0 (undefined/infinity). Watch for integer overflow when negating n.

4. Compare iterative vs. recursive approaches

Iterative: use a loop with a result variable and base, multiplying when the current bit of n is 1, then squaring base and halving n. Recursive: simpler but uses O(log n) stack space. Discuss trade-offs in space, readability, and potential stack overflow for large n.

5. Discuss numerical precision and performance

For floating-point x, repeated squaring can cause rounding errors; mention using Kahan summation or higher precision if needed. For integer x, discuss overflow and potential use of modular exponentiation. Confirm O(log |n|) time and O(1) space for iterative.

Key Points to Mention

Exponentiation by squaring reduces time complexity to O(log |n|) by halving the exponent each step.
Negative exponents: compute positive exponent then take reciprocal; handle overflow when n is INT_MIN.
Edge cases: n=0 returns 1 (including 0^0 as 1 by convention), x=0 with n>0 returns 0, x=0 with n<0 is undefined.
Iterative approach uses O(1) space and avoids recursion depth issues; recursive is cleaner but uses O(log n) stack space.
Numerical precision: floating-point repeated squaring can accumulate errors; consider compensated algorithms or higher precision.
Integer overflow: for large results, use modular exponentiation or big integers; for negative n, ensure no overflow when negating.

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

Given a string of lowercase letters and parentheses, remove the minimum number of parentheses to make the string valid. Return any valid result. Implement an O(n) solution and compare a stack-based approach to a two-pass marking approach, justifying why your removal count is minimal.

Algorithms & Data StructuresTechnical Trade-offs

Author's notes

Two-pass approach came to me first and I coded it up without much trouble.

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AI HintsAI Generated

Suggested Approach

First, clarify the problem and constraints, then present a stack-based solution that removes unmatched closing parentheses on the fly and unmatched opening parentheses at the end. Next, describe a two-pass marking approach that marks invalid parentheses in two scans, then removes them. Finally, compare both approaches in terms of time/space complexity and justify minimality by proving that each removed parenthesis is necessarily invalid.

Pro tip: Emphasize that both approaches achieve O(n) time, but the stack approach uses O(n) space while the two-pass marking can be done in O(1) extra space if we modify the string in place or use a boolean array. Highlight that the minimal removal count is exactly the number of unmatched parentheses, which both methods compute.

1. Clarify and Define Validity

Confirm that a valid string has balanced parentheses and that we can return any valid result. Define what makes a parenthesis invalid: a closing parenthesis without a matching opening before it, or an opening parenthesis without a matching closing after it.

2. Stack-Based Approach

Iterate through the string, using a stack to track indices of opening parentheses. When encountering a closing parenthesis, if the stack is non-empty, pop and mark as valid; otherwise, mark as invalid. After the scan, any remaining opening parentheses in the stack are invalid. Remove all marked invalid parentheses.

3. Two-Pass Marking Approach

First pass: left to right, count balance; mark any closing parenthesis that would make balance negative as invalid. Second pass: right to left, count balance; mark any opening parenthesis that would make balance negative as invalid. Remove all marked invalid parentheses.

4. Compare and Justify Minimality

Compare time and space: both O(n) time; stack uses O(n) space, two-pass can use O(1) extra space if using a mutable string or boolean array. Justify minimality: each removed parenthesis is unmatched and cannot be part of any valid string, so removing them is necessary; thus the count is minimal.

5. Discuss Trade-offs and Edge Cases

Mention trade-offs: stack is simpler to implement but uses extra space; two-pass is more space-efficient but requires two scans. Discuss edge cases: empty string, all parentheses, no parentheses, nested parentheses.

Key Points to Mention

Time complexity: both approaches are O(n) because each character is processed a constant number of times.
Space complexity: stack approach uses O(n) space for the stack; two-pass marking can be O(1) extra space if we modify the string in place or use a boolean array of size n (which is O(n) but can be optimized).
Minimality proof: every removed parenthesis is unmatched and thus cannot be part of any valid parentheses string; removing them is necessary and sufficient.
Stack approach: push indices of '('; on ')' pop if possible else mark for removal; after scan, mark all remaining '(' in stack for removal.
Two-pass approach: first pass removes excess ')', second pass removes excess '('; can be done by counting and marking.
Edge cases: empty string, string with only parentheses, string with no parentheses, deeply nested parentheses.

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