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Meta·Machine Learning Engineer·Technical Phone Screen·Senior

SeniorPrefer not to say
May 2026Remote

Summary

Two coding problems back to back for a Meta MLE screen. Both were algorithmic, no ML theory at all, which I half expected but still felt underprepared for the second one.

Questions Asked (2)

Q1

Implement a fast exponentiation function that computes x to the power n in O(log n) time using an iterative approach, handling negative exponents and edge cases like n being the minimum 32-bit integer.

Algorithms & Data StructuresTechnical Trade-offs
Author's notes

The core squaring logic came to me pretty fast but I fumbled on the edge cases.

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AI HintsAI Generated

Suggested Approach

Start by clarifying the problem constraints and edge cases, then explain the iterative binary exponentiation algorithm using bit manipulation. Emphasize how handling negative exponents and the minimum 32-bit integer requires careful type casting and overflow prevention.

Pro tip: Mention that converting n to a 64-bit integer before negation avoids overflow when n is INT_MIN, and use a long long for the exponent to safely handle the negation.

1. Clarify requirements and edge cases

Ask about input types (integer, float?), expected output precision, and constraints. Identify edge cases: n=0, x=0, negative n, and n = INT_MIN.

2. Explain the iterative binary exponentiation algorithm

Describe how to repeatedly square the base and halve the exponent, multiplying the result when the current exponent bit is 1. This achieves O(log n) time.

3. Handle negative exponents and INT_MIN

For negative n, compute x^(-n) as 1/(x^n). To avoid overflow when n = INT_MIN, cast n to a 64-bit integer before negation.

4. Implement the function with proper types

Use double for the base and result to handle fractional values, and long long for the exponent to safely negate INT_MIN. Write clean, iterative code with a while loop.

5. Test with edge cases and analyze complexity

Walk through examples: x=2, n=10; x=2, n=-2; x=0, n=0; x=1, n=INT_MIN. Confirm O(log n) time and O(1) space.

Key Points to Mention

  • Binary exponentiation (exponentiation by squaring) reduces time complexity to O(log n).
  • Iterative approach avoids recursion stack overhead and is more space-efficient.
  • Negative exponents require computing the reciprocal: x^(-n) = 1/(x^n).
  • INT_MIN cannot be negated directly in 32-bit; cast to 64-bit integer (long long) before negation.
  • Edge cases: n=0 returns 1 (including 0^0 convention), x=0 with negative n is undefined/infinity.
  • Use double for base and result to handle fractional powers and avoid integer overflow.

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

Q2

Given the root of a perfect binary tree where each node has a 'next' pointer, connect each node to its immediate right neighbor on the same level using O(1) extra space and no recursion.

Algorithms & Data Structures
Author's notes

I knew the recursive version cold so having to go iterative without a queue tripped me up for a bit.

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AI HintsAI Generated

Suggested Approach

Use the already-established next pointers of the current level to traverse and set the next pointers of the next level. Start with the root, and for each level, iterate through nodes using next pointers, linking the children of adjacent nodes. This achieves O(1) space and no recursion.

Pro tip: Emphasize that the algorithm leverages the next pointers of the current level to avoid any additional data structures, and mention that it works because the tree is perfect, ensuring all nodes have both children.

1. Initialize level start

Set a pointer to the root as the leftmost node of the current level.

2. Traverse current level

While the current node exists, connect its left child's next to its right child, and if the current node has a next, connect its right child's next to the next node's left child.

3. Move to next node

Advance the current node to its next pointer.

4. Move to next level

After finishing the current level, set the level start to its left child and repeat until the level start is null.

Key Points to Mention

  • O(1) space complexity by using existing next pointers
  • No recursion, iterative approach
  • Perfect binary tree property ensures all nodes have two children
  • Time complexity O(n) where n is number of nodes
  • Handling of edge cases: empty tree, single node
  • Comparison with level-order traversal using queue (O(n) space)

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