Meta·Machine Learning Engineer·Technical Phone Screen·Senior

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Jul 2026Remote

Summary

Meta MLE coding round, two algorithm problems back to back. Nothing behavioral, just pure coding. The second one was harder than I expected for the role.

Questions Asked (2)

Given a nested list that can contain integers or further nested lists, compute the weighted sum where each integer is multiplied by its depth in the nesting (top-level elements start at depth 1).

Algorithms & Data Structures

Author's notes

Recursive DFS, pretty clean once you see it.

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AI HintsAI Generated

Suggested Approach

Clarify the problem definition and edge cases, then present a recursive depth-first search solution that accumulates the weighted sum. Discuss iterative alternatives and analyze time and space complexity.

Pro tip: In ML engineering, nested structures are common in model configurations and data pipelines; mention how this problem relates to processing hierarchical data and emphasize writing clean, testable code.

1. Clarify the problem

Confirm the definition of depth (top-level = 1), input types (integers and lists), and expected output. Ask about edge cases like empty lists or negative integers.

2. Choose an approach

Decide between recursion and iteration. Recursion is natural for nested structures; iteration with a stack avoids recursion limits.

3. Implement the solution

Write a function that traverses the list, tracking depth. For each integer, add value * depth to the sum. For each sublist, recurse with depth + 1.

4. Test with examples

Walk through simple cases (e.g., [1, [2, 3]]) and edge cases (empty list, deeply nested) to verify correctness.

5. Analyze complexity

State that time complexity is O(n) where n is total number of elements (including nested), and space complexity is O(d) for recursion depth d.

Key Points to Mention

Depth definition: top-level elements have depth 1, increasing with each nesting level.
Recursive DFS: base case for integers, recursive case for lists.
Iterative alternative using a stack to avoid recursion depth limits.
Time complexity O(n) where n is total number of elements (integers and lists).
Space complexity O(d) for recursion depth or O(n) for iterative stack in worst case.
Handling edge cases: empty list, non-integer types, negative numbers.

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

On an m x n grid with buildings, empty land, and obstacles, find the empty cell whose total shortest-path distance to all buildings is minimized. Return that minimum distance, or -1 if no such cell exists.

Algorithms & Data Structures

Author's notes

BFS from each building, accumulating distances into a shared grid.

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AI HintsAI Generated

Suggested Approach

Model the grid as a graph and run BFS from each building to compute shortest distances to all reachable empty cells. Accumulate distances per cell and track the number of buildings that can reach it; the answer is the minimum total distance among cells reachable by all buildings, or -1 if none exist.

Pro tip: Early termination and pruning can drastically reduce runtime: stop BFS when a cell's accumulated distance already exceeds the current best, and skip buildings that are already unreachable from any candidate cell.

1. Parse the grid and identify buildings

Scan the grid to count total buildings and collect their coordinates. Also note the number of empty cells to gauge problem size.

2. Run BFS from each building

For each building, perform BFS to compute shortest distances to all reachable empty cells. Accumulate these distances in a total-distance matrix and increment a reach-count matrix for each visited cell.

3. Track reachability and minimum distance

After each BFS, update the minimum total distance among cells that have been reached by all buildings processed so far. Use pruning to avoid exploring cells that already exceed the current best.

4. Validate and return result

After processing all buildings, check if any cell was reached by every building. If so, return the minimum total distance; otherwise, return -1.

Key Points to Mention

BFS guarantees shortest path in unweighted grid.
Use a distance matrix to accumulate total distances and a reach matrix to count how many buildings can reach each cell.
Early termination: if a building cannot reach any empty cell, return -1 immediately.
Pruning: during BFS, skip cells whose accumulated distance already exceeds the current minimum total distance.
Time complexity: O(B * m * n) where B is number of buildings; space complexity O(m * n).
Edge cases: no empty cells, no buildings, or buildings isolated by obstacles.

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