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Meta·Data Scientist·Technical Phone Screen·Intermediate

Intermediate
Jun 2026Remote

Summary

Algorithmic screen for a Data Scientist role on Meta's VR/AR side, two back-to-back greedy problems with a clear theme once you spotted it. Nothing too wild but the second problem tripped me up longer than I'd like to admit.

Questions Asked (2)

Q1

Given an array of frame sizes and a bandwidth limit, split the array into the minimum number of contiguous segments such that the sum of each segment does not exceed the bandwidth limit. Return that minimum count.

Algorithms & Data Structures
Author's notes

Greedy running sum, pretty mechanical once you see it.

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AI HintsAI Generated

Suggested Approach

Use a greedy algorithm that scans the array once, accumulating segment sums and starting a new segment whenever adding the next element would exceed the bandwidth limit. This yields the minimum number of segments because any valid partition must cut at least at the same points. Clearly state the time and space complexity and discuss edge cases.

Pro tip: Mention that this is the classic 'minimum number of bins' problem where greedy is optimal due to the contiguous constraint, and proactively discuss how you would handle invalid inputs (e.g., a single frame larger than the limit).

1. Clarify the problem and constraints

Confirm that segments must be contiguous, that the sum of each segment must be ≤ bandwidth limit, and ask about edge cases like empty array or elements exceeding the limit.

2. Propose a greedy approach

Explain that you will iterate through the array, maintaining a running sum for the current segment, and start a new segment when adding the next element would exceed the limit.

3. Walk through an example

Use a small example (e.g., [1,2,3,4,5] with limit 6) to demonstrate how the greedy algorithm partitions the array and returns the minimum count.

4. Analyze complexity and prove optimality

State that the algorithm runs in O(n) time and O(1) extra space, and argue that greedy is optimal because any valid partition must make a cut whenever the cumulative sum exceeds the limit.

5. Discuss edge cases and extensions

Cover cases like an element larger than the limit (return -1 or error), empty array (return 0), and mention that if segments didn't need to be contiguous, the problem would be NP-hard (bin packing).

Key Points to Mention

  • Greedy algorithm: accumulate sum and start new segment when limit would be exceeded.
  • Optimality proof: any valid partition must cut at least at the points where the cumulative sum exceeds the limit.
  • Time complexity O(n) and space complexity O(1).
  • Edge cases: empty array, single element exceeding limit, all elements zero.
  • Contrast with non-contiguous version (bin packing) which is NP-hard.
  • Clear communication of assumptions and step-by-step reasoning.

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

Q2

Given n circles on a 1D line each represented by a center and radius, find the minimum number of arrows shot at a horizontal position p that would burst every circle, where a circle is burst if center minus radius is less than or equal to p and p is less than or equal to center plus radius.

Algorithms & Data StructuresTechnical Trade-offs
Author's notes

This one took me an embarrassingly long time to reframe.

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AI HintsAI Generated

Suggested Approach

Recognize this as the classic interval point cover problem: each circle corresponds to an interval [center - radius, center + radius] on the line. Sort intervals by their right endpoints, then greedily place an arrow at the right endpoint of the first interval, skip all intervals containing that point, and repeat. This yields the minimum number of arrows in O(n log n) time.

Pro tip: Explicitly connect the problem to the interval scheduling/point cover pattern, and mention that the greedy choice is optimal because placing an arrow at the earliest finishing interval's right endpoint maximizes coverage. Also, clarify edge cases like overlapping circles and identical intervals.

1. Model as intervals

Convert each circle (center, radius) into an interval [center - radius, center + radius]. This reduces the problem to covering all intervals with the minimum number of points.

2. Sort by right endpoint

Sort the intervals in ascending order of their right endpoints. This ordering is key for the greedy strategy.

3. Greedy placement

Initialize count = 0 and last_arrow = -infinity. Iterate through sorted intervals; if the current interval's left endpoint > last_arrow, place a new arrow at its right endpoint, increment count, and update last_arrow.

4. Return count

After processing all intervals, the count is the minimum number of arrows needed to burst all circles.

5. Analyze complexity and edge cases

State time complexity O(n log n) due to sorting, space O(1) or O(n) depending on sorting. Discuss edge cases: no circles, all circles overlapping, circles with zero radius.

Key Points to Mention

  • Interval representation: each circle becomes [c - r, c + r].
  • Greedy algorithm: sort by right endpoint and place arrows at right endpoints.
  • Proof of optimality: exchange argument or earliest deadline first.
  • Time complexity: O(n log n) for sorting, O(n) for iteration.
  • Edge cases: empty input, zero-radius circles, fully overlapping intervals.
  • Connection to classic problems: minimum points to cover intervals, activity selection.

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