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Meta·Software Engineer·Technical Phone Screen·Intermediate

Intermediate
May 2026

Summary

Meta SWE coding round, pretty much just the classic merge intervals problem. Nothing fancy, but it's one of those questions where you either know the sorting trick or you don't.

Questions Asked (1)

Q1

Given a collection of intervals as a 2D array, merge all overlapping intervals and return the resulting array.

Algorithms & Data Structures
Author's notes

Sort by start time first, that's the whole key.

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AI HintsAI Generated

Suggested Approach

Start by clarifying the problem and edge cases, then propose sorting the intervals by start time. After sorting, iterate through the intervals and merge overlapping ones by comparing the current interval's start with the previous merged interval's end. Return the merged list.

Pro tip: Mention that sorting is key to achieving O(n log n) time, and that you can optimize space by merging in-place if the input can be modified. Also, discuss how you would handle edge cases like empty input or single interval.

1. Clarify and Confirm

Ask clarifying questions: Are intervals sorted? Can they be modified? What is the expected output format? Confirm edge cases like empty input or single interval.

2. Sort Intervals

Sort the intervals by their start times. This ensures that any overlapping intervals are adjacent, simplifying the merging process.

3. Merge Overlapping

Initialize a result list with the first interval. Iterate through the sorted intervals, and if the current interval overlaps with the last interval in the result, merge them by updating the end time. Otherwise, add the current interval to the result.

4. Analyze Complexity

State the time complexity: O(n log n) due to sorting, and O(n) for the merge pass. Space complexity is O(n) for the output, or O(1) extra if merging in-place.

5. Test with Examples

Walk through a few test cases, including overlapping intervals, non-overlapping intervals, and edge cases like empty input, to verify correctness.

Key Points to Mention

  • Sorting by start time is crucial for O(n log n) efficiency.
  • Overlap condition: next.start <= current.end.
  • Merging by updating the end to max(current.end, next.end).
  • Handling edge cases: empty input, single interval, all overlapping.
  • Time and space complexity analysis.
  • Potential in-place merging to save space if allowed.

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