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Meta·Software Engineer·Technical Phone Screen·Intermediate

IntermediatePrefer not to say
May 2026Remote

Summary

Meta coding screen, pretty much just the classic interval merging problem. Nothing fancy, no behavioral, just code it up and explain your thinking.

Questions Asked (1)

Q1

Given an array of intervals, merge all overlapping ones and return the resulting non-overlapping set.

Algorithms & Data Structures
Author's notes

Sort by start time first, that part came to me quick.

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AI HintsAI Generated

Suggested Approach

Start by clarifying the problem: intervals are inclusive, input may be unsorted, and the output should be a list of merged intervals. Then propose sorting intervals by start time and iterating through them, merging when the current interval overlaps with the last merged interval. Analyze time and space complexity, and discuss edge cases like empty input or single interval.

Pro tip: At Optiver, emphasize robustness and efficiency: mention that sorting is O(n log n) and the merge is O(n), and proactively discuss how to handle edge cases like empty input or intervals that touch at endpoints (e.g., [1,2] and [2,3] should merge if inclusive).

1. Clarify requirements and edge cases

Ask whether intervals are inclusive, if input is sorted, and what to return for empty input. Confirm that overlapping includes touching endpoints.

2. Sort intervals by start time

Sort the intervals based on their start values. This ensures that any overlapping intervals are adjacent, simplifying the merge process.

3. Iterate and merge

Initialize a result list with the first interval. For each subsequent interval, if it overlaps with the last interval in the result (i.e., its start <= last end), merge them by updating the last interval's end to the maximum of both ends. Otherwise, add it to the result.

4. Analyze complexity and test

State that time complexity is O(n log n) due to sorting, and space is O(n) for the output. Walk through a few test cases, including overlapping, non-overlapping, and edge cases.

Key Points to Mention

  • Sorting intervals by start time is crucial for O(n log n) efficiency.
  • Merging condition: if current.start <= last.end, merge by updating last.end = max(last.end, current.end).
  • Time complexity: O(n log n) for sorting, O(n) for merging; overall O(n log n).
  • Space complexity: O(n) for the output list (or O(1) extra if modifying in place, but typically O(n)).
  • Edge cases: empty input, single interval, intervals that touch at endpoints, and intervals fully contained within others.
  • Potential follow-up: how to handle if intervals are given as a stream (requires different approach).

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