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Robinhood·Software Engineer·Technical Phone Screen·Intermediate

Intermediate
Apr 2026

Summary

Robinhood SWE interview with a classic interval merging problem. Pretty standard coding round, nothing too wild, but the kind of question where you either see it cleanly or you fumble the edge cases.

Questions Asked (1)

Q1

Given an array of intervals, merge all overlapping ones and return a list of non-overlapping intervals that cover the full input.

Algorithms & Data Structures
Author's notes

Classic problem but the edge cases get you if you're not careful.

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AI HintsAI Generated

Suggested Approach

Start by clarifying edge cases and constraints, then propose sorting the intervals by start time. Iterate through the sorted list, merging overlapping intervals by comparing the current interval's start with the previous merged interval's end. Return the merged list.

Pro tip: Mention that sorting is the key to achieving O(n log n) time, and explicitly handle edge cases like empty input or single interval. Also, discuss how you would test the solution with examples.

1. Clarify and Confirm

Ask clarifying questions about input format, interval inclusivity, and expected output. Confirm edge cases such as empty array or intervals with same start/end.

2. Sort Intervals

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

3. Merge Overlapping Intervals

Initialize a result list with the first interval. Iterate through the remaining intervals; if the current interval overlaps with the last interval in the result (i.e., its start <= last end), merge them by updating the end to the maximum of both ends. Otherwise, add the current interval to the result.

4. Return Result

After processing all intervals, return the result list containing non-overlapping intervals that cover the input.

5. Analyze Complexity and Test

State the time complexity O(n log n) due to sorting and space complexity O(n) for the output. Walk through a test case to verify correctness.

Key Points to Mention

  • Sorting by start time is crucial for O(n log n) efficiency.
  • Handle edge cases: empty input, single interval, intervals with same start/end.
  • Merge condition: current.start <= last.end (assuming inclusive intervals).
  • Update the end to max(last.end, current.end) when merging.
  • Time complexity: O(n log n) due to sorting; space complexity: O(n) for output.
  • Use a result list and compare with the last interval to avoid unnecessary checks.

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