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Maven Clinic·Software Engineer·Technical Phone Screen·Intermediate

Intermediate
May 2026

Summary

Maven Clinic coding round, came up in what felt like a follow-up to the classic meeting rooms problem family. One question, greedy solution, pretty standard stuff if you've seen interval problems before.

Questions Asked (1)

Q1

Given an array of meeting time intervals, what is the minimum number of intervals you need to remove so that no two remaining intervals overlap?

Algorithms & Data Structures
Author's notes

Knew this one from grinding interval problems.

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AI HintsAI Generated

Suggested Approach

Reframe the problem as finding the maximum number of non-overlapping intervals, then subtract that from the total number of intervals. Use a greedy algorithm: sort intervals by end time and iteratively select the next interval that starts after the last selected end time. The minimum removals equals total intervals minus the maximum non-overlapping intervals.

Pro tip: Mention that this is equivalent to the interval scheduling maximization problem, and that sorting by end time is optimal. Also, clarify edge cases like empty input or intervals with equal start/end times to show thoroughness.

1. Clarify the problem

Confirm that intervals are closed (e.g., [1,2] and [2,3] are considered overlapping) and that the goal is to remove the fewest intervals to eliminate all overlaps.

2. Reframe as maximization

Recognize that minimizing removals is equivalent to maximizing the number of non-overlapping intervals that can be kept.

3. Choose greedy strategy

Sort intervals by their end times. Iterate through them, selecting an interval if its start time is greater than or equal to the end time of the last selected interval.

4. Compute result

Count the selected intervals (max non-overlapping set). The minimum removals is total intervals minus this count.

5. Analyze complexity

State that sorting takes O(n log n) time and the iteration takes O(n) time, so overall O(n log n) time and O(1) extra space (if sorting in place).

Key Points to Mention

  • Greedy algorithm: sort by end time to maximize non-overlapping intervals.
  • Proof of optimality: exchange argument or earliest finish time property.
  • Edge cases: empty array, single interval, intervals with same start/end.
  • Time and space complexity: O(n log n) time, O(1) extra space.
  • Connection to interval scheduling maximization problem.
  • Handling of overlapping definition (e.g., [1,2] and [2,3] overlap or not).

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