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faire·Backend Engineer·Technical Phone Screen·Senior

Senior
May 2026

Summary

Faire backend interview with a line sweep problem that sounds easy until you actually have to defend your approach against alternatives. One question, but the discussion part was where they really dug in.

Questions Asked (1)

Q1

Given a list of events each with a start time, end time, and a capacity value, find the maximum total capacity required at any single point in time across all overlapping events.

Algorithms & Data StructuresTechnical Trade-offs
Author's notes

I went for a heap-based interval approach first because that's what my brain defaults to for overlap problems.

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AI HintsAI Generated

Suggested Approach

Clarify that each event contributes its capacity during the half-open interval [start, end), then use a sweep-line algorithm: create +capacity at start and -capacity at end, sort all events, and track the running sum to find the maximum. This yields O(n log n) time and O(n) space, which is optimal for comparison-based sorting.

Pro tip: Mention the half-open interval convention explicitly to avoid double-counting at boundaries, and note that if events are already sorted or times are bounded, you can achieve O(n) with counting sort or a difference array.

1. Clarify interval semantics and edge cases

Confirm whether events are inclusive/exclusive at boundaries (e.g., [start, end) vs [start, end]) and how to handle zero-duration events. This prevents off-by-one errors in the sweep.

2. Choose the sweep-line approach

Explain that you will convert each event into two points: (+capacity at start) and (-capacity at end). Sort all points by time, with end events processed before start events at the same timestamp if using half-open intervals.

3. Compute the running sum and track maximum

Iterate through sorted points, maintaining a running total of capacity. After each update, compare the running total to the current maximum and update if larger.

4. Analyze complexity and discuss trade-offs

State that sorting dominates at O(n log n) time and O(n) space. Discuss alternatives like difference arrays for bounded time ranges (O(n + T)) or if events are already sorted (O(n)).

5. Validate with examples and edge cases

Walk through a small example to verify correctness, and mention edge cases like no events, single event, all events overlapping, or events with zero capacity.

Key Points to Mention

  • Sweep-line algorithm with events sorted by time
  • Half-open interval convention to avoid double-counting at boundaries
  • Time complexity O(n log n) due to sorting, space O(n)
  • Alternative O(n) approach using difference array if time range is bounded
  • Handling of edge cases: empty list, zero-duration events, simultaneous start/end
  • Trade-off between sorting-based and counting-based approaches

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