Snowflake·Software Engineer·Technical Phone Screen·Intermediate

Intermediate

Jun 2026

Summary

Snowflake SWE interview, got a classic scheduling problem that I've seen before but still managed to overthink it a little. Two valid approaches came up and I ended up walking through both, which felt like either a flex or a waste of time depending on how you look at it.

Questions Asked (1)

Given a list of meeting time intervals, find the minimum number of conference rooms needed to accommodate all meetings.

Algorithms & Data Structures

Author's notes

I knew this problem but still second-guessed myself on the heap approach.

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AI HintsAI Generated

Suggested Approach

Clarify that intervals are half-open [start, end) to avoid false conflicts. Then present the sweep line algorithm: create events for starts (+1) and ends (-1), sort them, and track the running count to find the maximum. This gives O(n log n) time and O(n) space, which is optimal.

Pro tip: Mention that if the input is already sorted by start time, you can use a min-heap of end times to achieve O(n log n) without sorting all events, and that this approach naturally extends to finding the actual room assignments if needed.

1. Clarify assumptions

Confirm that intervals are half-open [start, end) so that a meeting ending at time t does not conflict with one starting at t. Also ask if the input is sorted or if we can modify it.

2. Explain the sweep line approach

Describe creating events: for each interval, add (start, +1) and (end, -1). Sort events by time, with ends before starts if times are equal (to handle half-open intervals). Then iterate, maintaining a running count and tracking the maximum.

3. Analyze complexity

State that sorting takes O(n log n) time and the sweep takes O(n), so overall O(n log n) time. Space is O(n) for the events array. This is optimal because we must at least read the input.

4. Discuss alternative approaches

Mention the min-heap method: sort intervals by start time, then iterate; if the earliest ending meeting is done, reuse its room; otherwise allocate a new room. The heap size at the end is the answer. This also runs in O(n log n).

5. Handle edge cases and extensions

Cover empty input (return 0), single meeting (return 1), and all meetings overlapping (return n). If asked, explain how to output the actual room assignments by tracking room IDs.

Key Points to Mention

Half-open interval assumption to avoid off-by-one errors at boundaries.
Sweep line algorithm with +1 for start and -1 for end events.
Sorting events with end events processed before start events when times are equal.
Time complexity O(n log n) and space complexity O(n).
Min-heap alternative that sorts by start time and tracks end times.
Edge cases: empty input, single meeting, all meetings overlapping.

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