Apple·Software Engineer·Technical Phone Screen·Intermediate

Intermediate

Jun 2026

Summary

Apple SWE interview with a scheduling problem that looks straightforward until you realize they want two answers out of one solution. Classic Apple move where the question has layers you don't see coming.

Questions Asked (1)

Given a list of meeting room booking intervals [start, end), first determine if all meetings can fit in a single room, then compute the minimum number of rooms needed if not.

Algorithms & Data Structures

Author's notes

I jumped straight to the min-rooms part and almost forgot they also wanted the single-room boolean as a separate output.

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AI HintsAI Generated

Suggested Approach

Start by clarifying the problem: intervals are half-open [start, end), and we need to first check if all meetings fit in one room (i.e., no overlaps), then compute the minimum rooms needed. For the single-room check, sort intervals by start time and verify each start is >= previous end. For the minimum rooms, use a min-heap to track end times of ongoing meetings, or use a sweep line with events; the heap size at any point gives the rooms needed.

Pro tip: Mention that the single-room check is a special case of the minimum rooms problem (if min rooms == 1, they fit). Also, discuss edge cases like empty input, zero-length meetings, and back-to-back meetings (end == start) which are allowed since intervals are half-open.

1. Clarify and confirm assumptions

Restate the problem: intervals are [start, end), so meetings ending at time t and starting at t do not conflict. Ask if input is sorted, if intervals are valid (start < end), and if we need to handle empty lists.

2. Check if all meetings fit in one room

Sort intervals by start time. Iterate through and check if each start is >= the previous end. If any overlap, they don't fit in one room. This is O(n log n) due to sorting.

3. Compute minimum rooms using a min-heap

Sort intervals by start time. Use a min-heap to store end times of ongoing meetings. For each interval, if the heap is not empty and the earliest end <= current start, pop it (room freed). Then push the current end. The heap size after processing all intervals is the minimum rooms needed.

4. Analyze complexity and alternatives

Time complexity: O(n log n) for sorting and heap operations. Space: O(n) for the heap. Mention an alternative sweep line approach: create events for starts (+1) and ends (-1), sort them, and track the running sum; the maximum sum is the answer.

5. Test with examples and edge cases

Walk through a simple example like [[0,30],[5,10],[15,20]] to show min rooms = 2. Test edge cases: empty list (0 rooms), single meeting (1 room), back-to-back meetings (1 room), and all overlapping (n rooms).

Key Points to Mention

Half-open intervals [start, end) mean meetings ending at t and starting at t do not conflict.
Sorting by start time is essential for both the single-room check and the heap approach.
Min-heap stores end times of ongoing meetings; pop when a room is freed (earliest end <= current start).
The heap size at any point represents the number of rooms currently in use; the maximum size is the answer.
Time complexity O(n log n) due to sorting and heap operations; space O(n) for the heap.
Alternative sweep line approach: events (+1 for start, -1 for end), sort, track running sum, max sum is min rooms.

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