Start by clarifying the problem: intervals are half-open [start, end), and we need the maximum number of overlapping meetings at any point. Present the sweep line algorithm: separate starts and ends, sort them, and use two pointers to count concurrent meetings, tracking the maximum. This gives O(n log n) time and O(n) space, which is optimal.
Pro tip: Mention that the minimum number of rooms equals the maximum depth of overlapping intervals, and that this can also be solved with a min-heap in O(n log n) time. Discussing trade-offs between sorting-based and heap-based approaches shows depth.
Confirm that intervals are half-open [start, end) so that a meeting ending at time t does not conflict with one starting at t. Ask about input size, whether intervals are sorted, and if we need to return the rooms or just the count.
The minimum number of rooms needed is the maximum number of meetings that overlap at any single point in time. This reduces the problem to finding the maximum overlap count.
Use the sweep line approach: extract all start and end times, sort them separately, then use two pointers to simulate time progression. Increment a counter on start, decrement on end, and track the maximum.
The algorithm runs in O(n log n) time due to sorting and O(n) space. Handle edge cases: empty input, single meeting, all meetings overlapping, and meetings that touch at endpoints.
Mention the min-heap approach: sort by start time, push end times into a heap, and pop when a meeting ends. Compare trade-offs: sweep line is simpler and uses less memory; heap approach may be more intuitive for some.
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