Clarify that intervals are half-open [start, end) so back-to-back meetings don't conflict. Then present the sweep line algorithm: separate starts and ends, sort both, and use two pointers to count maximum concurrent meetings. Alternatively, use a min-heap to track end times, adding a room when a new meeting starts before the earliest end.
Pro tip: Mention that the problem reduces to finding the maximum number of overlapping intervals at any point, and that the heap-based approach naturally extends to scheduling with room assignments if asked. Also, explicitly state the time and space complexity and compare with the sweep line method.
Confirm that intervals are half-open [start, end) and that start < end. Discuss edge cases: empty list, single meeting, all meetings overlapping, and back-to-back meetings.
State that the minimum number of rooms equals the maximum number of meetings happening at the same time. This transforms the problem into counting maximum overlaps.
Separate all start and end times into two arrays, sort both, and use two pointers to sweep through time. Increment a counter on start, decrement on end, and track the maximum.
Sort meetings by start time. Use a min-heap of end times. For each meeting, if the heap's minimum end <= current start, pop it (reuse room). Push the current end. The heap size at the end is the answer.
Both approaches are O(n log n) time and O(n) space. The sweep line is simpler and faster in practice; the heap approach is more intuitive for scheduling and can be extended to assign rooms.
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