I went with a min-heap on end times, which clicks once you realize you only care whether the earliest-ending meeting has freed up by the time the next one starts.
Clarify the problem and edge cases, then present the sweep line algorithm: separate start and end times, sort them, and use two pointers to count concurrent meetings, tracking the maximum. Alternatively, use a min-heap to simulate room allocation. Discuss time and space complexity, and compare with brute-force approaches.
Pro tip: Mention that the sweep line approach is optimal and can be implemented in O(n log n) time, and that the heap approach naturally handles the 'minimum rooms' requirement by reusing rooms as soon as they are free. Also, note that the problem is equivalent to finding the maximum number of overlapping intervals at any point.
Confirm the input format, whether intervals are inclusive/exclusive, and if meetings can be back-to-back (e.g., [1,2] and [2,3] do not overlap). Ask about constraints like input size.
Start with a brute-force O(n^2) method, then introduce the sweep line (sorting starts and ends) and min-heap approaches. Explain why they are more efficient.
For sweep line: create separate arrays for start and end times, sort both, use two pointers to count active meetings, and update max rooms. For heap: sort by start time, push end times into a min-heap, and pop when a room is free.
State that both approaches run in O(n log n) time due to sorting, and O(n) space. Compare with brute-force O(n^2) time and O(1) extra space.
Walk through a sample input like [[0,30],[5,10],[15,20]] to show the algorithm yields 2 rooms. Also test edge cases: empty list, single meeting, all overlapping.
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