Start by clarifying the problem and edge cases, then propose a solution using a min-heap to track meeting end times. Sort intervals by start time, iterate through them, and for each meeting, if the earliest ending meeting has ended, reuse that room; otherwise, allocate a new room. The heap size at the end gives the minimum number of rooms.
Pro tip: Mention that this problem is equivalent to finding the maximum number of overlapping intervals, and that the heap approach runs in O(n log n) time, which is optimal. Also, discuss how you would handle edge cases like empty input or back-to-back meetings.
Ask clarifying questions: Are intervals inclusive of start and exclusive of end? Can meetings be back-to-back? What if the list is empty? Confirm the expected input and output.
Mention that a brute force check for each meeting against all others is O(n^2). Then propose the optimal O(n log n) approach using sorting and a min-heap.
Sort intervals by start time. Initialize a min-heap for end times. For each interval, if the heap is not empty and the earliest end time is <= current start, pop the heap (reuse room). Then push the current end time. The heap size is the answer.
State time complexity O(n log n) due to sorting and heap operations, space O(n) for the heap. Discuss edge cases: empty input returns 0, single meeting returns 1, all meetings overlap returns n.
Walk through a small example, e.g., [[0,30],[5,10],[15,20]], showing how the heap updates and the final room count is 2. Verify correctness.
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