Got the core part fine, sorted by start time and used a min-heap to track end times.
Start by clarifying the problem: intervals are half-open, 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 a two-pointer technique to count active meetings. For the follow-up, discuss how to handle far-future meetings by using a priority queue or segment tree to efficiently query and update room availability over time.
Pro tip: Mention that the problem reduces to finding the maximum number of overlapping intervals, and that the sweep line approach is optimal O(n log n). For the follow-up, emphasize that if meetings are sparse over a long horizon, a segment tree with lazy propagation or an interval tree can answer queries in O(log n) while using O(n) space.
Confirm that intervals are half-open [start, end), and that we need the minimum number of rooms to avoid conflicts. Ask if the input is sorted or if we can modify it.
Recognize that the minimum number of rooms equals the maximum number of overlapping meetings at any time. This transforms the problem into finding the maximum overlap.
Explain the O(n log n) approach: create events for starts (+1) and ends (-1), sort them, and sweep through while tracking the current active meetings and the maximum.
For far-future meetings, propose using a priority queue to track end times of ongoing meetings, or a segment tree/interval tree to handle large time ranges efficiently. Mention that if meetings are added dynamically, a balanced BST or segment tree can support insertions and queries in O(log n).
Compare approaches: sweep line is simple and optimal for static input; for dynamic or far-future scenarios, segment trees offer better scalability but higher implementation complexity. Discuss space-time trade-offs.
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