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I knew the problem but fumbled the implementation a bit.
Clarify the problem and edge cases, then explain that the minimum number of rooms equals the maximum number of overlapping meetings at any point in time. Present the sweep line algorithm: separate start and end times, sort them, and use two pointers to count active meetings, tracking the maximum.
Pro tip: Mention that this is equivalent to finding the maximum clique in an interval graph, which can be solved in O(n log n) by sorting. Also, discuss how to handle edge cases like zero-length meetings and whether intervals are inclusive/exclusive.
Ask about input format, edge cases (empty array, zero-length meetings), and whether intervals are half-open or closed. Confirm that overlapping means any intersection, including touching endpoints if inclusive.
The minimum number of rooms needed is the maximum number of concurrent meetings. This can be found by sweeping through time and counting active intervals.
Extract all start and end times into separate arrays, sort both. Use two pointers to simulate a timeline: increment count when a start is encountered, decrement when an end is encountered, and track the maximum count.
Time complexity is O(n log n) due to sorting; space is O(n). Handle edge cases: empty input returns 0, meetings that end exactly when another starts may or may not overlap depending on definition.
Mention a min-heap approach: sort by start time, push end times into a heap, and for each meeting, if the earliest end time is <= current start, pop and reuse the room. This also runs in O(n log n) and may be more intuitive.
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