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My first instinct was a sweep line and I went with that, sorting events and using a min-heap of end times to track which servers free up.
Clarify the problem and edge cases, then present the sweep-line algorithm: sort events (starts and ends) and track active tasks with a min-heap for assignment. Explain the O(n log n) complexity and how to handle equal endpoints by processing ends before starts.
Pro tip: Mention that the minimum number of servers equals the maximum number of overlapping tasks at any point, and that a min-heap of server end times efficiently assigns tasks to the earliest available server, which is optimal.
Confirm interval semantics (half-open [start, end)), discuss empty input, duplicate intervals, and equal endpoints. State that tasks with end == start are non-overlapping.
Use a sweep-line approach: create events for starts and ends, sort them, and use a min-heap to track server availability. For assignment, store server IDs in the heap.
Iterate through sorted events: for a start event, if the heap is non-empty and the earliest end time <= current start, reuse that server; otherwise allocate a new server. For an end event, push the server back into the heap.
Sorting takes O(n log n), heap operations O(log n) each, total O(n log n) time and O(n) space. Prove optimality by relating to maximum overlap.
Mention alternative approaches (e.g., difference array for count only) and trade-offs. For assignment, the heap method naturally provides it.
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