Meeting Rooms II - Implementation

Here is the solution using event sweep:

function minMeetingRooms(intervals)
 events := []
 for each interval in intervals
 events.add((interval.start, +1))
 events.add((interval.end, -1))
 sort events by time (if tie, process -1 before +1)
 rooms := 0
 maxRooms := 0
 for each (time, delta) in events
 rooms := rooms + delta
 maxRooms := max(maxRooms, rooms)
 return maxRooms

Time: $O(n \log n)$. Space: $O(n)$ for events array.

The tie-breaking rule matters: if a meeting ends at time $t$ and another starts at $t$, the ending frees a room before the new one needs it.