How many rooms are needed at peak
A venue schedules sessions across the day. Find the minimum number of rooms so no two overlapping sessions share one.
- Each session has a start and an end in minutes.
- Intervals are half-open: [start, end).
- Sessions that merely touch (one ends exactly when the next starts) do not conflict.
sessionRooms(starts: list<int>, ends: list<int>) → int
Java needs a compiler and Drill does not host one yet, so this page is the reference rather than an exercise: the problem, worked examples, and the solution in full. To type it out, the same problem runs in Python.
Where you start
int sessionRooms(List<Integer> starts, List<Integer> ends) {
}Worked examples
| Call | Result |
|---|---|
sessionRooms(Main.<Integer>ls(10), Main.<Integer>ls(20)) | 1 |
sessionRooms(Main.<Integer>ls(10, 10, 10), Main.<Integer>ls(20, 20, 20)) | 3 |
sessionRooms(Main.<Integer>ls(10, 15), Main.<Integer>ls(20, 25)) | 2 |
sessionRooms(Main.<Integer>ls(10, 20, 30), Main.<Integer>ls(15, 25, 35)) | 1 |
Hint
Sort starts and ends independently. Walk both lists with two pointers; the gap between active starts and ends at each step is the current room count — track the peak.
Reference solution in Java
int sessionRooms(List<Integer> starts, List<Integer> ends) {
List<Integer> s = new ArrayList<>(starts);
Collections.sort(s);
List<Integer> e = new ArrayList<>(ends);
Collections.sort(e);
int peak = 0, active = 0, j = 0;
for (int i = 0; i < s.size(); i++) {
active++;
while (j < e.size() && e.get(j) <= s.get(i)) { active--; j++; }
if (active > peak) peak = active;
}
return peak;
}