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
C# 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
public int SessionRooms(List<int> starts, List<int> ends) {
}Worked examples
| Call | Result |
|---|---|
SessionRooms(new List<int> { 10 }, new List<int> { 20 }) | 1 |
SessionRooms(new List<int> { 10, 10, 10 }, new List<int> { 20, 20, 20 }) | 3 |
SessionRooms(new List<int> { 10, 15 }, new List<int> { 20, 25 }) | 2 |
SessionRooms(new List<int> { 10, 20, 30 }, new List<int> { 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 C#
public int SessionRooms(List<int> starts, List<int> ends) {
var s = new List<int>(starts);
s.Sort();
var e = new List<int>(ends);
e.Sort();
int peak = 0, active = 0, j = 0;
for (int i = 0; i < s.Count; i++) {
active++;
while (j < e.Count && e[j] <= s[i]) { active--; j++; }
if (active > peak) peak = active;
}
return peak;
}