Find the first free slot in a day
A scheduling assistant looks at a day of bookings and offers the earliest time a meeting of a given length would fit.
- All times are minutes since midnight. Bookings arrive sorted and do not overlap each other.
- The meeting must fit entirely between the start and end of the working day.
- A meeting may begin the moment a booking ends.
- A length of zero or less, or a day with no room, gives null.
nextFreeSlot(busy: list<Slot>, dayStart: int, dayEnd: int, minutes: 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
std::optional<int> nextFreeSlot(std::vector<Slot> busy, int dayStart, int dayEnd, int minutes) {
}Worked examples
| Call | Result |
|---|---|
nextFreeSlot(std::vector<Slot>{Slot{540, 600}}, 480, 1020, 30) | std::optional<int>(480) |
nextFreeSlot(std::vector<Slot>{Slot{480, 540}}, 480, 1020, 30) | std::optional<int>(540) |
nextFreeSlot(std::vector<Slot>{Slot{480, 1020}}, 480, 1020, 30) | std::nullopt |
nextFreeSlot(std::vector<Slot>{Slot{540, 600}, Slot{610, 700}}, 480, 1020, 90) | std::optional<int>(700) |
Hint
Sweep a cursor from the start of the day: before each booking, check whether the gap is wide enough, then jump the cursor past it.
Reference solution in C++
std::optional<int> nextFreeSlot(std::vector<Slot> busy, int dayStart, int dayEnd, int minutes) {
if (minutes <= 0) return std::nullopt;
int cursor = dayStart;
for (const auto& s : busy) {
if (s.start - cursor >= minutes) return cursor;
if (s.end > cursor) cursor = s.end;
}
if (dayEnd - cursor >= minutes) return cursor;
return std::nullopt;
}