Labour cost for the roster
A shift costs its staffed hours at ratePerHour (minor units per hour). Hours come from minute spans, prorated exactly.
- Pay = minutes ÷ 60 × rate, rounded down to whole minor units.
- A shift that lasts zero minutes costs nothing.
labourCost(shifts: list<Shift>, ratePerHour: 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
int labourCost(std::vector<Shift> shifts, int ratePerHour) {
}Worked examples
| Call | Result |
|---|---|
labourCost(std::vector<Shift>{Shift{480, 900}, Shift{900, 1080}}, 2000) | 20000 |
labourCost(std::vector<Shift>{Shift{0, 30}}, 2000) | 1000 |
labourCost(std::vector<Shift>{Shift{0, 0}}, 5000) | 0 |
labourCost(std::vector<Shift>{}, 1000) | 0 |
Hint
Sum (end - start) × rate, then divide by 60 once.
Reference solution in C++
int labourCost(std::vector<Shift> shifts, int ratePerHour) {
int mins = 0;
for (const auto& s : shifts) mins += s.end - s.start;
return (mins * ratePerHour) / 60;
}