How many machines are past calibration
Every machine has a last calibration date and a calibration interval in days. Count how many should have been recalibrated by today.
- A machine is overdue when today minus its last service is strictly greater than its interval.
- An interval of zero or less is a broken policy and counts as always overdue.
calibrationOverdue(machines: list<Machine>, today: 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 calibrationOverdue(std::vector<Machine> machines, int today) {
}Worked examples
| Call | Result |
|---|---|
calibrationOverdue(std::vector<Machine>{Machine{100, 10}, Machine{95, 10}, Machine{110, 10}}, 110) | 1 |
calibrationOverdue(std::vector<Machine>{Machine{0, 0}}, 100) | 1 |
calibrationOverdue(std::vector<Machine>{Machine{100, 20}, Machine{100, 20}}, 119) | 0 |
calibrationOverdue(std::vector<Machine>{}, 50) | 0 |
Hint
Run the strict comparison, with the broken-policy case short-circuiting.
Reference solution in C++
int calibrationOverdue(std::vector<Machine> machines, int today) {
int n = 0;
for (const auto& m : machines) {
if (m.intervalDays <= 0 || today - m.lastService > m.intervalDays) n++;
}
return n;
}