Rest-period violations in the hand-offs
Between any two shifts assigned to the same person there must be at least restMinutes of free time. Count the hand-offs in a day that break the rule.
- Each pair of consecutive shifts is one hand-off; consider them in the order listed.
- Sequential per person — here the day is one person's list, so every gap counts.
- Ending at 900 and starting at 1020 is exactly restMinutes if restMinutes is 120, which is fine.
restViolations(shifts: list<Shift>, restMinutes: int) → int
Go 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
func restViolations(shifts []Shift, restMinutes int) int {
}Worked examples
| Call | Result |
|---|---|
restViolations([]Shift{Shift{Start: 480, End: 900}, Shift{Start: 960, End: 1200}}, 120) | 1 |
restViolations([]Shift{Shift{Start: 480, End: 900}, Shift{Start: 1020, End: 1200}}, 120) | 0 |
restViolations([]Shift{Shift{Start: 0, End: 100}}, 100) | 0 |
restViolations([]Shift{Shift{Start: 0, End: 300}, Shift{Start: 400, End: 500}, Shift{Start: 700, End: 800}}, 300) | 2 |
Hint
Compare each next start to the previous end.
Reference solution in Go
func restViolations(shifts []Shift, restMinutes int) int {
bad := 0
for i := 1; i < len(shifts); i++ {
if shifts[i].Start-shifts[i-1].End < restMinutes {
bad++
}
}
return bad
}