Which station has the least work
Jobs are queued against stations on a line. Find the station currently carrying the fewest hours so the next job can be routed there.
- Each station is an index from 0 up to stationCount - 1.
- Sum the hours of every job queued at each in-range station.
- The answer is the index with the smallest total; ties go to the lowest index.
- Jobs pointing at an out-of-range station are ignored.
- With zero stations or no in-range jobs the answer is 0.
leastLoadedStation(jobs: list<Job>, stationCount: 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 leastLoadedStation(jobs []Job, stationCount int) int {
}Worked examples
| Call | Result |
|---|---|
leastLoadedStation([]Job{Job{Station: 0, Hours: 10}, Job{Station: 1, Hours: 5}, Job{Station: 0, Hours: 2}}, 2) | 1 |
leastLoadedStation([]Job{Job{Station: 0, Hours: 3}, Job{Station: 1, Hours: 3}}, 2) | 0 |
leastLoadedStation([]Job{Job{Station: 5, Hours: 10}}, 3) | 0 |
leastLoadedStation([]Job{}, 4) | 0 |
Hint
Tally per station in an array, then walk the array keeping a running best index.
Reference solution in Go
func leastLoadedStation(jobs []Job, stationCount int) int {
if stationCount <= 0 {
return 0
}
totals := make([]int, stationCount)
for _, j := range jobs {
if j.Station < 0 || j.Station >= stationCount {
continue
}
totals[j.Station] += j.Hours
}
best := 0
for i := 1; i < stationCount; i++ {
if totals[i] < totals[best] {
best = i
}
}
return best
}