Find the bottleneck
On an assembly line the slowest station sets the pace for everything else, so it is the one worth fixing first.
- The bottleneck is the station taking the most seconds per unit.
- If two are equally slow, the one earlier in the line wins.
- Stations quoting zero or fewer seconds are not real measurements and are ignored.
- With nothing measurable, return null.
slowestStation(stations: list<Station>) → string?
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<std::string> slowestStation(std::vector<Station> stations) {
}Worked examples
| Call | Result |
|---|---|
slowestStation(std::vector<Station>{Station{std::string("press"), 12}, Station{std::string("weld"), 30}, Station{std::string("pack"), 9}}) | std::optional<std::string>(std::string("weld")) |
slowestStation(std::vector<Station>{Station{std::string("a"), 20}, Station{std::string("b"), 20}}) | std::optional<std::string>(std::string("a")) |
slowestStation(std::vector<Station>{Station{std::string("broken"), 0}, Station{std::string("weld"), 5}}) | std::optional<std::string>(std::string("weld")) |
slowestStation(std::vector<Station>{Station{std::string("broken"), -1}}) | std::nullopt |
Hint
Track the best so far and only replace it on a strictly greater time, which gives the tie-break for free.
Reference solution in C++
std::optional<std::string> slowestStation(std::vector<Station> stations) {
std::optional<string> name;
int worst = 0;
for (const auto& s : stations) {
if (s.secondsPerUnit <= 0) continue;
if (s.secondsPerUnit > worst) { worst = s.secondsPerUnit; name = s.name; }
}
return name;
}