Pick an order off the shelves
A picker needs a quantity of one part, and it is spread across several shelves. Walk the shelves in the order given and take what you can from each until the order is filled.
- Take as much as a shelf holds, but never more than is still needed.
- Skip shelves that would contribute nothing.
- If the shelves cannot cover it, take everything they have and stop there.
- A request of zero or less picks nothing.
pickFromShelves(wanted: int, shelves: list<Shelf>) → list<Pick>
Java 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
List<Pick> pickFromShelves(int wanted, List<Shelf> shelves) {
}Worked examples
| Call | Result |
|---|---|
pickFromShelves(12, Main.<Shelf>ls(new Shelf("A1", 5), new Shelf("A2", 0), new Shelf("B3", 20))) | Main.<Pick>ls(new Pick("A1", 5), new Pick("B3", 7)) |
pickFromShelves(4, Main.<Shelf>ls(new Shelf("A1", 10))) | Main.<Pick>ls(new Pick("A1", 4)) |
pickFromShelves(30, Main.<Shelf>ls(new Shelf("A1", 5), new Shelf("B3", 6))) | Main.<Pick>ls(new Pick("A1", 5), new Pick("B3", 6)) |
pickFromShelves(0, Main.<Shelf>ls(new Shelf("A1", 5))) | Main.<Pick>ls() |
Hint
Carry a running "still needed" figure and stop as soon as it hits zero.
Reference solution in Java
List<Pick> pickFromShelves(int wanted, List<Shelf> shelves) {
List<Pick> out = new ArrayList<>();
int need = wanted;
for (Shelf s : shelves) {
if (need <= 0) break;
int take = Math.min(need, s.available);
if (take <= 0) continue;
out.add(new Pick(s.code, take));
need -= take;
}
return out;
}