Which pack size is the best value
The same product sits on the shelf in several pack sizes. A price-comparison badge needs the one with the lowest cost per unit.
- Compare price per unit, not sticker price.
- Skip any pack claiming zero or fewer units — that is bad data, not a bargain.
- On a tie, the larger pack wins.
- With nothing worth comparing, return null.
cheapestPack(packs: list<Pack>) → 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> cheapestPack(std::vector<Pack> packs) {
}Worked examples
| Call | Result |
|---|---|
cheapestPack(std::vector<Pack>{Pack{std::string("single"), 1, 300}, Pack{std::string("six"), 6, 1500}, Pack{std::string("crate"), 24, 6200}}) | std::optional<std::string>(std::string("six")) |
cheapestPack(std::vector<Pack>{Pack{std::string("a"), 2, 200}, Pack{std::string("b"), 4, 400}}) | std::optional<std::string>(std::string("b")) |
cheapestPack(std::vector<Pack>{Pack{std::string("broken"), 0, 100}, Pack{std::string("ok"), 3, 900}}) | std::optional<std::string>(std::string("ok")) |
cheapestPack(std::vector<Pack>{}) | std::nullopt |
Hint
Cross-multiply instead of dividing: a.price * b.units against b.price * a.units keeps it in integers.
Reference solution in C++
std::optional<std::string> cheapestPack(std::vector<Pack> packs) {
const Pack* best = nullptr;
for (const auto& p : packs) {
if (p.units <= 0) continue;
if (!best) { best = &p; continue; }
long long a = (long long) p.price * best->units, b = (long long) best->price * p.units;
if (a < b || (a == b && p.units > best->units)) best = &p;
}
if (!best) return std::nullopt;
return best->label;
}