How long to clear the order book
Each order has to be made in one piece on a single day, and the plant has a fixed capacity per day. Orders are taken in the order they were received.
- Fill each day with orders in sequence while they still fit; the first that does not fit starts the next day.
- An order bigger than a whole day of capacity can never be made — return -1.
- A capacity of zero or less is also -1.
- An empty book takes no days.
daysToClear(orders: list<int>, dailyCapacity: int) → int
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
int daysToClear(std::vector<int> orders, int dailyCapacity) {
}Worked examples
| Call | Result |
|---|---|
daysToClear(std::vector<int>{3, 4, 5}, 7) | 2 |
daysToClear(std::vector<int>{7, 7}, 7) | 2 |
daysToClear(std::vector<int>{8}, 7) | -1 |
daysToClear(std::vector<int>{1, 1, 1}, 10) | 1 |
Hint
Track how much of today is left. When the next order does not fit, start a new day rather than splitting it.
Reference solution in C++
int daysToClear(std::vector<int> orders, int dailyCapacity) {
if (dailyCapacity <= 0) return -1;
int days = 0, left = 0;
for (int o : orders) {
if (o > dailyCapacity) return -1;
if (o > left) { days++; left = dailyCapacity; }
left -= o;
}
return days;
}