Refund for partially used service
A customer paid upfront but only used part of the service. Refund the unused portion.
- Refund equals paid minus (usedUnits × unitPrice), clamped at zero.
- If usage cost meets or exceeds what was paid, no refund.
partialRefund(paid: int, usedUnits: int, unitPrice: 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 partialRefund(int paid, int usedUnits, int unitPrice) {
}Worked examples
| Call | Result |
|---|---|
partialRefund(1000, 40, 10) | 600 |
partialRefund(1000, 200, 10) | 0 |
partialRefund(500, 0, 50) | 500 |
partialRefund(1000, 100, 10) | 0 |
Hint
Multiply usage by price, subtract from what was paid, and clamp.
Reference solution in C++
int partialRefund(int paid, int usedUnits, int unitPrice) {
int cost = usedUnits * unitPrice;
return paid > cost ? paid - cost : 0;
}