Charge each request only once
The payment webhook is delivered at least once, sometimes more. Each delivery carries an idempotency key, and a replay must not charge the customer twice.
- Sum the amounts, but only the first event carrying a given key counts.
- A later event with a key already seen is a replay, whatever amount it claims.
applyOnce(events: list<Event>) → 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 applyOnce(std::vector<Event> events) {
}Worked examples
| Call | Result |
|---|---|
applyOnce(std::vector<Event>{Event{std::string("a"), 100}, Event{std::string("b"), 50}, Event{std::string("a"), 100}}) | 150 |
applyOnce(std::vector<Event>{Event{std::string("a"), 100}, Event{std::string("a"), -100}}) | 100 |
applyOnce(std::vector<Event>{Event{std::string("x"), 7}}) | 7 |
applyOnce(std::vector<Event>{}) | 0 |
Hint
Remember the keys you have already honoured.
Reference solution in C++
int applyOnce(std::vector<Event> events) {
std::set<string> seen;
int total = 0;
for (const auto& e : events) {
if (!seen.insert(e.key).second) continue;
total += e.amount;
}
return total;
}