Replay a log that can take it back
A stock adjustment screen records every change, and an undo command that cancels whichever change came last.
- A command is either a signed whole number to apply, or the word "undo".
- An undo cancels the most recent change that is still standing.
- An undo with nothing left to cancel does nothing.
- Start from zero and return the total once every command has run.
replayLog(commands: list<string>) → 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 replayLog(std::vector<std::string> commands) {
}Worked examples
| Call | Result |
|---|---|
replayLog(std::vector<std::string>{std::string("5"), std::string("3"), std::string("undo")}) | 5 |
replayLog(std::vector<std::string>{std::string("5"), std::string("undo"), std::string("undo")}) | 0 |
replayLog(std::vector<std::string>{std::string("10"), std::string("-4"), std::string("2")}) | 8 |
replayLog(std::vector<std::string>{}) | 0 |
Hint
Keep the applied changes on a stack. Undo pops the last one off and subtracts it back out.
Reference solution in C++
int replayLog(std::vector<std::string> commands) {
std::vector<int> applied;
int total = 0;
for (const std::string& command : commands) {
if (command == "undo") {
if (!applied.empty()) {
total -= applied.back();
applied.pop_back();
}
} else {
int amount = std::stoi(command);
applied.push_back(amount);
total += amount;
}
}
return total;
}