Answer a stack of range totals
A reporting screen asks for the total of many different day ranges over the same series, and asks a lot of them.
- Each query is a pair taken from the two lists by position: `starts[i]` to `ends[i]`.
- Both ends are inclusive.
- A query whose range falls outside the series, or runs backwards, totals 0.
- Return one total per query, in the order the queries came.
rangeTotals(series: list<int>, starts: list<int>, ends: list<int>) → list<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
std::vector<int> rangeTotals(std::vector<int> series, std::vector<int> starts, std::vector<int> ends) {
}Worked examples
| Call | Result |
|---|---|
rangeTotals(std::vector<int>{1, 2, 3, 4, 5}, std::vector<int>{0, 1, 0}, std::vector<int>{2, 3, 4}) | std::vector<int>{6, 9, 15} |
rangeTotals(std::vector<int>{1, 2, 3}, std::vector<int>{1}, std::vector<int>{1}) | std::vector<int>{2} |
rangeTotals(std::vector<int>{1, 2, 3}, std::vector<int>{2}, std::vector<int>{1}) | std::vector<int>{0} |
rangeTotals(std::vector<int>{1, 2, 3}, std::vector<int>{0}, std::vector<int>{99}) | std::vector<int>{0} |
Hint
Build the running total once, up front. Then any range is one subtraction — the total up to the end, less the total before the start.
Reference solution in C++
std::vector<int> rangeTotals(std::vector<int> series, std::vector<int> starts, std::vector<int> ends) {
std::vector<int> running{0};
for (int value : series) running.push_back(running.back() + value);
std::vector<int> out;
for (size_t i = 0; i < starts.size(); i++) {
int from = starts[i], to = ends[i];
if (from < 0 || to >= static_cast<int>(series.size()) || from > to) out.push_back(0);
else out.push_back(running[to + 1] - running[from]);
}
return out;
}