How often the load crossed the line
Capacity planning counts how many fixed-length stretches of the day carried at least a given amount of work.
- Every run of exactly `runLength` consecutive entries counts as one window, overlapping ones included.
- A window counts when its total is greater than or equal to the limit.
- If the window does not fit in the data, or the size is not positive, the answer is 0.
windowsOverLimit(load: list<int>, runLength: int, limit: 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 windowsOverLimit(std::vector<int> load, int runLength, int limit) {
}Worked examples
| Call | Result |
|---|---|
windowsOverLimit(std::vector<int>{1, 2, 3, 4, 5}, 2, 5) | 3 |
windowsOverLimit(std::vector<int>{1, 1, 1}, 2, 10) | 0 |
windowsOverLimit(std::vector<int>{5, 5, 5}, 1, 5) | 3 |
windowsOverLimit(std::vector<int>{1, 2}, 3, 1) | 0 |
Hint
Slide one total across the list rather than re-summing each window, and test it at every stop.
Reference solution in C++
int windowsOverLimit(std::vector<int> load, int runLength, int limit) {
if (runLength <= 0 || static_cast<int>(load.size()) < runLength) return 0;
int window = 0;
for (int i = 0; i < runLength; i++) window += load[i];
int hits = window >= limit ? 1 : 0;
for (size_t i = runLength; i < load.size(); i++) {
window += load[i] - load[i - runLength];
if (window >= limit) hits++;
}
return hits;
}