The peak in each window
A monitoring chart smooths readings by asking, for each fixed-size window, what the loudest moment inside it was.
- Each output entry is the largest value inside one contiguous window of size `window`.
- The windows slide by one position at a time.
- A window of zero or less, or one larger than the whole list, gives an empty result.
windowMax(values: list<int>, window: 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> windowMax(std::vector<int> values, int window) {
}Worked examples
| Call | Result |
|---|---|
windowMax(std::vector<int>{1, 3, 2, 5}, 2) | std::vector<int>{3, 3, 5} |
windowMax(std::vector<int>{1, 2, 3}, 3) | std::vector<int>{3} |
windowMax(std::vector<int>{5}, 1) | std::vector<int>{5} |
windowMax(std::vector<int>{4, 1, 3}, 2) | std::vector<int>{4, 3} |
Hint
For each starting index take the maximum of the next `window` values with an inner loop.
Reference solution in C++
std::vector<int> windowMax(std::vector<int> values, int window) {
std::vector<int> result;
if (window <= 0 || window > (int) values.size()) return result;
for (int i = 0; i + window <= (int) values.size(); i++) {
int m = values[i];
for (int j = i + 1; j < i + window; j++) if (values[j] > m) m = values[j];
result.push_back(m);
}
return result;
}