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
public List<int> WindowMax(List<int> values, int window) {
}Worked examples
| Call | Result |
|---|---|
WindowMax(new List<int> { 1, 3, 2, 5 }, 2) | new List<int> { 3, 3, 5 } |
WindowMax(new List<int> { 1, 2, 3 }, 3) | new List<int> { 3 } |
WindowMax(new List<int> { 5 }, 1) | new List<int> { 5 } |
WindowMax(new List<int> { 4, 1, 3 }, 2) | new List<int> { 4, 3 } |
Hint
For each starting index take the maximum of the next `window` values with an inner loop.
Reference solution in C#
public List<int> WindowMax(List<int> values, int window) {
var result = new List<int>();
if (window <= 0 || window > values.Count) return result;
for (int i = 0; i + window <= values.Count; i++) {
int m = values[i];
for (int j = i + 1; j < i + window; j++) if (values[j] > m) m = values[j];
result.Add(m);
}
return result;
}