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>
Go 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
func windowMax(values []int, window int) []int {
}Worked examples
| Call | Result |
|---|---|
windowMax([]int{1, 3, 2, 5}, 2) | []int{3, 3, 5} |
windowMax([]int{1, 2, 3}, 3) | []int{3} |
windowMax([]int{5}, 1) | []int{5} |
windowMax([]int{4, 1, 3}, 2) | []int{4, 3} |
Hint
For each starting index take the maximum of the next `window` values with an inner loop.
Reference solution in Go
func windowMax(values []int, window int) []int {
result := []int{}
if window <= 0 || window > len(values) {
return result
}
for i := 0; i+window <= len(values); i++ {
m := values[i]
for j := i + 1; j < i+window; j++ {
if values[j] > m {
m = values[j]
}
}
result = append(result, m)
}
return result
}