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>
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 rangeTotals(series []int, starts []int, ends []int) []int {
}Worked examples
| Call | Result |
|---|---|
rangeTotals([]int{1, 2, 3, 4, 5}, []int{0, 1, 0}, []int{2, 3, 4}) | []int{6, 9, 15} |
rangeTotals([]int{1, 2, 3}, []int{1}, []int{1}) | []int{2} |
rangeTotals([]int{1, 2, 3}, []int{2}, []int{1}) | []int{0} |
rangeTotals([]int{1, 2, 3}, []int{0}, []int{99}) | []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 Go
func rangeTotals(series []int, starts []int, ends []int) []int {
running := []int{0}
for _, value := range series {
running = append(running, running[len(running)-1]+value)
}
out := []int{}
for i := 0; i < len(starts); i++ {
from, to := starts[i], ends[i]
if from < 0 || to >= len(series) || from > to {
out = append(out, 0)
} else {
out = append(out, running[to+1]-running[from])
}
}
return out
}