The middle of a set of readings
A latency report quotes the median rather than the mean, because one slow request should not move the headline number.
- Sort first — the readings arrive in whatever order they were logged.
- An odd count has a single middle value.
- An even count takes the average of the two in the middle, which may be a half.
- No readings at all gives 0.
medianValue(values: list<int>) → float
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 medianValue(values []int) float64 {
}Worked examples
| Call | Result |
|---|---|
medianValue([]int{3, 1, 2}) | 2.0 |
medianValue([]int{1, 2, 3, 4}) | 2.5 |
medianValue([]int{7}) | 7.0 |
medianValue([]int{}) | 0.0 |
Hint
After sorting, the two middles for an even count sit at n/2 - 1 and n/2.
Reference solution in Go
func medianValue(values []int) float64 {
if len(values) == 0 {
return 0
}
s := append([]int{}, values...)
sort.Ints(s)
n := len(s)
mid := n / 2
if n%2 == 1 {
return float64(s[mid])
}
return float64(s[mid-1]+s[mid]) / 2
}