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
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
double medianValue(std::vector<int> values) {
}Worked examples
| Call | Result |
|---|---|
medianValue(std::vector<int>{3, 1, 2}) | 2.0 |
medianValue(std::vector<int>{1, 2, 3, 4}) | 2.5 |
medianValue(std::vector<int>{7}) | 7.0 |
medianValue(std::vector<int>{}) | 0.0 |
Hint
After sorting, the two middles for an even count sit at n/2 - 1 and n/2.
Reference solution in C++
double medianValue(std::vector<int> values) {
if (values.empty()) return 0.0;
std::vector<int> s = values;
std::sort(s.begin(), s.end());
int n = (int) s.size(), mid = n / 2;
if (n % 2 == 1) return s[mid];
return (s[mid - 1] + s[mid]) / 2.0;
}