Peak request burst in a sliding window
A rate limiter must know the worst-case burst: the most requests that ever land inside a fixed-size time window.
- The timestamps are sorted ascending and in whole seconds.
- A window spans [t, t + windowSeconds); a request at exactly t + windowSeconds is outside.
- For every possible window start, count how many timestamps fall inside it; return the largest count.
- When windowSeconds is zero or negative, return 0.
burstWindow(timestamps: list<int>, windowSeconds: int) → 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
int burstWindow(std::vector<int> timestamps, int windowSeconds) {
}Worked examples
| Call | Result |
|---|---|
burstWindow(std::vector<int>{1, 2, 5, 8, 10}, 5) | 3 |
burstWindow(std::vector<int>{0, 10, 20, 30}, 15) | 2 |
burstWindow(std::vector<int>{100}, 10) | 1 |
burstWindow(std::vector<int>{}, 5) | 0 |
Hint
Brute-force every starting index and count forward; the list is sorted, so stop at the first timestamp outside the window.
Reference solution in C++
int burstWindow(std::vector<int> timestamps, int windowSeconds) {
if (windowSeconds <= 0) return 0;
int best = 0;
for (size_t i = 0; i < timestamps.size(); i++) {
int count = 0;
for (size_t j = i; j < timestamps.size(); j++) {
if (timestamps[j] < timestamps[i] + windowSeconds) count++;
else break;
}
if (count > best) best = count;
}
return best;
}