Count the oversized gaps
A service watchdog flags any two consecutive events that arrived further apart than a threshold.
- Timestamps are ascending, so only neighbouring pairs need comparing.
- Count the pairs whose difference is strictly more than the threshold.
- The threshold is a positive gap above which a pair counts.
timestampsGaps(timestamps: list<int>, gapThreshold: 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 timestampsGaps(std::vector<int> timestamps, int gapThreshold) {
}Worked examples
| Call | Result |
|---|---|
timestampsGaps(std::vector<int>{0, 10, 11, 40}, 10) | 1 |
timestampsGaps(std::vector<int>{0, 10, 11, 40}, 5) | 2 |
timestampsGaps(std::vector<int>{0, 5, 10}, 5) | 0 |
timestampsGaps(std::vector<int>{1, 100}, 50) | 1 |
Hint
Compare each entry with the one before it and add to the count when the gap is too wide.
Reference solution in C++
int timestampsGaps(std::vector<int> timestamps, int gapThreshold) {
int count = 0;
for (size_t i = 1; i < timestamps.size(); i++) {
if (timestamps[i] - timestamps[i - 1] > gapThreshold) count++;
}
return count;
}