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.
timestamps_gaps(timestamps: list<int>, gap_threshold: int) → int
Where you start
def timestamps_gaps(timestamps: list[int], gap_threshold: int) -> int:
Worked examples
| Call | Result |
|---|---|
timestamps_gaps([0, 10, 11, 40], 10) | 1 |
timestamps_gaps([0, 10, 11, 40], 5) | 2 |
timestamps_gaps([0, 5, 10], 5) | 0 |
timestamps_gaps([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 Python
def timestamps_gaps(timestamps: list[int], gap_threshold: int) -> int:
count = 0
for i in range(1, len(timestamps)):
if timestamps[i] - timestamps[i - 1] > gap_threshold:
count += 1
return count