A server log has the times (in seconds) of every request in times, in no particular order. The on-call dashboard wants the widest quiet period: the biggest difference between two requests that are next to each other once the times are sorted.
Write widestGap(times) that returns that difference, or 0 if there are fewer than two requests. Aim for linear time: sorting first is too slow for the real log size.
Input: times = [3,6,9,1] Output: 3
Sorted: 1, 3, 6, 9. The gaps are 2, 3, and 3.
Input: times = [10] Output: 0
One request has no neighbor.
0 <= times.length <= 1000000 <= times[i] <= 10^9Run tries your code on the examples. Submit runs every test, including hidden edge cases, and solves the problem when they all pass.
Ctrl + Enter runs, Ctrl + Shift + Enter submits (⌘ on a Mac).
I'm here to help with #207 Widest Quiet Period. I give hints, never the solution, so you get to solve it yourself. Ask for a hint, select code to ask about it, or press Ask Tuto why on a failing test.
Your first 3 answers need no account. for free to keep asking.