A video platform logs each viewing session as [start, end] in whole minutes of the day: the viewer is watching during minutes start to end - 1 (the end minute is not included).
Write peakViewers(sessions) that returns [peak, minute]: the largest number of viewers watching during the same minute, and the earliest minute when that happens. If there are no sessions, return [0, 0].
Input: sessions = [[0,30],[10,40],[20,25],[35,60]] Output: [3,20]
From minute 20 to 24, three sessions overlap. That first happens at minute 20.
Input: sessions = [[5,10],[10,15]] Output: [1,5]
The first session ends just as the second starts, so they never overlap.
0 <= sessions.length <= 10^50 <= start < end <= 1440Run tries your code on the examples. Submit runs every test, including hidden edge cases, and solves the problem when they all pass.
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I'm here to help with #57 Peak Concurrent Viewers. 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.
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