Note - the original problem description is in the problem source section below đđ
Imagine youâre in charge of scheduling a new meeting for a group of employees who have already packed their day with various appointments. The day is measured in minutes, starting from 0 (midnight) and ending at 1440 minutes (the very last moment of the day). Each employeeâs busy schedule is laid out in a special array called schedules.
In this array:
Youâve been tasked with finding the earliest moment in this busy day where all employees are free to meet for a certain period of time, called length. But, thereâs a catch! The new meeting needs to fit within the same day, meaning it canât run past 1440 minutes (the end of the day).
Your job is to sift through the employeesâ schedules, find a time where everyone is available for this new meeting, and figure out the earliest possible time to set it. If thereâs no such moment when all are free for that meeting, youâll return -1, indicating that itâs impossible to schedule the meeting on this day.
Donât worry too much about finding the absolute most efficient solutionâjust make sure that your approach can handle the complexity of multiple busy schedules without taking too long.
schedules = [[[60, 150], [180, 240]], [[0, 210], [360, 420]]] length = 120 return = 240

schedules = [[[480, 510]], [[240, 330]], [[375, 400]]] length = 180 return = 0
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public int robloxScheduleMeeting(int[][][] schedules, int length) {
// write your code here
}