FastPrepEarliest Common Meeting Time

Earliest Common Meeting Time

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Problem statement

You are given one day's meeting schedules for several employees. schedules[i][j] = [start, finish] is a half-open busy interval in minutes from the start of the day.

Given a requested meeting length, return the earliest start minute at which every employee is free for the entire meeting. The meeting must finish by minute 1440. Return -1 if no common interval exists.

Function

earliestCommonMeeting(schedules: int[][][], length: int) → int

Examples

Example 1

schedules = [[[60,150],[180,240]],[[0,210],[360,420]]]length = 120return = 240

Every employee is free from minute 240 through minute 360, so 240 is the earliest valid start.

Example 2

schedules = [[[480,510]],[[240,330]],[[375,400]]]length = 180return = 0

All employees are free from minute 0 through minute 180.

Constraints

  • 1 <= schedules.length <= 100
  • 0 <= schedules[i].length <= 100
  • 0 <= start < finish <= 1440
  • 1 <= length <= 1440

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public int earliestCommonMeeting(int[][][] schedules, int length) {
    // Write your code here.
}
schedules[[[60,150],[180,240]],[[0,210],[360,420]]]
length120
expected240
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