474. Earliest Meeting Room Availability

You are given a list of scheduled meetings, where each meeting is represented as a pair of integers [start, end]. You need to find the earliest possible time a new meeting of duration 1 can be scheduled such that it does not overlap with any existing meetings. The earliest possible time you can start is 0. If there are no gaps between 0 and the end of the last meeting, return the end time of the last meeting.

Example:
Input: meetings = [[1, 3], [5, 6]]
Output: 0

(Explanation: Time 0 to 1 is free, so the earliest start time for a duration-1 meeting is 0.)

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