Meeting Rooms Solution

Sort by start time, then check if any meeting starts before the previous ends:

function canAttendMeetings(intervals)
    sort intervals by start time

    for i from 1 to length - 1
        if intervals[i].start < intervals[i-1].end then
            return false

    return true

After sorting, if no adjacent pairs overlap, then no pairs overlap at all. Why? If meeting $j$ overlaps with meeting $i$ (where $j > i + 1$), then meeting $i+1$ must also overlap with meeting $i$. After sorting, intervals[i+1].start <= intervals[j].start. So if intervals[i].end > intervals[j].start, then intervals[i].end > intervals[i+1].start too.

Time: $O(n \log n)$. Space: $O(1)$ if sorting in place.