Meeting Rooms

Check whether one person can attend every meeting.

Start with the concrete trace below. It shows the state the algorithm must carry as it runs.

Problem trace

Meeting Rooms: Sorting by start time makes the first overlap appear between adjacent meetings, so Python all can stop at the first failed boundary check.

Input and goalCheck whether one person can attend every meeting.

Start with unsorted meetingsUse intervals [[7,10],[2,4],[12,14],[9,11]]. Their input order does not reveal the chronological neighbor checks.

[7,10]

[2,4]

[12,14]

[9,11]

input[[7,10],[2,4],[12,14],[9,11]]

Recognize it

Use it when one resource must handle every interval and any temporal overlap makes the schedule invalid.

Keep true

Before checking index i, all adjacent boundaries before i are non-overlapping. In start-sorted order, any overlap must include an adjacent pair, so the first failed comparison proves impossibility.

Reuse it

Sort to make a global conflict local, then compare neighboring boundaries and stop at the first violation. This transfers to overlap validation, booking conflicts, and disjoint-range checks.

Pattern: Sort intervals by start.

Simple idea: After sorting, only neighboring meetings can reveal the first overlap. Each meeting must start at or after the previous meeting ends.

def can_attend_meetings(intervals: list[list[int]]) -> bool:
   intervals.sort()
   return all(
      intervals[index - 1][1] <= intervals[index][0]
      for index in range(1, len(intervals))
   )

Cost: $O(n\log n)$ time and $O(1)$ extra space, not counting sorting.