Meeting Rooms
The problem
Given meeting intervals with start < end, determine whether one person can attend every meeting. A meeting ending when another begins is allowed.
Example
[[1, 4], [4, 6], [8, 9]] → true
Need a hint?
After sorting, only neighboring meetings need comparison.
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Read the solution approach
Sort intervals by start. If any start is earlier than the previous interval’s end, return false. Otherwise return true. Empty and single-meeting schedules are conflict-free. Use a strict comparison so back-to-back meetings are accepted.
Complexity
O(n log n) time; sorting determines auxiliary space.
Before moving on, explain why the algorithm is correct and trace a boundary case without looking at the approach.