Meeting Rooms - Algorithm

Sort meetings by start time. Then scan through and check if any meeting starts before the previous one ends. If $meeting[i].start < meeting[i-1].end$, there is an overlap. You cannot attend both.

This is the simplest interval overlap check: sort, then linear scan comparing adjacent pairs. No need for complex data structures. Time: $O(n \log n)$. Space: $O(1)$ extra. Common bug: using <= instead of < when meetings can touch without conflicting.