We use analytics and advertising cookies to understand how the site is used and whether our ads on Facebook and Instagram work. They are set only if you accept. See our Privacy Policy for details.
Given an array of meeting time intervals where intervals[i] = [start, end], determine whether a person could attend all meetings.
Return true if no two meetings overlap, false otherwise.
A meeting that ends exactly when another begins does not count as an overlap - [1,2] and [2,3] are fine.
An easy intervals problem, graded against 6 test cases (3 of them hidden).
Sorting by start or end, then merging or scheduling overlapping ranges.
Reach for it when you see: Meetings, bookings, ranges, or anything with a start and an end.
More Intervals problems →The brute force compares all pairs, O(n^2). Sorting by start time buys a much stronger property: if any two meetings overlap at all, then some adjacent pair in sorted order overlaps. So a single pass over neighbours settles it.
For each consecutive pair, the conflict test is `next.start < current.end`. Getting the boundary right is the whole problem: with `<=` you would reject `[1,2]` and `[2,3]`, which are back-to-back rather than overlapping. Interviewers reliably probe this case, and it is worth stating your convention out loud before you write the comparison.
Sorting dominates the runtime; the scan is linear.
The full reference solution in every supported language stays in the editor above - reveal it there once you have had a real attempt.
Read off this problem's own test suite, so these are the cases a submission actually has to survive.
These apply to the pattern as a whole, not just this problem.