Meeting Rooms
Reported by candidates from Chainalysis's online assessment. Pattern, common pitfall, and the honest play if you blank under the timer.
The Chainalysis OA reported in April 2025 asks a question that looks too easy: can one person attend every meeting? The whole solution hinges on one data structure, a plain array of intervals sorted by start time. Once it's sorted, overlap checks only touch neighbors. If you've seen Meeting Rooms before, this is a five-minute problem. If you haven't, it's easy to overthink with nested loops that die on 10^5 intervals. StealthCoder sits invisible on your screen as a safety net if you blank during the live OA, but the idea here is small enough to hold in your head.
The problem
Each meeting is a half-open interval [start, end). Return true if one person can attend every meeting without overlap. Function canAttendMeetings(intervals: int[][]) → boolean Examples Example 1 intervals = [[0,30],[5,10],[15,20]] return = false The first meeting overlaps the others. Example 2 intervals = [[7,10],[10,12]] return = true Half-open intervals may touch at an endpoint. Constraints There are at most 10^5 intervals. start < end for every interval.
Reported by candidates. Source: FastPrep
Pattern and pitfall
Sort the intervals by start time, then walk through once. For each pair of neighbors, check whether the previous end is greater than the current start. If it is, return false. Otherwise keep going and return true at the end. The trap is the half-open rule. Intervals are [start, end), so [7,10] and [10,12] touch but don't overlap. Use a strict greater-than, not greater-or-equal. The other pitfall is the brute force pairwise comparison, which is O(n^2) and too slow with up to 10^5 intervals. Sorting gives O(n log n) time and tiny extra space. Also handle empty or single-interval input by returning true. If the comparison operator or the sort key slips away from you mid-assessment, StealthCoder can surface the clean version while you keep typing.
Memorize the pattern. If you can't, run StealthCoder. The proctor sees the IDE. They don't see what's behind it.
You can drill Meeting Rooms cold, or you can hedge it. StealthCoder runs invisibly during screen share and surfaces a working solution in under 2 seconds. The proctor sees the IDE. They don't see what's behind it. Made by an engineer who treats the OA as theater. If yours is tonight, you don't have time to grind. You have time to hedge.
Get StealthCoderRelated leaked OAs
This OA pattern shows up on LeetCode as meeting rooms. If you have time before the OA, drill that.
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Make sure you actually pass Chainalysis's OA.
Chainalysis reuses patterns across OAs. Made by an engineer who treats the OA as theater. If yours is tonight, you don't have time to grind. You have time to hedge. Works on HackerRank, CodeSignal, CoderPad, and Karat.
Meeting Rooms FAQ
How hard is the Chainalysis Meeting Rooms question really?+
It's easy. The core is sort by start, then compare each meeting's start to the previous meeting's end. Most candidates fail on the boundary condition, not the algorithm. Get the strict comparison right and you're done in a few minutes.
What's the trick to this problem?+
Sorting. Once intervals are ordered by start time, any overlap must show up between adjacent meetings. You never need to compare non-neighbors, which turns an O(n^2) check into a single pass after the sort.
Should the overlap check use > or >=?+
Use strict greater-than on previous end versus current start. The intervals are half-open, so [7,10] and [10,12] are fine. Using >= would wrongly return false on Example 2, which is exactly the case the problem is testing.
What's the time complexity I should state?+
O(n log n) for the sort, O(n) for the scan, so O(n log n) overall. Extra space depends on your language's sort, usually O(1) to O(n). With 10^5 intervals, that's comfortably fast, while pairwise comparison is not.
How do I prepare for this in 48 hours?+
Write the sorted-scan solution from scratch twice, then test it on touching endpoints, a single interval, and duplicates. Then try the follow-up of counting the minimum rooms needed, since interval problems often come in pairs.