Sort by start time, then just check if any meeting starts before the previous one ends.
First, clarify that intervals are half-open [start, end) to avoid ambiguity. Then sort the intervals by start time and check for any overlap between consecutive intervals. If no overlap is found, return true; otherwise, false.
Pro tip: Mention that sorting is O(n log n) and that this is optimal because the problem reduces to element uniqueness in the worst case. Also, discuss edge cases like empty input or single interval.
Confirm the definition of overlap (e.g., whether intervals are inclusive or exclusive) and handle edge cases like empty list or single interval.
Sort the intervals by start time. This allows a linear scan to detect overlaps.
Iterate through the sorted intervals and compare the end of the current interval with the start of the next. If the end is greater than the next start, there is an overlap.
If any overlap is found, return false; otherwise, return true after checking all intervals.
State that the time complexity is O(n log n) due to sorting, and space complexity is O(1) if sorting in-place or O(n) depending on the sorting algorithm.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.