The first thing I fumbled was not asking whether endpoints touching counts as overlapping.
First clarify the definition of non-overlapping (e.g., [1,2] and [2,3] are considered non-overlapping if endpoints touch). Then sort intervals by end time and use dynamic programming to count triplets: for each interval as the last one, count pairs of non-overlapping intervals before it. Alternatively, sort by start time and use binary search to find compatible intervals, then count triplets efficiently.
Pro tip: Mention that you can reduce the triplet count to O(n^2) by fixing the middle interval and counting non-overlapping intervals on both sides, or to O(n log n) with prefix sums and binary search. This shows you understand optimization beyond brute force.
Confirm whether intervals are closed or half-open, and whether touching endpoints count as non-overlapping. Also confirm if intervals are given as [start, end] and if they are sorted.
Sort the intervals by their end times (or start times) to enable efficient counting of non-overlapping intervals.
For each interval i, compute the number of intervals that end before i starts (using binary search). Then count triplets by iterating over the middle interval or using prefix sums.
Use the precomputed counts to sum over all possible triplets. For example, for each j, add (number of non-overlapping intervals before j) * (number of non-overlapping intervals after j).
State the time complexity (e.g., O(n log n) or O(n^2)) and space complexity, and discuss potential optimizations or edge cases.
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