Happy to document it, but I don't know the answer.

Haha, okay, got it. Well, hopefully someone else who is algorithmically inclined will come along and see this :)

Haha, okay, got it. Well, hopefully someone else who is algorithmically inclined will come along and see this :)

If arrayOne's length is a and arrayTwo's length is b, isn't it the time complexity is O(ab)?

1. We need at least iterate arrayOne a times and generally iterate arrayTwo b/2 times as an average situation.
2. The iteration of arrayTwo is in each arrayOne's iteration, so the time complexity is O(a * b/2).
3. The constant level factor should be ignored.
4. As a result, the answer is O(ab)

If you call arrayDiffer(a, b) and m = a.length, n = b.length, then turning b into a set takes O(n * log n) because inserting a single element into the set takes O(n), and filtering a takes O(m * log n). To summarize, the complexity is O(max(m, n) * log n). But that's a worst-case analysis which might be irrelevant in real life.