[teo_zero]

With real numbers, I have the same intuition. But with integers, where 2 or more elements can be exactly the same, and with the two sets defined as they are defined in TFA, that is one "less than" and one "greater or equal", then I'd argue that the second set will be bigger than the former.

In the pathological case where all the elements are the same value, one set will always be empty and the algorithm will not even terminate.

In a less extreme case where nearly all the items are the same except a few ones, then the algorithm will slowly advance, but not with the progression n, n/2, n/4, etc. that is needed to prove it's O(n).

Please note that the "less extreme case" I depicted above is quite common in significant real-world statistics. For example, how many times a site is visited by unique users per day: a long sequence of 1s with some sparse numbers>1. Or how many children/cars/pets per family: many repeated small numbers with a few sparse outliers. Etc.