[menaerus]

> It is not O(N^4/3), it is O(N * M^1/3). There's a good argument to be made that M^(1/3) should be considered a constant

Math isn't mathing here. If M^1/3 >> N, like in memory-bound algorithms, then why should we consider it a constant?

> The point of Big-O is to have some reasonable understanding of how much worse it will get if you need to run this algorithm on 10x or 100x as much data

And this also isn't true and can be easily proved by contradiction. O(N) linear search over array is sometimes faster than O(1) search in a hash-map.