[BearsAreCool]

With a linear time O(N) first pass you can identify the minimum element in the list. This reduces M from being the largest element to being the difference between the largest and the smallest element.

[rileymat2]

If we are playing with this, why not take one pass to find the largest element, then another pass to divide each element in the list by that number, leaving it 3N+M/M.

Edit:

Why divide by m, when you can divide by M*C. Where C will speed this portion up by a factor!

[david_allison]

Let's do better.

You can find the largest element in a list in O(1) time with n^2 processors[0]*.

Division of the list by that number is O(1) with n processors

Therefore the operation is O(1)

[0]* on a CRCW PRAM https://www.cpp.edu/~gsyoung/CS535/CS535Notes/Part2PRAM.pdf#...

[lupire]

You're ignoring communication overhead.