The optimal algorithm will scan the input array once; indeed, you can show by a standard adversary argument that no faster algorithm exists.

Your current solution does two passes over the input array (the two invocations of `np.where`) which as stated seems wasteful. Given that we know there's a linear lower bound, the natural question is: how do we solve the problem in a *single* pass? The answer is to use `np.argmax`:

    import numpy as np
    arr = np.array([-1, -2, -3, 4, 5, 6])
    np.argmax(arr > 0) # Return (index) 3

At this point, if we can assume (as you seem to imply) that there is a single position of interest (and/or that we only care about the first one, read from left to right) we are done: just decrement one from the answer and you have the other cut-off point (but do check whether we are at a corner case so as not to return a non-valid index).