I know this isn't supposed to necessarily be an efficient solution, but - the result of arr[1:][::2] is equivalent to arr[1::2] and arr[0:][::2] is equivalent to arr[::2]. And if I'm not mistaken the latter options are also more efficient as you're not having to take two separate slices each time. Also inside your sum you can just use a generator rather than a list comprehension, ie. you don't need the surrounding [].

It's called an absolute majority.

Shortest O(n) solution that I can think of. Using c.update([x]) inside a list comprehension is a bit horrid though!

It'd be better to replace all([(n ** .5) in array1 for n in array2]) with all((n ** .5) in array1 for n in array2). The first unnecessarily constructs the whole list of booleans before passing it to all. The second uses a generator expression, which means it gets the booleans lazily. If the the first value is False, then all will return False without ever bothering to get the rest of the values.

Only just had a chance to look at this - looks like the ideal solution. Thanks for the recommendation on iterators - I haven't used itertools all that much so forget about how useful it can be.

Yes, Unnamed, you're entirely correct and this definitely should come under 'clever' rather than 'best practice'.

You're correct. I used just [None]*len(keys) for the sake of brevity, but ideally you'd use [None]*(len(keys) - len(values)) in practice.