honestly you could say you opened my eyes. xD
my ansewr is way simpler compared to this xD

I mean the best way to do it wound be O(nlog(n)) which would be a hastle to implement (or you use fancy libraries as other solutions do);

Even though I prefer a branchless solution, this code is great to let the compiler optimise things. Good job.

Nice way of checking for odd numbers, but the complexity of this algorithm is O(n^2) which is far from optimal (though it is space effecient).

I think the problem is that it would return 1 for <2,1,1>, which is wrong. It's always wrong if the odd one is in the first position.

But we have to return 2 for {2,1,1}, don't we?

             [1,2,2]   this way will return 2 ?

   <2,1,1> you will return 2 !

wrong