I assume most of you are familiar with the ancient legend of the rice (but I see wikipedia suggests wheat, for some reason) problem, but a quick recap for you: a young man asks as a compensation only 1 grain of rice for the first square, 2 grains for the second, 4 for the third, 8 for the fourth and so on, always doubling the previous.
Your task is pretty straightforward (but not necessarily easy): given an amount of grains, you need to return up to which square of the chessboard one should count in order to get at least as many.
As usual, a few examples might be way better than thousands of words from me:
squaresNeeded(0) === 0
squaresNeeded(1) === 1
squaresNeeded(2) === 2
squaresNeeded(3) === 2
squaresNeeded(4) === 3
Input is always going to be valid/reasonable: ie: a non negative number; extra cookie for not using a loop to compute square-by-square (at least not directly) and instead trying a smarter approach [hint: some peculiar operator]; a trick converting the number might also work: impress me!
for (let i=0;i