Problem: Find the second rightmost bit in a number (1 line code)

Presented with the integer n, find the 0-based position of the second rightmost zero bit in its binary representation (it is guaranteed that such a bit exists), counting from right to left.

Return the value of 2position_of_the_found_bit.


For n = 37, the output should be
secondRightmostZeroBit(n) = 8.

3710 = 1001012. The second rightmost zero bit is at position 3 (0-based) from the right in the binary representation of n.
Thus, the answer is 23 = 8.

def secondRightmostZeroBit(n):
    return 2**(len(bin(n)[2:])-1-bin(n)[2:].rfind('0',0,len(bin(n)[2:])-(len(bin(n)[2:]) - bin(n)[2:].rfind('0'))))