Suppose we have a number n. We have to check whether this is power of 2 or not.
So, if the input is like n = 2048, then the output will be True as 2048 is 2^11.
To solve this, we will follow these steps −
if n is same as 0, then
return true when (n AND (n - 1)) is same as 0 otherwise false
Let us see the following implementation to get better understanding
def solve(n): if n == 0: return False return (n & (n - 1)) == 0 n = 2048 print(solve(n))