**Solution**

**Power of Two** https://leetcode.com/problems/power-of-two/

**Algorithm 1**

- n & (n - 1) drops the** lowest set bit.** It's a neat little bit trick. https://discuss.leetcode.com/topic/20120/c-solution-n-n-1/2

Let's use n = 00101100 as an example. This binary representation has three 1s.

- If n = 00101100, then n - 1 = 00101011, so n & (n - 1) = 00101100 & 00101011 = 00101000. Count = 1.
- If n = 00101000, then n - 1 = 00100111, so n & (n - 1) = 00101000 & 00100111 = 00100000. Count = 2.
- If n = 00100000, then n - 1 = 00011111, so n & (n - 1) = 00100000 & 00011111 = 00000000. Count = 3.
- n is now zero, so the while loop ends, and the final count (the numbers of set bits) is returned.
- Test for power of two:
`(n & (n-1)) == 0`

```
class Solution(object):
def isPowerOfTwo(self, n):
"""
:type n: int
:rtype: bool
"""
if n <= 0:
return False
return (n & (n-1)) == 0
```