Austin Powers in Python

Suppose we have a number greater than 0, we have to check whether the number is a power of two or not. A power of two is any number that can be expressed as 2ⁿ where n is a non-negative integer (1, 2, 4, 8, 16, 32, etc.).

So, if the input is like 1024, then the output will be True because 1024 = 2¹⁰.

Approach

To solve this, we will follow these steps ?

• While n > 1, do

  • n := n / 2

• Return true when n is same as 1, otherwise false

Let us see the following implementation to get better understanding ?

Example

class Solution:
    def solve(self, n):
        while n > 1:
            n /= 2
        return n == 1

ob = Solution()
print(ob.solve(1024))
print(ob.solve(18))
print(ob.solve(16))

True
False
True

Alternative Approach Using Bit Operations

A more efficient way is to use the bit manipulation trick. A power of two has only one bit set in its binary representation ?

def is_power_of_two(n):
    if n <= 0:
        return False
    return (n & (n - 1)) == 0

# Test with different numbers
numbers = [1, 2, 4, 8, 16, 32, 64, 1024, 18, 100]

for num in numbers:
    result = is_power_of_two(num)
    print(f"{num} is power of two: {result}")

1 is power of two: True
2 is power of two: True
4 is power of two: True
8 is power of two: True
16 is power of two: True
32 is power of two: True
64 is power of two: True
1024 is power of two: True
18 is power of two: False
100 is power of two: False

How It Works

The bit manipulation approach works because:

• Powers of two have exactly one bit set: 8 = 1000, 16 = 10000

• When we subtract 1, all bits after the set bit become 1: 8-1 = 0111

• The AND operation between n and (n-1) gives 0 for powers of two

Comparison

Conclusion

Use the bit manipulation approach n & (n-1) == 0 for the most efficient solution. The division method is easier to understand but slower for large numbers.