Python Program to Check if Number is Power of 2

You can check if a number is a power of 2 in Python using n > 0 and (n & (n - 1)) == 0, which returns True if n is a power of 2.

Examples

Input16

OutputTrue

Input18

OutputFalse

Input1

OutputTrue

How to Think About It

To check if a number is a power of 2, think about its binary form. A power of 2 has exactly one '1' bit in binary. For example, 8 is 1000 in binary. If you subtract 1 from it, you get 7, which is 0111. When you do a bitwise AND between the number and one less than it, the result is zero only if the number is a power of 2.

Algorithm

1

Get the input number n.

2

Check if n is greater than zero.

3

Perform bitwise AND between n and n-1.

4

If the result is zero, return True (n is power of 2).

5

Otherwise, return False.

Code

python

def is_power_of_two(n):
    return n > 0 and (n & (n - 1)) == 0

# Test examples
print(is_power_of_two(16))  # True
print(is_power_of_two(18))  # False
print(is_power_of_two(1))   # True

Output

True
False
True

Dry Run

Let's trace the number 16 through the code

1

Check if n > 0

n = 16, which is greater than 0, so continue

2

Calculate n & (n - 1)

16 in binary is 10000, 15 in binary is 01111 Bitwise AND: 10000 & 01111 = 00000

3

Compare result to zero

Result is 0, so return True

n	n-1	n & (n-1)	Result
16	15	0	True

Why This Works

Step 1: Check positive number

The number must be greater than zero because powers of 2 are positive.

Step 2: Bitwise AND operation

Using n & (n - 1) clears the lowest set bit. If the result is zero, only one bit was set.

Step 3: Result means power of 2

If the bitwise AND result is zero, the number has exactly one '1' bit, so it is a power of 2.

Alternative Approaches

Using math.log2 and is_integer

python

import math
def is_power_of_two(n):
    return n > 0 and math.log2(n).is_integer()

print(is_power_of_two(16))  # True
print(is_power_of_two(18))  # False

This method uses logarithms but can be slower and less precise for large numbers.

Using a loop to divide by 2

python

def is_power_of_two(n):
    if n <= 0:
        return False
    while n % 2 == 0:
        n //= 2
    return n == 1

print(is_power_of_two(16))  # True
print(is_power_of_two(18))  # False

This method is simple but slower for large numbers because it uses a loop.

Complexity: O(1) time, O(1) space

Time Complexity

The bitwise operation runs in constant time because it uses a fixed number of CPU instructions.

Space Complexity

No extra memory is needed; the check is done in-place with constant space.

Which Approach is Fastest?

The bitwise method is fastest and simplest compared to logarithm or loop methods.

Approach	Time	Space	Best For
Bitwise AND	O(1)	O(1)	Fastest and simplest
Logarithm	O(1)	O(1)	Readable but slower, may have precision issues
Loop division	O(log n)	O(1)	Easy to understand, slower for large numbers