This notebook was prepared by Donne Martin. Source and license info is on GitHub.

Solution Notebook

Problem: Determine if a number is a power of two.

Constraints

  • Is the input number an int?
    • Yes
  • Can we assume the inputs are valid?
    • No
  • Is the output a boolean?
    • Yes
  • Can we assume this fits memory?
    • Yes

Test Cases

  • None -> TypeError
  • 0 -> False
  • 1 -> True
  • 2 -> True
  • 15 -> False
  • 16 -> True

Algorithm

We can use bit manipulation to determine if a number is a power of two.

For a number to be a power of two, there must only be one bit that is a 1.

We can use the following bit manipulation trick to determine this:

n & (n - 1)

Here's an example why:

0000 1000 = n
0000 0001 = 1
0000 0111 = n-1

0000 1000 = n
0000 0111 = n-1
0000 0000 = n & n-1, result = 0

Complexity:

  • Time: O(1)
  • Space: O(1)

Code

In [1]:
class Solution(object):

    def is_power_of_two(self, n):
        if n is None:
            raise TypeError('n cannot be None')
        if n <= 0:
            return False
        return (n & (n - 1)) == 0

Unit Test

In [2]:
%%writefile test_is_power_of_two.py
import unittest


class TestSolution(unittest.TestCase):

    def test_is_power_of_two(self):
        solution = Solution()
        self.assertRaises(TypeError, solution.is_power_of_two, None)
        self.assertEqual(solution.is_power_of_two(0), False)
        self.assertEqual(solution.is_power_of_two(1), True)
        self.assertEqual(solution.is_power_of_two(2), True)
        self.assertEqual(solution.is_power_of_two(15), False)
        self.assertEqual(solution.is_power_of_two(16), True)
        print('Success: test_is_power_of_two')


def main():
    test = TestSolution()
    test.test_is_power_of_two()


if __name__ == '__main__':
    main()
Overwriting test_is_power_of_two.py
In [3]:
%run -i test_is_power_of_two.py
Success: test_is_power_of_two