231. Power of Two

EasyLeetCode

Given an integer n, return true if it is a power of two. Otherwise, return false.

An integer n is a power of two, if there exists an integer x such that n == 2^x.

Example 1:

Input: n = 1

Output: true

Explanation: 2⁰ = 1

Example 2:

Input: n = 16

Output: true

Explanation: 2⁴ = 16

Example 3:

Input: n = 3

Output: false

Constraints:

-2³¹ <= n <= 2³¹ - 1

How to solve the problem

1. Using while loop.

2. Using logarithm, if a number is a power of 2, then its log₂(n) must be an integer.

Code

• Loop

class Solution:
    def isPowerOfTwo(self, n: int) -> bool:
        if n <= 0: return False
        while n > 1:
            if n % 2 != 0: return False
            n //= 2
        return True

• Logarithm

class Solution:
    def isPowerOfTwo(self, n: int) -> bool:
        if n <= 0: return False
        return math.log2(n).is_integer()

Complexity

Time complexity: O(log n)

The number of iterations is equal to the number of digits in n, which is ⌊log₁₀(n)⌋ + 1.

Space complexity: O(1)