PowerOfThree

Problem

https://leetcode.com/problems/power-of-three/

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

An integer n is a power of three, if there exists an integer x such that n == 3:sup:`x`.

Example 1:

Input: n = 27
Output: true
Explanation: 27 = 33

Example 2:

Input: n = 0
Output: false
Explanation: There is no x where 3x = 0.

Example 3:

Input: n = -1
Output: false
Explanation: There is no x where 3x = (-1).

Constraints:

• -2:sup:`31`<= n <= 2:sup:`31`- 1

Follow up: Could you solve it without loops/recursion?

Solution

Note that, if \(n = 3^x\), \(n\) can never be negative. For positive \(n`\), check if \(n\) is divisble by 3. If it is, divide \(n\) by 3 and check again. If it is not, then \(n\) is not a power of 3.

Pattern

Math, Recursion

Code

def isPowerOfThree(n: int) -> bool:
    """Check if ``n`` is a power of 3.
    """
    if n <= 0:
        return False
    elif n == 1:
        return True
    elif n % 3 == 0:
        return isPowerOfThree(n / 3)
    else:
        return False

Test

>>> from PowerOfThree import isPowerOfThree
>>> isPowerOfThree(27)
True
>>> isPowerOfThree(0)
False
>>> isPowerOfThree(-1)
False

PowerOfThree.isPowerOfThree(n: int) → bool

Check if n is a power of 3.