Power of three

Time: O(1); Space: O(1); easy

Given an integer, write a function to determine if it is a power of three.

Example 1:

Input: num = 27

Output: True

Example 2:

Input: num = 0

Output: False

Example 3:

Input: num = 9

Output: True

Example 4:

Input: num = 45

Output: False

Follow up:

  1. Could you do it without using any loop / recursion?

1. Loop Iteration [O(log[b]N), O(1)]

2. Base Conversion [O(log[3]N, O(log[3]N]

3. Mathematics [Unknown, O(1)]

4. Integer Limitations [O(1), O(1)]

See also: https://leetcode.com/problems/power-of-three/solution

[2]:

import math

class Solution1(object):
    def __init__(self):
        # print(int(math.log(0x7fffffff)))  # 21
        # print(int(math.log(3)))           # 1
        self.__max_log3 = int(math.log(0x7fffffff) // math.log(3))
        self.__max_pow3 = 3 ** self.__max_log3

    def isPowerOfThree(self, num) -> bool:
        """
        :type num: int
        :rtype: bool
        """
        return num > 0 and self.__max_pow3 % num == 0

[3]:

s = Solution1()
num = 27
assert s.isPowerOfThree(num) == True
num = 0
assert s.isPowerOfThree(num) == False
num = 9
assert s.isPowerOfThree(num) == True
num = 45
assert s.isPowerOfThree(num) == False