Description
Given an integer n, return true if it is a power of four. Otherwise, return false.
An integer n is a power of four, if there exists an integer x such that n == 4x.
Example 1:
Input: n = 16 Output: true
Example 2:
Input: n = 5 Output: false
Example 3:
Input: n = 1 Output: true
Constraints:
-231 <= n <= 231 - 1
Follow up: Could you solve it without loops/recursion?
Solutions
Solution 1: Bit Manipulation
If a number is a power of 4, then it must be greater than 0. Suppose this number is 4^x, which is 2^{2x}. Therefore, its binary representation has only one 1, and this 1 appears at an even position.
First, we check if the number is greater than 0. Then, we verify if the number is 2^{2x} by checking if the bitwise AND of n and n-1 is 0. Finally, we check if the 1 appears at an even position by verifying if the bitwise AND of n and 0xAAAAAAAA is 0. If all three conditions are met, then the number is a power of 4.
The time complexity is O(1), and the space complexity is O(1).
class Solution: def isPowerOfFour(self, n: int) -> bool: return n > 0 and (n & (n - 1)) == 0 and (n & 0xAAAAAAAA) == 0(code-box)
