342. Power of Four
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 $\textit{0xAAAAAAAA}$ 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 $O(1)$.
Python3
class Solution:
def isPowerOfFour(self, n: int) -> bool:
return n > 0 and (n & (n - 1)) == 0 and (n & 0xAAAAAAAA) == 0
Java
class Solution {
public boolean isPowerOfFour(int n) {
return n > 0 && (n & (n - 1)) == 0 && (n & 0xaaaaaaaa) == 0;
}
}
C++
class Solution {
public:
bool isPowerOfFour(int n) {
return n > 0 && (n & (n - 1)) == 0 && (n & 0xaaaaaaaa) == 0;
}
};
Go
func isPowerOfFour(n int) bool {
return n > 0 && (n&(n-1)) == 0 && (n&0xaaaaaaaa) == 0
}
TypeScript
function isPowerOfFour(n: number): boolean {
return n > 0 && (n & (n - 1)) == 0 && (n & 0xaaaaaaaa) == 0;
}
JavaScript
/**
* @param {number} n
* @return {boolean}
*/
var isPowerOfFour = function (n) {
return n > 0 && (n & (n - 1)) == 0 && (n & 0xaaaaaaaa) == 0;
};