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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 $O(1)$.

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class Solution:
    def isPowerOfFour(self, n: int) -> bool:
        return n > 0 and (n & (n - 1)) == 0 and (n & 0xAAAAAAAA) == 0

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class Solution {
    public boolean isPowerOfFour(int n) {
        return n > 0 && (n & (n - 1)) == 0 && (n & 0xaaaaaaaa) == 0;
    }
}

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class Solution {
public:
    bool isPowerOfFour(int n) {
        return n > 0 && (n & (n - 1)) == 0 && (n & 0xaaaaaaaa) == 0;
    }
};

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func isPowerOfFour(n int) bool {
    return n > 0 && (n&(n-1)) == 0 && (n&0xaaaaaaaa) == 0
}

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function isPowerOfFour(n: number): boolean {
    return n > 0 && (n & (n - 1)) == 0 && (n & 0xaaaaaaaa) == 0;
}

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/**
 * @param {number} n
 * @return {boolean}
 */
var isPowerOfFour = function (n) {
    return n > 0 && (n & (n - 1)) == 0 && (n & 0xaaaaaaaa) == 0;
};

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