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# Power of Four LeetCode Solution

## Problem – Power of Four

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?

### Power of Four LeetCode Solution in Java

``````    public boolean isPowerOfFour(int num) {
return num > 0 && (num&(num-1)) == 0 && (num & 0x55555555) != 0;
//0x55555555 is to get rid of those power of 2 but not power of 4
//so that the single 1 bit always appears at the odd position
}``````

### Power of Four LeetCode Solution in C++

``````bool isPowerOfFour(int num) {
return num > 0 && (num & (num - 1)) == 0 && (num - 1) % 3 == 0;
}``````

### Power of Four LeetCode Solution in Python

``````def isPowerOfFour(self, num):
return num != 0 and num &(num-1) == 0 and num & 1431655765== num``````
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