## 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 == 4`^{x}

.

**Example 1:**

```
Input: n = 16
Output: true
```

**Example 2:**

```
Input: n = 5
Output: false
```

**Example 3:**

```
Input: n = 1
Output: true
```

**Constraints:**

`-2`^{31} <= n <= 2^{31} - 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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