## Problem – Power of Two LeetCode Solution

Given an integer `n`

, return `true`

if it is a power of two. Otherwise, return `false`

.

An integer `n`

is a power of two, if there exists an integer `x`

such that `n == 2`^{x}

.

**Example 1:**

**Input:** n = 1
**Output:** true
**Explanation: **2^{0} = 1

**Example 2:**

**Input:** n = 16
**Output:** true
**Explanation: **2^{4} = 16

**Example 3:**

**Input:** n = 3
**Output:** false

**Constraints:**

`-2`^{31} <= n <= 2^{31} - 1

**Follow up:** Could you solve it without loops/recursion?

## Power of Two LeetCode Solution in Python

```
class Solution(object):
def isPowerOfTwo(self, n):
return n and not (n & n - 1)
```

## Power of Two LeetCode Solution in Java

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

## Power of Two LeetCode Solution in C++

```
class Solution {
public:
bool isPowerOfTwo(int n) {
return n > 0 && not (n & n - 1);
}
};
```

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