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# Power of Two LeetCode Solution

## 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 == 2x`.

Example 1:

``````Input: n = 1
Output: true
Explanation: 20 = 1
``````

Example 2:

``````Input: n = 16
Output: true
Explanation: 24 = 16
``````

Example 3:

``````Input: n = 3
Output: false
``````

Constraints:

• `-231 <= n <= 231 - 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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