# Solving Leetcode Interviews in Seconds with AI: Power of Two


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "231" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like [Chatmagic](https://www.chatmagic.app), we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

	# Problem Statement
	> 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?

	# Explanation
	Here's the breakdown of the solution and the Python code:

*   **Approach:**
    *   Handle non-positive numbers: Powers of two are always positive, so immediately return `False` if `n` is zero or negative.
    *   Bitwise AND: A power of two has only one bit set to 1 in its binary representation. The expression `n & (n - 1)` will be zero if and only if `n` is a power of two. This works because subtracting 1 from a power of two flips all the bits from the rightmost set bit to the end.

*   **Complexity:**
    *   Runtime: O(1)
    *   Storage: O(1)

	
	# Code
	```python
	class Solution:
    def isPowerOfTwo(self, n: int) -> bool:
        if n <= 0:
            return False
        return (n & (n - 1)) == 0
	```