Power of Two

This repository contains a solution to the Power of Two problem, where the goal is to determine if a given integer is a power of two. The problem is solved using efficient bitwise operations in C++.

Problem Description

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 Input and Output

Input	Output	Explanation
n = 1	true	( 2^0 = 1 )
n = 16	true	( 2^4 = 16 )
n = 3	false	3 is not a power of two

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Constraints

• ( -2^{31} \leq n \leq 2^{31} - 1 )

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Solution

Approach

The solution leverages the following properties of numbers that are powers of two:

1. A power of two has exactly one bit set in its binary representation.
2. The bitwise operation ( n & (n - 1) = 0 ) holds true only for powers of two.

Algorithm

1. If n <= 0, return false (negative numbers and zero are not powers of two).
2. Use the condition ( n & (n - 1) == 0 ) to determine if n is a power of two.

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Implementation

C++ Solution

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

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Usage

Input

Provide an integer ( n ) as input.

Output

The function returns true if the input is a power of two; otherwise, it returns false.

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Test Cases

Test Case	Expected Output
1	true
16	true
3	false
0	false
-2	false

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Complexity

Time Complexity

• ( O(1) ): The bitwise operation is performed in constant time.

Space Complexity

• ( O(1) ): No extra space is used.

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License

This repository is available under the MIT License.