When N is a power of two, it has only one bit set to 1, which is the leftmost bit. Now, if we subtract 1 from Kth power of two, we get a number where all bits from position 0 up to the position (but not including) position k are set to 1 and the kth bit is 0.

This is because subtracting 1 from a number flips all the bits from the rightmost set bit to the least significant bit. Hence, when we perform a bitwise AND operation between N and N-1, all bits become 0.

Code

#include <bits/stdc++.h>
using namespace std;

// Solution class
class Solution {
public:
    // Function to check if a number is power of two
    bool isPowerOfTwo(int n) {
        // Return false if n is less than or equal to 0
        if (n <= 0) return false;

        // Check if n has only one set bit
        return (n & (n - 1)) == 0;
    }
};

// Main function
int main() {
    int n = 16;

    // Create Solution object
    Solution sol;

    // Call the function and print result
    if (sol.isPowerOfTwo(n))
        cout << "True\n";
    else
        cout << "False\n";

    return 0;
}