Java Program to find whether a no is power of two

Given a positive integer, write a function to find if it is a power of two or not.

Examples :

Input : n = 4
Output : Yes
22 = 4

Input : n = 7
Output : No

Input : n = 32
Output : Yes
25 = 32

1. A simple method for this is to simply take the log of the number on base 2 and if you get an integer then number is power of 2. 

Java

// Java Program to find whether a
// no is power of two
class GFG {
    /* Function to check if x is power of 2*/
    static boolean isPowerOfTwo(int n)
    {
        return (int)(Math.ceil((Math.log(n) / Math.log(2))))
            == (int)(Math.floor(((Math.log(n) / Math.log(2)))));
    }

    // Driver Code
    public static void main(String[] args)
    {
        if (isPowerOfTwo(31))
            System.out.println("Yes");
        else
            System.out.println("No");

        if (isPowerOfTwo(64))
            System.out.println("Yes");
        else
            System.out.println("No");
    }
}

// This code is contributed by mits

Output

No
Yes

Time Complexity: O(log₂n)
Auxiliary Space: O(1)

2. Another solution is to keep dividing the number by two, i.e, do n = n/2 iteratively. In any iteration, if n%2 becomes non-zero and n is not 1 then n is not a power of 2. If n becomes 1 then it is a power of 2.

Java

// Java program to find whether
// a no is power of two
import java.io.*;

class GFG {

    // Function to check if
    // x is power of 2
    static boolean isPowerOfTwo(int n)
    {
        if (n == 0)
            return false;

        while (n != 1) {
            if (n % 2 != 0)
                return false;
            n = n / 2;
        }
        return true;
    }

    // Driver program
    public static void main(String args[])
    {
        if (isPowerOfTwo(31))
            System.out.println("Yes");
        else
            System.out.println("No");

        if (isPowerOfTwo(64))
            System.out.println("Yes");
        else
            System.out.println("No");
    }
}

// This code is contributed by Nikita tiwari.

Output

No
Yes

Time Complexity: O(log2n)
Auxiliary Space: O(1)

3. All power of two numbers have only one bit set. So count the no. of set bits and if you get 1 then number is a power of 2. Please see Count set bits in an integer for counting set bits.
4. If we subtract a power of 2 numbers by 1 then all unset bits after the only set bit become set; and the set bit become unset.
For example for 4 ( 100) and 16(10000), we get following after subtracting 1 
3 –> 011 
15 –> 01111
So, if a number n is a power of 2 then bitwise & of n and n-1 will be zero. We can say n is a power of 2 or not based on value of n&(n-1). The expression n&(n-1) will not work when n is 0. To handle this case also, our expression will become n& (!n&(n-1)) (thanks to https://www.geeksforgeeks.org/program-to-find-whether-a-no-is-power-of-two/Mohammad for adding this case). 

Below is the implementation of this method.  

Java

// Java program to efficiently
// check for power for 2

class Test {
    /* Method to check if x is power of 2*/
    static boolean isPowerOfTwo(int x)
    {
        /* First x in the below expression is
        for the case when x is 0 */
        return x != 0 && ((x & (x - 1)) == 0);
    }

    // Driver method
    public static void main(String[] args)
    {
        System.out.println(isPowerOfTwo(31) ? "Yes" : "No");
        System.out.println(isPowerOfTwo(64) ? "Yes" : "No");
    }
}
// This program is contributed by Gaurav Miglani

Output

No
Yes

Time Complexity: O(1)
Auxiliary Space: O(1)

Please refer complete article on Program to find whether a no is power of two for more details!