Program to find whether a given number is power of 2

Given a positive integer n, write a function to find if it is a power of 2 or not

Examples:

Input : n = 4
Output : Yes
Explanation: 2² = 4

Input : n = 32
Output : Yes
Explanation: 2⁵ = 32

To solve the problem follow the below idea:

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

Below is the implementation of the above approach:

C++

// C++ Program to find whether a
// no is a power of two
#include <bits/stdc++.h>
using namespace std;

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

    return (ceil(log2(n)) == floor(log2(n)));
}

// Driver code
int main()
{
    // Function call
    isPowerOfTwo(31) ? cout << "Yes" << endl
                     : cout << "No" << endl;
    isPowerOfTwo(64) ? cout << "Yes" << endl
                     : cout << "No" << endl;

    return 0;
}

// This code is contributed by Surendra_Gangwar

C

// C Program to find whether a
// no is power of two
#include <math.h>
#include <stdbool.h>
#include <stdio.h>

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

    return (ceil(log2(n)) == floor(log2(n)));
}

// Driver code
int main()
{
    // Function call
    isPowerOfTwo(31) ? printf("Yes\n") : printf("No\n");
    isPowerOfTwo(64) ? printf("Yes\n") : printf("No\n");
    return 0;
}

// This code is contributed by bibhudhendra

Java

// Java Program to find whether a
// no is power of two
import java.lang.Math;

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

        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)
    {
        // Function call
        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

Python3

# Python3 Program to find
# whether a no is
# power of two
import math

# Function to check
# Log base 2


def Log2(x):
    if x == 0:
        return false

    return (math.log10(x) /
            math.log10(2))

# Function to check
# if x is power of 2


def isPowerOfTwo(n):
    return (math.ceil(Log2(n)) ==
            math.floor(Log2(n)))


# Driver Code
if __name__ == "__main__":

    # Function call
    if(isPowerOfTwo(31)):
        print("Yes")
    else:
        print("No")

    if(isPowerOfTwo(64)):
        print("Yes")
    else:
        print("No")

# This code is contributed
# by mits

C#

// C# Program to find whether
// a no is power of two
using System;

class GFG {

    /* Function to check if
       x is power of 2*/
    static bool isPowerOfTwo(int n)
    {

        if (n == 0)
            return false;

        return (int)(Math.Ceiling(
                   (Math.Log(n) / Math.Log(2))))
            == (int)(Math.Floor(
                ((Math.Log(n) / Math.Log(2)))));
    }

    // Driver Code
    public static void Main()
    {

        // Function call
        if (isPowerOfTwo(31))
            Console.WriteLine("Yes");
        else
            Console.WriteLine("No");

        if (isPowerOfTwo(64))
            Console.WriteLine("Yes");
        else
            Console.WriteLine("No");
    }
}

// This code is contributed
// by Akanksha Rai(Abby_akku)

PHP

<?php
// PHP Program to find
// whether a no is
// power of two

// Function to check
// Log base 2
function Log2($x)
{
    return (log10($x) /
            log10(2));
}


// Function to check
// if x is power of 2
function isPowerOfTwo($n)
{
    return (ceil(Log2($n)) ==
            floor(Log2($n)));
}

// Driver Code

// Function call
if(isPowerOfTwo(31))
echo "Yes\n";
else
echo "No\n";

if(isPowerOfTwo(64))
echo "Yes\n";
else
echo "No\n";

// This code is contributed
// by Sam007
?>

Javascript

<script>
// javascript Program to find whether a
// no is power of two

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

        return parseInt( (Math.ceil((Math.log(n) / Math.log(2))))) == parseInt( (Math.floor(((Math.log(n) / Math.log(2))))));
    }

    // Driver Code

    if (isPowerOfTwo(31))
        document.write("Yes<br/>");
    else
        document.write("No<br/>");

    if (isPowerOfTwo(64))
        document.write("Yes<br/>");
    else
        document.write("No<br/>");

// This code is contributed by shikhasingrajput.
</script>

Output

No
Yes

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

Find whether a given number is a power of 2 using the division operator:

To solve the problem follow the below idea:

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. 

Below is the implementation of the above approach:

C++

// C++ program for the above approach

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

/* Function to check if x is power of 2*/
bool isPowerOfTwo(int n)
{
    if (n == 0)
        return 0;
    while (n != 1) {
        if (n % 2 != 0)
            return 0;
        n = n / 2;
    }
    return 1;
}

// Driver code
int main()
{
    // Function call
    isPowerOfTwo(31) ? cout << "Yes\n" : cout << "No\n";
    isPowerOfTwo(64) ? cout << "Yes\n" : cout << "No\n";
    return 0;
}

// This code is contributed by rathbhupendra

C

// C program for the above approach

#include <stdbool.h>
#include <stdio.h>

/* Function to check if x is power of 2*/
bool isPowerOfTwo(int n)
{
    if (n == 0)
        return 0;
    while (n != 1) {
        if (n % 2 != 0)
            return 0;
        n = n / 2;
    }
    return 1;
}

// Driver code
int main()
{
    // Function call
    isPowerOfTwo(31) ? printf("Yes\n") : printf("No\n");
    isPowerOfTwo(64) ? printf("Yes\n") : printf("No\n");
    return 0;
}

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 code
    public static void main(String args[])
    {

        // Function call
        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.

Python3

# Python program to check if given
# number is power of 2 or not

# Function to check if x is power of 2


def isPowerOfTwo(n):
    if (n == 0):
        return False
    while (n != 1):
        if (n % 2 != 0):
            return False
        n = n // 2

    return True


# Driver code
if __name__ == "__main__":

    # Function call
    if(isPowerOfTwo(31)):
        print('Yes')
    else:
        print('No')
    if(isPowerOfTwo(64)):
        print('Yes')
    else:
        print('No')

# This code is contributed by Danish Raza

C#

// C# program to find whether
// a no is power of two
using System;

class GFG {

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

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

            n = n / 2;
        }
        return true;
    }

    // Driver code
    public static void Main()
    {
        // Function call
        Console.WriteLine(isPowerOfTwo(31) ? "Yes" : "No");
        Console.WriteLine(isPowerOfTwo(64) ? "Yes" : "No");
    }
}

// This code is contributed by Sam007

PHP

<?php
// php program for the above approach

// Function to check if
// x is power of 2
function isPowerOfTwo($n)
{
if ($n == 0)
    return 0;
while ($n != 1)
{
    if ($n % 2 != 0)
        return 0;
    $n = $n / 2;
}
return 1;
}

// Driver Code

// Function call
if(isPowerOfTwo(31))
    echo "Yes\n";
else
    echo "No\n";

if(isPowerOfTwo(64))
    echo "Yes\n";
else
    echo "No\n";

// This code is contributed
// by Sam007
?>

Javascript

<script>

    /* Function to check if x is power of 2*/
    function isPowerOfTwo(n)
    {
        if (n == 0)
            return 0;
        while (n != 1)
        {
            if (n%2 != 0)
                return 0;
            n = n/2;
        }
        return 1;
    }

    isPowerOfTwo(31)? document.write("Yes" + "</br>"): document.write("No" + "</br>");
    isPowerOfTwo(64)? document.write("Yes"): document.write("No");

</script>

Output

No
Yes

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

Below is the recursive implementation of the above approach:

C++

// C++ program for above approach
#include <bits/stdc++.h>
using namespace std;

// Function which checks whether a number is a power of 2
bool powerOf2(int n)
{
    // base cases
    // '1' is the only odd number which is a power of 2(2^0)
    if (n == 1)
        return true;

    // all other odd numbers are not powers of 2
    else if (n % 2 != 0 || n == 0)
        return false;

    // recursive function call
    return powerOf2(n / 2);
}

// Driver Code
int main()
{
    int n = 64; // True
    int m = 12; // False

    // Function call
    if (powerOf2(n) == 1)
        cout << "True" << endl;

    else
        cout << "False" << endl;

    if (powerOf2(m) == 1)
        cout << "True" << endl;

    else
        cout << "False" << endl;
}

// This code is contributed by Aditya Kumar (adityakumar129)

C

// C program for above approach
#include <stdbool.h>
#include <stdio.h>

// Function which checks whether a number is a power of 2
bool powerOf2(int n)
{
    // base cases
    // '1' is the only odd number which is a power of 2(2^0)
    if (n == 1)
        return true;

    // all other odd numbers are not powers of 2
    else if (n % 2 != 0 || n == 0)
        return false;

    // recursive function call
    return powerOf2(n / 2);
}

// Driver Code
int main()
{
    int n = 64; // True
    int m = 12; // False

    // Function call
    if (powerOf2(n) == 1)
        printf("True\n");

    else
        printf("False\n");

    if (powerOf2(m) == 1)
        printf("True\n");

    else
        printf("False\n");
}

// This code is contributed by Aditya Kumar (adityakumar129)

Java

// Java program for
// the above approach
import java.util.*;
class GFG {

    // Function which checks
    // whether a number is a
    // power of 2
    static boolean powerOf2(int n)
    {
        // base cases
        // '1' is the only odd number
        // which is a power of 2(2^0)
        if (n == 1)
            return true;

        // all other odd numbers are
        // not powers of 2
        else if (n % 2 != 0 || n == 0)
            return false;

        // recursive function call
        return powerOf2(n / 2);
    }

    // Driver Code
    public static void main(String[] args)
    {
        // True
        int n = 64;

        // False
        int m = 12;

        // Function call
        if (powerOf2(n) == true)
            System.out.print("True"
                             + "\n");
        else
            System.out.print("False"
                             + "\n");

        if (powerOf2(m) == true)
            System.out.print("True"
                             + "\n");
        else
            System.out.print("False"
                             + "\n");
    }
}

// This code is contributed by Princi Singh

Python3

# Python program for above approach

# function which checks whether a
# number is a power of 2


def powerof2(n):

    # base cases
    # '1' is the only odd number
    # which is a power of 2(2^0)
    if n == 1:
        return True

    # all other odd numbers are not powers of 2
    elif n % 2 != 0 or n == 0:
        return False

    # recursive function call
    return powerof2(n/2)


# Driver Code
if __name__ == "__main__":

        # Function call
    print(powerof2(64))  # True
    print(powerof2(12))  # False

# code contributed by Moukthik a.k.a rowdyninja

C#

// C# program for above approach
using System;

class GFG {

    // Function which checks whether a
    // number is a power of 2
    static bool powerOf2(int n)
    {

        // Base cases
        // '1' is the only odd number
        // which is a power of 2(2^0)
        if (n == 1)
            return true;

        // All other odd numbers
        // are not powers of 2
        else if (n % 2 != 0 || n == 0)
            return false;

        // Recursive function call
        return powerOf2(n / 2);
    }

    // Driver code
    static void Main()
    {

        int n = 64; // True
        int m = 12; // False

        // Function call
        if (powerOf2(n)) {
            Console.Write("True"
                          + "\n");
        }
        else {
            Console.Write("False"
                          + "\n");
        }

        if (powerOf2(m)) {
            Console.Write("True");
        }
        else {
            Console.Write("False");
        }
    }
}

// This code is contributed by rutvik_56

Javascript

<script>

// javascript program for
// the above approach

// Function which checks
// whether a number is a
// power of 2

function powerOf2(n)
{
  // base cases
  // '1' is the only odd number
  // which is a power of 2(2^0)
  if (n == 1)
    return true;

  // all other odd numbers are
  // not powers of 2
  else if (n % 2 != 0 ||
           n ==0)
    return false;

  // recursive function call
  return powerOf2(n / 2);
}

// Driver Code
//True
var n = 64;

//False
var m = 12;

if (powerOf2(n) == true)
  document.write("True" + "\n");
else document.write("False" + "\n");

if (powerOf2(m) == true)
  document.write("True" + "\n");
else
  document.write("False" + "\n");

// This code contributed by shikhasingrajput

</script>

Output

True
False

Time Complexity: O(log N)
Auxiliary Space: O(log N)

Find whether a given number is a power of 2 by checking the count of set bits:

To solve the problem follow the below idea:

All power of two numbers has only a one-bit set. So count the no. of set bits and if you get 1 then the number is a power of 2. Please see Count set bits in an integer for counting set bits.

Below is the implementation of the above approach:

C++

// C++ program of the above approach

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

/* Function to check if x is power of 2*/
bool isPowerOfTwo(int n)
{
    /* First x in the below expression is for the case when
     * x is 0 */
    int cnt = 0;
    while (n > 0) {
        if ((n & 1) == 1) {
            cnt++;
        }
        n = n >> 1; // keep dividing n by 2 using right
                    // shift operator
    }

    if (cnt == 1) { // if cnt = 1 only then it is power of 2
        return true;
    }
    return false;
}

// Driver code
int main()
{
    // Function call
    isPowerOfTwo(31) ? cout << "Yes\n" : cout << "No\n";
    isPowerOfTwo(64) ? cout << "Yes\n" : cout << "No\n";
    return 0;
}

// This code is contributed by devendra salunke

Java

// Java program of the above approach
import java.io.*;

class GFG {

    // Function to check if x is power of 2
    static boolean isPowerofTwo(int n)
    {
        int cnt = 0;
        while (n > 0) {
            if ((n & 1) == 1) {
                cnt++; // if n&1 == 1 keep incrementing cnt
                // variable
            }
            n = n >> 1; // keep dividing n by 2 using right
                        // shift operator
        }
        if (cnt == 1) {
            // if cnt = 1 only then it is power of 2
            return true;
        }
        return false;
    }

    // Driver code
    public static void main(String[] args)
    {
        // Function call
        if (isPowerofTwo(30) == true)
            System.out.println("Yes");
        else
            System.out.println("No");

        if (isPowerofTwo(128) == true)
            System.out.println("Yes");
        else
            System.out.println("No");
    }
}

// This code is contributed by devendra salunke.

C#

// C# program to check for power for 2
using System;

class GFG {

    // Method to check if x is power of 2
    static bool isPowerOfTwo(int n)
    {
        int cnt = 0; // initialize count to 0
        while (n > 0) {

            // run loop till n > 0
            if ((n & 1) == 1) {

                // if n&1 == 1 keep incrementing cnt
                // variable
                cnt++;
            }
            n = n >> 1; // keep dividing n by 2 using right
                        // shift operator
        }

        if (cnt
            == 1) // if cnt = 1 only then it is power of 2
            return true;
        return false;
    }

    // Driver code
    public static void Main()
    {
        // Function call
        Console.WriteLine(isPowerOfTwo(31) ? "Yes" : "No");
        Console.WriteLine(isPowerOfTwo(64) ? "Yes" : "No");
    }
}

// This code is contributed by devendra salunke

Python3

# Python3 program to check if given
# number is power of 2 or not

# Function to check if x is power of 2


def isPowerOfTwo(n):
    cnt = 0
    while n > 0:
        if n & 1 == 1:
            cnt = cnt + 1
        n = n >> 1

    if cnt == 1:
        return 1
    return 0


# Driver code
if __name__ == "__main__":

    # Function call
    if(isPowerOfTwo(31)):
        print('Yes')
    else:
        print('No')

    if(isPowerOfTwo(64)):
        print('Yes')
    else:
        print('No')

# This code is contributed by devendra salunke

Javascript

<script>
    // JavaScript code for the above approach

    // Function to check if x is power of 2
    function isPowerofTwo(n)
    {
        let cnt = 0;
        while (n > 0) {
            if ((n & 1) == 1) {
                cnt++; // if n&1 == 1 keep incrementing cnt
                // variable
            }
            n = n >> 1; // keep dividing n by 2 using right
                        // shift operator
        }
        if (cnt == 1) {
            // if cnt = 1 only then it is power of 2
            return true;
        }
        return false;
    }

    // Driver code

    if (isPowerofTwo(30) == true)
            document.write("Yes" + "<br/>");
        else
            document.write("No" + "<br/>");

        if (isPowerofTwo(128) == true)
            document.write("Yes" + "<br/>");
        else
            document.write("No" + "<br/>");

            // This code is contributed by sanjoy_62.
  </script>

Output

No
Yes

Time complexity: O(log N)
Auxiliary Space: O(1)

Find whether a given number is a power of 2 using the AND(&) operator:

To solve the problem follow the below idea:

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 becomes unset.
For example for 4 ( 100) and 16(10000), we get the 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 the 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 the above approach:

C++

// C++ program for the above approach

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

/* Function to check if x is power of 2*/
bool isPowerOfTwo(int x)
{
    /* First x in the below expression is for the case when
     * x is 0 */
    return x && (!(x & (x - 1)));
}

// Driver code
int main()
{
    // Function call
    isPowerOfTwo(31) ? cout << "Yes\n" : cout << "No\n";
    isPowerOfTwo(64) ? cout << "Yes\n" : cout << "No\n";
    return 0;
}

// This code is contributed by rathbhupendra

C

// C program for the above approach

#include <stdio.h>
#define bool int

/* Function to check if x is power of 2*/
bool isPowerOfTwo(int x)
{
    /* First x in the below expression is for the case when
     * x is 0 */
    return x && (!(x & (x - 1)));
}

// Driver code
int main()
{
    // Function call
    isPowerOfTwo(31) ? printf("Yes\n") : printf("No\n");
    isPowerOfTwo(64) ? printf("Yes\n") : printf("No\n");
    return 0;
}

Java

// Java program for the above approach

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

Python3

# Python3 program for the above approach

# Function to check if x is power of 2


def isPowerOfTwo(x):

    # First x in the below expression
    # is for the case when x is 0
    return (x and (not(x & (x - 1))))


# Driver code
if __name__ == "__main__":

    # Function call
    if(isPowerOfTwo(31)):
        print('Yes')
    else:
        print('No')

    if(isPowerOfTwo(64)):
        print('Yes')
    else:
        print('No')

# This code is contributed by Danish Raza

C#

// C# program for the above approach
using System;

class GFG {
    // Method to check if x is power of 2
    static bool 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 code
    public static void Main()
    {
        // Function call
        Console.WriteLine(isPowerOfTwo(31) ? "Yes" : "No");
        Console.WriteLine(isPowerOfTwo(64) ? "Yes" : "No");
    }
}

// This code is contributed by Sam007

PHP

<?php
// php program for the above approach

// Function to check if
// x is power of 2
function isPowerOfTwo ($x)
{
// First x in the below expression
// is for the case when x is 0
return $x && (!($x & ($x - 1)));
}

// Driver Code

// Function call
if(isPowerOfTwo(31))
    echo "Yes\n" ;
else
    echo "No\n";

if(isPowerOfTwo(64))
    echo "Yes\n" ;
else
    echo "No\n";

// This code is contributed by Sam007
?>

Javascript

<script>

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

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

    }

// Driver method
document.write(isPowerOfTwo(31) ? "Yes" : "No");
document.write("<br>"+(isPowerOfTwo(64) ? "Yes" : "No"));

// This code is contributed by 29AjayKumar

</script>

Output

No
Yes

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

Find whether a given number is a power of 2 using the AND(&) and NOT(~) operator:

To solve the problem follow the below idea:

Another way is to use the logic to find the rightmost bit set of a given number and then check if (n & (~(n-1))) is equal to n or not

Below is the implementation of the above approach:

C++

// C++ program of the above approach

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

/* Function to check if x is power of 2*/
bool isPowerofTwo(long long n)
{
    if (n == 0)
        return 0;
    if ((n & (~(n - 1))) == n)
        return 1;
    return 0;
}

// Driver code
int main()
{
    // Function call
    isPowerofTwo(30) ? cout << "Yes\n" : cout << "No\n";
    isPowerofTwo(128) ? cout << "Yes\n" : cout << "No\n";
    return 0;
}
// This code is contributed by Sachin

Java

// Java program of the above approach
import java.io.*;

class GFG {

    // Function to check if x is power of 2
    static boolean isPowerofTwo(int n)
    {
        if (n == 0)
            return false;
        if ((n & (~(n - 1))) == n)
            return true;
        return false;
    }

    // Driver code
    public static void main(String[] args)
    {
        // Function call
        if (isPowerofTwo(30) == true)
            System.out.println("Yes");
        else
            System.out.println("No");

        if (isPowerofTwo(128) == true)
            System.out.println("Yes");
        else
            System.out.println("No");
    }
}

// This code is contributed by rajsanghavi9.

Python3

# Python program of the above approach

# Function to check if x is power of 2*/


def isPowerofTwo(n):

    if (n == 0):
        return 0
    if ((n & (~(n - 1))) == n):
        return 1
    return 0


# Driver code
if __name__ == "__main__":

    # Function call
    if(isPowerofTwo(30)):
        print('Yes')
    else:
        print('No')

    if(isPowerofTwo(128)):
        print('Yes')
    else:
        print('No')

# This code is contributed by shivanisinghss2110

C#

// C# program of the above approach

using System;
public class GFG {

    // Function to check if x is power of 2
    static bool isPowerofTwo(int n)
    {
        if (n == 0)
            return false;
        if ((n & (~(n - 1))) == n)
            return true;
        return false;
    }

    // Driver code
    public static void Main(String[] args)
    {

        // Function call
        if (isPowerofTwo(30) == true)
            Console.WriteLine("Yes");
        else
            Console.WriteLine("No");

        if (isPowerofTwo(128) == true)
            Console.WriteLine("Yes");
        else
            Console.WriteLine("No");
    }
}

// This code contributed by gauravrajput1

Javascript

<script>
// javascript program of the above approach

    // Function to check if x is power of 2
    function isPowerofTwo(n)
    {
        if (n == 0)
            return false;
        if ((n & (~(n - 1))) == n)
            return true;
        return false;
    }

        if (isPowerofTwo(30) == true)
            document.write("Yes<br/>");
        else
            document.write("No<br/>");

          if (isPowerofTwo(128) == true)
            document.write("Yes<br/>");
        else
            document.write("No<br/>");

// This code is contributed by umadevi9616
</script>

Output

No
Yes

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

Find whether a given number is a power of 2 using Brian Kernighan’s algorithm:

To solve the problem follow the below idea:

As we know that the number which will be the power of two have only one set bit , therefore when we do bitwise AND with the number which is just less than the number which can be represented as the power of (2) then the result will be 0 . 

Example : 4 can be represented as (2^2 ) , 
                (4 & 3)=0  or in binary (100 & 011=0)

Below is the implementation of the above approach:

C++

// C++ program of the above approach

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

/* Function to check if x is power of 2*/
bool isPowerofTwo(long long n)
{
    return (n != 0) && ((n & (n - 1)) == 0);
}

// Driver code
int main()
{
    // Function call
    isPowerofTwo(30) ? cout << "Yes\n" : cout << "No\n";
    isPowerofTwo(128) ? cout << "Yes\n" : cout << "No\n";
    return 0;
}
// This code is contributed by Suruchi Kumari

Java

// Java program of the above approach

import java.io.*;
class GFG {

    /* Function to check if x is power of 2*/
    public static boolean isPowerofTwo(long n)
    {
        return (n != 0) && ((n & (n - 1)) == 0);
    }

    // Driver code
    public static void main(String[] args)
    {
        // Function call
        if (isPowerofTwo(30)) {
            System.out.println("Yes");
        }
        else {
            System.out.println("No");
        }

        if (isPowerofTwo(128)) {
            System.out.println("Yes");
        }
        else {
            System.out.println("No");
        }
    }
}

// This code is contributed by akashish__

Python3

# Python3 program of the above approach

# Function to check if x is power of 2


def isPowerofTwo(n):

    return (n != 0) and ((n & (n - 1)) == 0)


# Driver code
if __name__ == "__main__":

        # Function call
    if isPowerofTwo(30):
        print("Yes")
    else:
        print("No")

    if isPowerofTwo(128):
        print("Yes")
    else:
        print("No")

# this code is contributed by aditya942003patil

C#

// C# program of the above approach

using System;

public class GFG {
    /* Function to check if x is power of 2*/
    static public bool isPowerofTwo(ulong n)
    {
        return (n != 0) && ((n & (n - 1)) == 0);
    }

    // Driver code
    static public void Main()
    {

        // Function call
        if (isPowerofTwo(30)) {
            System.Console.WriteLine("Yes");
        }
        else {
            System.Console.WriteLine("No");
        }

        if (isPowerofTwo(128)) {
            System.Console.WriteLine("Yes");
        }
        else {
            System.Console.WriteLine("No");
        }
    }
}

// This code is contributed by akashish__

Javascript

<script>
function isPowerofTwo(n){
     return (n != 0) && ((n & (n - 1)) == 0);
}

  /* Function to check if x is power of 2*/
 if(isPowerofTwo(30))
      {
        document.write("Yes");
      }
      else
      {
         document.write("No");
      }

       if(isPowerofTwo(128))
      {
        document.write("Yes");
      }
      else
      {
         document.write("No");
      }

</script>

Output

No
Yes

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

Find whether a given number is a power of 2 using a floating point bit hack:

We can also harness a unique property of IEEE Standard 754 to infer if the given integer is a power of 2 using the following bit hack that only works in a few languages that allow pointer casting.

This works because we know that a power of 2 only has 1 bit high and all the other bits low. Therefore if we represent such a number in scientific notation, we’ll always be left with a mantissa of 1. But in IEEE Standard 754 the 1 is discarded from the mantissa as it is redundant. Now we should be left with 0, if not, then the number must not be a power of 2. We will be using double precision.

Example:

Let’s take 23 first –
23 = 00010111
     =1.0111000 x 2^4

Biased Exponent 1023+4=1027
1027 = 10000000011
Normalised Mantissa = 01110000
We will add 0’s to complete the 52 bits

The IEEE 754 Double Precision is:
= 0 10000000011 0111000000000000000000000000000000000000000000000000

Notice that the mantissa is not 0.

——————————————————————————————

Now let’s take a power of 2, say 16 –
16 = 00010000
      =1.0000000 x 2^4

Biased Exponent 1023+4=1027
1027 = 10000000011
Normalised Mantissa = 00000000
We will add 0’s to complete the 52 bits

The IEEE 754 Double Precision is:
= 0 10000000011 0000000000000000000000000000000000000000000000000000

Now the mantissa is strictly 0.

——————————————————————————————

Below is the implementation of the above approach:

C

#include <stdio.h>
#include <stdint.h>

// Function to check if x is power of 2.
int isPowerOfTwo(int x)
{
    // Power of 2 can't be less than 1.
    if (x < 1)
      return 0;
    // Converting the number to double precision floating point.
    double n = x;
    /*
     * Storing the binary representation of double to an unsigned 64 bit integer.
     * This is not the same as direct casting to integer, as the binary representation changes.
     */
    uint64_t r = *((uint64_t*)(&n));
    // Discarding 1 sign bit and 11 exponent bits.
    r = (r << 12);
    return r == 0;
}

// Driver code
int main()
{
    // Function calls
    isPowerOfTwo(31) ? printf("Yes\n") : printf("No\n");
    isPowerOfTwo(32) ? printf("Yes\n") : printf("No\n");
    isPowerOfTwo(33) ? printf("Yes\n") : printf("No\n");
    isPowerOfTwo(64) ? printf("Yes\n") : printf("No\n");
    return 0;
}
// This code is contributed by Aryan Rai.

C++

#include <iostream>

using namespace std;

// Function to check if x is power of 2.
bool isPowerOfTwo(int x)
{
    // Power of 2 can't be less than 1.
    if (x < 1)
      return false;
    // Converting the number to double precision floating point.
    double n = x;
    /*
     * Storing the binary representation of double to an unsigned 64 bit integer.
     * This is not the same as direct casting to integer, as the binary representation changes.
     */
    uint64_t r = *((uint64_t*)(&n));
    // Discarding 1 sign bit and 11 exponent bits.
    r = (r << 12);
    return r == 0;
}

// Driver code
int main()
{
    // Function calls

    cout << ( isPowerOfTwo(31) ? "Yes" : "No" ) << endl;
    cout << ( isPowerOfTwo(32) ? "Yes" : "No" ) << endl;
    cout << ( isPowerOfTwo(33) ? "Yes" : "No" ) << endl;
    cout << ( isPowerOfTwo(64) ? "Yes" : "No" ) << endl;
    return 0;
}
// This code is contributed by Aryan Rai.

Output

No
Yes
No
Yes

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

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