Home Data Structures and Algorithms Program to find whether a given number is power of 2

Program to find whether a given number is power of 2

Find the Number Occurring Odd Number of Times

Find position of the only set bit

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 = 42
Output : No
Explanation: 42 is not a power of 2

Input : n = 1
Output : Yes
Explanation: 20 = 1

Recommended Practice
Power of 2
Try It!

Finding whether a given number is a power of 2 using Log operator:

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
Python 
# 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)
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>
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
?>

Output
No
Yes

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

Finding whether a given number is a power of 2 using the modulo & division operator:

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.
Python 
# 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
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>
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
?>

Output
No
Yes

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

Finding 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.
Python 
# 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
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
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)

Finding 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.w3wiki.net/program-to-find-whether-a-no-is-power-of-two/Mohammad for adding this case). 

Note: With any method which involves subtraction by 1 (just as below, ‘n-1’), some online platforms may give an error if n is given the minimum value of integer or -2147483648. Subtraction by 1 gives integer overflow error in C++.

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)
{
    if(x<0) return false;
    /* 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
Python 
# 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
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>
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
?>

Output
No
Yes

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

Finding 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

Note: With any method which involves subtraction by 1 (just as below, ‘n-1’), some online platforms may give an error if n is given the minimum value of integer or -2147483648. Subtraction by 1 gives integer overflow error in C++.

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.
Python 
# 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) 

 



Tags:
#FactSet #Qualcomm #Samsung #SAP Labs #Bit Magic #DSA #Mathematical #FactSet #Qualcomm #Samsung #SAP Labs #Bit Magic #Mathematical

Find the Number Occurring Odd Number of Times

Find position of the only set bit

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