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.

C

#include<stdio.h>
#include<stdbool.h>
#include<math.h>
  
/* Function to check if x is power of 2*/
bool isPowerOfTwo(int n)
{
   return (ceil(log2(n)) == floor(log2(n)));
}
  
// Driver program
int main()
{
    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
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

Python3

# Python3 Program to find 
# whether a no is 
# power of two
import math
  
# Function to check
# Log base 2
def Log2(x):
    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(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)
{
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()
{
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
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

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.

C

#include<stdio.h>
#include<stdbool.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 program to test above function*/
int main()
{
  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 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.

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(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 program 
    public static void Main()
    {
        Console.WriteLine(isPowerOfTwo(31) ? "Yes" : "No");
        Console.WriteLine(isPowerOfTwo(64) ? "Yes" : "No");
  
    }
}
  
// This code is contributed by Sam007

PHP

<?php
  
// 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
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

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.

C

#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 program to test above function*/
int main()
{
  isPowerOfTwo(31)? printf("Yes\n"): printf("No\n");
  isPowerOfTwo(64)? printf("Yes\n"): printf("No\n");
  return 0;
}

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    

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 (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(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 efficiently 
// check for power for 2
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 method
    public static void Main() 
    {
        Console.WriteLine(isPowerOfTwo(31) ? "Yes" : "No");
        Console.WriteLine(isPowerOfTwo(64) ? "Yes" : "No");
          
    }
}
  
// This code is contributed by Sam007

PHP

<?php
// PHP program to efficiently 
// check for power for 2
  
// 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
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

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Improved By : Sam007, Mithun Kumar, Abby_akku



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