Tutorialspoint.dev

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++

// C++ Program to find whether a
// no is 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 program
int main()
{
    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<stdio.h>
#include<stdbool.h>
#include<math.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 program
int main()
{
    isPowerOfTwo(31)? printf("Yes "): printf("No ");
    isPowerOfTwo(64)? printf("Yes "): printf("No ");
    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)
{
    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)
{
    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

/div>

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(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()
{
    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 ";
else
echo "No ";
  
if(isPowerOfTwo(64))
echo "Yes ";
else
echo "No ";
      
// 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 "): printf("No ");
  isPowerOfTwo(64)? printf("Yes "): printf("No ");
  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 ";
else
    echo "No ";
  
if(isPowerOfTwo(64))
    echo "Yes ";
else
    echo "No ";
  
// 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://tutorialspoint.dev/slugresolver/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 "): printf("No ");
  isPowerOfTwo(64)? printf("Yes "): printf("No ");
  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 " ;
else
    echo "No ";
  
if(isPowerOfTwo(64))
    echo "Yes " ;
else
    echo "No ";
          
// This code is contributed by Sam007
?>


Output :

No
Yes

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above



This article is attributed to GeeksforGeeks.org

You Might Also Like

leave a comment

code

0 Comments

load comments

Subscribe to Our Newsletter