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

Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

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) 
{ 
   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) 
{ 
   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 
?> 