Find whether a given number is a power of 4 or not

We strongly recommend that you click here and practice it, before moving on to the solution.

1. A simple method is to take log of the given number on base 4, and if we get an integer then number is power of 4.

2. Another solution is to keep dividing the number by 4, i.e, do n = n/4 iteratively. In any iteration, if n%4 becomes non-zero and n is not 1 then n is not a power of 4, otherwise n is a power of 4.

C

#include<stdio.h>
#define bool int

/* Function to check if x is power of 4*/
bool isPowerOfFour(int n)
{
  if(n == 0)
    return 0;
  while(n != 1)
  {    
   if(n % 4 != 0)
      return 0;
    n = n / 4;      
  }
  return 1;
} 

/*Driver program to test above function*/
int main()
{
  int test_no = 64;
  if(isPowerOfFour(test_no))
    printf("%d is a power of 4", test_no);
  else
    printf("%d is not a power of 4", test_no);
  getchar();
}

Java

// Java code to check if given
// number is power of 4 or not

class GFG {

    // Function to check if 
    // x is power of 4
    static int isPowerOfFour(int n)
    {
        if(n == 0)
        return 0;
        while(n != 1)
        { 
            if(n % 4 != 0)
            return 0;
            n = n / 4;     
        }
        return 1;
    } 

    // Driver program 
    public static void main(String[] args)
    {
        int test_no = 64;
        if(isPowerOfFour(test_no) == 1)
         System.out.println(test_no + 
                           " is a power of 4");
        else
         System.out.println(test_no + 
                           "is not a power of 4");
    }
}

// This code is contributed
// by  prerna saini

Python3

# Python3 program to check if given
# number is power of 4 or not 

# Function to check if x is power of 4
def isPowerOfFour(n):
    if (n == 0):
        return False
    while (n != 1):
            if (n % 4 != 0):
                return False
            n = n // 4
            
    return True

# Driver code
test_no = 64
if(isPowerOfFour(64)):
    print(test_no, 'is a power of 4')
else:
    print(test_no, 'is not a power of 4')

# This code is contributed by Danish Raza

C#

// C# code to check if given
// number is power of 4 or not
using System;

class GFG {
    
    // Function to check if
    // x is power of 4
    static int isPowerOfFour(int n)
    {
        if (n == 0)
            return 0;
        while (n != 1) {
            if (n % 4 != 0)
                return 0;
            n = n / 4;
        }
        return 1;
    }

    // Driver code
    public static void Main()
    {
        int test_no = 64;
        if (isPowerOfFour(test_no) == 1)
            Console.Write(test_no + 
                    " is a power of 4");
        else
            Console.Write(test_no + 
                " is not a power of 4");
    }
}

// This code is contributed by Sam007

Output:

64 is a power of 4

3. A number n is a power of 4 if following conditions are met.
a) There is only one bit set in the binary representation of n (or n is a power of 2)
b) The count of zero bits before the (only) set bit is even.

For example: 16 (10000) is power of 4 because there is only one bit set and count of 0s before the set bit is 4 which is even.

Thanks to Geek4u for suggesting the approach and providing the code.

C

#include<stdio.h>
#define bool int

bool isPowerOfFour(unsigned int n)
{
  int count = 0;

  /*Check if there is only one bit set in n*/
  if ( n && !(n&(n-1)) )
  {
     /* count 0 bits before set bit */
     while(n > 1)
     {
       n  >>= 1;
       count += 1;
     }      

    /*If count is even then return true else false*/
    return (count%2 == 0)? 1 :0;
  }

  /* If there are more than 1 bit set
    then n is not a power of 4*/
  return 0;
}    

/*Driver program to test above function*/
int main()
{
   int test_no = 64;
   if(isPowerOfFour(test_no))
     printf("%d is a power of 4", test_no);
   else
     printf("%d is not a power of 4", test_no);
   getchar();
}

Python3

# Python3 program to check if given
# number is power of 4 or not 

# Function to check if x is power of 4
def isPowerOfFour(n):
    
    count = 0
    
    # Check if there is only one 
    # bit set in n
    if (n and (not(n & (n - 1)))):
        
        # count 0 bits before set bit 
        while(n > 1):
            n >>= 1
            count += 1
        
        # If count is even then return
        # true else false
        if(count % 2 == 0):
            return True
        else:
            return False

# Driver code
test_no = 64
if(isPowerOfFour(64)):
    print(test_no, 'is a power of 4')
else:
    print(test_no, 'is not a power of 4')

# This code is contribued by Danish Raza

Output:

64 is a power of 4

Please write comments if you find any of the above codes/algorithms incorrect, or find other ways to solve the same problem.

We strongly recommend that you click here and practice it, before moving on to the solution.

C

Java

Python3

C#

C

Python3

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