Given a positive integer, write a function to find if it is a power of two or not.

**Examples :**

Input : n = 4 Output : Yes 2^{2}= 4 Input : n = 7 Output : No Input : n = 32 Output : Yes 2^{5}= 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

## 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 ?>

**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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