# C# Program to find whether a no is power of two

• Last Updated : 31 May, 2022

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` `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)`

Output:

```No
Yes```

Time Complexity: O(log2n)

Auxiliary Space: O(1)

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#

 `// 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`

Output:

```No
Yes```

Time Complexity: O(log2n)

Auxiliary Space: O(1)

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#

 `// 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`

Output:

```No
Yes```

Time Complexity: O(1)

Auxiliary Space: O(1)

Please refer complete article on Program to find whether a no is power of two for more details!

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