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# Java 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.

## 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`

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.

## 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.`

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.

## 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`

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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