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C Program to find whether a no is power of two

• Last Updated : 19 Jul, 2021

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``#include ``#include ``#include ` `/* 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`
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

 `#include ``#include ` `/* 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;``}`
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 ``#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;``}`
Output:
```No
Yes```

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

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