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

**Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.**

**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.

**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.

**Output:**

No No Yes Yes 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 http://www.geeksforgeeks.org/?p=1176 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 Mohammad for adding this case).

Below is the implementation of this method.

**Output:**

No No Yes Yes No Yes

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