/*

Explaination: If number equals or less than 0 then it cant be any power of 2. Now for positive numbers,

it should have only one binary bit as 1 (becaus power of 2 will come as 2^0,2^1,2^2.......).

for example,

1 -> 2^0 -> 0000001

2 -> 2^1 -> 0000010

4 -> 2^2 -> 0000100

8 -> 2^3 -> 0001000

16-> 2^4 -> 0010000

....... and so on.

So here we can take a bitwise and (&) operation between the number and a number less than it,

if it equal to 0 then it means that no other position has high bit, except at one place.

Thus return true, else false.

for ex: take the number, n=16

n (in bits) -> 1 0 0 0 0 (16 is a power of 2 and thus has only one high bit)

n-1 (in bits) -> 0 1 1 1 1 ( n-1 i.e. 15 will make all bits high excpet the 5th bit)

n & n-1 -> 0 0 0 0 0 (& operation will make all the bits to 0, thus its power of 2)

*/

class Solution {

public boolean isPowerOfTwo(int n) {

if(n<=0)

return false;

return ((n&(n-1))==0);

}

}

// Iterative

class Solution {

public boolean isPowerOfTwo(int n) {

if(n==0)

return false;

while(n!=1)

{

if(n%2!=0)

return false;

n = n/2;

}

return true;

}

}

// Recursive

class Solution {

public boolean isPowerOfTwo(int n) {

if(n==1)

return true;

if(n%2!=0 || n==0)

return false;

return isPowerOfTwo(n/2);

}

}

// Bitwise Solution

class Solution {

public boolean isPowerOfTwo(int n) {

int count=0;

while(n>0)

{

if((n & 1) == 1)

count++;

n = n >> 1;

}

return count == 1 ? true : false;

}

}

// One Liner

public class Solution {

public boolean isPowerOfTwo(int n) {

return ((n & n - 1) == 0 && n > 0);

}

}