Power of Two

Given an integer, write a function to determine if it is a power of two.

Example 1:

Input: 1
Output: true 
Explanation: 20 = 1

Example 2:

Input: 16
Output: true
Explanation: 24 = 16

Example 3:

Input: 218
Output: false

METHOD 1:

 class Solution {

public boolean isPowerOfTwo(int n) {
    if(n<=0)
        return false;
 
    while(n>2){
        int t = n>>1;
        int c = t<<1;
 
        if(n-c != 0)
            return false;
 
        n = n>>1;
    }
 
    return true;
}
}


If a number is power of 2, it's binary form should be 10...0. So if we right shift a bit of the number and then left shift a bit, the value should be the same when the number >= 10(i.e.,2).

Check below method for more optimized solution.

METHOD 2:

 class Solution {

public boolean isPowerOfTwo(int n) {
  return n>0 && (n&n-1)==0;
}
}


If a number is power of 2, then its highly bit is 1 and there is only one 1. Therefore, n & (n-1) is 0.

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.


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