E

bit: XOR, &, shift>>

```
/*
The Hamming distance between two integers is the number of positions at which the corresponding bits are different.

Given two integers x and y, calculate the Hamming distance.

Note:
0 ≤ x, y < 231.

Example:

Input: x = 1, y = 4

Output: 2

Explanation:
1   (0 0 0 1)
4   (0 1 0 0)
       ↑   ↑

The above arrows point to positions where the corresponding bits are different.
*/

/*
Thoughts:
^XOR: gives out all the different bits.
Count the 1's by shiftting right and & 1
*/
class Solution {
    public int hammingDistance(int x, int y) {
        int xor = x ^ y;
        int count = 0;
        while (xor != 0) {
            count += xor & 1;
            xor = xor >> 1;
        }
        return count;
    }
}
```