461. Hamming Distance

https://leetcode.com/problems/hamming-distance/#/description

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.

Sol:

The hamming distance is the number of different digits of two numbers in binary digit.   

Get every mod of x and y divided by 2, and add one to the answer when they are different.

class Solution(object):
    def hammingDistance(self, x, y):
        """
        :type x: int
        :type y: int
        :rtype: int
        """
        
        # We can find the i-th bit (from the right) of a number by dividing by 2 i times, then taking the number mod 2.

        # Using this, lets compare each of the i-th bits, adding 1 to our answer when they are different.
        ans = 0
        while x or y:
            ans += ( x%2 ) ^ ( y%2 )
            x /= 2
            y /= 2
        return ans

Note:

1 x ^ y,  where ^ means exclusive or.

return 1 only when two bits are different.

1 0 ==> 1

0 1 ==> 1

0 0 ==> 0

1 1 ==> 0

2 x & y, where & means and.

return 1 only when two bits are 1.

1 1 ==> 1

1 0 ==> 0

0 1 ==> 0

0 0 ==> 0

3 This problem is clearly about bit manipulation. Think of  bit operators. 

4 The end condition of division is the last dividend goes down to zero.