Hamming distance between 2 integers

Time: O(1); Space: O(1); easy

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

Example 1:

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.

Constraints:

• 0 < x, y < 231

class Solution1(object):
    def hammingDistance(self, x, y) -> int:
        """
        :type x: int
        :type y: int
        :rtype: int
        """
        distance = 0
        z = x ^ y
        while z:
            distance += 1
            z &= z - 1
        return distance

s = Solution1()
assert s.hammingDistance(1, 4) == 2

class Solution2(object):
    def hammingDistance(self, x, y) -> int:
        """
        :type x: int
        :type y: int
        :rtype: int
        """
        return bin(x ^ y).count('1')

s = Solution2()
assert s.hammingDistance(1, 4) == 2

See also:

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

https://www.lintcode.com/problem/hamming-distance