Hamming Distance

Problem

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.

Solution

public class Solution {
    public int hammingDistance(int x, int y) {
        int diff = x ^ y, count = 0;
        while (diff != 0) {
            count += diff & 1;
            diff = diff >> 1;
        }
        return count;
    }
}

One line built-in ;)

public class Solution {
    public int hammingDistance(int x, int y) {
        return Integer.bitCount(x ^ y);
    }
}

Analysis

This problem asks to count the difference of bits in each position
We first use ^ (bitwise XOR) to get the difference in bit format
Then we need to count how many 1 are there to get the result
We can either use built-in function Integer.bitCount() or implement it by ourselves

Remember the difference between | bitwise or, & bitwise and, and ^ bitwise XOR
In bit operations, | returns 1 if there is just one or more 1 in the bit position
& returns 1 if both bit positions have 1
^ returns 1 if values in two positions are different