The Hamming distance between two integers is the number of positions at which the corresponding bits are different.
Given two integers x and y, return the Hamming distance between them.
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.
Example 2:
Input: x = 3, y = 1
Output: 1
Constraints:
- 0 <= x, y <= 231 - 1
My Solution
We can solve this problem by XOR-ing x and y together and then counting the number of set bits in the result.
Time complexity would be O(1) because it only requires a constant 31 iterations. Space complexity also would be O(1) since it only requires constant extra space.
class Solution {
public int hammingDistance(int x, int y) {
int num = x ^ y;
int out = 0;
while (num > 0) {
out += (num & 1);
num >>= 1;
}
return out;
}
}