Count Differing Positions in Two Binary Integers
Company: Google
Role: Software Engineer
Category: Coding & Algorithms
Difficulty: easy
Interview Round: Other
# Count Differing Positions in Two Binary Integers
Given nonnegative integers `x` and `y`, return the number of binary bit positions where their bits differ. Leading zero bits are included when aligning the representations and do not change the count. This count is their Hamming distance.
### Interface
`class Solution { public: int hammingDistance(int x, int y); };`
### Example 1
```text
Input: x = 10, y = 5
Output: 4
```
The aligned representations are `1010` and `0101`; all four positions differ.
### Example 2
```text
Input: x = 0, y = 8
Output: 1
```
The representations `0000` and `1000` differ at one position.
### Constraints
- `0 <= x, y <= 2147483647` (`2^31 - 1`).
### What a Strong Answer Covers
- Counting positions rather than the arithmetic difference of the values.
- Zero inputs, equal values and the highest permitted bit.
- Safe bit operations for the full nonnegative input domain.
```hint Compare aligned positions
Consider one corresponding binary position at a time. What should it contribute when the bits agree or disagree?
```
Overview: Count the binary positions at which two nonnegative integers differ. Preserve zero inputs, the nonnegative 31-bit range and the original Hamming-distance callable interface.