# 461. Hamming Distance

### Problem statement

The [Hamming distance](https://en.wikipedia.org/wiki/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

```plaintext
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

```plaintext
Input: x = 3, y = 1
Output: 1
```

#### Constraints

* `0 <= x, y <= 2^31`.
    

### Solution: Using bitwise operator XOR

You could use bit operator ^ (XOR) to get the bit positions where `x` and `y` are different. Then use bit operator & (AND) at each position to count them.

#### Code

```cpp
#include <iostream>
int hammingDistance(int x, int y) {
    int z = x ^ y;
    int count = 0;
    while (z) {
        count += z & 1;
        z = z >> 1;
    }
    return count;
}
int main() {
    std::cout << hammingDistance(1,4) << std::endl;
    std::cout << hammingDistance(1,3) << std::endl;
}
```

```plaintext
Output:
2
1
```

#### Complexity

* Runtime: `O(1)`.
    
* Extra space: `O(1)`.
    

### References

* [https://leetcode.com/problems/hamming-distance/](https://leetcode.com/problems/hamming-distance/)
    

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