### Problem statement

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 <= 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

```
#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;
}
```

```
Output:
2
1
```

#### Complexity

Runtime:

`O(1)`

.Extra space:

`O(1)`

.

### References

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