#   # 461. Hamming Distance

## Yet another example of the bitwise XOR operator

### 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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