### View *Hamming Distance* on LeetCode

## Statistics

**Time Complexity**

O(n + m + k) - The algorithm iterates through all characters within both the integer inputs and their XOR representation, resulting in the O(n + m + k) time complexity.

**Space Complexity**

O(k) - The algorithm performs XOR on both input integers and then converts the result into its binary representation; although only an intermediate value, the computer still stores this binary representation temporarily, resulting in the O(k) space complexity.

**Runtime Beats**

91.65% of other submissions

**Memory Beats**

96.31% of other sumbissions

## Explanation

The algorithm converts the XOR of x and y into binary and counts the number of `1`

s. This results in the correct output because `1`

only appears if a `1`

exists in one binary number and not the other at the same index.

## Solution

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class Solution:
def hammingDistance(self, x: int, y: int) -> int:
return bin(x^y).count("1")