LeetCode 0461. Hamming Distance Solution in Java, C++, Python & More | Explanation + Code

Description

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 <= 231 - 1

 

Note: This question is the same as 2220: Minimum Bit Flips to Convert Number.

Solutions

Solution 1

PythonJavaC++GoTypeScriptJavaScript

class Solution: def hammingDistance(self, x: int, y: int) -> int: return (x ^ y).bit_count()(code-box)

class Solution { public int hammingDistance(int x, int y) { return Integer.bitCount(x ^ y); } }(code-box)

class Solution { public: int hammingDistance(int x, int y) { return __builtin_popcount(x ^ y); } };(code-box)

func hammingDistance(x int, y int) int { return bits.OnesCount(uint(x ^ y)) }(code-box)

function hammingDistance(x: number, y: number): number { x ^= y; let ans = 0; while (x) { x -= x & -x; ++ans; } return ans; }(code-box)

/** * @param {number} x * @param {number} y * @return {number} */ var hammingDistance = function (x, y) { x ^= y; let ans = 0; while (x) { x -= x & -x; ++ans; } return ans; };(code-box)