Hamming Distance

Easy Bit Manipulation XOR Basics Original on LeetCode

Hamming Distance is a easy bit manipulation problem solved with the xor basics pattern. The best approach, optimal (xor, then count set bits), runs in O(number of differing bits) time and O(1) space. Below are 2 approaches in Java, from compare bit by bit up.

Problem

The Hamming distance between two integers is the number of bit positions where they differ. Return it.

Examples

Example 1

Input: x = 1, y = 4
Output: 2
Why: 001 vs 100: two positions differ.

Example 2

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

Constraints

0 <= x, y <= 2^31 - 1

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Solutions

Try it yourself first. Then compare: each approach lists its idea, the steps, its time and space, and the Java code.

Approach	Time	Space
Compare bit by bit	`O(32)`	`O(1)`
Optimal (XOR, then count set bits)	`O(number of differing bits)`	`O(1)`

1Compare bit by bit

TimeO(32)

SpaceO(1)

Check each of the 32 positions.

For i in 0..31: if ((x >> i) & 1) != ((y >> i) & 1), count++.

Java

class Solution {
    public int hammingDistance(int x, int y) {
        int count = 0;
        for (int i = 0; i < 32; i++) if (((x >> i) & 1) != ((y >> i) & 1)) count++;
        return count;
    }
}

2Optimal (XOR, then count set bits)

TimeO(number of differing bits)

SpaceO(1)

XOR marks the differing bits; Kernighan's trick counts them.

d = x ^ y; while d != 0: d &= d - 1; count++.

Java

class Solution {
    public int hammingDistance(int x, int y) {
        int d = x ^ y, count = 0;
        while (d != 0) {
            d &= d - 1;
            count++;
        }
        return count;
    }
}

Edge cases to test

x == y (0)

Hints

Hint 1

x ^ y has a 1 exactly where the bits differ. Count its set bits.

FAQ

What is the best time complexity for Hamming Distance?

Optimal (XOR, then count set bits) runs in O(number of differing bits) time and O(1) extra space.

Which pattern does Hamming Distance use?

It is a bit manipulation problem that uses the xor basics pattern. Other problems with the same pattern: Swap two numbers.

Is there a brute force solution for Hamming Distance?

Yes. Compare bit by bit takes O(32) time and O(1) space. Check each of the 32 positions.

Which edge cases should I test for Hamming Distance?

x == y (0).