Hamming Distance

Hamming Distance - Problem

Bit Manipulation Easy

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: For x = 1 (binary: 0001) and y = 4 (binary: 0100), the Hamming distance is 2 because bits differ at positions 0 and 2.

Input & Output

Example 1 — Basic Case

$ Input: x = 1, y = 4

› Output: 2

💡 Note: x = 1 (binary: 0001), y = 4 (binary: 0100). Bits differ at positions 0 and 2, so Hamming distance = 2

Example 2 — Same Numbers

$ Input: x = 3, y = 3

› Output: 0

💡 Note: x = 3 (binary: 11), y = 3 (binary: 11). No bits differ, so Hamming distance = 0

Example 3 — Adjacent Numbers

$ Input: x = 7, y = 8

› Output: 4

💡 Note: x = 7 (binary: 0111), y = 8 (binary: 1000). All 4 bit positions differ, so Hamming distance = 4

Constraints

0 ≤ x, y ≤ 2³¹ - 1

Visualization

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Asked in

f Facebook 15 G Google 12

The key insight is to use XOR operation which gives 1 where bits differ, then count the 1s. Best approach uses Brian Kernighan's algorithm for efficient bit counting. Time: O(k), Space: O(1) where k is number of differing bits.

Common Approaches

✓ Brian Kernighan's Algorithm

⏱️ Time: O(k) Space: O(1)

After XORing the numbers, use Brian Kernighan's algorithm which efficiently counts set bits by repeatedly clearing the lowest set bit until the number becomes zero.

Brute Force - Binary String Comparison

⏱️ Time: O(log n) Space: O(log n)

Convert both integers to their binary string representations, pad them to equal length, then iterate through each position counting differences.

XOR and Bit Counting

⏱️ Time: O(log n) Space: O(1)

XOR the two numbers to get a result where 1s indicate positions with different bits, then count the number of 1s in the XOR result.

Brian Kernighan's Algorithm — Algorithm Steps

Compute XOR of x and y
Use n & (n-1) repeatedly to clear lowest set bit
Count iterations until XOR result becomes 0

Visualization

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Step-by-Step Walkthrough

XOR Result

Get XOR result with differing bits marked as 1

Clear Lowest Bit

Repeatedly use n & (n-1) to clear lowest set bit

Count Operations

Number of operations equals Hamming distance

Code -

solution.c — C

#include <stdio.h>

int solution(int x, int y) {
    // XOR to find differing bits
    int xorResult = x ^ y;
    
    // Brian Kernighan's algorithm to count set bits
    int count = 0;
    while (xorResult) {
        xorResult &= xorResult - 1;  // Clear the lowest set bit
        count++;
    }
    
    return count;
}

int main() {
    int x, y;
    scanf("%d", &x);
    scanf("%d", &y);
    
    printf("%d\n", solution(x, y));
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(k)

Where k is the number of differing bits (set bits in XOR result), much faster than O(log n)

⚡ Linearithmic

Space Complexity

O(1)

Only uses a few integer variables

✓ Linear Space

125.0K Views

Medium Frequency

~10 min Avg. Time

3.4K Likes

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Smart Actions

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Input & Output

Constraints

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Related Problems

Common Approaches

Brian Kernighan's Algorithm — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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