Hamming Distance

In this lesson, we find the number of positions where the bits are different for the given input.

Introduction

In this question, we will find the number of positions at which the corresponding bits are different.

Problem Statement

Given integers x, y finds the positions where the corresponding bits are different.

Input: x = 1, y = 8
Output: 2

Explanation:
1   (0 0 0 1)
8   (1 0 0 0)
     ↑     ↑

Input: x = 12, y = 15
Output: 2
Explanation:
12   (1 1 0 0)
15   (1 1 1 1)
          ↑ ↑

Solution

We solve this using shifting operation and then move to solve it more optimally.

Bit Shifting

This approach is better as it takes O(1) time and complexity. We shift the bits to left or right and then check whether the bit is one.

Algorithm

We use the right shift operation, where each bit would have its turn to be shifted to the rightmost position.

Once shifted, we use either modulo % (i.e., i % 2) or & operation (i.e., i & 1).

Hint: you can check if a number does not equal 0 by the ^ operator.

Java

class HammingDistance {

    public static int hammingDistance(int a, int b) {
        int xor = a ^ b;
        int distance = 0;

        while (xor != 0) {
            if (xor % 2 == 1) {
                distance += 1;
            }
            xor >>= 1;
        }

        return distance;
    }

    public static void main(String[] args) {
        int a = 1;
        int b = 8;
        System.out.println("Hamming Distance between two integers is " + hammingDistance(a, b));
    }
}

Python

def HammingDistance(a, b):
   xor = a ^ b
   distance = 0

   while (xor ^ 0) :
    if (xor % 2 == 1) :
        distance += 1
    
    xor >>= 1
    
   return distance

a = 1
b = 8
print("Hamming Distance between two integers is", HammingDistance(a, b))

JavaScript

function HammingDistance(a, b) {
  let xor = a ^ b;
  let distance = 0;

  while (xor ^ 0) {
    if (xor % 2 === 1) {
        distance += 1;
    }
    xor >>= 1;
  }

  return distance;
}

let a = 1;
let b = 8;
console.log("Hamming Distance between two integers is", HammingDistance(a, b));

C++

#include <iostream>
using namespace std;

void hammingDistance(int a, int b){
  int xorVal = a ^ b;
  int distance = 0;

  while(xorVal ^ 0){
    if(xorVal % 2 == 1){
      distance += 1;
    }
    xorVal >>= 1;
  }
  cout << "Hamming Distance between two integers is " << distance << endl;
}

int main() {
  int a = 1;
  int b = 8;
  hammingDistance(a, b);
  return 0;
}

TypeScript

export const HammingDistance = (a: number, b: number): number => {
  let xor: number = a ^ b;
  let distance: number = 0;

  while (xor ^ 0) {
    if (xor % 2 === 1) {
        distance += 1;
    }
    xor >>= 1;
  }

  return distance;
}

let a: number= 1;
let b: number = 8;
console.log("Hamming Distance between two integers is", HammingDistance(a, b));

Complexity Analysis

Time complexity: O(1). For a 32-bit integer, the algorithm would take at most 32 iterations.

Space complexity: O(1). Memory is constant irrespective of the input.

Brian Kernighan’s Algorithm

In the above approach, we shifted each bit one by one. So, is there a better approach to finding the hamming distance? Yes.

Algorithm

When we do & bit operation between number n and (n-1), the rightmost bit of one in the original number n would be cleared.

      n       = 40  => 00101000
    n - 1     = 39  => 00100111
----------------------------------
(n & (n - 1)) = 32  => 00100000   
----------------------------------

Based on the above idea, we can count the distance in 2 iterations rather than all the shifting iterations we did earlier. Let’s see the code in action.

Java

class HammingDistance {
  public static int hammingDistance(int a, int b) {
    int xor = a ^ b;
    int distance = 0;

    while (xor != 0) {
      distance += 1;
      xor &= ( xor - 1); // equals to `xor = xor & ( xor - 1);`
    }

    return distance;
  }

  public static void main(String[] args) {
    int a = 1;
    int b = 8;
    System.out.println("Hamming Distance between two integers is " + hammingDistance(a, b));
  }
}

Python

def HammingDistance(a, b):
   xor = a ^ b
   distance = 0

   while (xor != 0) :
     distance += 1
     xor &= ( xor - 1)
  
   return distance

a = 1
b = 8
print("Hamming Distance between two integers is", HammingDistance(a, b))

JavaScript

function HammingDistance(a, b) {
  let xor = a ^ b;
  let distance = 0;

  while (xor != 0) {
    distance += 1;
    xor &= ( xor - 1); // equals to `xor = xor & ( xor - 1);`
  }

  return distance;
}

let a = 1;
let b = 8;
console.log("Hamming Distance between two integers is", HammingDistance(a, b));

C++

#include <iostream>
using namespace std;

int hammingDistance(int a, int b){
  int xorVal = a ^ b;
  int distance = 0;

  while (xorVal != 0) {
    distance += 1;
    xorVal &= ( xorVal - 1); // equals to `xorVal = xorVal & ( xorVal - 1);`
  }

  return distance;
}

int main() {
  int a = 1;
  int b = 8;
  cout << "Hamming Distance between two integers is " << hammingDistance(a, b);
  return 0;
}

TypeScript

export const HammingDistance = (a: number, b: number): number => {
  let xor: number = a ^ b;
  let distance: number = 0;

  while (xor != 0) {
    distance += 1;
    xor &= ( xor - 1); // equals to `xor = xor & ( xor - 1);`
  }

  return distance;
}

let a: number= 1;
let b: number = 8;
console.log("Hamming Distance between two integers is", HammingDistance(a, b));

Complexity Analysis

Time complexity: O(1). The input size of the integer is fixed, and we have a constant time complexity.

Space complexity: O(1). Memory is constant irrespective of the input.