Hamming Distance - C++ Solution

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.

Example 01:

Input: x = 1, y = 8
Output: 2
Explanation:
1   (0 0 0 1)
8   (1 0 0 0)
     ↑     ↑

Example 02:

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 we move to solve it in a more optimal way.

Bit Shifting

This approach is better as it takes O(1) time complexity. We shift the bits to left or right and then check if the bit is one or not.

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).

Code

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

#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;
}

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 in 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   
----------------------------------

Code

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.

#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;
}

Complexity Analysis

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

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

Extras

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