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 , finds the positions where the corresponding bits are different. x y 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 time complexity. We shift the bits to left or right and then check if the bit is one or not. O(1) 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 you can check if a number does not equal 0 by the operator. Hint: ^ #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 . For a integer, the algorithm would take at most 32 iterations. Time complexity: O(1) 32-bit . Memory is constant irrespective of the input. Space complexity: O(1) 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 and , the rightmost bit of one in the original number would be cleared. n (n-1) n 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 . The input size of the is fixed, we have a constant time complexity. Time complexity: O(1) integer . Memory is constant irrespective of the input. Space complexity: O(1) First Published here

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How to Solve the Hamming Distance Problem in C++, A Google Interview Question

How to Solve the Hamming Distance Problem in C++, A Google Interview Question

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