Hamming Distance - Java Solution
Tags: algorithms, java, beginners, faang
Published: 2022-03-01 04:19:46

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:**

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

**Example 02:**
```javascript
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. 

```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));
  }
}
```

### 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.

```javascript
      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.

```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));
  }
}
```

### 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.

---

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