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.

```
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.

```
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.

```
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.

## Extras

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