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Hamming distance between two Integers

  • Difficulty Level : Easy
  • Last Updated : 02 Aug, 2021

Given two integers, the task is to find the hamming distance between two integers. Hamming Distance between two integers is the number of bits that are different at the same position in both numbers. 
Examples: 
 

Input: n1 = 9, n2 = 14
Output: 3
9 = 1001, 14 = 1110
No. of Different bits = 3

Input: n1 = 4, n2 = 8
Output: 2

 

Approach: 
 

  1. Calculate the XOR of two numbers.
  2. Count the number of set bits.

Below is the implementation of above approach: 
 

C++




// C++ implementation of above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to calculate hamming distance
int hammingDistance(int n1, int n2)
{
    int x = n1 ^ n2;
    int setBits = 0;
 
    while (x > 0) {
        setBits += x & 1;
        x >>= 1;
    }
 
    return setBits;
}
 
// Driver code
int main()
{
    int n1 = 9, n2 = 14;
    cout << hammingDistance(9, 14) << endl;
 
    return 0;
}

Java




// Java implementation of above approach
class GFG
{
 
// Function to calculate hamming distance
static int hammingDistance(int n1, int n2)
{
    int x = n1 ^ n2;
    int setBits = 0;
 
    while (x > 0)
    {
        setBits += x & 1;
        x >>= 1;
    }
 
    return setBits;
}
 
// Driver code
public static void main(String[] args)
{
    int n1 = 9, n2 = 14;
    System.out.println(hammingDistance(n1, n2));
}
}
 
// This code is contributed by Bilal

Python3




# Python3 implementation of above approach
 
# Function to calculate hamming distance
def hammingDistance(n1, n2) :
 
    x = n1 ^ n2
    setBits = 0
 
    while (x > 0) :
        setBits += x & 1
        x >>= 1
     
    return setBits
 
if __name__=='__main__':
    n1 = 9
    n2 = 14
    print(hammingDistance(9, 14))
 
# this code is contributed by Smitha Dinesh Semwal

C#




// C# implementation of above approach
class GFG
{
 
// Function to calculate
// hamming distance
static int hammingDistance(int n1, int n2)
{
    int x = n1 ^ n2;
    int setBits = 0;
 
    while (x > 0)
    {
        setBits += x & 1;
        x >>= 1;
    }
 
    return setBits;
}
 
// Driver code
static void Main()
{
    int n1 = 9, n2 = 14;
    System.Console.WriteLine(hammingDistance(n1, n2));
}
}
 
// This code is contributed by mits

PHP




<?PHP
// PHP implementation of above approach
 
// Function to calculate hamming distance
function hammingDistance($n1, $n2)
{
    $x = $n1 ^ $n2;
    $setBits = 0;
 
    while ($x > 0)
    {
        $setBits += $x & 1;
        $x >>= 1;
    }
 
    return $setBits;
}
 
// Driver code
$n1 = 9;
$n2 = 14;
echo(hammingDistance(9, 14));
 
// This code is contributed by Smitha
?>

Javascript




<script>
 
// Javascript implementation of above approach
 
// Function to calculate hamming distance
function hammingDistance(n1, n2)
{
    let x = n1 ^ n2;
    let setBits = 0;
 
    while (x > 0) {
        setBits += x & 1;
        x >>= 1;
    }
 
    return setBits;
}
 
// Driver code
    let n1 = 9, n2 = 14;
    document.write(hammingDistance(9, 14));
 
</script>
Output



3

Note: No. of set bits can be count using __builtin_popcount() function.

Approach 2:

1. Calculate the maximum of both the numbers.

2.Check the set bits for both at each position.

C++




#include <bits/stdc++.h>
using namespace std;
int hammingDistance(int x, int y)
{
    int ans = 0;
    int m = max(x, y);
    while (m) {
        int c1 = x & 1;
        int c2 = y & 1;
        if (c1 != c2)
            ans += 1;
        m = m >> 1;
        x = x >> 1;
        y = y >> 1;
    }
    return ans;
}
int main()
{
    int n1 = 4, n2 = 8;
    int hdist = hammingDistance(n1, n2);
    cout << hdist << endl;
    return 0;
}
Output
2

 

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