Hamming distance between two Integers

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;
}

// This code is contributed by Sania Kumari Gupta (kriSania804)

C

// C implementation of above approach
#include <stdio.h>

// 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;
    printf("%d\n", hammingDistance(9, 14));

    return 0;
}

// This code is contributed by Sania Kumari Gupta (kriSania804)

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 counted using __builtin_popcount() function.

Time Complexity: O(log₂x), where x is n1 ^ n2, this can be interpreted as O(1), since an int can hold a maximum of 32 bits

Auxiliary Space: O(1)

Approach 2:

1. Calculate the maximum of both 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;
}

Java

/*package whatever //do not write package name here */

import java.io.*;

class GFG {
    static int hammingDistance(int x, int y)
{
    int ans = 0;
    int m = Math.max(x, y);
    while (m>0) {
        int c1 = x & 1;
        int c2 = y & 1;
        if (c1 != c2)
            ans += 1;
        m = m >> 1;
        x = x >> 1;
        y = y >> 1;
    }
    return ans;
}

// Drivers code
public static void main(String args[])
{
    int n1 = 4, n2 = 8;
    int hdist = hammingDistance(n1, n2);
    System.out.println(hdist);
}
}

// This code is contributed by shinjanpatra

Python3

# Python3 code for the same approach
def hammingDistance(x, y):

   ans = 0
   m = max(x, y)
   while (m):
      c1 = x & 1
      c2 = y & 1
      if (c1 != c2):
         ans += 1
      m = m >> 1
      x = x >> 1
      y = y >> 1
   return ans

# driver program
n1 = 4
n2 = 8
hdist = hammingDistance(n1, n2)
print(hdist)

# This code is contributed by shinjanpatra

C#

// C# code to implement the approach

using System;

public class GFG
{

    // Function that return the hamming distance
    // between x and y
    static int hammingDistance(int x, int y)
    {
        int ans = 0;
        int m = Math.Max(x, y);

        // checking the set bits
        while (m > 0) {
            int c1 = x & 1;
            int c2 = y & 1;
            if (c1 != c2)
                ans += 1;
            m = m >> 1;
            x = x >> 1;
            y = y >> 1;
        }
        return ans;
    }

    // Driver code
    public static void Main(string[] args)
    {
        int n1 = 4, n2 = 8;
        // Function call
        int hdist = hammingDistance(n1, n2);
        Console.WriteLine(hdist);
    }
}

// This code is contributed by phasing17

Javascript

<script>

// JavaScript code for the same approach

function hammingDistance(x,y)
{
    let ans = 0;
    let m = Math.max(x, y);
    while (m) {
        let c1 = x & 1;
        let c2 = y & 1;
        if (c1 != c2)
            ans += 1;
        m = m >> 1;
        x = x >> 1;
        y = y >> 1;
    }
    return ans;
}

// driver program
let n1 = 4,n2 = 8;
let hdist = hammingDistance(n1, n2);
document.write(hdist,"</br>");

// This code is contributed by shinjanpatra

</script>

Output

2

Time Complexity: O(log₂(max(n1, n2)).
Auxiliary Space: O(1)

Approach#3:  Using string manipulation

Algorithm

1. Convert the two given integers to binary strings.
2. Pad the shorter binary string with leading zeros to make both strings of equal length.
3. Compare the two binary strings bit by bit and count the number of differences.
4. The count obtained in step 3 represents the Hamming distance between the two given integers.

C++

// c++ code implementation
#include <iostream>
#include <bitset>
#include <string>

using namespace std;

int hammingDistance(int n1, int n2) {
    string binStr1 = bitset<32>(n1).to_string();
    string binStr2 = bitset<32>(n2).to_string();
    int lenDiff = abs(static_cast<int>(binStr1.length() - binStr2.length()));
    if (binStr1.length() < binStr2.length()) {
        binStr1 = string(lenDiff, '0') + binStr1;
    } else {
        binStr2 = string(lenDiff, '0') + binStr2;
    }
    int count = 0;
    for (int i = 0; i < binStr1.length(); i++) {
        if (binStr1[i] != binStr2[i]) {
            count++;
        }
    }
    return count;
}

int main() {
    int n1 = 4;
    int n2 = 8;
    cout << hammingDistance(n1, n2) << endl;
    return 0;
}

Java

public class Main {

    public static int hammingDistance(int n1, int n2) {
        String binStr1 = Integer.toBinaryString(n1);
        String binStr2 = Integer.toBinaryString(n2);
        int lenDiff = Math.abs(binStr1.length() - binStr2.length());
        if (binStr1.length() < binStr2.length()) {
            binStr1 = "0".repeat(lenDiff) + binStr1;
        } else {
            binStr2 = "0".repeat(lenDiff) + binStr2;
        }
        int count = 0;
        for (int i = 0; i < binStr1.length(); i++) {
            if (binStr1.charAt(i) != binStr2.charAt(i)) {
                count++;
            }
        }
        return count;
    }

    public static void main(String[] args) {
        int n1 = 4;
        int n2 = 8;
        System.out.println(hammingDistance(n1, n2));
    }
}

Python3

def hamming_distance(n1, n2):
    bin_str1 = bin(n1)[2:]
    bin_str2 = bin(n2)[2:]
    len_diff = abs(len(bin_str1) - len(bin_str2))
    if len(bin_str1) < len(bin_str2):
        bin_str1 = '0' * len_diff + bin_str1
    else:
        bin_str2 = '0' * len_diff + bin_str2
    count = 0
    for i in range(len(bin_str1)):
        if bin_str1[i] != bin_str2[i]:
            count += 1
    return count
n1 = 4
n2 = 8
print(hamming_distance(n1, n2))

Javascript

function hamming_distance(n1, n2) {
    let bin_str1 = n1.toString(2);
    let bin_str2 = n2.toString(2);
    let len_diff = Math.abs(bin_str1.length - bin_str2.length);
    if (bin_str1.length < bin_str2.length) {
        bin_str1 = '0'.repeat(len_diff) + bin_str1;
    } else {
        bin_str2 = '0'.repeat(len_diff) + bin_str2;
    }
    let count = 0;
    for (let i = 0; i < bin_str1.length; i++) {
        if (bin_str1[i] !== bin_str2[i]) {
            count += 1;
        }
    }
    return count;
}

let n1 = 4;
let n2 = 8;
console.log(hamming_distance(n1, n2));

Output

2

Time Complexity: O(log n), where n is the maximum of the two integers.
Space Complexity: O(log n), where n is the maximum of the two integers.