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Count of matching set and unset bits in given integers.

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  • Last Updated : 30 Jun, 2022
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For two integers X and Y are the tasks is to find the number of bits that are the same in their binary representation by not considering any leading zeroes after the leftmost set bit of the greater number in binary form.

Examples:

Input:  X = 10, Y = 6
Output:  2
Explanation: The binary representation of 10 is 1010 and 6 is 0110. So, the number of same bits are 2.

Input : X = 3, Y = 16
Output : 2

 

Approach: 

Intuition:

Calculate least significant bit (LSB) by using (i % 2), where i is any integer and Check if the least significant bit (LSB) of the given integer X, Y are same or not. If same then, increment the answer and right shift the both integer by 1 for checking the LSB are same or not for next bit.

Algorithm:

  • Keep doing the following steps until either X or Y is not 0.
    • Calculate the LSB of given integer X and Y by X % 2 and Y % 2 and check if both are same or not.
      • If same, then increment the answer
    • Right shift the both given integer by 1

Below is the implementation of the above approach:

C++




// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to find the number of same bit between
// two integer.
int solve(int X, int Y)
{
    int ans = 0;
 
    while (X != 0 || Y != 0) {
        // Check if both LSB are same or not
        if (X % 2 == Y % 2)
            ans++;
 
        // Right shift the both given integer by 1
        X >>= 1;
        Y >>= 1;
    }
 
    return ans;
}
 
// Driver code
int main()
{
    int X = 10, Y = 6;
 
    // Find number of same bits
    cout << solve(X, Y);
 
    return 0;
}

Java




// Java implementation of the approach
import java.io.*;
 
class GFG
{
 
  // Function to find the number of same bit between
  // two integer.
  public static int solve(int X, int Y)
  {
    int ans = 0;
 
    while (X != 0 || Y != 0)
    {
       
      // Check if both LSB are same or not
      if (X % 2 == Y % 2)
        ans++;
 
      // Right shift the both given integer by 1
      X >>= 1;
      Y >>= 1;
    }
 
    return ans;
  }
 
  public static void main(String[] args)
  {
    int X = 10, Y = 6;
 
    // Find number of same bits
    System.out.print(solve(X, Y));
  }
}
 
// This code is contributed by Rohit Pradhan

Python3




# Python3 implementation of the approach
 
# Function to find the number of same bit between
# two integer.
def solve(X, Y):
    ans = 0
    while (X != 0 or Y != 0):
       
        # Check if both LSB are same or not
        if X % 2 == Y % 2:
            ans += 1
 
        # Right shift the both given integer by 1
        X >>= 1
        Y >>= 1
 
    return ans
 
# Driver code
X = 10
Y = 6
 
# function call
print(solve(X, Y))
 
# This code is contributed by phasing17

C#




// C# implementation of the approach
using System;
 
public class GFG{
 
  // Function to find the number of same bit between
  // two integer.
  public static int solve(int X, int Y)
  {
    int ans = 0;
 
    while (X != 0 || Y != 0)
    {
 
      // Check if both LSB are same or not
      if (X % 2 == Y % 2)
        ans++;
 
      // Right shift the both given integer by 1
      X >>= 1;
      Y >>= 1;
    }
 
    return ans;
  }
 
  static public void Main (){
    int X = 10, Y = 6;
 
    // Find number of same bits
    Console.Write(solve(X, Y));
  }
}
 
// This code is contributed by hrithikgarg03188.

Javascript




<script>
// JS implementation of the approach
 
// Function to find the number of same bit between
// two integer.
function solve(X, Y)
{
    var ans = 0;
 
    while (X != 0 || Y != 0)
    {
     
        // Check if both LSB are same or not
        if (X % 2 == Y % 2)
            ans++;
 
        // Right shift the both given integer by 1
        X >>= 1;
        Y >>= 1;
    }
 
    return ans;
}
 
// Driver code
var X = 10;
var Y = 6;
 
// Find number of same bits
document.write(solve(X, Y));
 
// This code is contributed by phasing17
</script>

Output

2

Time Complexity: O(1)
Auxiliary Space: O(1)

Another Approach: Counting unset bits in XOR of the numbers

The XOR of two bits is 1/set if they are unmatched (ie, one is unset and the other is set), and the XOR of two bits is 0/unset if they are matched (ie both are set or unset).

Therefore, the total number of matching bits in two numbers is equal to the number of unmatched bits in their XOR, or the difference between total bits and unmatched bits in their XOR.

C++




// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
  
// compute number of matching bits
int countUnmatchedBits(int A, int B)
{
    //calculate XOR
    int XOR = (A ^ B);
    //count total number of bits using log2(XOR) + 1
    int totalBits = (int)log2(XOR) + 1;
    // Check for 1's in the binary form using
    // Brian Kerninghan's Algorithm
    int setCount = 0;
    while (XOR) {
        XOR = XOR & (XOR - 1);
        setCount++;
    }
    //the number of unset bits is equal to
    //the difference between total bits
    //and the set bits count
    return totalBits - setCount;
}
  
// Driver code
int main()
{
    int A = 10, B = 6;
    int result = countUnmatchedBits(A, B);
    cout << "Number of matching bits : " << result;
  
    return 0;
}
 
//this code is contributed by phasing17

Java




import java.util.*;
import java.io.*;
 
// Java program for the above approach
class GFG{
 
    // compute number of matching bits
    static int countUnmatchedBits(int A, int B)
    {
        //calculate XOR
        int XOR = (A ^ B);
        //count total number of bits using log2(XOR) + 1
        int totalBits = (int)Math.floor(Math.log(XOR)/Math.log(2)) + 1;
        // Check for 1's in the binary form using
        // Brian Kerninghan's Algorithm
        int setCount = 0;
        while (XOR > 0) {
            XOR = XOR & (XOR - 1);
            setCount++;
        }
        //the number of unset bits is equal to
        //the difference between total bits
        //and the set bits count
        return totalBits - setCount;
    }
 
    // Driver code
    public static void main(String args[])
    {
        int A = 10, B = 6;
        int result = countUnmatchedBits(A, B);
        System.out.println("Number of matching bits : " + result);
    }
}
 
// This code is contributed by subhamgoyal2014.

Python




# Python program for the above approach
import math
# compute number of matching bits
 
 
def countUnmatchedBits(A, B):
    # calculate XOR
    XOR = (A ^ B)
    # count total number of bits using log2(XOR) + 1
    totalBits = int(math.floor(math.log(XOR) / math.log(2)) + 1)
    # Check for 1's in the binary form using Brian Kerninghan's Algorithm
    setCount = 0
    while XOR > 0:
        XOR = XOR & (XOR - 1)
        setCount += 1
    # the number of unset bits is equal to the difference between
    # total bits and the set bits count
    return totalBits - setCount
 
 
A, B = 10, 6
result = countUnmatchedBits(A, B)
print 'Number of matching bits : ', result
# This code is contributed by KaaL-EL.

C#




// C# program for the above approach
 
using System;
 
class GFG {
 
    // compute number of matching bits
    static int countUnmatchedBits(int A, int B)
    {
        // calculate XOR
        int XOR = (A ^ B);
        // count total number of bits using log2(XOR) + 1
        int totalBits
            = (int)Math.Floor(Math.Log(XOR) / Math.Log(2))
              + 1;
        // Check for 1's in the binary form using
        // Brian Kerninghan's Algorithm
        int setCount = 0;
        while (XOR > 0) {
            XOR = XOR & (XOR - 1);
            setCount++;
        }
        // the number of unset bits is equal to
        // the difference between total bits
        // and the set bits count
        return totalBits - setCount;
    }
 
    // Driver code
    public static void Main(string[] args)
    {
        int A = 10, B = 6;
        // Function call
        int result = countUnmatchedBits(A, B);
        Console.WriteLine("Number of matching bits : "
                          + result);
    }
}
 
// This code is contributed by phasing17

Javascript




// JavaScript implementation of the approach
  
// compute number of matching bits
function countUnmatchedBits(A, B)
{
    // calculate XOR
    let XOR = (A ^ B);
     
    // count total number of bits using log2(XOR) + 1
    let totalBits = Math.floor(Math.log2(XOR)) + 1;
     
    // Check for 1's in the binary form using
    // Brian Kerninghan's Algorithm
    let setCount = 0;
    while (XOR) {
        XOR = XOR & (XOR - 1);
        setCount++;
    }
     
    // the number of unset bits is equal to
    // the difference between total bits
    // and the set bits count
    return totalBits - setCount;
}
  
// Driver code
let A = 10;
let B = 6;
 
// Function Call
let result = countUnmatchedBits(A, B);
console.log("Number of matching bits :", result);
 
// this code is contributed by phasing17

Output

Number of matching bits : 2

Time Complexity: O(1)
Auxiliary Space: O(1)


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