Count number of common elements between two arrays by using Bitset and Bitwise operation

Given two arrays a[] and b[], the task is to find the count of common elements in both the given arrays. Note that both the arrays contain distinct (individually) positive integers.

Examples:

Input: a[] = {1, 2, 3}, b[] = {2, 4, 3}
Output: 2
2 and 3 are common to both the arrays.

Input: a[] = {1, 4, 7, 2, 3}, b[] = {2, 11, 7, 4, 15, 20, 24}
Output: 3

Approach: We will use 3 bitset of same size. First we will traverse first array and set the bit 1 to position a[i] in first bitset.
After that we will traverse second array and set the bit 1 to position b[i] in second bitset.
At last we will find the bitwise AND of both the bitsets and if the ith position of the resultant bitset is 1 then it implies that ith position of first and second bitsets are also 1 and i is the common element in both the arrays.

Below is the implementation of the above approach:

C++

filter_none

edit
close

play_arrow

link
brightness_4
code

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
#define MAX 100000
bitset<MAX> bit1, bit2, bit3;
  
// Function to return the count of common elements
int count_common(int a[], int n, int b[], int m)
{
  
    // Traverse the first array
    for (int i = 0; i < n; i++) {
  
        // Set 1 at position a[i]
        bit1.set(a[i]);
    }
  
    // Traverse the second array
    for (int i = 0; i < m; i++) {
  
        // Set 1 at position b[i]
        bit2.set(b[i]);
    }
  
    // Bitwise AND of both the bitsets
    bit3 = bit1 & bit2;
  
    // Find the count of 1's
    int count = bit3.count();
  
    return count;
}
  
// Driver code
int main()
{
  
    int a[] = { 1, 4, 7, 2, 3 };
    int b[] = { 2, 11, 7, 4, 15, 20, 24 };
    int n = sizeof(a) / sizeof(a[0]);
    int m = sizeof(b) / sizeof(b[0]);
  
    cout << count_common(a, n, b, m);
  
    return 0;
}

chevron_right


Python3

filter_none

edit
close

play_arrow

link
brightness_4
code

# Python3 implementation of the approach 
  
MAX = 100000
bit1 , bit2, bit3 = 0, 0, 0
  
# Function to return the count of common elements 
def count_common(a, n, b, m) : 
  
    # Traverse the first array 
    for i in range(n) :
          
        global bit1, bit2, bit3
          
        # Set 1 at (index)position a[i]
        bit1 = bit1 | (1<<a[i])
  
    # Traverse the second array 
    for i in range(m) :
  
        # Set 1 at (index)position b[i] 
        bit2 = bit2 | (1<<b[i])
  
    # Bitwise AND of both the bitsets 
    bit3 = bit1 & bit2; 
  
    # Find the count of 1's 
    count = bin(bit3).count('1'); 
  
    return count; 
  
# Driver code 
if __name__ == "__main__"
  
    a = [ 1, 4, 7, 2, 3 ]; 
    b = [ 2, 11, 7, 4, 15, 20, 24 ]; 
    n = len(a); 
    m = len(b); 
  
    print(count_common(a, n, b, m)); 
  
# This code is contributed by AnkitRai01

chevron_right


Output:

3


My Personal Notes arrow_drop_up

Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.



Improved By : AnkitRai01