# 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

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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++

 `// C++ implementation of the approach ` `#include ` `using` `namespace` `std; ` `#define MAX 100000 ` `bitset 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; ` `} `

## Python3

 `# 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``<

Output:

```3
```

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