# Bitwise XOR of Bitwise AND of all pairs from two given arrays

Given two arrays arr1[] and arr2[] consisting of N and M integers respectively, the task is to print the Bitwise XOR of Bitwise AND of all pairs possible by selecting an element from arr1[] and arr2[].

Examples:

Input: arr1[] = {1, 2, 3}, arr2[] = {6, 5}
Output: 0
Explanation:
Bitwise AND of the pair (arr1[0], arr2[]) = 1 & 6 = 0.
Bitwise AND of the pair (arr1[0], arr2[1]) = 1 & 5 = 1.
Bitwise AND of the pair (arr1[1], arr2[0]) = 2 & 6 = 2.
Bitwise AND of the pair (arr1[1], arr2[1]) = 2 & 5 = 0.
Bitwise AND of the pair (arr1[2], arr2[0]) = 3 & 6 = 2.
Bitwise AND of the pair (arr1[2], arr2[1]) = 3 & 5 = 1.
Therefore, the Bitwise XOR of the obtained Bitwise AND values = 0 ^ 1 ^ 2 ^ 0^ 2 ^ 1 = 0.

Input: arr1[] = {12}, arr2[] = {4}
Output: 4

Naive Approach: The simplest approach is to find Bitwise AND of all possible pairs possible by selecting an element from arr1[] and another element from arr2[] and then, calculating the Bitwise XOR of all Bitwise AND of resultant pairs.

Below is the implementation of the above approach:

## C++

 `// C++ program for the above approach` `#include ` `using` `namespace` `std;`   `// Function to find the Bitwise XOR` `// of Bitwise AND of all pairs from` `// the arrays arr1[] and arr2[]` `int` `findXORS(``int` `arr1[], ``int` `arr2[], ``int` `N, ``int` `M)` `{` `    ``// Stores the result` `    ``int` `res = 0;`   `    ``// Iterate over the range [0, N - 1]` `    ``for` `(``int` `i = 0; i < N; i++) {`   `        ``// Iterate over the range [0, M - 1]` `        ``for` `(``int` `j = 0; j < M; j++) {`   `            ``// Stores Bitwise AND of` `            ``// the pair {arr1[i], arr2[j]}` `            ``int` `temp = arr1[i] & arr2[j];`   `            ``// Update res` `            ``res ^= temp;` `        ``}` `    ``}` `    ``// Return the res` `    ``return` `res;` `}`   `// Driver Code` `int` `main()` `{` `    ``// Input` `    ``int` `arr1[] = { 1, 2, 3 };` `    ``int` `arr2[] = { 6, 5 };` `    ``int` `N = ``sizeof``(arr1) / ``sizeof``(arr1[0]);` `    ``int` `M = ``sizeof``(arr2) / ``sizeof``(arr2[0]);`   `    ``cout << findXORS(arr1, arr2, N, M);`   `    ``return` `0;` `}`

## Java

 `// Java program for the above approach` `import` `java.util.*;`   `class` `GFG{`   `// Function to find the Bitwise XOR` `// of Bitwise AND of all pairs from` `// the arrays arr1[] and arr2[]` `static` `int` `findXORS(``int` `arr1[], ``int` `arr2[],` `                    ``int` `N, ``int` `M)` `{` `    `  `    ``// Stores the result` `    ``int` `res = ``0``;`   `    ``// Iterate over the range [0, N - 1]` `    ``for``(``int` `i = ``0``; i < N; i++) ` `    ``{` `        `  `        ``// Iterate over the range [0, M - 1]` `        ``for``(``int` `j = ``0``; j < M; j++) ` `        ``{` `            `  `            ``// Stores Bitwise AND of` `            ``// the pair {arr1[i], arr2[j]}` `            ``int` `temp = arr1[i] & arr2[j];`   `            ``// Update res` `            ``res ^= temp;` `        ``}` `    ``}` `    `  `    ``// Return the res` `    ``return` `res;` `}`   `// Driver Code` `public` `static` `void` `main(String[] args)` `{` `    `  `    ``// Input` `    ``int` `arr1[] = { ``1``, ``2``, ``3` `};` `    ``int` `arr2[] = { ``6``, ``5` `};` `    ``int` `N = arr1.length;` `    ``int` `M = arr2.length;`   `    ``System.out.print(findXORS(arr1, arr2, N, M));` `}` `}`   `// This code is contributed by 29AjayKumar`

## Python3

 `# Python 3 program for the above approach`   `# Function to find the Bitwise XOR` `# of Bitwise AND of all pairs from` `# the arrays arr1[] and arr2[]` `def` `findXORS(arr1, arr2, N, M):` `  `  `    ``# Stores the result` `    ``res ``=` `0`   `    ``# Iterate over the range [0, N - 1]` `    ``for` `i ``in` `range``(N):` `      `  `        ``# Iterate over the range [0, M - 1]` `        ``for` `j ``in` `range``(M):` `            ``# Stores Bitwise AND of` `            ``# the pair {arr1[i], arr2[j]}` `            ``temp ``=` `arr1[i] & arr2[j]`   `            ``# Update res` `            ``res ^``=` `temp` `    ``# Return the res` `    ``return` `res`   `# Driver Code` `if` `__name__ ``=``=` `'__main__'``:` `  `  `    ``# Input` `    ``arr1 ``=` `[``1``, ``2``, ``3``]` `    ``arr2 ``=` `[``6``, ``5``]` `    ``N ``=` `len``(arr1)` `    ``M ``=` `len``(arr2)` `    ``print``(findXORS(arr1, arr2, N, M))` `    `  `    ``# This code is contributed by ipg2016107.`

## C#

 `// C# program for the above approach` `using` `System;` `class` `GFG` `{` `  `  `    ``// Function to find the Bitwise XOR` `    ``// of Bitwise AND of all pairs from` `    ``// the arrays arr1[] and arr2[]` `    ``static` `int` `findXORS(``int``[] arr1, ``int``[] arr2, ``int` `N,` `                        ``int` `M)` `    ``{` `        ``// Stores the result` `        ``int` `res = 0;`   `        ``// Iterate over the range [0, N - 1]` `        ``for` `(``int` `i = 0; i < N; i++) {`   `            ``// Iterate over the range [0, M - 1]` `            ``for` `(``int` `j = 0; j < M; j++)` `            ``{`   `                ``// Stores Bitwise AND of` `                ``// the pair {arr1[i], arr2[j]}` `                ``int` `temp = arr1[i] & arr2[j];`   `                ``// Update res` `                ``res ^= temp;` `            ``}` `        ``}` `      `  `        ``// Return the res` `        ``return` `res;` `    ``}`   `    ``// Driver Code` `    ``public` `static` `void` `Main()` `    ``{` `      `  `        ``// Input` `        ``int``[] arr1 = { 1, 2, 3 };` `        ``int``[] arr2 = { 6, 5 };` `        ``int` `N = arr1.Length;` `        ``int` `M = arr2.Length;`   `        ``Console.Write(findXORS(arr1, arr2, N, M));` `    ``}` `}`   `// This code is contributed by ukasp.`

## Javascript

 ``

Output:

`0`

Time Complexity: O(N * M)
Auxiliary Space: O(1)

Efficient Approach: The above approach can be optimized based on the following observations:

• The Bitwise Xor and Bitwise And operation have Additive and Distributive properties.
• Therefore, considering the arrays as arr1[] = {A, B} and arr2[] = {X, Y}:
• (A AND X) XOR (A AND Y) XOR (B AND X) XOR (B AND Y)
• (A AND ( X XOR Y)) XOR (B AND ( X XOR Y))
• (A XOR B) AND (X XOR Y)
• Hence, from the above steps, the task is reduced to finding the bitwise And of bitwise XOR of arr1[] and arr2[].

Follow the steps below to solve the problem:

Below is the implementation of the above approach:

## C++

 `// C++ program for the above approach` `#include ` `using` `namespace` `std;`   `// Function to find the Bitwise XOR` `// of Bitwise AND of all pairs from` `// the arrays arr1[] and arr2[]` `int` `findXORS(``int` `arr1[], ``int` `arr2[],` `             ``int` `N, ``int` `M)` `{` `    ``// Stores XOR of array arr1[]` `    ``int` `XORS1 = 0;`   `    ``// Stores XOR of array arr2[]` `    ``int` `XORS2 = 0;`   `    ``// Traverse the array arr1[]` `    ``for` `(``int` `i = 0; i < N; i++) {` `        ``XORS1 ^= arr1[i];` `    ``}`   `    ``// Traverse the array arr2[]` `    ``for` `(``int` `i = 0; i < M; i++) {` `        ``XORS2 ^= arr2[i];` `    ``}`   `    ``// Return the result` `    ``return` `XORS1 and XORS2;` `}`   `// Driver Code` `int` `main()` `{` `    ``// Input` `    ``int` `arr1[] = { 1, 2, 3 };` `    ``int` `arr2[] = { 6, 5 };` `    ``int` `N = ``sizeof``(arr1) / ``sizeof``(arr1[0]);` `    ``int` `M = ``sizeof``(arr2) / ``sizeof``(arr2[0]);`   `    ``cout << findXORS(arr1, arr2, N, M);`   `    ``return` `0;` `}`

## Java

 `// Java program for the above approach` `import` `java.io.*;` `import` `java.lang.*;` `import` `java.util.*;`   `class` `GFG{`   `// Function to find the Bitwise XOR` `// of Bitwise AND of all pairs from` `// the arrays arr1[] and arr2[]` `static` `int` `findXORS(``int` `arr1[], ``int` `arr2[],` `                    ``int` `N, ``int` `M)` `{` `    `  `    ``// Stores XOR of array arr1[]` `    ``int` `XORS1 = ``0``;` ` `  `    ``// Stores XOR of array arr2[]` `    ``int` `XORS2 = ``0``;` ` `  `    ``// Traverse the array arr1[]` `    ``for``(``int` `i = ``0``; i < N; i++) ` `    ``{` `        ``XORS1 ^= arr1[i];` `    ``}` ` `  `    ``// Traverse the array arr2[]` `    ``for``(``int` `i = ``0``; i < M; i++)` `    ``{` `        ``XORS2 ^= arr2[i];` `    ``}` ` `  `    ``// Return the result` `    ``return` `(XORS1 & XORS2);` `}`   `// Driver Code` `public` `static` `void` `main(String[] args)` `{` `    `  `    ``// Input` `    ``int` `arr1[] = { ``1``, ``2``, ``3` `};` `    ``int` `arr2[] = { ``6``, ``5` `};` `    ``int` `N = arr1.length;` `    ``int` `M = arr2.length;` `    `  `    ``System.out.println(findXORS(arr1, arr2, N, M));` `}` `}`   `// This code is contributed by susmitakundugoaldanga`

## Python3

 `# Python3 program for the above approach`   `# Function to find the Bitwise XOR` `# of Bitwise AND of all pairs from` `# the arrays arr1[] and arr2[]` `def` `findXORS(arr1, arr2, N, M):` `    `  `    ``# Stores XOR of array arr1[]` `    ``XORS1 ``=` `0`   `    ``# Stores XOR of array arr2[]` `    ``XORS2 ``=` `0`   `    ``# Traverse the array arr1[]` `    ``for` `i ``in` `range``(N):` `        ``XORS1 ^``=` `arr1[i]`   `    ``# Traverse the array arr2[]` `    ``for` `i ``in` `range``(M):` `        ``XORS2 ^``=` `arr2[i]`   `    ``# Return the result` `    ``return` `XORS1 ``and` `XORS2`   `# Driver Code` `if` `__name__ ``=``=` `'__main__'``:` `    `  `    ``# Input` `    ``arr1 ``=` `[ ``1``, ``2``, ``3` `]` `    ``arr2 ``=` `[ ``6``, ``5` `]` `    ``N ``=` `len``(arr1)` `    ``M ``=` `len``(arr2)` `    `  `    ``print``(findXORS(arr1, arr2, N, M))`   `# This code is contributed by bgangwar59`

## C#

 `// C# program for the above approach` `using` `System;`   `class` `GFG{`   `// Function to find the Bitwise XOR` `// of Bitwise AND of all pairs from` `// the arrays arr1[] and arr2[]` `static` `int` `findXORS(``int` `[]arr1, ``int` `[]arr2,` `                    ``int` `N, ``int` `M)` `{` `    `  `    ``// Stores XOR of array arr1[]` `    ``int` `XORS1 = 0;` ` `  `    ``// Stores XOR of array arr2[]` `    ``int` `XORS2 = 0;` ` `  `    ``// Traverse the array arr1[]` `    ``for``(``int` `i = 0; i < N; i++) ` `    ``{` `        ``XORS1 ^= arr1[i];` `    ``}` ` `  `    ``// Traverse the array arr2[]` `    ``for``(``int` `i = 0; i < M; i++)` `    ``{` `        ``XORS2 ^= arr2[i];` `    ``}` ` `  `    ``// Return the result` `    ``return` `(XORS1 & XORS2);` `}`   `// Driver Code` `public` `static` `void` `Main(String[] args)` `{` `    `  `    ``// Input` `    ``int` `[]arr1 = { 1, 2, 3 };` `    ``int` `[]arr2 = { 6, 5 };` `    ``int` `N = arr1.Length;` `    ``int` `M = arr2.Length;` `    `  `    ``Console.WriteLine(findXORS(arr1, arr2, N, M));` `}` `}`   `// This code is contributed by 29AjayKumar`

## Javascript

 ``

Output:

`0`

Time Complexity: O(N + M)
Auxiliary Space: O(1)

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