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Bitwise XOR of Bitwise AND of all pairs from two given arrays
• Difficulty Level : Medium
• Last Updated : 15 Jun, 2021

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, arr2[]) = 1 & 6 = 0.
Bitwise AND of the pair (arr1, arr2) = 1 & 5 = 1.
Bitwise AND of the pair (arr1, arr2) = 2 & 6 = 2.
Bitwise AND of the pair (arr1, arr2) = 2 & 5 = 0.
Bitwise AND of the pair (arr1, arr2) = 3 & 6 = 2.
Bitwise AND of the pair (arr1, arr2) = 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);``    ``int` `M = ``sizeof``(arr2) / ``sizeof``(arr2);` `    ``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`

## 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.`

## 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.`

## 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);``    ``int` `M = ``sizeof``(arr2) / ``sizeof``(arr2);` `    ``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`

## 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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