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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[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 <bits/stdc++.h>
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

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




<script>
// Javascript program for the above approach
 
 
// Function to find the Bitwise XOR
// of Bitwise AND of all pairs from
// the arrays arr1[] and arr2[]
function findXORS(arr1, arr2, N, M) {
    // Stores the result
    let res = 0;
 
    // Iterate over the range [0, N - 1]
    for (let i = 0; i < N; i++) {
 
        // Iterate over the range [0, M - 1]
        for (let j = 0; j < M; j++) {
 
            // Stores Bitwise AND of
            // the pair {arr1[i], arr2[j]}
            let temp = arr1[i] & arr2[j];
 
            // Update res
            res ^= temp;
        }
    }
    // Return the res
    return res;
}
 
// Driver Code
 
// Input
let arr1 = [1, 2, 3];
let arr2 = [6, 5];
let N = arr1.length;
let M = arr2.length;
 
document.write(findXORS(arr1, arr2, N, M));
 
// This code is contributed by _saurabh_jaiswal
</script>
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 <bits/stdc++.h>
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

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




<script>
// JavaScript program for the above approach
 
// Function to find the Bitwise XOR
// of Bitwise AND of all pairs from
// the arrays arr1[] and arr2[]
function findXORS(arr1, arr2, N, M)
{
    // Stores XOR of array arr1[]
    let XORS1 = 0;
 
    // Stores XOR of array arr2[]
    let XORS2 = 0;
 
    // Traverse the array arr1[]
    for (let i = 0; i < N; i++) {
        XORS1 ^= arr1[i];
    }
 
    // Traverse the array arr2[]
    for (let i = 0; i < M; i++) {
        XORS2 ^= arr2[i];
    }
 
    // Return the result
    return XORS1 && XORS2;
}
 
// Driver Code
 
    // Input
    let arr1 = [ 1, 2, 3 ];
    let arr2 = [ 6, 5 ];
    let N = arr1.length;
    let M = arr2.length;
 
    document.write(findXORS(arr1, arr2, N, M));
     
       // This code is contributed by Dharanendra L V.
 
</script>
Output: 
0

 

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

 

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