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 Codeint 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 approachimport 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 Codepublic 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 Codeif __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 approachusing 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
<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// Inputlet 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:
- Find the bitwise Xor of every array element of an array arr1[] and store it in a variable, say XORS1.
- Find the bitwise Xor of every array element of an array arr2[] and store it in a variable, say XORS2.
- Finally, print the result as bitwise AND of XORS1 and XORS2.
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 Codeint 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 approachimport 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 Codepublic 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 Codeif __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 approachusing 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 Codepublic 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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