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++ 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 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 |
# 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# 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. |
<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> |
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++ 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 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 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# 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 |
<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> |
0
Time Complexity: O(N + M)
Auxiliary Space: O(1)