来自两个给定数组的所有对的按位与的按位异或

给定两个数组arr1[]和arr2[] 分别由N和M 个整数组成，任务是通过从arr1[]和arr2[] 中选择一个元素来打印所有可能对的按位与的按位异或。

例子：

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, Bitwise XOR of the obtained Bitwise AND values = 0 ^ 1 ^ 2 ^ 0^ 2 ^ 1 = 0.

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

朴素的方法：最简单的方法是通过从arr1[] 中选择一个元素和从arr2[] 中选择另一个元素，然后计算结果对的所有按位与的按位异或，找到所有可能对的按位与。

下面是上述方法的实现：

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

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.

输出：

0

时间复杂度： O(N * M)
辅助空间： O(1)

高效的方法：上述方法可以基于以下观察进行优化：

• Bitwise Xor 和 Bitwise And 操作具有 Additive 和 Distributive 属性。
• 因此，将数组视为 arr1[] = {A, B} 和 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)
• 因此，从上述步骤，任务被简化为找到arr1[]和arr2[]的按位异或的按位与。

请按照以下步骤解决问题：

• 找到数组 arr1[] 的每个数组元素的按位异或并将其存储在一个变量中，比如XORS1。
• 找到数组 arr2[] 的每个数组元素的按位异或并将其存储在一个变量中，比如XORS2 。
• 最后，将结果打印为XORS1 和 XORS2 的按位与。

下面是上述方法的实现：

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

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

输出：

0

时间复杂度： O(N + M)
辅助空间： O(1)

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