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Find XOR sum of Bitwise AND of all pairs from given two Arrays

  • Last Updated : 16 Sep, 2021

Given two arrays A and B of sizes N and M respectively, the task is to calculate the XOR sum of bitwise ANDs of all pairs of A and B

Examples:

Input: A={3, 5}, B={2, 3}, N=2, M=2
Output:
0
Explanation:
The answer is (3&2)^(3&3)^(5&2)^(5&3)=1^3^0^2=0.

Input: A={1, 2, 3}, B={5, 6}, N=3, M=2
Output:
0

Naive Approach: The naive approach would be to use nested loops to calculate the bitwise ANDs of all pairs and then find their XOR sum. Follow the steps below to solve the problem:



  1. Initialize a variable ans to -1, which will store that final answer.
  2. Traverse array A, and do the following:
    1. For each current element, traverse the array B, and do the following:
      1. If ans is equal to -1, store the bitwise AND of the elements in ans.
      2. Otherwise, store the bitwise XOR of ans and the bitwise AND of the elements in ans.
  3. Return ans.

Below is the implementation of the above approach:

C++




// C++ algorithm for the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to calculate the XOR sum of all ANDS of all
// pairs on A[] and B[]
int XorSum(int A[], int B[], int N, int M)
{
    // variable to store anshu
    int ans = -1;
    for (int i = 0; i < N; i++) {
        for (int j = 0; j < M; j++) {
            // when there has been no
            // AND of pairs before this
            if (ans == -1)
                ans = (A[i] & B[j]);
            else
                ans ^= (A[i] & B[j]);
        }
    }
    return ans;
}
 
// Driver code
int main()
{
    // Input
    int A[] = { 3, 5 };
    int B[] = { 2, 3 };
    int N = sizeof(A) / sizeof(A[0]);
    int M = sizeof(B) / sizeof(B[0]);
 
    // Function call
    cout << XorSum(A, B, N, M) << endl;
}

Java




// Java program for the above approach
import java.io.*;
class GFG
{
   
  // Function to calculate the XOR sum of all ANDS of all
// pairs on A[] and B[]
public static int XorSum(int A[], int B[], int N, int M)
{
   
    // variable to store anshu
    int ans = -1;
    for (int i = 0; i < N; i++)
    {
        for (int j = 0; j < M; j++)
        {
           
            // when there has been no
            // AND of pairs before this
            if (ans == -1)
                ans = (A[i] & B[j]);
            else
                ans ^= (A[i] & B[j]);
        }
    }
    return ans;
}
 
// Driver code
    public static void main (String[] args)
    {
       
         // Input
    int A[] = { 3, 5 };
    int B[] = { 2, 3 };
    int N = A.length;
    int M =B.length;
 
    // Function call
    System.out.println(XorSum(A, B, N, M));
       
    }
}
 
// This code is contributed by Potta Lokesh

Python3




# Python3 algorithm for the above approach
 
# Function to calculate the XOR sum of all ANDS of all
# pairs on A and B
def XorSum(A, B, N, M):
   
    # variable to store anshu
    ans = -1
    for i in range(N):
        for j in range(M):
           
            # when there has been no
            # AND of pairs before this
            if (ans == -1):
                ans = (A[i] & B[j])
            else:
                ans ^= (A[i] & B[j])
 
    return ans
   
# Driver code
if __name__ == '__main__':
   
    # Input
    A = [3, 5]
    B = [2, 3]
 
    N = len(A)
    M = len(B)
 
    # Function call
    print (XorSum(A, B, N, M))
 
# This code is contributed by mohit kumar 29.

C#




// C# algorithm for the above approach
using System;
using System.Collections.Generic;
 
class GFG{
  
// Function to calculate the XOR sum of all ANDS of all
// pairs on A[] and B[]
static int XorSum(int []A, int []B, int N, int M)
{
   
    // variable to store anshu
    int ans = -1;
    for (int i = 0; i < N; i++)
    {
        for (int j = 0; j < M; j++)
        {
           
            // when there has been no
            // AND of pairs before this
            if (ans == -1)
                ans = (A[i] & B[j]);
            else
                ans ^= (A[i] & B[j]);
        }
    }
    return ans;
}
 
// Driver code
public static void Main()
{
    // Input]
    int []A = {3, 5};
    int []B = {2, 3};
    int N = A.Length;
    int M = B.Length;
 
    // Function call
    Console.Write(XorSum(A, B, N, M));
}
}
 
// This code is contributed by SURENDER_GANGWAR.

Javascript




<script>
// JavaScript program for the above approach
 
        // Function to calculate the XOR sum of all ANDS of all
        // pairs on A[] and B[]
        function XorSum(A, B, N, M)
        {
         
            // variable to store anshu
            let ans = -1;
            for (let i = 0; i < N; i++)
            {
                for (let j = 0; j < M; j++)
                {
                 
                    // when there has been no
                    // AND of pairs before this
                    if (ans == -1)
                        ans = (A[i] & B[j]);
                    else
                        ans ^= (A[i] & B[j]);
                }
            }
            return ans;
        }
 
        // Driver code
 
        // Input
        let A = [3, 5];
        let B = [2, 3];
        let N = A.length;
        let M = B.length;
 
        // Function call
        document.write(XorSum(A, B, N, M));
 
  // This code is contributed by Potta Lokesh
    </script>
Output
0

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

Efficient Approach:

Observation: The distributive property of XOR over AND can be used to solve the problem. 

let A[] = [a, b, c] 
let B[] = [d, e] 
Result : 
(a & d) ^ (a & e) ^ (b & d) ^ (b & e) ^ (c & d) ^ (c & e) 
By applying distributive property, 
[(a ^ b ^ c) & d] ^ [(a ^ b ^ e) & e] 
(a ^ b ^ c) & (d ^ e) 
 

Follow the steps below to solve the problem: 

  • Initialize two variables ans1 and ans2 to 0, which will store the bitwise XOR sums of the first array and the second array respectively.
  • Traverse A and for each current element:
    • Store the bitwise XOR of ans1 and the current element in ans1.
  • Traverse B and for each current element:
    • Store the bitwise XOR of ans2 and the current element in ans2.
  • Return ans1&ans2.

Below is the implementation of the above approach: 

C++14




// C++ algorithm for the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to calculate the XOR sum of all ANDS of all
// pairs on A[] and B[]
int XorSum(int A[], int B[], int N, int M)
{
    // variable to store xor sums
    // of first array and second
    // array respectively.
    int ans1 = 0, ans2 = 0;
    // Xor sum of first array
    for (int i = 0; i < N; i++)
        ans1 = ans1 ^ A[i];
    // Xor sum of second array
    for (int i = 0; i < M; i++)
        ans2 = ans2 ^ B[i];
    // required answer
    return (ans1 & ans2);
}
// Driver code
int main()
{
    // Input
    int A[] = { 3, 5 };
    int B[] = { 2, 3 };
    int N = sizeof(A) / sizeof(A[0]);
    int M = sizeof(B) / sizeof(B[0]);
 
    // Function call
    cout << XorSum(A, B, N, M) << endl;
}

Java




// Java algorithm for the above approach
import java.io.*;
 
class GFG{
 
// Function to calculate the XOR sum of
// all ANDS of all pairs on A[] and B[]
static int XorSum(int A[], int B[], int N, int M)
{
     
    // Variable to store xor sums
    // of first array and second
    // array respectively.
    int ans1 = 0, ans2 = 0;
     
    // Xor sum of first array
    for(int i = 0; i < N; i++)
        ans1 = ans1 ^ A[i];
         
    // Xor sum of second array
    for(int i = 0; i < M; i++)
        ans2 = ans2 ^ B[i];
         
    // Required answer
    return (ans1 & ans2);
}
 
// Driver code
public static void main(String[] args)
{
     
    // Input
    int A[] = { 3, 5 };
    int B[] = { 2, 3 };
    int N = A.length;
    int M = B.length;
 
    // Function call
    System.out.print(XorSum(A, B, N, M));
}
}
 
// This code is contributed by subham348

Python3




# python 3 algorithm for the above approach
 
# Function to calculate the XOR sum of all ANDS of all
# pairs on A[] and B[]
def XorSum(A, B, N, M):
   
    # variable to store xor sums
    # of first array and second
    # array respectively.
    ans1 = 0
    ans2 = 0
     
    # Xor sum of first array
    for i in range(N):
        ans1 = ans1 ^ A[i]
         
    # Xor sum of second array
    for i in range(M):
        ans2 = ans2 ^ B[i]
         
    # required answer
    return (ans1 & ans2)
 
# Driver code
if __name__ == '__main__':
   
    # Input
    A = [3, 5]
    B = [2, 3]
    N = len(A)
    M = len(B)
 
    # Function call
    print(XorSum(A, B, N, M))
     
    # This code is contributed by bgangwar59.

C#




// C# program for the above approach
 
using System;
 
class GFG {
 
// Function to calculate the XOR sum of
// all ANDS of all pairs on A[] and B[]
static int XorSum(int[] A, int[] B, int N, int M)
{
     
    // Variable to store xor sums
    // of first array and second
    // array respectively.
    int ans1 = 0, ans2 = 0;
     
    // Xor sum of first array
    for(int i = 0; i < N; i++)
        ans1 = ans1 ^ A[i];
         
    // Xor sum of second array
    for(int i = 0; i < M; i++)
        ans2 = ans2 ^ B[i];
         
    // Required answer
    return (ans1 & ans2);
}
 
  // Driver code
  public static void Main (String[] args)
  {
 
    // Input
    int[] A = { 3, 5 };
    int[] B = { 2, 3 };
    int N = A.Length;
    int M = B.Length;
 
    // Function call
    Console.Write(XorSum(A, B, N, M));
  }
}
 
// This code is contributed by code_hunt.

Javascript




<script>
 
// JavaScript algorithm for the above approach
 
// Function to calculate the XOR sum of all ANDS of all
// pairs on A[] and B[]
function XorSum(A, B, N, M)
{
    // variable to store xor sums
    // of first array and second
    // array respectively.
    let ans1 = 0, ans2 = 0;
    // Xor sum of first array
    for (let i = 0; i < N; i++)
        ans1 = ans1 ^ A[i];
    // Xor sum of second array
    for (let i = 0; i < M; i++)
        ans2 = ans2 ^ B[i];
    // required answer
    return (ans1 & ans2);
}
// Driver code
    // Input
    let A = [ 3, 5 ];
    let B = [ 2, 3 ];
    let N = A.length;
    let M = B.length;
 
    // Function call
    document.write(XorSum(A, B, N, M));
 
</script>
Output
0

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

 

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