Generate an array of minimum sum whose XOR of same-indexed elements with given array are Prime Numbers

Given an array Arr[] of N ( 1 ? N ? 10⁵)integers, the task is to generate an array B[] consisting of N non-zero elements, such that XOR of A_i ^ B_i always results in a prime number.

Note: The sum of XORs obtained should be minimized.

Examples:

Input: arr[] = {5, 4, 7, 6} 
Output: {7, 6, 5, 4} 
Explanation:
2 is the smallest prime number. Therefore, XORing A[i] with (A[i] ^ 2) 
gives us the smallest number which is prime.
A[i] ^ (A[i] ^ 2) = (A[i] ^ A[i]) ^ 2 = 0 ^ 2 = 2
because 
1. XOR of 5 ^ 7 = 2, which is prime 
2. XOR of 4 ^ 6 = 2, which is prime. 
3. XOR of 7 ^ 5 = 2, which is prime. 
4. XOR of 6 ^ 4 = 2, which is prime. 
The resultant sum is – 2 + 2 + 2 + 2 = 8, which is the minimum possible

Input: arr[] = {10, 16} 
Output: {8, 18}

Approach: This problem can be solved using a Greedy technique. Follow the steps below to solve the problem:

1. Since 2 is the smallest prime number possible, XOR of Arr[i] with B[i] = (Arr[i] ^ 2) will give us a prime number 2.
2. The contradiction arises when any of the array elements itself is Arr[i] = 2. In this case, B[i] = 2 ^ 2 results in 0.
3. Therefore, if Arr[i] = 2, set B[i] = (2 ^ 3) = 1, such that Arr[i] ^ K = 3, next smallest prime number.

Below is the implementation of the above approach:

C++

// C++ implementation of the above approach

#include <bits/stdc++.h>
using namespace std;

// Function to generate an array whose XOR
// with same-indexed elements of the given
// array is always a prime
void minXOR(vector<int>& Arr, int N)
{
    // Traverse the array
    for (int i = 0; i < N; i++) {

        // If current array element is 2
        if (Arr[i] == 2) {

            // Print its XOR with 3
            cout << (Arr[i] ^ 3) << " ";
        }
        // Otherwise
        else {

            // Print its XOR with 2
            cout << (Arr[i] ^ 2) << " ";
        }
    }
}

// Driver Code
int main()
{
    // Given array
    vector<int> Arr = { 5, 4, 7, 6 };

    // Size of the array
    int N = Arr.size();

    // Prints the required array
    minXOR(Arr, N);
    return 0;
}

Java

// Java implementation of the above approach
class GFG{

// Function to generate an array whose XOR
// with same-indexed elements of the given
// array is always a prime
private static void minXOR(int Arr[], int N)
{

    // Traverse the array
    for(int i = 0; i < N; i++)
    {

        // If current array element is 2
        if (Arr[i] == 2)
        {

            // Print its XOR with 3
            System.out.print((Arr[i] ^ 3) + " ");
        }

        // Otherwise
        else
        {

            // Print its XOR with 2
            System.out.print((Arr[i] ^ 2) + " ");
        }
    }
}

// Driver code
public static void main(String[] args)
{

    // Given array
    int Arr[] = { 5, 4, 7, 6 };

    // Size of the array
    int N = Arr.length;

    // Prints the required array
    minXOR(Arr, N);
}
}

// This code is contributed by MuskanKalra1

Python3

# Python3 implementation of the above approach

# Function to generate an array whose XOR
# with same-indexed elements of the given
# array is always a prime
def minXOR(Arr, N):

    # Traverse the array
    for i in range(N):

        # If current array element is 2
        if (Arr[i] == 2):

            # Print its XOR with 3
            print(Arr[i] ^ 3,end=" ")
        # Otherwise
        else:

            # Print its XOR with 2
            print(Arr[i] ^ 2,end=" ")

# Driver Code
if __name__ == '__main__':

    # Given array
    Arr = [5, 4, 7, 6 ]

    # Size of the array
    N = len(Arr)

    # Prints the required array
    minXOR(Arr, N)

# This code is contributed by mohit kumar 29

C#

// C# program to implement
// the above approach
using System;

class GFG{

// Function to generate an array whose XOR
// with same-indexed elements of the given
// array is always a prime
private static void minXOR(int[] Arr, int N)
{

    // Traverse the array
    for(int i = 0; i < N; i++)
    {

        // If current array element is 2
        if (Arr[i] == 2)
        {

            // Print its XOR with 3
            Console.Write((Arr[i] ^ 3) + " ");
        }

        // Otherwise
        else
        {

            // Print its XOR with 2
            Console.Write((Arr[i] ^ 2) + " ");
        }
    }
}

// Driver code
public static void Main()
{

    // Given array
    int[] Arr = { 5, 4, 7, 6 };

    // Size of the array
    int N = Arr.Length;

    // Prints the required array
    minXOR(Arr, N);
}
}

Javascript

<script>

// JavaScript implementation of the above approach

// Function to generate an array whose XOR
// with same-indexed elements of the given
// array is always a prime
function minXOR(Arr, N)
{

    // Traverse the array
    for(let i = 0; i < N; i++)
    {

        // If current array element is 2
        if (Arr[i] == 2)
        {

            // Print its XOR with 3
            document.write((Arr[i] ^ 3) + " ");
        }

        // Otherwise
        else
        {

            // Print its XOR with 2
            document.write((Arr[i] ^ 2) + " ");
        }
    }
}

// Driver code

    // Given array
    let Arr = [ 5, 4, 7, 6 ];

    // Size of the array
    let N = Arr.length;

    // Prints the required array
    minXOR(Arr, N);

</script>

Output:

7 6 5 4

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