Form an Array so that their Bitwise XOR sum satisfies given conditions

Given an array arr[] of size N ( N is even ), the task is to construct an array B[] such that the sum of B[i] XOR PrefixXor[i], where 1 ? i ? N is even and is divisible by the first N/2 elements of arr[]. 

PrefixXor[] array is an array where each element will indicate the XOR of all elements from 1st element to the current index element.

Examples:

Input: N = 4, arr[] = {2, 6, 1, 5}
Output: B[] = {0, 6, 3, 6}
Explanation: Since, prefix array is prefixXor[] = {2, 4, 5, 0}, (2 ^ 0) + (4 ^ 6) + (5 ^ 3) + (0^6) = 2 + 2 + 6 + 6 = 16, which is even and divisible by 8.

Input: N = 6, arr[] = {2,  5, 3,  4, 6, 1}
Output: B[] = {0, 5, 1, 5, 5, 4}
Explanation: Since  prefix, the array is prefixXor[]={2, 7, 4, 0, 6, 7}, So, (2 ^ 0) + (7 ^ 5) + (4 ^ 1) + (0 ^ 5) + (6 ^ 5) + (7 ^ 4) = 2 + 2 + 5 + 5 + 3 + 3 = 20, which is even and divisible by 10.

Approach: Implement the idea below to solve the problem:

Let's first analyze the question. The given sum should be even and should be divisible by the first N/2 elements. So more formally it means :

if array arr[] = {a, b, c, d},  then its prefix xor is Prefix_xor[]=[a, a^b, a^b^c, a^b^c^d} 
Let assume array B[] = {x, y, z, w} then S= (x^a) + (y^(a^b)) + (z^(a^b^c)) + ( w^(a^b^c^d)). So according to the question:

• S%2 == 0 and
• S%(a+b) == 0

So on combining we get S % (2*(a + b)) == 0

Steps were taken to solve the problem:

• Initialize the pointer = 0 variable to iterate over array arr[].
• Calculate prefixXor[] array.
• Initialize array B[] of size N.
• While iterating over the array, if the difference == 2, increment the pointer, otherwise:
  • Store arr[pointer] ^ prefixXor[i] in B[i].
  • Increment difference by 1. 

Below is the implementation of the above approach:

// C++ code to implement the approach

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

// Function to generates valid array
// according to the given conditions
void valid_array_formation(int N, int arr[])
{
    int prefix_xor[N] = { arr[0] };

    // Calculating the prefix xor array
    for (int i = 1; i < N; i++) {
        prefix_xor[i] = prefix_xor[i - 1] ^ arr[i];
    }

    int B[N] = { 0 };

    // Pointer to extract which element
    // is to be extracted from array a
    int pointer = 0;

    int difference = 0;

    for (int i = 0; i < N; i++) {

        // If difference becomes 2 then we
        // increment the pointer
        if (difference == 2) {
            pointer++;
            difference = 0;
        }

        // B[i] i calculated accordingly
        B[i] = prefix_xor[i] ^ arr[pointer];
        difference++;
    }

    for (int i = 0; i < N; i++) {

        // Printing the required array b
        cout << B[i] << " ";
    }

    cout << endl;
}

// Driver code
int main()
{

    // Test case 1
    int N = 4;
    int arr1[] = { 2, 6, 1, 5 };

    // Function call
    valid_array_formation(N, arr1);

    // Test case 2
    N = 6;
    int arr2[] = { 2, 5, 3, 4, 6, 1 };

    // Function call
    valid_array_formation(N, arr2);

    return 0;
}

// Java code to implement the approach
import java.util.*;

class GFG {

  // Function to generates valid array
  // according to the given conditions
  public static void valid_array_formation(int N,
                                           int arr[])
  {
    int prefix_xor[] = new int[N];
    prefix_xor[0] = arr[0];

    // Calculating the prefix xor array
    for (int i = 1; i < N; i++) {
      prefix_xor[i] = prefix_xor[i - 1] ^ arr[i];
    }

    int B[] = new int[N];

    // Pointer to extract which element
    // is to be extracted from array a
    int pointer = 0;

    int difference = 0;

    for (int i = 0; i < N; i++) {

      // If difference becomes 2 then we
      // increment the pointer
      if (difference == 2) {
        pointer++;
        difference = 0;
      }

      // B[i] i calculated accordingly
      B[i] = prefix_xor[i] ^ arr[pointer];
      difference++;
    }

    for (int i = 0; i < N; i++) {

      // Printing the required array b
      System.out.print(B[i] + " ");
    }

    System.out.println();
  }

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

    // Test case 1
    int N = 4;
    int arr1[] = { 2, 6, 1, 5 };

    // Function call
    valid_array_formation(N, arr1);

    // Test case 2
    N = 6;
    int arr2[] = { 2, 5, 3, 4, 6, 1 };

    // Function call
    valid_array_formation(N, arr2);
  }
}

// This Code is Contributed by Prasad Kandekar(prasad264)

// Function to generates valid array
// according to the given conditions
function validArrayFormation(N, arr) {
  let prefixXor = [arr[0]];

  // Calculating the prefix xor array
  for (let i = 1; i < N; i++) {
    prefixXor.push(prefixXor[i - 1] ^ arr[i]);
  }

  let B = Array(N).fill(0);

  // Pointer to extract which element
  // is to be extracted from array a
  let pointer = 0;

  let difference = 0;

  for (let i = 0; i < N; i++) {
    // If difference becomes 2 then we
    // increment the pointer
    if (difference == 2) {
      pointer++;
      difference = 0;
    }

    // B[i] i calculated accordingly
    B[i] = prefixXor[i] ^ arr[pointer];
    difference++;
  }

  // Printing the required array b
  console.log(B.join(" "));
}

// Test case 1
let N = 4;
let arr1 = [2, 6, 1, 5];

// Function call
validArrayFormation(N, arr1);

// Test case 2
N = 6;
let arr2 = [2, 5, 3, 4, 6, 1];

// Function call
validArrayFormation(N, arr2);

# Function to generates valid array
# according to the given conditions
def valid_array_formation(N, arr):
    prefix_xor = [arr[0]]

    # Calculating the prefix xor array
    for i in range(1, N):
        prefix_xor.append(prefix_xor[i-1] ^ arr[i])

    B = [0] * N

    # Pointer to extract which element
    # is to be extracted from array a
    pointer = 0

    difference = 0

    for i in range(N):
        # If difference becomes 2 then we
        # increment the pointer
        if difference == 2:
            pointer += 1
            difference = 0

        # B[i] i calculated accordingly
        B[i] = prefix_xor[i] ^ arr[pointer]
        difference += 1

    # Printing the required array b
    print(*B)

# Test case 1
N = 4
arr1 = [2, 6, 1, 5]

# Function call
valid_array_formation(N, arr1)

# Test case 2
N = 6
arr2 = [2, 5, 3, 4, 6, 1]

# Function call
valid_array_formation(N, arr2)

using System;

public class Program
{
    // Function to generates valid array
    // according to the given conditions
    public static void ValidArrayFormation(int N, int[] arr)
    {
        int[] prefixXor = new int[N];
        prefixXor[0] = arr[0];

        // Calculating the prefix xor array
        for (int i = 1; i < N; i++)
        {
            prefixXor[i] = prefixXor[i - 1] ^ arr[i];
        }

        int[] B = new int[N];

        // Pointer to extract which element
        // is to be extracted from array a
        int pointer = 0;

        int difference = 0;

        for (int i = 0; i < N; i++)
        {
            // If difference becomes 2 then we
            // increment the pointer
            if (difference == 2)
            {
                pointer++;
                difference = 0;
            }

            // B[i] i calculated accordingly
            B[i] = prefixXor[i] ^ arr[pointer];
            difference++;
        }

        // Printing the required array b
        Console.WriteLine(string.Join(" ", B));
    }

    public static void Main()
    {
        // Test case 1
        int N = 4;
        int[] arr1 = new int[] { 2, 6, 1, 5 };

        // Function call
        ValidArrayFormation(N, arr1);

        // Test case 2
        N = 6;
        int[] arr2 = new int[] { 2, 5, 3, 4, 6, 1 };

        // Function call
        ValidArrayFormation(N, arr2);
    }
}

0 6 3 6 
0 5 1 5 5 4 

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