Construct original array starting with K from an array of XOR of all elements except elements at same index

Given an array A[] consisting of N integers and first element of the array B[] as K, the task is to construct the array B[] from A[] such that for any index i, A[i] is the Bitwise XOR of all the array elements of B[] except B[i].

Examples:

Input: A[] = {13, 14, 10, 6}, K = 2
Output: 2 1 5 9
Explanation:
For any index i, A[i] is the Bitwise XOR of all elements of B[] except B[i]. 

1. B[1] ^ B[2] ^ B[3] = 1 ^ 5 ^ 9 = 13 = A[0]
2. B[0] ^ B[2] ^ B[3] = 2 ^ 5 ^ 9 = 14 = A[1]
3. B[0] ^ B[1] ^ B[3] = 2 ^ 1 ^ 9 = 10 = A[2]
4. B[0] ^ B[1] ^ B[2] = 2 ^ 1 ^ 5 = 6 = A[3]

Input: A[] = {3, 5, 0, 2, 4}, K = 2
Output: 2 4 1 3 5

Approach: The idea is based on the observation that Bitwise XOR of the same value calculated even number of times is 0.

For any index i, 
A[i] = B[0] ^ B[1] ^ … B[i-1] ^ B[i+1] ^ … B[n-1] 
Therefore, XOR of all elements of B[], totalXor = B[0] ^ B[1] ^ … B[i – 1] ^ B[i] ^ B[i + 1] ^ … ^ B[N – 1].
Therefore, B[i] = totalXor ^ A[i]. (Since every element occurs twice except B[i])

Follow the below steps to solve the problem:

• Store the Bitwise XOR of all the elements present in the array B[] in a variable, say totalXOR, where totalXOR = A[0] ^ K.
• Traverse the given array A[] for each array element A[i], store the value of B[i] as totalXOR ^ A[i].
• After completing the above steps, print the element stored in the array B[].

Below is the implementation of the above approach:

C++

// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;

// Function to construct an array
// with each element equal to XOR
// of all array elements except
// the element at the same index
void constructArray(int A[], int N,
                    int K)
{
    // Original array
    int B[N];

    // Stores Bitwise XOR of array
    int totalXOR = A[0] ^ K;

    // Calculate XOR of all array elements
    for (int i = 0; i < N; i++)
        B[i] = totalXOR ^ A[i];

    // Print the original array B[]
    for (int i = 0; i < N; i++) {
        cout << B[i] << " ";
    }
}

// Driver Code
int main()
{
    int A[] = { 13, 14, 10, 6 }, K = 2;
    int N = sizeof(A) / sizeof(A[0]);

    // Function Call
    constructArray(A, N, K);

    return 0;
}

Java

// Java program for the above approach
class GFG{

// Function to construct an array
// with each element equal to XOR
// of all array elements except
// the element at the same index
static void constructArray(int A[], int N,
                           int K)
{

    // Original array
    int B[] = new int[N];

    // Stores Bitwise XOR of array
    int totalXOR = A[0] ^ K;

    // Calculate XOR of all array elements
    for(int i = 0; i < N; i++)
        B[i] = totalXOR ^ A[i];

    // Print the original array B[]
    for(int i = 0; i < N; i++)
    {
        System.out.print(B[i] + " ");
    }
}

// Driver Code
public static void main(String[] args)
{
    int A[] = { 13, 14, 10, 6 }, K = 2;
    int N = A.length;

    // Function Call
    constructArray(A, N, K);
}
}

// This code is contributed by divyeshrabadiya07

Python3

# Python program for the above approach

# Function to construct an array
# with each element equal to XOR
# of all array elements except
# the element at the same index
def constructArray(A, N, K):

    # Original array
    B = [0] * N;

    # Stores Bitwise XOR of array
    totalXOR = A[0] ^ K;

    # Calculate XOR of all array elements
    for i in range(N):
        B[i] = totalXOR ^ A[i];

    # Print the original array B
    for i in range(N):
        print(B[i], end = " ");

# Driver Code
if __name__ == '__main__':
    A = [13, 14, 10, 6];
    K = 2;
    N = len(A);

    # Function Call
    constructArray(A, N, K);

# This code is contributed by Princi Singh

C#

// C# program for the above approach
using System;
using System.Collections;
class GFG {

    // Function to construct an array
    // with each element equal to XOR
    // of all array elements except
    // the element at the same index
    static void constructArray(int[] A, int N,
                               int K)
    {

        // Original array
        int[] B = new int[N];

        // Stores Bitwise XOR of array
        int totalXOR = A[0] ^ K;

        // Calculate XOR of all array elements
        for(int i = 0; i < N; i++)
            B[i] = totalXOR ^ A[i];

        // Print the original array B[]
        for(int i = 0; i < N; i++)
        {
            Console.Write(B[i] + " ");
        }
    }

  static void Main() {
    int[] A = { 13, 14, 10, 6 };
    int K = 2;
    int N = A.Length;

    // Function Call
    constructArray(A, N, K);
  }
}

// This code is contributed by divyesh072019

Javascript

<script>

// JavaScript program for the above approach

// Function to construct an array
// with each element equal to XOR
// of all array elements except
// the element at the same index
function constructArray(A, N, K)
{
    // Original array
    let B = new Array(N);

    // Stores Bitwise XOR of array
    let totalXOR = A[0] ^ K;

    // Calculate XOR of all array elements
    for (let i = 0; i < N; i++)
        B[i] = totalXOR ^ A[i];

    // Print the original array B[]
    for (let i = 0; i < N; i++) {
        document.write(B[i] + " ");
    }
}

// Driver Code
    let A = [ 13, 14, 10, 6 ], K = 2;
    let N = A.length;

    // Function Call
    constructArray(A, N, K);

// This code is contributed by Surbhi Tyagi.

</script>

Output:

2 1 5 9

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