Construct an array from XOR of all elements of array except element at same index

Given an array A[] having n positive elements. The task to create another array B[] such as B[i] is XOR of all elements of array A[] except A[i].
Examples :

Input : A[] = {2, 1, 5, 9}
Output : B[] = {13, 14, 10, 6}

Input : A[] = {2, 1, 3, 6}
Output : B[] = {4, 7, 5, 0}

Naive Approach : 
We can simple calculate B[i] as XOR of all elements of A[] except A[i], as 

for (int i = 0; i < n; i++)
{
    B[i] = 0;
    for (int j = 0; j < n; j++)
        if ( i != j)
            B[i] ^= A[j];
}

Time complexity for this naive approach is O (n^2). 
Auxiliary Space for this naive approach is O (n).
Optimized Approach : 
First calculate XOR of all elements of array A[] say ‘xor’, and for each element of array A[] calculate A[i] = xor ^ A[i] . 

int xor = 0;
for (int i = 0; i < n; i++)
    xor ^= A[i];

for (int i = 0; i < n; i++)
        A[i] = xor ^ A[i];

Time complexity for this approach is O (n). 
Auxiliary Space for this approach is O (1). 

C++

// C++ program to construct array from
// XOR of elements of given array
#include <bits/stdc++.h>
using namespace std;

// function to construct new array
void constructXOR(int A[], int n)
{
    // calculate xor of array
    int XOR = 0;
    for (int i = 0; i < n; i++)
        XOR ^= A[i];

    // update array
    for (int i = 0; i < n; i++)
        A[i] = XOR ^ A[i];
}

// Driver code
int main()
{
    int A[] = { 2, 4, 1, 3, 5};
    int n = sizeof(A) / sizeof(A[0]);
    constructXOR(A, n);

    // print result
    for (int i = 0; i < n; i++)
        cout << A[i] << " ";
    return 0;
}

Java

// Java program to construct array from
// XOR of elements of given array
class GFG
{

    // function to construct new array
    static void constructXOR(int A[], int n)
    {

        // calculate xor of array
        int XOR = 0;
        for (int i = 0; i < n; i++)
            XOR ^= A[i];

        // update array
        for (int i = 0; i < n; i++)
            A[i] = XOR ^ A[i];
    }

    // Driver code
    public static void main(String[] args)
    {
        int A[] = { 2, 4, 1, 3, 5};
        int n = A.length;
        constructXOR(A, n);

        // print result
        for (int i = 0; i < n; i++)
            System.out.print(A[i] + " ");
    }
}

// This code is contributed by Anant Agarwal.

Python3

# Python 3 program to construct
# array from XOR of elements
# of given array

# function to construct new array
def constructXOR(A, n):

    # calculate xor of array
    XOR = 0
    for i in range(0, n):
        XOR ^= A[i]

    # update array
    for i in range(0, n):
        A[i] = XOR ^ A[i]

# Driver code
A = [ 2, 4, 1, 3, 5 ]
n = len(A)
constructXOR(A, n)

# print result
for i in range(0,n):
    print(A[i], end =" ")

# This code is contributed by Smitha Dinesh Semwal

C#

// C# program to construct array from
// XOR of elements of given array
using System;

class GFG
{

    // function to construct new array
    static void constructXOR(int []A, int n)
    {

        // calculate xor of array
        int XOR = 0;
        for (int i = 0; i < n; i++)
            XOR ^= A[i];

        // update array
        for (int i = 0; i < n; i++)
            A[i] = XOR ^ A[i];
    }

    // Driver code
    public static void Main()
    {
        int []A = { 2, 4, 1, 3, 5};
        int n = A.Length;
        constructXOR(A, n);

        // print result
        for (int i = 0; i < n; i++)
        Console.Write(A[i] + " ");
    }
}

// This code is contributed by nitin mittal

PHP

<?php
// Program to construct array from
// XOR of elements of given array

// function to construct new array
function constructXOR(&$A, $n)
{
    // calculate xor of array
    $XOR = 0;
    for ($i = 0; $i < $n; $i++)
        $XOR ^= $A[$i];

    // update array
    for ($i = 0; $i < $n; $i++)
        $A[$i] = $XOR ^ $A[$i];
}

// Driver code
$A = array( 2, 4, 1, 3, 5);
$n = sizeof($A);
constructXOR($A, $n);

// print result
for ($i = 0; $i < $n; $i++)
    echo $A[$i] ." ";

// This code is contributed
// by ChitraNayal
?>

Javascript

<script>

// JavaScript program to construct array from
// XOR of elements of given array

// function to construct new array
function constructXOR(A, n)
{
    // calculate xor of array
    let XOR = 0;
    for (let i = 0; i < n; i++)
        XOR ^= A[i];

    // update array
    for (let i = 0; i < n; i++)
        A[i] = XOR ^ A[i];
}

// Driver code
    let A = [ 2, 4, 1, 3, 5];
    let n = A.length;
    constructXOR(A, n);

    // print result
    for (let i = 0; i < n; i++)
        document.write(A[i] + " ");

// This code is contributed by Surbhi Tyagi.

</script>

Output: 

3 5 0 2 4

Time Complexity : O(n)
Auxiliary Space : O(1)
Related Problem : 
A Product Array Puzzle