Replace every array element by Bitwise Xor of previous and next element

Given an array of integers, replace every element with xor of previous and next elements with following exceptions.
a) First element is replaced by sum of first and second.
b) Last element is replaced by sum of last and second last.

Examples:

Input: arr[] = { 2, 3, 4, 5, 6}
Output: 1 6 6 2 3 

We get the following array as {2^3, 2^4, 3^5, 4^6, 5^6}

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

We get the following array as {1^2, 1^1, 2^5, 1^5}

A Simple Solution is to create an auxiliary array, copy contents of given array to auxiliary array. Finally traverse the auxiliary array and update given array using copied values. Time complexity of this solution is O(n), but it requires O(n) extra space.

An efficient solution can solve the problem in O(n) time and O(1) space. The idea is to keep track of previous element in loop. Xor the previous element using the extra variable and the next element to get each element.

Below is the implementation of the above approach:

C++

// C++ program to update every array element with
// sum of previous and next numbers in array
#include <iostream>
using namespace std;

void ReplaceElements(int arr[], int n)
{
    // Nothing to do when array size is 1
    if (n <= 1)
        return;

    // store current value of arr[0] and update it
    int prev = arr[0];
    arr[0] = arr[0] ^ arr[1];

    // Update rest of the array elements
    for (int i = 1; i < n - 1; i++) {
        // Store current value of next interation
        int curr = arr[i];

        // Update current value using previos value
        arr[i] = prev ^ arr[i + 1];

        // Update previous value
        prev = curr;
    }

    // Update last array element separately
    arr[n - 1] = prev ^ arr[n - 1];
}

// Driver program
int main()
{
    int arr[] = { 2, 3, 4, 5, 6 };
    int n = sizeof(arr) / sizeof(arr[0]);

    ReplaceElements(arr, n);

    // Print the modified array
    for (int i = 0; i < n; i++)
        cout << arr[i] << " ";
    return 0;
}

Java

// Java  program to update every array
// element with sum of previous and
// next numbers in array
import  java .io.*;

class GFG
{
static void ReplaceElements(int[] arr,
                            int n)
{
    // Nothing to do when array size is 1
    if (n <= 1)
        return;

    // store current value of arr[0]
    // and update it
    int prev = arr[0];
    arr[0] = arr[0] ^ arr[1];

    // Update rest of the array elements
    for (int i = 1; i < n - 1; i++)
    {
        // Store current value of
        // next interation
        int curr = arr[i];

        // Update current value using
        // previous value
        arr[i] = prev ^ arr[i + 1];

        // Update previous value
        prev = curr;
    }

    // Update last array element separately
    arr[n - 1] = prev ^ arr[n - 1];
}

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

{
    int[] arr = { 2, 3, 4, 5, 6 };
    int n = arr.length;

    ReplaceElements(arr, n);

    // Print the modified array
    for (int i = 0; i < n; i++)
        System.out.print(arr[i] + " ");
}
}

// This code is contributed
// by anuj_67..

Python3

# Python3 program to update every
# array element with sum of previous
# and next numbers in array
def ReplaceElements(arr, n):

    # Nothing to do when array
    # size is 1
    if n <= 1:
        return

    # store current value of arr[0]
    # and update it
    prev = arr[0]
    arr[0] = arr[0] ^ arr[1]

    # Update rest of the array elements
    for i in range(1, n - 1):

        # Store current value of
        # next interation
        curr = arr[i]

        # Update current value using
        # previos value
        arr[i] = prev ^ arr[i + 1]

        # Update previous value
        prev = curr

    # Update last array element separately
    arr[n - 1] = prev ^ arr[n - 1]

# Driver Code
arr = [2, 3, 4, 5, 6]
n = len(arr)
ReplaceElements(arr, n)
for i in range(n):
    print(arr[i], end = " ")

# This code is contributed
# by Shrikant13

C#

// C# program to update every array
// element with sum of previous and
// next numbers in array
using System;

class GFG
{
static void ReplaceElements(int[] arr,
                            int n)
{
    // Nothing to do when array size is 1
    if (n <= 1)
        return;

    // store current value of arr[0]
    // and update it
    int prev = arr[0];
    arr[0] = arr[0] ^ arr[1];

    // Update rest of the array elements
    for (int i = 1; i < n - 1; i++)
    {
        // Store current value of
        // next interation
        int curr = arr[i];

        // Update current value using
        // previous value
        arr[i] = prev ^ arr[i + 1];

        // Update previous value
        prev = curr;
    }

    // Update last array element separately
    arr[n - 1] = prev ^ arr[n - 1];
}

// Driver Code
public static void Main()
{
    int[] arr = { 2, 3, 4, 5, 6 };
    int n = arr.Length;

    ReplaceElements(arr, n);

    // Print the modified array
    for (int i = 0; i < n; i++)
        Console.Write(arr[i] + " ");
}
}

// This code is contributed
// by Akanskha Rai(Abby_akku)

PHP

<?php
// PHP program to update every array
// element with sum of previous and
// next numbers in array
function ReplaceElements(&$arr, $n)
{
    // Nothing to do when array size is 1
    if ($n <= 1)
        return;

    // store current value of arr[0]
    // and update it
    $prev = $arr[0];
    $arr[0] = $arr[0] ^ $arr[1];

    // Update rest of the array elements
    for ($i = 1; $i < $n - 1; $i++)
    {
        // Store current value of next
        // interation
        $curr = $arr[$i];

        // Update current value using
        // previos value
        $arr[$i] = $prev ^ $arr[$i + 1];

        // Update previous value
        $prev = $curr;
    }

    // Update last array element separately
    $arr[$n - 1] = $prev ^ $arr[$n - 1];
}

// Driver Code
$arr = array( 2, 3, 4, 5, 6 );
$n = sizeof($arr);

ReplaceElements($arr, $n);

// Print the modified array
for ($i = 0; $i < $n; $i++)
    echo $arr[$i] . " ";

// This code is contributed by ita_c
?>

Output:

1 6 6 2 3

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