Find elements of array using XOR of consecutive elements

Given an array arr[] in which XOR of every 2 consecutive elements of the original array is given i.e if the total number of elements in the original array is then the size of this XOR array would be n-1. The first element in the original array is also given. The task is to find out the rest of n-1 elements of the original array.

• Let a, b, c, d, e, f are the original elements, and the xor of every 2 consecutive elements is given, i.e a^b = k1, b ^ c = k2, c ^ d = k3, d ^ e = k4, e ^ f = k5 (where k1, k2, k3, k4, k5 are the elements that are given us along with the first element a), and we have to find the value of b, c, d, e, f.

Examples:

Input : arr[] = {13, 2, 6, 1}, a = 5
Output : 5 8 10 12 13
5^8=13, 8^10=2, 10^12=6, 12^13=1

Input : arr[] = {12, 5, 26, 7}, a = 6
Output : 6 10 15 21 18 

Approach: We can find all the elements one by one with the help of (first elements), and to find the next element i.e we have to xor a by arr[0], similarly for xor arr[1] with b and so on.

This works by following the properties of XOR as stated below:  

• XOR of a number to itself is zero.
• XOR of a number with zero given the number itself.

So, as arr[0] contains a^b. Therefore,  

a ^ arr[0] = a ^ a ^ b
           = 0 ^ b
           = b

Similarly, arr[i] contains XOR of a_i and a_i+1. Therefore, 

ai ^ arr[i] = ai ^ ai ^ ai+1
            = 0 ^ ai+1
            = ai+1

Algorithm:

Step 1: Start
Step 2: Let’s start with creating a function named getElements with three parameters: an integer a, an array arr of size n, and an                  integer n.
Step 3: Now let’s form a new integer array of size n+1 in which we will store the value of the original array.
Step 4: Now set the value of the 0th index to a.
Step 5: Start a for loop which traverses through the given array from index 0 to n-1.
Step 6: Calculate the XOR of the current element of arr with the previous element of elements for each iteration, then store the                    result in the subsequent index of elements.
Step 7: Now print all elements with another for a loop.
Step 8: End 

Below is the implementation of the above approach

C++

// C++ program to find the array elements
// using XOR of consecutive elements

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

// Function to find the array elements
// using XOR of consecutive elements
void getElements(int a, int arr[], int n)
{
    // array to store the original
    // elements
    int elements[n + 1];

    // first element a i.e elements[0]=a
    elements[0] = a;

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

        /*  To get the next elements we have to calculate
            xor of previous elements with given xor of 2
            consecutive elements.
            e.g. if a^b=k1 so to get b xor a both side.
            b = k1^a
        */
        elements[i + 1] = arr[i] ^ elements[i];
    }

    // Printing the original array elements
    for (int i = 0; i < n + 1; i++)
        cout << elements[i] << " ";
}

// Driver Code
int main()
{
    int arr[] = { 13, 2, 6, 1 };

    int n = sizeof(arr) / sizeof(arr[0]);

    int a = 5;

    getElements(a, arr, n);

    return 0;
}

Java

// Java  program to find the array elements
// using XOR of consecutive elements

import java.io.*;

class GFG {


// Function to find the array elements
// using XOR of consecutive elements
static void getElements(int a, int arr[], int n)
{
    // array to store the original
    // elements
    int elements[] = new int[n + 1];

    // first element a i.e elements[0]=a
    elements[0] = a;

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

        /* To get the next elements we have to calculate
            xor of previous elements with given xor of 2
            consecutive elements.
            e.g. if a^b=k1 so to get b xor a both side.
            b = k1^a
        */
        elements[i + 1] = arr[i] ^ elements[i];
    }

    // Printing the original array elements
    for (int i = 0; i < n + 1; i++)
        System.out.print( elements[i] + " ");
}

// Driver Code

    public static void main (String[] args) {
            int arr[] = { 13, 2, 6, 1 };

    int n = arr.length;

    int a = 5;

    getElements(a, arr, n);
    }
}
// This code is contributed by anuj_67..

Python3

# Python3 program to find the array
# elements using xor of consecutive elements

# Function to find the array elements
# using XOR of consecutive elements

def getElements(a, arr, n):

    # array to store the original elements
    elements = [1 for i in range(n + 1)]

    # first elements a i.e elements[0]=a
    elements[0] = a

    for i in range(n):

        # To get the next elements we have to
        # calculate xor of previous elements
        # with given xor of 2 consecutive elements.
        # e.g. if a^b=k1 so to get b xor a both side.
        # b = k1^a
        elements[i + 1] = arr[i] ^ elements[i]

    # Printing the original array elements
    for i in range(n + 1):
        print(elements[i], end = " ")

# Driver code
arr = [13, 2, 6, 1]
n = len(arr)
a = 5
getElements(a, arr, n)

# This code is contributed by Mohit Kumar

C#

// C# program to find the array elements
// using XOR of consecutive elements

using System;

class GFG {
    // Function to find the array elements
    // using XOR of consecutive elements
    static void getElements(int a, int []arr, int n)
    {
        // array to store the original
        // elements
        int []elements = new int[n + 1];

        // first element a i.e elements[0]=a
        elements[0] = a;

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

            /* To get the next elements we have to calculate
                xor of previous elements with given xor of 2
                consecutive elements.
                e.g. if a^b=k1 so to get b xor a both side.
                b = k1^a
            */
            elements[i + 1] = arr[i] ^ elements[i];
        }

        // Printing the original array elements
        for (int i = 0; i < n + 1; i++)
            Console.Write( elements[i] + " ");
    }

    // Driver Code
    public static void Main () {
            int []arr = { 13, 2, 6, 1 };

            int n = arr.Length;

            int a = 5;

            getElements(a, arr, n);
    }
        // This code is contributed by Ryuga
}

PHP

<?php
// PHP program to find the array elements
// using XOR of consecutive elements

// Function to find the array elements
// using XOR of consecutive elements
function getElements($a, &$arr, &$n)
{
    // array to store the original
    // elements

    // first element a i.e elements[0]=a
    $elements[0] = $a;

    for ($i = 0; $i < $n; $i++)
    {

        /*  To get the next elements we have to
            calculate xor of previous elements
            with given xor of 2 consecutive elements.
            e.g. if a^b=k1 so to get b xor a both side.
            b = k1^a
        */
        $elements[$i + 1] = $arr[$i] ^ $elements[$i];
    }

    // Printing the original array elements
    for ($i = 0; $i < $n + 1; $i++)
    {
        echo($elements[$i] . " ");
    }
}

// Driver Code
$arr = array(13, 2, 6, 1);

$n = sizeof($arr);

$a = 5;

getElements($a, $arr, $n);

// This code is contributed by Shivi_Aggarwal
?>

Javascript

<script>

// Javascript program to find the array elements
// using XOR of consecutive elements

// Function to find the array elements
// using XOR of consecutive elements
function getElements(a, arr, n)
{

    // Array to store the original
    // elements
    let elements = new Array(n + 1);
    for(let i = 0; i < n + 1; i++)
    {
        elements[i] = 0;
    }

    // first element a i.e elements[0]=a
    elements[0] = a;

    for(let i = 0; i < n; i++)
    {

        /* To get the next elements we have to calculate
            xor of previous elements with given xor of 2
            consecutive elements.
            e.g. if a^b=k1 so to get b xor a both side.
            b = k1^a
        */
        elements[i + 1] = arr[i] ^ elements[i];
    }

    // Printing the original array elements
    for(let i = 0; i < n + 1; i++)
        document.write( elements[i] + " ");
}

// Driver Code
let arr = [ 13, 2, 6, 1 ];
let n = arr.length;
let a = 5;

getElements(a, arr, n);

// This code is contributed by unknown2108

</script>

Output

5 8 10 12 13 

Complexity Analysis:

• Time Complexity: O(N), since there runs a loop for N times.
• Auxiliary Space: O(N), since N extra space has been taken.