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 n 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 as 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 a (first elements), and to find the next element i.e b we have to xor a by arr[0], similarly for c 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 as arr[i] contains XOR of ai and ai+1. Therefore,

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

Below is the implementation of above approach

C++

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// CPP 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 orginal
    // 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;
}

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Java

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// 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 orginal
    // 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..

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Python3

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# 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

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C#

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// 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 orginal 
        // 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

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PHP

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<?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 orginal
    // 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
?>

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Output:

5 8 10 12 13


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