Find array using different XORs of elements in groups of size 4

Given an array q[] of XOR queries of size N (N is a multiple of 4) which describe an array of the same size as follows:

q[0 – 3] describe arr[0 – 3], q[4 – 7] describe arr[4 – 7], and so on…
If arr[0 – 3] = {a1, a2, a3, a4} then
q[0 – 3] = {a1 ⊕ a2 ⊕ a3, a1 ⊕ a2 ⊕ a4, a1 ⊕ a3 ⊕ a4, a2 ⊕ a3 ⊕ a4}

The task is to find the values of the original array arr[] when only q[] is given.

Example:

Input: q[] = {4, 1, 7, 0}
Output: 2 5 3 6
a1 ⊕ a2 ⊕ a3 = 4 ⊕ 1 ⊕ 7 = 2
a1 ⊕ a2 ⊕ a4 = 4 ⊕ 1 ⊕ 0 = 5
a1 ⊕ a3 ⊕ a4 = 4 ⊕ 7 ⊕ 0 = 3
a2 ⊕ a3 ⊕ a4 = 1 ⊕ 7 ⊕ 0 = 6

Input: q[] = {4, 1, 7, 0, 8, 5, 1, 4}
Output: 2 5 3 6 12 9 13 0



Approach: From the properties of xor, a a = 0 and a 0 = a.
(abc) (bcd) = ad (As (bc) (bc) = 0)
So we will divide array in groups of 4 elements and for each group(a, b, c, d) we will
be given results of the following queries:

  1. abc
  2. abd
  3. acd
  4. bcd

As (abc) (bcd) = ad,
using this (ad) we can get b and c from query 2 and 3 in the following manner:
(abd) (ad) = b
(acd) (ad) = c
Then using b and c we can get a from query 1 and d from query 4 in the following way:
(abc) (b) (c) = a
(bcd) (b) (c) = d
This process will be repeated (iteratively) for all groups of 4 elements and we will get elements of all indices(ex.(a_{1}, a_{2}, a_{3}, a_{4}) then (a_{5}, a_{6}, a_{7}, a_{8}) etc.)

Below is the implementation of the above approach:

C++

filter_none

edit
close

play_arrow

link
brightness_4
code

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
  
// Utility function to print the contents of the array
void printArray(int arr[], int n)
{
    for (int i = 0; i < n; i++)
        cout << arr[i] << " ";
}
  
// Function to find the required array
void findArray(int q[], int n)
{
    int arr[n], ans;
    for (int k = 0, j = 0; j < n / 4; j++) {
        ans = q[k] ^ q[k + 3];
        arr[k + 1] = q[k + 1] ^ ans;
        arr[k + 2] = q[k + 2] ^ ans;
        arr[k] = q[k] ^ ((arr[k + 1]) ^ (arr[k + 2]));
        arr[k + 3] = q[k + 3] ^ (arr[k + 1] ^ arr[k + 2]);
        k += 4;
    }
  
    // Print the array
    printArray(arr, n);
}
  
// Driver code
int main()
{
    int q[] = { 4, 1, 7, 0 };
    int n = sizeof(q) / sizeof(q[0]);
    findArray(q, n);
    return 0;
}

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// Java implementation of the approach 
class GFG 
      
    // Utility function to print
    // the contents of the array 
    static void printArray(int []arr, int n) 
    
        for (int i = 0; i < n; i++) 
            System.out.print(arr[i] + " "); 
    
      
    // Function to find the required array 
    static void findArray(int []q, int n) 
    
        int ans;
        int []arr = new int[n];
        for (int k = 0, j = 0
                        j < n / 4; j++)
        
            ans = q[k] ^ q[k + 3]; 
            arr[k + 1] = q[k + 1] ^ ans; 
            arr[k + 2] = q[k + 2] ^ ans; 
            arr[k] = q[k] ^ ((arr[k + 1]) ^ 
                             (arr[k + 2])); 
            arr[k + 3] = q[k + 3] ^ (arr[k + 1] ^
                                     arr[k + 2]); 
            k += 4
        
      
        // Print the array 
        printArray(arr, n); 
    
      
    // Driver code 
    public static void main(String args[]) 
    
        int []q = { 4, 1, 7, 0 }; 
        int n = q.length;
        findArray(q, n); 
    
  
// This code is contributed
// by Akanksha Rai

chevron_right


Python3

filter_none

edit
close

play_arrow

link
brightness_4
code

# Python 3 implementation of the approach
  
# Utility function to print the 
# contents of the array
def printArray(arr, n):
    for i in range(n):
        print(arr[i], end = " ")
  
# Function to find the required array
def findArray(q, n):
    arr = [None] * n
    k = 0
    for j in range(int(n / 4)):
        ans = q[k] ^ q[k + 3]
        arr[k + 1] = q[k + 1] ^ ans
        arr[k + 2] = q[k + 2] ^ ans
        arr[k] = q[k] ^ ((arr[k + 1]) ^ 
                         (arr[k + 2]))
        arr[k + 3] = q[k + 3] ^ (arr[k + 1] ^
                                 arr[k + 2])
        k += 4
  
    # Print the array
    printArray(arr, n)
  
# Driver code
if __name__ == '__main__':
    q = [4, 1, 7, 0]
    n = len(q)
    findArray(q, n)
  
# This code is contributed by
# Surendra_Gangwar

chevron_right


C#

filter_none

edit
close

play_arrow

link
brightness_4
code

// C# implementation of the approach 
using System;
  
class GFG 
      
    // Utility function to print
    // the contents of the array 
    static void printArray(int []arr, int n) 
    
        for (int i = 0; i < n; i++) 
            Console.Write(arr[i] + " "); 
    
      
    // Function to find the required array 
    static void findArray(int []q, int n) 
    
        int ans;
        int []arr = new int[n] ;
        for (int k = 0, j = 0; j < n / 4; j++)
        
            ans = q[k] ^ q[k + 3]; 
            arr[k + 1] = q[k + 1] ^ ans; 
            arr[k + 2] = q[k + 2] ^ ans; 
            arr[k] = q[k] ^ ((arr[k + 1]) ^ (arr[k + 2])); 
            arr[k + 3] = q[k + 3] ^ (arr[k + 1] ^ arr[k + 2]); 
            k += 4; 
        
      
        // Print the array 
        printArray(arr, n); 
    
      
    // Driver code 
    public static void Main() 
    
        int []q = { 4, 1, 7, 0 }; 
        int n = q.Length ;
        findArray(q, n); 
    
  
// This code is contributed by Ryuga

chevron_right


PHP

filter_none

edit
close

play_arrow

link
brightness_4
code

<?php
// PHP implementation of the approach 
  
// Utility function to print
// the contents of the array 
function printArray($arr, $n
    for ($i = 0; $i < $n; $i++) 
        echo($arr[$i] ." "); 
      
// Function to find the required array 
function findArray($q, $n)
    $ans;
    $arr = array($n);
    for ($k = 0, $j = 0; $j < $n / 4; $j++)
    
        $ans = $q[$k] ^ $q[$k + 3]; 
        $arr[$k + 1] = $q[$k + 1] ^ $ans
        $arr[$k + 2] = $q[$k + 2] ^ $ans
        $arr[$k] = $q[$k] ^ (($arr[$k + 1]) ^ 
                            ($arr[$k + 2])); 
        $arr[$k + 3] = $q[$k + 3] ^ ($arr[$k + 1] ^
                                    $arr[$k + 2]); 
        $k += 4; 
    
      
    // Print the array 
    printArray($arr, $n); 
      
// Driver code 
    $q = array( 4, 1, 7, 0 ); 
    $n = sizeof($q);
    findArray($q, $n); 
  
// This code is contributed
// by Code_Mech.

chevron_right


Output:

2 5 3 6

Time Complexity: O(N)



My Personal Notes arrow_drop_up

I am a second year BTech (CSE) engineering student at GB Pant Goverment Engineering College Delhi(IPU)

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.