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

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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 , a , a , a ) then (a , a , a , a ) etc.)

Below is the implementation of the above approach:

C++

 // C++ implementation of the approach #include 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);     findArray(q, n);     return 0; }

Java

 // 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

Python3

 # 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

C#

 // 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

PHP



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)

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