Find the original Array using XOR values of all adjacent elements

Given a sequence arr[] of N-1 elements which is xor of all adjacent pairs in an array, the task is to find that original array from the arr[].

Note: It is given that the N is always odd and arr[] contains the permutation of N natural number.

Examples:

Input: arr[] = {3, 1}
Output: 1 2 3
Explanation:
The XOR of the output array will lead to the given array that is:
1 ^ 2 = 3
2 ^ 3 = 1

Input: arr[] = {7, 5, 3, 7}
Output: 3 4 1 2 5
Explanation:
The XOR of the output array will lead to the given array that is:
3 ^ 4 = 7
4 ^ 1 = 5
1 ^ 2 = 3
2 ^ 5 = 7



Approach:

  1. The idea is to find the XOR of all the elements of the 1 to N and the xor of the adjacent elements of the given array to find the last element of the expected array.
  2. As the XOR of adjacent elements will contain all the elements except the last element, then the XOR of this with all the numbers from 1 to N will give the last element of the expected permutation.
    For Example:
    Let's the expected array be - {a, b, c, d, e}
    Then the XOR array for this array will be - 
    {a^b, b^c, c^d, d^e}
    
    Now XOR of all the element from 1 to N -
    xor_all => a ^ b ^ c ^ d ^ e
    
    XOR of the adjacent elements -
    xor_adjacent => ((a ^ b) ^ (c ^ d))
    
    Now the XOR of the both the array will be the 
    last element of the expected permutation 
    => (a ^ b ^ c ^ d ^ e) ^ ((a ^ b) ^ (c ^ d))
    => As all elements are in pair except the last element.
    => (a ^ a ^ b ^ b ^ c ^ c ^ d ^ d ^ e)
    => (0 ^ 0 ^ 0 ^ 0 ^ e)
    => e
    
  3. Now for the rest of the element, continuously, on xor of this last element we will get last second element, i.e, d.
  4. Repeatedly, Update the last element and finally get the first element, i.e, a.

Below is the implementation of the above approach:

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// C++ implementation to find the
// Array from the XOR array
// of the adjacent elements of array
  
#include <bits/stdc++.h>
using namespace std;
  
// XOR of all elements from 1 to N
int xor_all_elements(int n)
{
  
    switch (n & 3) {
  
    case 0:
        return n;
    case 1:
        return 1;
    case 2:
        return n + 1;
    case 3:
        return 0;
    }
}
  
// Function to find the Array
// from the XOR Array
vector<int> findArray(int xorr[], int n)
{
    // Take a vector to store
    // the permuatation
    vector<int> arr;
  
    // XOR of N natural numbers
    int xor_all = xor_all_elements(n);
    int xor_adjacent = 0;
  
    // Loop to find the XOR of
    // adjacent elements of the XOR Array
    for (int i = 0; i < n - 1; i += 2) {
        xor_adjacent = xor_adjacent ^ xorr[i];
    }
    int last_element = xor_all ^ xor_adjacent;
    arr.push_back(last_element);
  
    // Loop to find the other
    // elements of the permuatation
    for (int i = n - 2; i >= 0; i--) {
        // Finding the next and next elements
        last_element = xorr[i] ^ last_element;
        arr.push_back(last_element);
    }
  
    return arr;
}
  
// Driver Code
int main()
{
    vector<int> arr;
  
    int xorr[] = { 7, 5, 3, 7 };
    int n = 5;
  
    arr = findArray(xorr, n);
  
    // Required Permutation
    for (int i = n - 1; i >= 0; i--) {
        cout << arr[i] << " ";
    }
}
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// Java implementation to find the
// Array from the XOR array
// of the adjacent elements of array
import java.util.*;
  
class GFG{
  
// XOR of all elements from 1 to N
static int xor_all_elements(int n)
{
  
    switch (n & 3) {
  
    case 0:
        return n;
    case 1:
        return 1;
    case 2:
        return n + 1
    }
    return 0;
}
  
// Function to find the Array
// from the XOR Array
static Vector<Integer> findArray(int xorr[], int n)
{
    // Take a vector to store
    // the permuatation
    Vector<Integer> arr = new Vector<Integer>();
  
    // XOR of N natural numbers
    int xor_all = xor_all_elements(n);
    int xor_adjacent = 0;
  
    // Loop to find the XOR of
    // adjacent elements of the XOR Array
    for (int i = 0; i < n - 1; i += 2) {
        xor_adjacent = xor_adjacent ^ xorr[i];
    }
    int last_element = xor_all ^ xor_adjacent;
    arr.add(last_element);
  
    // Loop to find the other
    // elements of the permuatation
    for (int i = n - 2; i >= 0; i--)
    {
        // Finding the next and next elements
        last_element = xorr[i] ^ last_element;
        arr.add(last_element);
    }
  
    return arr;
}
  
// Driver Code
public static void main(String[] args)
{
    Vector<Integer> arr = new Vector<Integer>();
  
    int xorr[] = { 7, 5, 3, 7 };
    int n = 5;
  
    arr = findArray(xorr, n);
  
    // Required Permutation
    for (int i = n - 1; i >= 0; i--) 
    {
        System.out.print(arr.get(i)+ " ");
    }
}
}
  
// This code is contributed by PrinciRaj1992
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# Python3 implementation to find the
# Array from the XOR array
# of the adjacent elements of array
  
# XOR of all elements from 1 to N
def xor_all_elements(n):
  
    if n & 3 == 0:
        return n
    elif n & 3 == 1:
        return 1
    elif n & 3 == 2:
        return n + 1
    else:
        return 0
  
# Function to find the Array
# from the XOR Array
def findArray(xorr, n):
      
    # Take a vector to store
    # the permuatation
    arr = []
  
    # XOR of N natural numbers
    xor_all = xor_all_elements(n)
    xor_adjacent = 0
  
    # Loop to find the XOR of
    # adjacent elements of the XOR Array
    for i in range(0, n - 1, 2):
        xor_adjacent = xor_adjacent ^ xorr[i]
  
    last_element = xor_all ^ xor_adjacent
    arr.append(last_element)
  
    # Loop to find the other
    # elements of the permuatation
    for i in range(n - 2, -1, -1):
          
        # Finding the next and next elements
        last_element = xorr[i] ^ last_element
        arr.append(last_element)
  
    return arr
  
# Driver Code
xorr = [7, 5, 3, 7]
n = 5
  
arr = findArray(xorr, n)
  
# Required Permutation
for i in range(n - 1, -1, -1):
    print(arr[i], end=" ")
      
# This code is contributed by mohit kumar 29    
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// C# implementation to find the
// Array from the XOR array
// of the adjacent elements of array
using System;
using System.Collections.Generic;
  
class GFG{
   
// XOR of all elements from 1 to N
static int xor_all_elements(int n)
{
   
    switch (n & 3) {
   
    case 0:
        return n;
    case 1:
        return 1;
    case 2:
        return n + 1; 
    }
    return 0;
}
   
// Function to find the Array
// from the XOR Array
static List<int> findArray(int []xorr, int n)
{
    // Take a vector to store
    // the permuatation
    List<int> arr = new List<int>();
   
    // XOR of N natural numbers
    int xor_all = xor_all_elements(n);
    int xor_adjacent = 0;
   
    // Loop to find the XOR of
    // adjacent elements of the XOR Array
    for (int i = 0; i < n - 1; i += 2) {
        xor_adjacent = xor_adjacent ^ xorr[i];
    }
    int last_element = xor_all ^ xor_adjacent;
    arr.Add(last_element);
   
    // Loop to find the other
    // elements of the permuatation
    for (int i = n - 2; i >= 0; i--)
    {
        // Finding the next and next elements
        last_element = xorr[i] ^ last_element;
        arr.Add(last_element);
    }
   
    return arr;
}
   
// Driver Code
public static void Main(String[] args)
{
    List<int> arr = new List<int>();
   
    int []xorr = { 7, 5, 3, 7 };
    int n = 5;
   
    arr = findArray(xorr, n);
   
    // Required Permutation
    for (int i = n - 1; i >= 0; i--) 
    {
        Console.Write(arr[i]+ " ");
    }
}
}
  
// This code contributed by Rajput-Ji
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Output:
3 4 1 2 5

Performance Analysis:

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