# 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

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

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:

 `// C++ implementation to find the ` `// Array from the XOR array ` `// of the adjacent elements of array ` ` `  `#include ` `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] << ``" "``; ` `    ``} ` `} `

 `// 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 findArray(``int` `xorr[], ``int` `n) ` `{ ` `    ``// Take a vector to store ` `    ``// the permuatation ` `    ``Vector arr = ``new` `Vector(); ` ` `  `    ``// 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 arr = ``new` `Vector(); ` ` `  `    ``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 `

 `# 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     `

 `// 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 `

Output:
```3 4 1 2 5
```

Performance Analysis:

• Time Complexity: In the above approach we iterate over the entire xor array to find the XOR of the Adjacent elements, then the complexity in the worst case will be O(N)
• Space Complexity: In the above approach there is a vector array used to store the Permutation of the numbers from 1 to N, then the space complexity will be O(N)

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