Adjacent xor code in python

View Discussion

Improve Article

Save Article

View Discussion

Improve Article

Save Article

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

1. Now for the rest of the element, continuously, on xor of this last element we will get last second element, i.e, d.
2. Repeatedly, Update the last element and finally get the first element, i.e, a.

Below is the implementation of the above approach:
 

C++

#include

using namespace std;

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;

    }

}

vector<int> findArray(int xorr[], int n)

{

    vector<int> arr;

    int xor_all = xor_all_elements(n);

    int xor_adjacent = 0;

    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);

    for (int i = n - 2; i >= 0; i--) {

        last_element = xorr[i] ^ last_element;

        arr.push_back(last_element);

    }

    return arr;

}

int main()

{

    vector<int> arr;

    int xorr[] = { 7, 5, 3, 7 };

    int n = 5;

    arr = findArray(xorr, n);

    for (int i = n - 1; i >= 0; i--) {

        cout << arr[i] << " ";

    }

}

Java

import java.util.*;

class GFG{

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;

}

static Vector findArray(int xorr[], int n)

{

    Vector arr = new Vector();

    int xor_all = xor_all_elements(n);

    int xor_adjacent = 0;

    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);

    for (int i = n - 2; i >= 0; i--)

    {

        last_element = xorr[i] ^ last_element;

        arr.add(last_element);

    }

    return arr;

}

public static void main(String[] args)

{

    Vector arr = new Vector();

    int xorr[] = { 7, 5, 3, 7 };

    int n = 5;

    arr = findArray(xorr, n);

    for (int i = n - 1; i >= 0; i--)

    {

        System.out.print(arr.get(i)+ " ");

    }

}

}

Python3

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

def findArray(xorr, n):

    arr = []

    xor_all = xor_all_elements(n)

    xor_adjacent = 0

    for i in range(0, n - 1, 2):

        xor_adjacent = xor_adjacent ^ xorr[i]

    last_element = xor_all ^ xor_adjacent

    arr.append(last_element)

    for i in range(n - 2, -1, -1):

        last_element = xorr[i] ^ last_element

        arr.append(last_element)

    return arr

xorr = [7, 5, 3, 7]

n = 5

arr = findArray(xorr, n)

for i in range(n - 1, -1, -1):

    print(arr[i], end=" ")

C#

using System;

using System.Collections.Generic;

class GFG{

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;

}

static List<int> findArray(int []xorr, int n)

{

    List<int> arr = new List<int>();

    int xor_all = xor_all_elements(n);

    int xor_adjacent = 0;

    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);

    for (int i = n - 2; i >= 0; i--)

    {

        last_element = xorr[i] ^ last_element;

        arr.Add(last_element);

    }

    return arr;

}

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);

    for (int i = n - 1; i >= 0; i--)

    {

        Console.Write(arr[i]+ " ");

    }

}

}

Javascript

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)