Find elements of array using XOR of consecutive elements

• Difficulty Level : Basic
• Last Updated : 03 Aug, 2021

Given an array arr[] in which XOR of every 2 consecutive elements of the original array is given i.e if the total number of elements in the original array is then the size of this XOR array would be n-1. The first element in the original array is also given. The task is to find out the rest of n-1 elements of the original array.

• Let a, b, c, d, e, f are the original elements, and the xor of every 2 consecutive elements is given, i.e a^b = k1, b ^ c = k2, c ^ d = k3, d ^ e = k4, e ^ f = k5 (where k1, k2, k3, k4, k5 are the elements that are given us along with the first element a), and we have to find the value of b, c, d, e, f.

Examples:

Input : arr[] = {13, 2, 6, 1}, a = 5
Output : 5 8 10 12 13
5^8=13, 8^10=2, 10^12=6, 12^13=1

Input : arr[] = {12, 5, 26, 7}, a = 6
Output : 6 10 15 21 18

Approach: We can find all the elements one by one with the help of (first elements), and to find the next element i.e we have to xor a by arr, similarly for xor arr with b and so on.

This works by following the properties of XOR as stated below:

• XOR of a number to itself is zero.
• XOR of a number with zero given the number itself.

So, as arr contains a^b. Therefore,

a ^ arr = a ^ a ^ b
= 0 ^ b
= b

Similarly, arr[i] contains XOR of ai and ai+1. Therefore,

ai ^ arr[i] = ai ^ ai ^ ai+1
= 0 ^ ai+1
= ai+1

Below is the implementation of the above approach

C++

 // C++ program to find the array elements// using XOR of consecutive elements #include using namespace std; // Function to find the array elements// using XOR of consecutive elementsvoid getElements(int a, int arr[], int n){    // array to store the original    // elements    int elements[n + 1];     // first element a i.e elements=a    elements = a;     for (int i = 0; i < n; i++) {         /*  To get the next elements we have to calculate            xor of previous elements with given xor of 2            consecutive elements.            e.g. if a^b=k1 so to get b xor a both side.            b = k1^a        */        elements[i + 1] = arr[i] ^ elements[i];    }     // Printing the original array elements    for (int i = 0; i < n + 1; i++)        cout << elements[i] << " ";} // Driver Codeint main(){    int arr[] = { 13, 2, 6, 1 };     int n = sizeof(arr) / sizeof(arr);     int a = 5;     getElements(a, arr, n);     return 0;}

Java

 // Java  program to find the array elements// using XOR of consecutive elements import java.io.*; class GFG {     // Function to find the array elements// using XOR of consecutive elementsstatic void getElements(int a, int arr[], int n){    // array to store the original    // elements    int elements[] = new int[n + 1];     // first element a i.e elements=a    elements = a;     for (int i = 0; i < n; i++) {         /* To get the next elements we have to calculate            xor of previous elements with given xor of 2            consecutive elements.            e.g. if a^b=k1 so to get b xor a both side.            b = k1^a        */        elements[i + 1] = arr[i] ^ elements[i];    }     // Printing the original array elements    for (int i = 0; i < n + 1; i++)        System.out.print( elements[i] + " ");} // Driver Code     public static void main (String[] args) {            int arr[] = { 13, 2, 6, 1 };     int n = arr.length;     int a = 5;     getElements(a, arr, n);    }}// This code is contributed by anuj_67..

Python3

 # Python3 program to find the array# elements using xor of consecutive elements # Function to find the array elements# using XOR of consecutive elements def getElements(a, arr, n):         # array to store the original elements    elements = [1 for i in range(n + 1)]         # first elements a i.e elements=a    elements = a         for i in range(n):                 # To get the next elements we have to        # calculate xor of previous elements        # with given xor of 2 consecutive elements.        # e.g. if a^b=k1 so to get b xor a both side.        # b = k1^a        elements[i + 1] = arr[i] ^ elements[i]                 # Printing the original array elements    for i in range(n + 1):        print(elements[i], end = " ") # Driver codearr = [13, 2, 6, 1]n = len(arr)a = 5getElements(a, arr, n) # This code is contributed by Mohit Kumar

C#

 // C# program to find the array elements// using XOR of consecutive elements using System; class GFG {    // Function to find the array elements    // using XOR of consecutive elements    static void getElements(int a, int []arr, int n)    {        // array to store the original        // elements        int []elements = new int[n + 1];             // first element a i.e elements=a        elements = a;             for (int i = 0; i < n; i++) {                 /* To get the next elements we have to calculate                xor of previous elements with given xor of 2                consecutive elements.                e.g. if a^b=k1 so to get b xor a both side.                b = k1^a            */            elements[i + 1] = arr[i] ^ elements[i];        }             // Printing the original array elements        for (int i = 0; i < n + 1; i++)            Console.Write( elements[i] + " ");    }     // Driver Code    public static void Main () {            int []arr = { 13, 2, 6, 1 };             int n = arr.Length;             int a = 5;             getElements(a, arr, n);    }        // This code is contributed by Ryuga}



Javascript


Output:
5 8 10 12 13

Time Complexity: O(N)
Auxiliary Space: O(N)

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