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# Count of elements which are equal to the XOR of the next two elements

• Difficulty Level : Medium
• Last Updated : 14 Sep, 2022

Given an array arr[] of n elements. The task is to find the count of elements that are equal to the XOR of the next two elements.

Examples:

```Input: arr[] = {4, 2, 1, 3, 7, 8}
Output: 1
2 is the only valid element as 1 ^ 3 = 2```
```Input: arr[] = {23, 1, 7, 8, 6}
Output: 0 ```

Approach: Initialize count = 0 and for every element of the array such that it has at least two elements appearing after it in the array, if it is equal to the XOR of the next two elements then increment the count.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach` `#include ` `using` `namespace` `std;` `// Function to return the count of elements``// which are equal to the XOR``// of the next two elements``int` `cntElements(``int` `arr[], ``int` `n)``{` `    ``// To store the required count``    ``int` `cnt = 0;` `    ``// For every element of the array such that``    ``// it has at least two elements appearing``    ``// after it in the array``    ``for` `(``int` `i = 0; i < n - 2; i++) {` `        ``// If current element is equal to the XOR``        ``// of the next two elements in the array``        ``if` `(arr[i] == (arr[i + 1] ^ arr[i + 2])) {``            ``cnt++;``        ``}``    ``}` `    ``return` `cnt;``}` `// Driver code``int` `main()``{``    ``int` `arr[] = { 4, 2, 1, 3, 7, 8 };``    ``int` `n = ``sizeof``(arr) / ``sizeof``(``int``);` `    ``cout << cntElements(arr, n);` `    ``return` `0;``}`

## Java

 `// Java implementation of the approach``import` `java.io.*;` `class` `GFG``{``    ` `// Function to return the count of elements``// which are equal to the XOR``// of the next two elements``static` `int` `cntElements(``int` `arr[], ``int` `n)``{` `    ``// To store the required count``    ``int` `cnt = ``0``;` `    ``// For every element of the array such that``    ``// it has at least two elements appearing``    ``// after it in the array``    ``for` `(``int` `i = ``0``; i < n - ``2``; i++)``    ``{` `        ``// If current element is equal to the XOR``        ``// of the next two elements in the array``        ``if` `(arr[i] == (arr[i + ``1``] ^ arr[i + ``2``]))``        ``{``            ``cnt++;``        ``}``    ``}``    ``return` `cnt;``}` `// Driver code``public` `static` `void` `main (String[] args)``{``    ``int` `arr[] = { ``4``, ``2``, ``1``, ``3``, ``7``, ``8` `};``    ``int` `n = arr.length;` `    ``System.out.println (cntElements(arr, n));``}``}` `// This code is contributed by jit_t`

## Python3

 `# Python3 implementation of the approach` `# Function to return the count of elements``# which are equal to the XOR``# of the next two elements``def` `cntElements(arr, n):` `    ``# To store the required count``    ``cnt ``=` `0` `    ``# For every element of the array such that``    ``# it has at least two elements appearing``    ``# after it in the array``    ``for` `i ``in` `range``(n ``-` `2``):` `        ``# If current element is equal to the XOR``        ``# of the next two elements in the array``        ``if` `(arr[i] ``=``=` `(arr[i ``+` `1``] ^ arr[i ``+` `2``])):``            ``cnt ``+``=` `1` `    ``return` `cnt` `# Driver code``arr ``=` `[``4``, ``2``, ``1``, ``3``, ``7``, ``8``]``n ``=` `len``(arr)` `print``(cntElements(arr, n))` `# This code is contributed by Mohit Kumar`

## C#

 `// C# implementation of the approach``using` `System;``using` `System.Collections.Generic;``    ` `class` `GFG``{``    ` `// Function to return the count of elements``// which are equal to the XOR``// of the next two elements``static` `int` `cntElements(``int` `[]arr, ``int` `n)``{` `    ``// To store the required count``    ``int` `cnt = 0;` `    ``// For every element of the array such that``    ``// it has at least two elements appearing``    ``// after it in the array``    ``for` `(``int` `i = 0; i < n - 2; i++)``    ``{` `        ``// If current element is equal to the XOR``        ``// of the next two elements in the array``        ``if` `(arr[i] == (arr[i + 1] ^ arr[i + 2]))``        ``{``            ``cnt++;``        ``}``    ``}``    ``return` `cnt;``}` `// Driver code``public` `static` `void` `Main (String[] args)``{``    ``int` `[]arr = { 4, 2, 1, 3, 7, 8 };``    ``int` `n = arr.Length;` `    ``Console.WriteLine(cntElements(arr, n));``}``}``    ` `// This code is contributed by Rajput-Ji`

## Javascript

 ``

Output:

`1`

Time complexity: O(n) where n is no of elements of the given array.

Auxiliary space: O(1)

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