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

Given an array arr[] of n elements. The task is to find the count of elements which 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: Initialise 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++

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// C++ implementation of the approach
#include <bits/stdc++.h>
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;
}

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Java

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// 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

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Python3

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# 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

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C#

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// 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

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Output:

1


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