等于下两个元素的XOR的元素计数

给定n个元素的数组arr [] 。任务是找到等于下两个元素的XOR的元素数。
例子：

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 
 

方法：初始化count = 0 ，对于数组的每个元素，使其至少有两个元素出现在数组之后，如果它等于接下来的两个元素的XOR，则增加count 。
下面是上述方法的实现：

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

 
// Javascript implementation of the approach
 
// Function to return the count of elements
// which are equal to the XOR
// of the next two elements
function cntElements(arr, n)
{
 
    // To store the required count
    let cnt = 0;
 
    // For every element of the array such that
    // it has at least two elements appearing
    // after it in the array
    for (let 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
    let arr = [ 4, 2, 1, 3, 7, 8 ];
    let n = arr.length;
 
    document.write(cntElements(arr, n));
 

输出：

1