通过删除数组元素来使奇数和偶数索引元素的按位XOR相等的计数方法

给定长度为N的数组arr [] ，任务是查找数组索引的数量，以便从这些索引中删除元素使奇数索引元素和偶数索引(基于1的索引)元素的按位异或相等。

例子：

Input: arr[] = {1, 0, 1, 0, 1}, N = 5
Output: 3
Explanation: 

1. Removing an element from index 3 modifies arr[] to {1, 0, 0, 1}. Therefore, xor of odd and even indexed elements is 0.
2. Removing an element from index 1 modifies arr[] to {0, 1, 0, 1}. Therefore, xor of odd and even indexed elements is 0.
3. Removing an element from index 5 modifies arr[] to {1, 0, 1, 0}. Therefore, xor of odd and even indexed elements is 0.

Input: arr[] = {1, 0, 0, 0, 1}, N=5
Output: 3 

1. Removing an element from index 3 modifies arr[] to {1, 0, 0, 1}. Therefore, xor of odd and even indexed elements is 0.
2. Removing an element from index 2 modifies arr[] to {1, 0, 0, 1}. Therefore, xor of odd and even indexed elements is 1.
3. Removing an element from index 4 modifies arr[] to {1, 0, 0, 0}. Therefore, xor of odd and even indexed elements is 1.

天真的方法：解决此问题的最简单方法是遍历数组，对于每个数组元素，检查从数组中删除元素是否使偶数索引和奇数索引数组元素的按位XOR相等或不相等。如果发现为真，则增加计数。最后，打印计数。

时间复杂度： O(N ² )
辅助空间： O(1)

高效方法：可以基于以下观察条件来优化上述方法：从给定数组中删除任何元素，使后续元素的偶数索引为奇数，而后续元素的奇数索引为偶数。请按照以下步骤解决问题：

1. 用0初始化变量curr_odd，curr_even，post_odd，post_even和res 。
2. 反向遍历数组并执行以下操作：
   • 如果当前元素为奇数，则对post_odd进行XOR。
   • 否则，将当前元素与post_even进行XOR。
3. 现在，遍历给定的数组并执行以下操作：
   • 如果当前索引为奇数，则通过使用post_odd和当前元素的按位XOR更新post_odd来从post_odd中删除当前元素。
   • 否则，请类似地从post_even中删除当前元素。
   • 初始化变量X和Y。
   • 在X中分配curr_odd和post_even的XOR。因此， X存储所有奇数索引元素的异或。
   • 在Y中指定curr_even和post_odd的xor。因此， Y存储所有偶数索引元素的异或。
   • 检查X是否等于Y。如果确定为true，则将res增加1 。
   • 如果当前索引为奇数，则将当前元素与curr_odd进行XOR。
   • 否则，将当前元素与curr_even进行XOR。
4. 最后，打印res 。

下面是上述方法的实现：

C++

// C++ program for the above approach
 
#include 
using namespace std;
 
// Function to count ways to make Bitwise
// XOR of odd and even indexed elements
// equal by removing an array element
void Remove_one_element(int arr[], int n)
{
    // Stores xor of odd and even
    // indexed elements from the end
    int post_odd = 0, post_even = 0;
 
    // Stores xor of odd and even
    // indexed elements from the start
    int curr_odd = 0, curr_even = 0;
 
    // Stores the required count
    int res = 0;
 
    // Traverse the array in reverse
    for (int i = n - 1; i >= 0; i--) {
 
        // If i is odd
        if (i % 2)
            post_odd ^= arr[i];
 
        // If i is even
        else
            post_even ^= arr[i];
    }
 
    // Traverse the array
    for (int i = 0; i < n; i++) {
 
        // If i is odd
        if (i % 2)
            post_odd ^= arr[i];
 
        // If i is even
        else
            post_even ^= arr[i];
 
        // Removing arr[i], post_even stores
        // XOR of odd indexed elements
        int X = curr_odd ^ post_even;
 
        // Removing arr[i], post_odd stores
        // XOR of even indexed elements
        int Y = curr_even ^ post_odd;
 
        // Check if they are equal
        if (X == Y)
            res++;
 
        // If i is odd, xor it
        // with curr_odd
        if (i % 2)
            curr_odd ^= arr[i];
 
        // If i is even, xor it
        // with curr_even
        else
            curr_even ^= arr[i];
    }
 
    // Finally print res
    cout << res << endl;
}
 
// Drivers Code
int main()
{
 
    // Given array
    int arr[] = { 1, 0, 1, 0, 1 };
 
    // Given size
    int N = sizeof(arr) / sizeof(arr[0]);
 
    // Function call
    Remove_one_element(arr, N);
 
    return 0;
}

Java

// Java program for the above approach
class GFG {
     
    // Function to count ways to make Bitwise
    // XOR of odd and even indexed elements
    // equal by removing an array element
    static void Remove_one_element(int arr[], int n)
    {
        // Stores xor of odd and even
        // indexed elements from the end
        int post_odd = 0, post_even = 0;
     
        // Stores xor of odd and even
        // indexed elements from the start
        int curr_odd = 0, curr_even = 0;
     
        // Stores the required count
        int res = 0;
     
        // Traverse the array in reverse
        for (int i = n - 1; i >= 0; i--)
        {
     
            // If i is odd
            if (i % 2 != 0)
                post_odd ^= arr[i];
     
            // If i is even
            else
                post_even ^= arr[i];
        }
     
        // Traverse the array
        for (int i = 0; i < n; i++)
        {
     
            // If i is odd
            if (i % 2 != 0)
                post_odd ^= arr[i];
     
            // If i is even
            else
                post_even ^= arr[i];
     
            // Removing arr[i], post_even stores
            // XOR of odd indexed elements
            int X = curr_odd ^ post_even;
     
            // Removing arr[i], post_odd stores
            // XOR of even indexed elements
            int Y = curr_even ^ post_odd;
     
            // Check if they are equal
            if (X == Y)
                res++;
     
            // If i is odd, xor it
            // with curr_odd
            if (i % 2 != 0)
                curr_odd ^= arr[i];
     
            // If i is even, xor it
            // with curr_even
            else
                curr_even ^= arr[i];
        }
     
        // Finally print res
        System.out.println(res);
    }
     
    // Drivers Code
    public static void main (String[] args)
    {
     
        // Given array
        int arr[] = { 1, 0, 1, 0, 1 };
     
        // Given size
        int N = arr.length;
     
        // Function call
        Remove_one_element(arr, N);   
    }
 
}
 
// This code is contributed by AnkitRai01

Python3

# Python3 program for the above approach
 
# Function to count ways to make Bitwise
# XOR of odd and even indexed elements
# equal by removing an array element
def Remove_one_element(arr, n):
 
    # Stores xor of odd and even
    # indexed elements from the end
    post_odd = 0
    post_even = 0
 
    # Stores xor of odd and even
    # indexed elements from the start
    curr_odd = 0
    curr_even = 0
 
    # Stores the required count
    res = 0
 
    # Traverse the array in reverse
    for i in range(n - 1, -1, -1):
 
        # If i is odd
        if (i % 2):
            post_odd ^= arr[i]
 
        # If i is even
        else:
            post_even ^= arr[i]
 
    # Traverse the array
    for i in range(n):
 
        # If i is odd
        if (i % 2):
            post_odd ^= arr[i]
 
        # If i is even
        else:
            post_even ^= arr[i]
 
        # Removing arr[i], post_even stores
        # XOR of odd indexed elements
        X = curr_odd ^ post_even
 
        # Removing arr[i], post_odd stores
        # XOR of even indexed elements
        Y = curr_even ^ post_odd
 
        # Check if they are equal
        if (X == Y):
            res += 1
 
        # If i is odd, xor it
        # with curr_odd
        if (i % 2):
            curr_odd ^= arr[i]
 
        # If i is even, xor it
        # with curr_even
        else:
            curr_even ^= arr[i]
 
    # Finally print res
    print(res)
 
# Drivers Code
if __name__ == "__main__" :
 
    # Given array
    arr = [ 1, 0, 1, 0, 1 ]
 
    # Given size
    N = len(arr)
 
    # Function call
    Remove_one_element(arr, N)
 
# This code is contributed by AnkitRai01

C#

// C# program to implement
// the above approach 
using System;
 
class GFG {
      
    // Function to count ways to make Bitwise
    // XOR of odd and even indexed elements
    // equal by removing an array element
    static void Remove_one_element(int[] arr, int n)
    {
        // Stores xor of odd and even
        // indexed elements from the end
        int post_odd = 0, post_even = 0;
      
        // Stores xor of odd and even
        // indexed elements from the start
        int curr_odd = 0, curr_even = 0;
      
        // Stores the required count
        int res = 0;
      
        // Traverse the array in reverse
        for (int i = n - 1; i >= 0; i--)
        {
      
            // If i is odd
            if (i % 2 != 0)
                post_odd ^= arr[i];
      
            // If i is even
            else
                post_even ^= arr[i];
        }
      
        // Traverse the array
        for (int i = 0; i < n; i++)
        {
      
            // If i is odd
            if (i % 2 != 0)
                post_odd ^= arr[i];
      
            // If i is even
            else
                post_even ^= arr[i];
      
            // Removing arr[i], post_even stores
            // XOR of odd indexed elements
            int X = curr_odd ^ post_even;
      
            // Removing arr[i], post_odd stores
            // XOR of even indexed elements
            int Y = curr_even ^ post_odd;
      
            // Check if they are equal
            if (X == Y)
                res++;
      
            // If i is odd, xor it
            // with curr_odd
            if (i % 2 != 0)
                curr_odd ^= arr[i];
      
            // If i is even, xor it
            // with curr_even
            else
                curr_even ^= arr[i];
        }
      
        // Finally print res
        Console.WriteLine(res);
    }
      
    // Drivers Code
    public static void Main ()
    {
      
        // Given array
        int[] arr = { 1, 0, 1, 0, 1 };
      
        // Given size
        int N = arr.Length;
      
        // Function call
        Remove_one_element(arr, N);   
    }
  
}
 
// This code is contributed by susmitakundugoaldanga

输出：

3

时间复杂度： O(N)
辅助空间： O(N)