通过相邻对的异或对数组元素进行 M 次替换后的第 K 个数组元素

给定一个大小为N的数组arr[]和两个整数M和K ，任务是在对给定数组执行以下M 次操作后找到第K^个索引处的数组元素。

In a single operation, a new array is formed whose elements have the Bitwise XOR values of the adjacent elements of the current array.

如果操作次数M或M 次操作后的K值无效，则打印-1 。

例子：

Input: arr[] = {1, 4, 5, 6, 7}, M = 1, K = 2 
Output: 3 
Explanation: 
Since M = 1, therefore, the operation has to be performed only once on the array. 
The array is modified to {1^4, 4^5, 5^6, 6^7} = {5, 1, 3, 1}. 
The value of the element at index K = 2 in the updated array is 3.

Input: arr[] = {3, 2, 6}, M = 2, K = 1 
Output: -1 
Explanation: 
Since M = 2, operation has be performed twice. 
On performing 1st operation on the array {3, 2, 6} it is modified as {3^2, 2^6} = {1, 4}. 
On performing 2nd operation on the array {1, 4}, it is modified as {1^4} = {5}. 
After performing 2 operations, the size of the array is reduced to 1. This implies that the index K = 1 doesn’t exist after 2 operations. 
Hence, the given input is invalid. So the output is -1.

方法：想法是在每次操作后生成数组，并在每次迭代中签入是否可以进一步进行操作。如果不可能，则返回“-1” 。以下是步骤：

1. 检查是否可以对数组执行给定数量的操作。
2. 每次迭代后，数组的长度减 1，这意味着如果M大于或等于N ，则无法执行该操作。因此，打印“-1” 。
3. 如果M的值有效，则检查给定的索引K是否有效。
4. 经过M 次操作后，数组中还剩下(N – M) 个元素。因此，如果索引值大于或等于(N – M)的值，则打印“-1” 。
5. 如果M和K值都有效，则迭代循环M次并执行以下操作：
   • 在每次迭代中，创建一个临时数组temp[] ，用于存储通过对原始数组的相邻值执行操作获得的值。
   • 遍历数组arr[]并计算相邻元素的按位异或值，并将这些计算值保存在临时数组temp[] 中。
   • 使用temp[]更新当前数组arr [] 。
6. 在M次迭代之后，返回arr[K]的值，这是所需的输出。

下面是上述方法的实现：

C++

// C++ program for the above approach
#include
using namespace std;
 
// Method that returns the
// corresponding output by
// taking the given inputs.
int xor_operations(int N, int arr[], 
                   int M, int K)
{
     
    // If this condition is satisfied,
    // value of M is invalid
    if (M < 0 or M >= N)
        return -1;
     
    // Check if index K is valid
    if (K < 0 or K >= N - M)
        return -1;
     
    // Loop to perform M operations
    for(int p = 0; p < M; p++)
    {
         
        // Creating a temporary list
        vectortemp;
     
        // Traversing the array
        for(int i = 0; i < N; i++)
        {
             
            // Calculate XOR values
            // of adjacent elements
            int value = arr[i] ^ arr[i + 1];
             
            // Adding this value to
            // the temporary list
            temp.push_back(value);
             
            // Update the original array
            arr[i] = temp[i];
        }
    }
     
    // Getting value at index K
    int ans = arr[K];
     
    return ans;
}
 
// Driver Code
int main()
{
    // Number of elements
    int N = 5;
     
    // Given array arr[]
    int arr[] = {1, 4, 5, 6, 7};
    int M = 1, K = 2;
     
    // Function call
    cout << xor_operations(N, arr, M, K);
    return 0;
}
 
// This code is contributed by gauravrajput1

Java

// Java program for the above approach
import java.util.*;
 
class GFG{
 
// Method that returns the
// corresponding output by
// taking the given inputs.
static int xor_operations(int N, int arr[],
                          int M, int K)
{
     
    // If this condition is satisfied,
    // value of M is invalid
    if (M < 0 || M >= N)
        return -1;
     
    // Check if index K is valid
    if (K < 0 || K >= N - M)
        return -1;
     
    // Loop to perform M operations
    for(int p = 0; p < M; p++)
    {
         
        // Creating a temporary list
        Vectortemp = new Vector();
     
        // Traversing the array
        for(int i = 0; i < N - 1; i++)
        {
             
            // Calculate XOR values
            // of adjacent elements
            int value = arr[i] ^ arr[i + 1];
             
            // Adding this value to
            // the temporary list
            temp.add(value);
             
            // Update the original array
            arr[i] = temp.get(i);
        }
    }
     
    // Getting value at index K
    int ans = arr[K];
     
    return ans;
}
 
// Driver Code
public static void main(String[] args)
{
     
    // Number of elements
    int N = 5;
     
    // Given array arr[]
    int arr[] = { 1, 4, 5, 6, 7 };
    int M = 1, K = 2;
     
    // Function call
    System.out.print(xor_operations(N, arr, M, K));
}
}
 
// This code is contributed by Amit Katiyar

Python3

# Python3 program for the above approach
 
# Method that returns the
# corresponding output by
# taking the given inputs.
def xor_operations(N, arr, M, K):
     
    # If this condition is satisfied,
    # value of M is invalid
    if M < 0 or M >= N:
        return -1
     
    # Check if index K is valid
    if K < 0 or K >= N-M:
        return -1
     
 
    # Loop to perform M operations
    for _ in range(M):
     
        # Creating a temporary list
        temp = []
     
        # Traversing the array
        for i in range(len(arr)-1):
             
            # Calculate XOR values
            # of adjacent elements
            value = arr[i]^arr[i + 1]
             
            # Adding this value to
            # the temporary list
            temp.append(value)
         
        # Update the original array
        arr = temp[:]
     
    # Getting value at index K
    ans = arr[K]
    return ans
 
# Driver Code
 
# Number of elements
N = 5
 
# Given array arr[]
arr = [1, 4, 5, 6, 7]
M = 1
K = 2
 
# Function Call
print(xor_operations(N, arr, M, K))

C#

// C# program for the above approach
using System;
using System.Collections.Generic;
class GFG{
 
// Method that returns the
// corresponding output by
// taking the given inputs.
static int xor_operations(int N, int []arr,
                          int M, int K)
{
  // If this condition is satisfied,
  // value of M is invalid
  if (M < 0 || M >= N)
    return -1;
 
  // Check if index K is valid
  if (K < 0 || K >= N - M)
    return -1;
 
  // Loop to perform M operations
  for(int p = 0; p < M; p++)
  {
    // Creating a temporary list
    Listtemp = new List();
 
    // Traversing the array
    for(int i = 0; i < N - 1; i++)
    {
      // Calculate XOR values
      // of adjacent elements
      int value = arr[i] ^ arr[i + 1];
 
      // Adding this value to
      // the temporary list
      temp.Add(value);
 
      // Update the original array
      arr[i] = temp[i];
    }
  }
 
  // Getting value at index K
  int ans = arr[K];
 
  return ans;
}
 
// Driver Code
public static void Main(String[] args)
{
  // Number of elements
  int N = 5;
 
  // Given array []arr
  int []arr = {1, 4, 5, 6, 7};
  int M = 1, K = 2;
 
  // Function call
  Console.Write(xor_operations(N, arr,
                               M, K));
}
}
 
// This code is contributed by Rajput-Ji

Javascript

 
    // Javascript program for the above approach
     
    // Method that returns the
    // corresponding output by
    // taking the given inputs.
    function xor_operations(N, arr, M, K)
    {
 
        // If this condition is satisfied,
        // value of M is invalid
        if (M < 0 || M >= N)
            return -1;
 
        // Check if index K is valid
        if (K < 0 || K >= N - M)
            return -1;
 
        // Loop to perform M operations
        for(let p = 0; p < M; p++)
        {
 
            // Creating a temporary list
            let temp = [];
 
            // Traversing the array
            for(let i = 0; i < N; i++)
            {
 
                // Calculate XOR values
                // of adjacent elements
                let value = arr[i] ^ arr[i + 1];
 
                // Adding this value to
                // the temporary list
                temp.push(value);
 
                // Update the original array
                arr[i] = temp[i];
            }
        }
 
        // Getting value at index K
        let ans = arr[K];
 
        return ans;
    }
     
    // Number of elements
    let N = 5;
      
    // Given array arr[]
    let arr = [1, 4, 5, 6, 7];
    let M = 1, K = 2;
      
    // Function call
    document.write(xor_operations(N, arr, M, K));
 
     

输出：

3

时间复杂度： O(M * N)
辅助空间： O(N)

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