查找具有最小XOR的大小为K的子数组

给定数组arr []和整数K ，任务是找到给定数组中大小为K的任何子数组的最小按位XOR和。

例子：

Input: arr[] = {3, 7, 90, 20, 10, 50, 40}, K = 3
Output: 16
Explanation:
The subarray {10, 50, 40} has the minimum XOR

Input: arr[] = {3, 7, 5, 20, -10, 0, 12}, K = 2
Output: 17
Explanation:
The subarray {5, 20} has the minimum XOR

天真的方法：一个简单的解决方案是将每个元素视为大小为k的子数组的开始，并计算从该元素开始的子数组的XOR。
时间复杂度： O(N * K)

高效方法：想法是使用大小为K的滑动窗口技术并跟踪当前K个元素的XOR和。要计算当前窗口的XOR，请与前一个窗口的第一个元素进行XOR以丢弃该元素，并与当前元素进行异或以将该元素添加到窗口中。同样，滑动窗口以找到大小为K的子数组的最小XOR。

下面是上述方法的实现：

C++

// C++ implementation to find the
// subarray with minimum XOR
  
#include 
  
using namespace std;
      
// Function to find the minimum 
// XOR of the subarray of size K
void findMinXORSubarray(int arr[], 
                     int n, int k)
{
    // K must be smaller 
    // than or equal to n
    if (n < k)
        return;
  
    // Initialize beginning 
    // index of result
    int res_index = 0;
  
    // Compute XOR sum of first 
    // subarray of size K
    int curr_xor = 0;
    for (int i = 0; i < k; i++)
        curr_xor ^= arr[i];
  
    // Initialize minimum XOR 
    // sum as current xor
    int min_xor = curr_xor;
  
    // Traverse from (k+1)'th 
    // element to n'th element
    for (int i = k; i < n; i++) {
          
        // XOR with current item 
        // and first item of
        // previous subarray
        curr_xor ^= (arr[i] ^ arr[i - k]);
  
        // Update result if needed
        if (curr_xor < min_xor) {
            min_xor = curr_xor;
            res_index = (i - k + 1);
        }
    }
  
    cout << min_xor << "\n";
}
  
// Driver Code
int main()
{
    int arr[] = { 3, 7, 90, 20, 10, 50, 40 };
    int k = 3; // Subarray size
    int n = sizeof arr / sizeof arr[0];
      
    // Function Call
    findMinXORSubarray(arr, n, k);
    return 0;
}

Java

// Java implementation to find the
// subarray with minimum XOR
class GFG{
      
// Function to find the minimum 
// XOR of the subarray of size K
static void findMinXORSubarray(int arr[], 
                               int n, int k)
{
      
    // K must be smaller 
    // than or equal to n
    if (n < k)
        return;
  
    // Initialize beginning 
    // index of result
    int res_index = 0;
  
    // Compute XOR sum of first 
    // subarray of size K
    int curr_xor = 0;
    for(int i = 0; i < k; i++)
       curr_xor ^= arr[i];
  
    // Initialize minimum XOR 
    // sum as current xor
    int min_xor = curr_xor;
  
    // Traverse from (k+1)'th 
    // element to n'th element
    for(int i = k; i < n; i++)
    {
         
       // XOR with current item 
       // and first item of
       // previous subarray
       curr_xor ^= (arr[i] ^ arr[i - k]);
         
       // Update result if needed
       if (curr_xor < min_xor)
       {
           min_xor = curr_xor;
           res_index = (i - k + 1);
       }
    }
    System.out.print(min_xor + "\n");
}
  
// Driver Code
public static void main(String[] args)
{
    int arr[] = { 3, 7, 90, 20, 10, 50, 40 };
      
    // Subarray size
    int k = 3; 
    int n = arr.length;
      
    // Function Call
    findMinXORSubarray(arr, n, k);
}
}
  
// This code is contributed by Amit Katiyar

Python3

# Python3 implementation to find the 
# subarray with minimum XOR 
      
# Function to find the minimum 
# XOR of the subarray of size K 
def findMinXORSubarray(arr, n, k):
  
    # K must be smaller 
    # than or equal to n 
    if (n < k): 
        return
  
    # Initialize beginning 
    # index of result 
    res_index = 0
  
    # Compute XOR sum of first 
    # subarray of size K 
    curr_xor = 0
    for i in range(0, k): 
        curr_xor = curr_xor ^ arr[i] 
  
    # Initialize minimum XOR 
    # sum as current xor 
    min_xor = curr_xor
  
    # Traverse from (k+1)'th 
    # element to n'th element 
    for i in range(k, n):
          
        # XOR with current item 
        # and first item of 
        # previous subarray 
        curr_xor ^= (arr[i] ^ arr[i - k])
  
        # Update result if needed 
        if (curr_xor < min_xor): 
            min_xor = curr_xor
            res_index = (i - k + 1) 
  
    print(min_xor, end = '\n')
  
# Driver Code 
arr = [ 3, 7, 90, 20, 10, 50, 40 ] 
  
# Subarray size 
k = 3 
n = len(arr)
  
# Function Call 
findMinXORSubarray(arr, n, k)
  
# This code is contributed by PratikBasu

C#

// C# implementation to find the
// subarray with minimum XOR
using System;
  
class GFG{
      
// Function to find the minimum 
// XOR of the subarray of size K
static void findMinXORSubarray(int []arr, 
                               int n, int k)
{
      
    // K must be smaller 
    // than or equal to n
    if (n < k)
        return;
  
    // Initialize beginning 
    // index of result
    int res_index = 0;
  
    // Compute XOR sum of first 
    // subarray of size K
    int curr_xor = 0;
    for(int i = 0; i < k; i++)
       curr_xor ^= arr[i];
  
    // Initialize minimum XOR 
    // sum as current xor
    int min_xor = curr_xor;
  
    // Traverse from (k+1)'th 
    // element to n'th element
    for(int i = k; i < n; i++)
    {
         
       // XOR with current item 
       // and first item of
       // previous subarray
       curr_xor ^= (arr[i] ^ arr[i - k]);
         
       // Update result if needed
       if (curr_xor < min_xor)
       {
           min_xor = curr_xor;
           res_index = (i - k + 1);
       }
    }
    Console.Write(min_xor + "\n");
}
  
// Driver Code
public static void Main(String[] args)
{
    int []arr = { 3, 7, 90, 20, 10, 50, 40 };
      
    // Subarray size
    int k = 3; 
    int n = arr.Length;
      
    // Function Call
    findMinXORSubarray(arr, n, k);
}
}
  
// This code is contributed by Amit Katiyar

输出：

16

时间复杂度： O(N)
辅助空间： O(1)