给定一个大小为N的数组arr [] ，任务是检查数组中是否存在任何大小为K的子数组，其按位XOR等于其余数组元素的按位XOR。如果发现是真的，则打印“是” 。否则，打印“否” 。
例子：
Input: arr[] = { 2, 3, 3, 5, 7, 7, 3, 4 }, K = 5
Output: YES
Explanation:
Bitwise XOR of the subarray { 3, 3, 5, 7, 7 } is equal to 5
Bitwise XOR of { 2, 3, 4 } is equal to 5.
Therefore, the required output is YES.
Input: arr[] = { 2, 3, 4, 5, 6, 7, 4 }, K = 2
Output: NO
天真的方法：解决此问题的最简单方法是生成大小为K的所有子数组。对于每个子数组，检查子数组的按位XOR是否等于其余元素的按位XOR。如果发现是真的，则打印“是” 。否则，打印“否” 。
时间复杂度： O(N 2 )
辅助空间： O(1)
高效的方法：可以使用滑动窗口技术来优化上述方法，以下是观察结果：
If X ^ Y = Z, then X ^ Z = Y
SubarrayXOR = arr[i] ^ arr[i + 1] ^ … ^ arr[j]
totalXOR = arr[0] ^ arr[1] ^ arr[2] ….. ^ arr[N – 1]
Bitwise XOR of the remaining array elements = totalXOR ^ SubarrayXOR
- 计算所有数组元素的按位XOR，例如totalXOR 。
- 计算数组的前K个元素的按位XOR ，例如SubarrayXOR 。
- 使用滑动窗技术，遍历大小为K的每个子阵列，并检查子阵列的位异或等于剩余的数组元素或不属于位异或。如果发现是真的，则打印“是” 。
- 否则，打印“否” 。
下面是上述方法的实现：
C++
// C++ program to implement
// the above approach
#include
using namespace std;
// Utility function to check if subarray
// of size K exits whose XOR of elements
// equal to XOR ofremaning array elements
bool isSubarrayExistUtil(int arr[], int K, int N)
{
int totalXOR = 0;
int SubarrayXOR = 0;
// Find XOR of whole array
for (int i = 0; i < N; i++)
totalXOR ^= arr[i];
// Find XOR of first K elements
for (int i = 0; i < K; i++)
SubarrayXOR ^= arr[i];
if (SubarrayXOR == (totalXOR ^ SubarrayXOR))
return true;
for (int i = K; i < N; i++) {
// Adding XOR of next element
SubarrayXOR ^= arr[i];
// Removing XOR of previous element
SubarrayXOR ^= arr[i - 1];
// Check if XOR of current subarray matches
// with the XOR of remaining elements or not
if (SubarrayXOR == (totalXOR ^ SubarrayXOR))
return true;
}
return false;
}
// Function to check if subarray of size
// K exits whose XOR of elements equal
// to XOR ofremaning array elements
void isSubarrayExist(int arr[], int K, int N)
{
if (isSubarrayExistUtil(arr, K, N))
cout << "YES\n";
else
cout << "NO\n";
}
// Driver Code
int32_t main()
{
// Given array
int arr[] = { 2, 3, 3, 5, 7, 7, 3, 4 };
// Size of the array
int N = sizeof(arr) / sizeof(arr[0]);
// Given K
int K = 5;
// Function Call
isSubarrayExist(arr, K, N);
} Java
// Java program to implement
// the above approach
import java.util.*;
class GFG{
// Utility function to check if subarray
// of size K exits whose XOR of elements
// equal to XOR ofremaning array elements
static boolean isSubarrayExistUtil(int arr[],
int K, int N)
{
int totalXOR = 0;
int SubarrayXOR = 0;
// Find XOR of whole array
for(int i = 0; i < N; i++)
totalXOR ^= arr[i];
// Find XOR of first K elements
for(int i = 0; i < K; i++)
SubarrayXOR ^= arr[i];
if (SubarrayXOR == (totalXOR ^ SubarrayXOR))
return true;
for(int i = K; i < N; i++)
{
// Adding XOR of next element
SubarrayXOR ^= arr[i];
// Removing XOR of previous element
SubarrayXOR ^= arr[i - 1];
// Check if XOR of current subarray matches
// with the XOR of remaining elements or not
if (SubarrayXOR == (totalXOR ^ SubarrayXOR))
return true;
}
return false;
}
// Function to check if subarray of size
// K exits whose XOR of elements equal
// to XOR ofremaning array elements
static void isSubarrayExist(int arr[],
int K, int N)
{
if (isSubarrayExistUtil(arr, K, N))
System.out.print("YES\n");
else
System.out.print("NO\n");
}
// Driver Code
public static void main(String[] args)
{
// Given array
int arr[] = { 2, 3, 3, 5, 7, 7, 3, 4 };
// Size of the array
int N = arr.length;
// Given K
int K = 5;
// Function Call
isSubarrayExist(arr, K, N);
}
}
// This code is contributed by 29AjayKumarPython3
# Python3 program to implement
# the above approach
# Utility function to check if subarray
# of size K exits whose XOR of elements
# equal to XOR ofremaning array elements
def isSubarrayExistUtil(arr, K, N):
totalXOR = 0
SubarrayXOR = 0
# Find XOR of whole array
for i in range(N):
totalXOR ^= arr[i]
# Find XOR of first K elements
for i in range(K):
SubarrayXOR ^= arr[i]
if (SubarrayXOR == (totalXOR ^ SubarrayXOR)):
return True
for i in range(K, N):
# Adding XOR of next element
SubarrayXOR ^= arr[i]
# Removing XOR of previous element
SubarrayXOR ^= arr[i - 1]
# Check if XOR of current subarray matches
# with the XOR of remaining elements or not
if (SubarrayXOR == (totalXOR ^ SubarrayXOR)):
return True
return False
# Function to check if subarray of size
# K exits whose XOR of elements equal
# to XOR ofremaning array elements
def isSubarrayExist(arr, K, N):
if (isSubarrayExistUtil(arr, K, N)):
print("YES")
else:
print("NO")
# Driver Code
if __name__ == '__main__':
# Given array
arr = [2, 3, 3, 5, 7, 7, 3, 4]
# Size of the array
N = len(arr)
# Given K
K = 5
# Function Call
isSubarrayExist(arr, K, N)
# This code is contributed by mohit kumar 29C#
// C# program to implement
// the above approach
using System;
class GFG
{
// Utility function to check if subarray
// of size K exits whose XOR of elements
// equal to XOR ofremaning array elements
static bool isSubarrayExistUtil(int []arr,
int K, int N)
{
int totalXOR = 0;
int SubarrayXOR = 0;
// Find XOR of whole array
for(int i = 0; i < N; i++)
totalXOR ^= arr[i];
// Find XOR of first K elements
for(int i = 0; i < K; i++)
SubarrayXOR ^= arr[i];
if (SubarrayXOR == (totalXOR ^ SubarrayXOR))
return true;
for(int i = K; i < N; i++)
{
// Adding XOR of next element
SubarrayXOR ^= arr[i];
// Removing XOR of previous element
SubarrayXOR ^= arr[i - 1];
// Check if XOR of current subarray matches
// with the XOR of remaining elements or not
if (SubarrayXOR == (totalXOR ^ SubarrayXOR))
return true;
}
return false;
}
// Function to check if subarray of size
// K exits whose XOR of elements equal
// to XOR ofremaning array elements
static void isSubarrayExist(int []arr,
int K, int N)
{
if (isSubarrayExistUtil(arr, K, N))
Console.Write("YES\n");
else
Console.Write("NO\n");
}
// Driver Code
public static void Main(String[] args)
{
// Given array
int []arr = { 2, 3, 3, 5, 7, 7, 3, 4 };
// Size of the array
int N = arr.Length;
// Given K
int K = 5;
// Function Call
isSubarrayExist(arr, K, N);
}
}
// This code is contributed by 29AjayKumarJavascript
输出：
YES时间复杂度： O(N)
辅助空间： O(N)