检查 Array 是否可以拆分为子数组，使得这些子数组的最长递减子序列的长度的 XOR 为 0

给定一个大小为N的整数数组arr[] ，任务是检查是否可以将arr[]拆分为不同的子数组，使得所有子数组的 LDS（最长递减子序列）长度的 XOR 等于0 .如果可以拆分则打印' YES '，否则打印' NO '。

例子：

Input: arr[] = {1, 0, 3, 4, 5}
Output: YES
Explanation: {1}, {0}, {3}, {4, 5} length of LDS of subarrays: 1, 1, 1, 1, XOR of lengths = 0. So answer is Yes.

Input: arr[] = {5, 4, 3}
Output: NO

方法：偶数个1 的XOR 操作为0。因此，如果数组长度为偶数，则每个元素都可以视为大小为1的 LDS，这使得偶数1 的XOR 等于0 ，并且对于奇数长度的数组具有偶数 LDS 1 的任何大小为2的子数组都可以与 LDS 一起考虑，即子数组应该严格增加，因此 LDS 将为1 。请按照以下步骤解决问题：

• 初始化一个发现为假的变量。
• 如果N甚至打印'YES'并返回。
• 否则，使用变量i迭代范围(0, N – 1]并执行以下任务：
  • 通过arr[i]>arr[i-1]检查是否有连续增加的元素
  • 如果arr[i]>arr[i-1] make发现为真并退出循环
• 如果找到为真打印“YES”，否则打印“NO”

下面是上述方法的实现。

C++

// C++ code for the above approach
#include 
using namespace std;
 
// Function to find XOR of LDS's
// of subarrays
void xor_of_lds(int arr[], int n)
{
 
    // If length is even each element
    // can be considered as lds of length
    // 1 which makes even 1's and xor is 0
    if (n % 2 == 0) {
        cout << "YES" << endl;
        return;
    }
    else {
 
        // For odd length we need to find
        // even subarray of length 2 which
        // is strictly increasing so that it
        // makes a subarray with lds=1
 
        bool found = 0;
        for (int i = 1; i < n; i++) {
 
            // Check if there are 2
            // consecutive increasing elements
            if (arr[i] > arr[i - 1]) {
                found = 1;
                break;
            }
        }
        if (found == 1)
            cout << "YES" << endl;
        else
            cout << "NO" << endl;
    }
}
 
// Driver Code
int main()
{
 
    // Initializing array of arr
    int arr[] = { 1, 0, 3, 4, 5 };
    int N = sizeof(arr) / sizeof(arr[0]);
 
    // Call the function
    xor_of_lds(arr, N);
 
    return 0;
}

Java

// Java code for the above approach
import java.util.*;
class GFG{
 
// Function to find XOR of LDS's
// of subarrays
static void xor_of_lds(int arr[], int n)
{
 
    // If length is even each element
    // can be considered as lds of length
    // 1 which makes even 1's and xor is 0
    if (n % 2 == 0) {
        System.out.print("YES" +"\n");
        return;
    }
    else {
 
        // For odd length we need to find
        // even subarray of length 2 which
        // is strictly increasing so that it
        // makes a subarray with lds=1
 
        boolean found = false;
        for (int i = 1; i < n; i++) {
 
            // Check if there are 2
            // consecutive increasing elements
            if (arr[i] > arr[i - 1]) {
                found = true;
                break;
            }
        }
        if (found == true)
            System.out.print("YES" +"\n");
        else
            System.out.print("NO" +"\n");
    }
}
 
// Driver Code
public static void main(String[] args)
{
 
    // Initializing array of arr
    int arr[] = { 1, 0, 3, 4, 5 };
    int N = arr.length;
 
    // Call the function
    xor_of_lds(arr, N);
 
}
}
 
// This code is contributed by shikhasingrajput

Python3

# Python code for the above approach
 
# Function to find XOR of LDS's
# of subarrays
def xor_of_lds  (arr, n) :
 
    # If length is even each element
    # can be considered as lds of length
    # 1 which makes even 1's and xor is 0
    if (n % 2 == 0):
        print("YES")
        return
    else:
 
        # For odd length we need to find
        # even subarray of length 2 which
        # is strictly increasing so that it
        # makes a subarray with lds=1
 
        found = 0
        for i in range(1, n):
            # Check if there are 2
            # consecutive increasing elements
            if (arr[i] > arr[i - 1]):
                found = 1
                break
        if (found == 1):
            print("YES")
        else:
            print("NO")
 
# Driver Code
# Initializing array of arr
arr = [1, 0, 3, 4, 5]
N = len(arr)
 
# Call the function
xor_of_lds(arr, N)
 
# This code is contributed by Saurabh Jaiswal

C#

// C# code for the above approach
using System;
 
class GFG
{
   
// Function to find XOR of LDS's
// of subarrays
static void xor_of_lds(int []arr, int n)
{
 
    // If length is even each element
    // can be considered as lds of length
    // 1 which makes even 1's and xor is 0
    if (n % 2 == 0) {
        Console.Write("YES" + "\n");
        return;
    }
    else {
 
        // For odd length we need to find
        // even subarray of length 2 which
        // is strictly increasing so that it
        // makes a subarray with lds=1
 
        bool found = false;
        for (int i = 1; i < n; i++) {
 
            // Check if there are 2
            // consecutive increasing elements
            if (arr[i] > arr[i - 1]) {
                found = true;
                break;
            }
        }
        if (found == true)
            Console.Write("YES" +"\n");
        else
            Console.Write("NO" +"\n");
    }
}
 
// Driver Code
public static void Main()
{
   
    // Initializing array of arr
    int []arr = { 1, 0, 3, 4, 5 };
    int N = arr.Length;
 
    // Call the function
    xor_of_lds(arr, N);
}
}
 
// This code is contributed by Samim Hossain Mondal.

Javascript

    // JavaScript code for the above approach
 
    // Function to find XOR of LDS's
    // of subarrays
    const xor_of_lds = (arr, n) => {
 
        // If length is even each element
        // can be considered as lds of length
        // 1 which makes even 1's and xor is 0
        if (n % 2 == 0) {
            document.write("YES<br/>");
            return;
        }
        else {
 
            // For odd length we need to find
            // even subarray of length 2 which
            // is strictly increasing so that it
            // makes a subarray with lds=1
 
            let found = 0;
            for (let i = 1; i < n; i++) {
 
                // Check if there are 2
                // consecutive increasing elements
                if (arr[i] > arr[i - 1]) {
                    found = 1;
                    break;
                }
            }
            if (found == 1)
                document.write("YES<br/>");
            else
                document.write("NO<br/>");
        }
    }
 
    // Driver Code
    // Initializing array of arr
    let arr = [1, 0, 3, 4, 5];
    let N = arr.length;
 
    // Call the function
    xor_of_lds(arr, N);
 
    // This code is contributed by rakeshsahni

输出

YES

时间复杂度： O(N)
辅助空间： O(1)