Problem:
Score : 700 points

Problem Statement

Snuke has a permutation (P_0,P_1,\cdots,P_{N-1}) of (0,1,\cdots,N-1).

Now, he will perform the following operation exactly once:

Choose K consecutive elements in P and sort them in ascending order.

Find the number of permutations that can be produced as P after the operation.

Constraints

2 \leq N \leq 200000

2 \leq K \leq N

0 \leq P_i \leq N-1

P_0,P_1,\cdots,P_{N-1} are all different.

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N K
P_0 P_1 \cdots P_{N-1}

Output

Print the number of permutations that can be produced as P after the operation.

Sample Input 1

5 3
0 2 1 4 3

Sample Output 1

2

Two permutations can be produced as P after the operation: (0,1,2,4,3) and (0,2,1,3,4).

Sample Input 2

4 4
0 1 2 3

Sample Output 2

1

Sample Input 3

10 4
2 0 1 3 7 5 4 6 8 9

Sample Output 3

6

Solution in Java (OpenJDK 11.0.6) (judge verdict: Wrong Answer)

import java.util.*;
public class Main{
    
    static final int INF = (int)1e9;
    
    public static void main(String[] args){
        Scanner sc = new Scanner(System.in);
        
        int n = Integer.parseInt(sc.next());
        int k = Integer.parseInt(sc.next());
        int[] a = new int[n];
        for(int i=0; i<n; i++){
            a[i] = Integer.parseInt(sc.next());
        }
        
        int[] b = new int[n];
        for(int i=0; i<n-1; i++){
            b[i] = (a[i]<a[i+1])? 1 : 0;
        }
        for(int i=0; i<n-1; i++){
            b[i+1] += b[i];
        }
        
        TreeSet<Integer> set = new TreeSet<>();
        for(int i=0; i<k; i++){
            set.add(a[i]);
        }
        
        boolean ct = (b[k-1]==k-1);
        int ans = (ct)? 0 : 1;
        
        for(int i=k; i<n; i++){
            set.add(a[i]);
            boolean ok = (set.first()!=a[i-k]) || (set.last()!=a[i]);
            set.remove(a[i-k]);
            
            if(b[i] - b[i-k+1] == k-1){
                ct = true;
            }else if(ok){
                ans++;
            }
        }
        
        if(ct){
            ans++;
        }
        System.out.println(ans);
    }
    
}