Problem:
Score : 200 points

Problem Statement

There are N + 1 squares arranged in a row, numbered 0, 1, ..., N from left to right.

Initially, you are in Square X.
You can freely travel between adjacent squares. Your goal is to reach Square 0 or Square N.
However, for each i = 1, 2, ..., M, there is a toll gate in Square A_i, and traveling to Square A_i incurs a cost of 1.
It is guaranteed that there is no toll gate in Square 0, Square X and Square N.

Find the minimum cost incurred before reaching the goal.

Constraints

1 \leq N \leq 100

1 \leq M \leq 100

1 \leq X \leq N - 1

1 \leq A_1 < A_2 < ... < A_M \leq N

A_i \neq X

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N M X
A_1 A_2 ... A_M

Output

Print the minimum cost incurred before reaching the goal.

Sample Input 1

5 3 3
1 2 4

Sample Output 1

1

The optimal solution is as follows:

First, travel from Square 3 to Square 4. Here, there is a toll gate in Square 4, so the cost of 1 is incurred.

Then, travel from Square 4 to Square 5. This time, no cost is incurred.

Now, we are in Square 5 and we have reached the goal.

In this case, the total cost incurred is 1.

Sample Input 2

7 3 2
4 5 6

Sample Output 2

0

We may be able to reach the goal at no cost.

Sample Input 3

10 7 5
1 2 3 4 6 8 9

Sample Output 3

3

Solution in Java8 (OpenJDK 1.8.0) (judge verdict: Wrong Answer)



import java.io.*;
import java.util.*;
 class Main {

	public static void main(String[] args) throws IOException{
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		String s[]=br.readLine().split(" ");
		int  n =Integer.parseInt(s[0]);
		int  m =Integer.parseInt(s[1]);
		int  x =Integer.parseInt(s[2]);
		int arr[] = new int[m];
		String s1[] = br.readLine().split(" ");
		for (int i = 0; i < s1.length; i++) {
			arr[i]=Integer.parseInt(s1[i]);
		}
		/*int res=0;
		int left = 0;
		int right = m-1;
		if(x > arr[m-1] || x< arr[0]) {
			System.out.println(0);
			
		}
		else {
		while(left<right) {
			int mid = (left+right)/2;
			//System.out.println("mid"+mid);
			if(arr[mid] < x) {
				res = mid;
				left = mid+1;
			}
			else {
				right = mid-1;
			}
		}
		if(arr[m-1] == n) {
			System.out.println(Math.min(res+1, m-res-2));
		}
		else
		System.out.println(Math.min(res+1, m-res-1));
		}*/
		int res=0;
		for (int i = 0; i < s1.length; i++) {
			if(arr[i]<x) {
				res++;
			}
			else {
				break;
			}
		}
		System.out.println(Math.min(res+1, m-1-res));
	}

}