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 C++14 (Clang 3.8.0) (judge verdict: Accepted)

#include <algorithm>
#include <iostream>
#include <vector>
using namespace std;

int main() {
  int N = 0;
  int M = 0;
  int X = 0;
  std::cin >> N >> M >> X;
  std::vector<int> A(0, 0);

  for (int i = 0; i < M; i++) {
    int a = 0;
    cin >> a;
    A.push_back(a);
    //std::cout << A[i] << std::endl;
  }

  int right_cost = 0;
   int left_cost = 0 ;
  int i = 0;
   if(X < A[0]){
   left_cost=0;
     right_cost=M;
   }else if(X > A[M-1]){
   right_cost=M;
     left_cost=0;
   }else{
  for (;;) {
    if (A[i] < X && A[i + 1] > X) {
      break;
    }
    i += 1;
  }
      right_cost = M - (i+1 );
  left_cost = i+1 ;
   }
 
 //std::cout << left_cost << std::endl;
 //  std::cout << right_cost << std::endl;
  std::cout << min(left_cost, right_cost) << std::endl;
  return 0;
}