Problem:
Score : 300 points

Problem Statement

Given are an integer X and an integer sequence of length N: p_1, \ldots, p_N.

Among the integers not contained in the sequence p_1, \ldots, p_N (not necessarily positive), find the integer nearest to X, that is, find the integer whose absolute difference with X is the minimum. If there are multiple such integers, report the smallest such integer.

Constraints

1 \leq X \leq 100

0 \leq N \leq 100

1 \leq p_i \leq 100

p_1, \ldots, p_N are all distinct.

All values in input are integers.

Input

Input is given from Standard Input in the following format:

X N
p_1 ... p_N

Output

Print the answer.

Sample Input 1

6 5
4 7 10 6 5

Sample Output 1

8

Among the integers not contained in the sequence 4, 7, 10, 6, 5, the one nearest to 6 is 8.

Sample Input 2

10 5
4 7 10 6 5

Sample Output 2

9

Among the integers not contained in the sequence 4, 7, 10, 6, 5, the ones nearest to 10 are 9 and 11. We should print the smaller one, 9.

Sample Input 3

100 0

Sample Output 3

100

When N = 0, the second line in the input will be empty. Also, as seen here, X itself can be the answer.

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

import java.util.*;

class Main {
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int x = sc.nextInt();
    int n = sc.nextInt();
    ArrayList<Integer> p = new ArrayList<Integer>();
    boolean f = true;
    if(n == 0) f = false;

    for(int i=0; i<n; i++) {
      int tmp = sc.nextInt();
      p.add(tmp);
    }

    Collections.sort(p);
    int ans = 101;
    for(int i=0; i<101; i++) {
      if(p.contains(i)) continue;
      //System.out.println(Math.abs(x-i));
      ans = Math.min(ans, Math.abs(x-i));
      //System.out.println(ans);
    }

    if(f = true) {
      if(p.contains(x-ans)) {
        System.out.println(x+ans);
      } else {
        System.out.println(x-ans);
      }
    } else {
      System.out.println(x);
    }

	}
}