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: Accepted)

import java.util.*;
public class Main {
	public static void main(String[] args){
		Scanner sc = new Scanner(System.in);
		int x = sc.nextInt();
		int n = sc.nextInt();
		if(n==0){
			System.out.println(x);
			System.exit(0);
		}
		int ans1=0;
		int ans2=0;
		List<Integer> l = new ArrayList<Integer>();
		for(int i = 0;i<n;i++){
			l.add(sc.nextInt());
		}
		int count = 0;
		boolean flg1 = false;
		boolean flg2 = false;
		while(!flg1&&!flg2){

			if(!l.contains(x+count)){
				flg1 = true;
				ans1 = x+count;
			}
			if(!l.contains(x-count)){
				flg2 = true;
				ans2 = x-count;
			}
          count++;
		}
		if(flg2){
			System.out.println(ans2);
		}else{
			System.out.println(ans1);
		}		
	}
}