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 scan = new Scanner(System.in);
		int x = scan.nextInt();
		int n = scan.nextInt();
		HashSet<Integer> set = new HashSet<>();
		int min = Integer.MAX_VALUE;
		int max = -1;
		for(int i = 0;i < n; ++i){
			int p = scan.nextInt();
			set.add(p);
			min = Math.min(min, p);
			max = Math.max(max, p);
		}
		if (set.size() == 0){
			System.out.print(x);
			return;
		}
		int diff = Integer.MAX_VALUE;
		int ans = max + 1;
		for(int i = max + 1; i >= min-1; --i){
			if(!set.contains(i) && Math.abs(x - i) <= diff){
				ans = Math.min(ans,i);
				diff = Math.abs(x - i);
			}
		}
		System.out.print(ans);

	}
}