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.Scanner;
import java.util.Arrays;

public class Main {
    public static void main(String[] args) {
        Scanner scan = new Scanner(System.in);
        int x = scan.nextInt();
        int n = scan.nextInt();
        
        int minDiff = 10000;
        int closest = -2;
        boolean[] vals = new boolean[101];
        for(int i = 0 ; i < n ; i++) {
            vals[scan.nextInt()] = true;
            
        }

        
        for(int i = -1 ; i <= 101 ;i++) {
            if(i == -1 || i == 101 || !vals[i]) {
                if(minDiff > Math.abs(x-i)) {
                    minDiff = Math.abs(x-i);
                    closest = i;
                }    
            }
        }
        System.out.println(closest);
    }
}