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.Arrays;
import java.util.Scanner;

public class Main {

	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		
		int k = Integer.parseInt(sc.next());
		int n = Integer.parseInt(sc.next());
		
		int []a = new int [n];
		boolean []b = new boolean [200];
		
		for(int i = 0 ; i < n  ;i++) {
			a[i] = Integer.parseInt(sc.next());
		}

		Arrays.fill(b, false);
		
		int min = 1000000000;
		int ans = k;
		for(int i = 0 ; i < n  ;i++) {
			b[a[i]] = true;
		}
		
		for(int i = 1 ; i < 200 ;i++) {
			if(min > Math.abs(i-k) && !b[i]) {
				min = Math.abs(i-k);
				ans = i;
			}
		}
		System.out.println(ans);

	}
}