Problem:
Score : 500 points

Problem Statement

In CODE FESTIVAL XXXX, there are N+1 participants from all over the world, including Takahashi.

Takahashi checked and found that the time gap (defined below) between the local times in his city and the i-th person's city was D_i hours.
The time gap between two cities is defined as follows. For two cities A and B, if the local time in city B is d o'clock at the moment when the local time in city A is 0 o'clock, then the time gap between these two cities is defined to be min(d,24-d) hours.
Here, we are using 24-hour notation.
That is, the local time in the i-th person's city is either d o'clock or 24-d o'clock at the moment when the local time in Takahashi's city is 0 o'clock, for example.

Then, for each pair of two people chosen from the N+1 people, he wrote out the time gap between their cities. Let the smallest time gap among them be s hours.

Find the maximum possible value of s.

Constraints

1 \leq N \leq 50

0 \leq D_i \leq 12

All input values are integers.

Input

Input is given from Standard Input in the following format:

N
D_1 D_2 ... D_N

Output

Print the maximum possible value of s.

Sample Input 1

3
7 12 8

Sample Output 1

4

For example, consider the situation where it is 7, 12 and 16 o'clock in each person's city at the moment when it is 0 o'clock in Takahashi's city. In this case, the time gap between the second and third persons' cities is 4 hours.

Sample Input 2

2
11 11

Sample Output 2

2

Sample Input 3

1
0

Sample Output 3

0

Note that Takahashi himself is also a participant.

Solution in Java8 (OpenJDK 1.8.0) (judge verdict: Wrong Answer)

import java.util.*;

public class Main {
    public void main(Scanner sc) {
        int n = sc.nextInt();
        int d[] = new int[13];
        for (int i = 0; i < n; i++) {
            d[sc.nextInt()]++;
        }

        if (d[0] >= 1 || d[12] >= 2 || Arrays.stream(d).anyMatch(cnt -> cnt >= 3)) {
            System.out.println(0);
            return;
        }

        int len = 0;
        int one[] = new int[12];
        boolean time[] = new boolean[24];
        for (int i = 0; i <= 12; i++) {
            if (d[i] == 2) {
                time[d[i]] = true;
                time[(24 - d[i]) % 24] = true;
            } else if (d[i] == 1) {
                one[len++] = i;
            }
        }
        time[0] = true;

        int ans = 0;
        for (int i = 0; i < (1 << len); i++) {
            boolean tmp[] = new boolean[24];
            for (int j = 0; j < 24; j++) {
                tmp[j] = time[j];
            }
            for (int j = 0; j < len; j++) {
                if ((i >> j) % 2 == 0) {
                    tmp[one[j]] = true;
                } else {
                    tmp[24 - one[j]] = true;
                }
            }
            ans = Math.max(ans, calc(tmp));
        }

        System.out.println(ans);
    }

    private int calc(boolean time[]) {
        int prv = 0;
        int ans = 25;
        for (int i = 1; i < 24; i++) {
            if (time[i]) {
                ans = Math.min(ans, i - prv);
                prv = i;
            }
        }

        return ans;
    }

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        new Main().main(sc);
        sc.close();
    }
}