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 C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>

#define fr first
#define sc second
#define pb push_back
#define ll long long

using namespace std;

int b[55];
vector<int>a;
vector<int>p[55];
vector<int>vec;
pair<int, int>ans;
int s[15];
int cnt;
int u[15];
void make(int b, int n)
{
    int m = n;
    while (b)
    {
        s[n - 1] = b % 2;
        b /= 2;
        n--;
    }
}
main()
{
    int n, mn;
    mn = 25;
    cin >> n;
    mn = INT_MAX;
    for (int i = 0; i < n; i++)
    {
        cin >> b[i];
        u[b[i]]++;
    }
    for (int i = 0; i < n; i++)
    {
        if (u[b[i]] > 2)
        {
            cout << 0;
            return 0;
        }
    }
    u[0]++;
    b[n] =  0;
    n ++;

    sort(b, b + n);

    for (int i = 0; i < n - 1; i++)
    {
        if (b[i] != b[i + 1])
            a.pb(b[i]);
    }

    a.pb(b[n - 1]);

    if (n == 1)
    {
        cout << b[0];
        return 0;
    }
    if (n > 25)
    {
        cout << 0;
        return 0;
    }
    if (n >= 24)
    {
        cout << 1;
        return 0;
    }
    n = a.size();
    if (n == 1)
    {
        cout << abs((24 - a[0]) - a[0]);
        return 0;
    }

    for (int i = 0; i < n; i++)
    {
        if (a[i])
        {
            p[i].pb(a[i]);
            p[i].pb(24 - a[i]);
        }
        else
        {
            cout << 0;
            return 0;
        }
    }

    for (int i = 0; i < (1 << n); i++)
    {
        make(i, n);
        for (int j = 0; j < n - 1; j++)
        {
            for (int k = j + 1; k < n; k++)
            {
                mn = min(mn, abs(p[j][s[j]] - p[k][s[k]]));
            }
        }
        for (int j = 0; j < n; j++)
        {
            int mm = min(p[j][s[j]], 24 - p[j][s[j]]);
            mn = min(mn, mm);
        }
        if(mn < 1e9)
            vec.pb(mn);
        mn = 1e9 + 1;
    }
    sort(vec.begin(), vec.end());

    cout << vec.back();
}