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++ (GCC 9.2.1) (judge verdict: Accepted)

#include <bits/stdc++.h>

using namespace std;

int main(){
  int n, d, mn=12;
  map<int, int> mp;
  cin >> n;
  for(int i=0;i<n;i++){
    cin >> d;
    mn = min(mn, d);
    mp[d]++;
  }
  int mxl=0, mxr=0;
  for(auto p: mp){
    if(p.second==1){
      if(mxl<=mxr){
        mn = min(mn, p.first-mxl);
        mxl = p.first;
      }
      else{
        mn = min(mn, p.first-mxr);
        mxr = p.first;
      }
    }
    else if(p.second==2){
      mn = min({mn, p.first-mxl, p.first-mxr});
      mxl=p.first;
      mxr=p.first;
    }
    else{
      mn = 0;
    }
  }
  mn = min(mn, (12-mxl)+(12-mxr));
  cout << mn << endl;
}