Problem:
Score : 300 points

Problem Statement

Takahashi made N problems for competitive programming.
The problems are numbered 1 to N, and the difficulty of Problem i is represented as an integer d_i (the higher, the harder).

He is dividing the problems into two categories by choosing an integer K, as follows:

A problem with difficulty K or higher will be for ARCs.

A problem with difficulty lower than K will be for ABCs.

How many choices of the integer K make the number of problems for ARCs and the number of problems for ABCs the same?

Problem Statement

2 \leq N \leq 10^5

N is an even number.

1 \leq d_i \leq 10^5

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N
d_1 d_2 ... d_N

Output

Print the number of choices of the integer K that make the number of problems for ARCs and the number of problems for ABCs the same.

Sample Input 1

6
9 1 4 4 6 7

Sample Output 1

2

If we choose K=5 or 6, Problem 1, 5, and 6 will be for ARCs, Problem 2, 3, and 4 will be for ABCs, and the objective is achieved.
Thus, the answer is 2.

Sample Input 2

8
9 1 14 5 5 4 4 14

Sample Output 2

0

There may be no choice of the integer K that make the number of problems for ARCs and the number of problems for ABCs the same.

Sample Input 3

14
99592 10342 29105 78532 83018 11639 92015 77204 30914 21912 34519 80835 100000 1

Sample Output 3

42685

Solution in C# (Mono 4.6.2.0) (judge verdict: Accepted)

using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Threading.Tasks;

class Program
{
    static void Main(string[] args)
    {
        int iCount = int.Parse(Console.ReadLine());
        List<int> N = new List<int>();
        string[] P = Console.ReadLine().Split(' ');
        for (int i = 0; i < P.Length; i++)
        {
            N.Add(int.Parse(P[i]));
        }

        N.Sort();

        int st =N[0];
        int oCount = 0;
        for (int i = 1; i < N.Count; i++)
        {
            if(N[i] != st)
            {
                if(N.Count/2 == i)
                {
                    oCount = N[i] - N[i - 1]; 
                    break;
                }
                else if(N.Count / 2 < i)
                {
                    break;
                }
            }
            else
            {
                st = N[i];
            }
        }

        Console.WriteLine(oCount);
        Console.ReadKey();

    }
}