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.Linq;

namespace C
{
    class Program
    {
        static void Main(string[] args)
        {
            var n = int.Parse(Console.ReadLine());
            var num = Array.ConvertAll(Console.ReadLine().Split(' '), int.Parse).ToList();
            var ans = 0;
            if (n % 2 != 0)
            {
                Console.WriteLine(0);
                return;
            }
            else
            {
                num.Sort((a,b) => b -a);
                for(var i=0;i<n;i++)
                {
                    if(i+1 == n/2)
                    {
                        if(num[i] == num[i+1])
                        {
                            Console.WriteLine(0);
                            return;
                        }
                        ans = num[i] - num[i + 1];
                    }
                }
            }
            Console.WriteLine(ans);
        }
    }
}