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;
using System.Collections.Generic;

namespace ConsoleApp2
{
    public class Program
    {
        public static void Main(string[] args)
        {
            int N = int.Parse(Console.ReadLine());
            int[] dN = Read(Console.ReadLine());

            List<int> dn = dN.ToList();
            dn.Sort();

            int center = N / 2;

            Console.WriteLine(dn[center] - dn[center-1]);


        }


        // 最小公倍数
        public static int Lcm(int a, int b)
        {
            return a * b / Gcd(a, b);
        }

        // ユークリッドの互除法 
        public static int Gcd(int a, int b)
        {
            if (a < b)
                // 引数を入替えて自分を呼び出す
                return Gcd(b, a);
            while (b != 0)
            {
                var remainder = a % b;
                a = b;
                b = remainder;
            }
            return a;
        }


        private static bool S(int[] data,int K ,int l ,int x)
        {
            int sum = 0;
            int max = Math.Min(x+1, data.Length);
            for (int i = l; i < max; i++)
            {
                sum += data[i];
            }
            //Console.WriteLine(sum);
            return sum>=K;
        }

        private static int Stair(int N)
        {
            if (N < 0) return 0;
            int twoM = N / 2;
            int ans = 0;
            for (int two = 0; two < twoM+1; two++)
            {
                int one = N - 2 * two;
                ans += Combination(one + two,two);
            }
            return ans;
        }

        private static int Combination(int N,int M)
        {
            
            int ans = Product(N) / (Product(M) * Product(N - M));
            return ans;

        }

        private static int Product(int N)
        {
            if (N == 0||N ==1) return 1;
            int ans = N;
            for (int i = 1; i < N; i++)
            {
                ans *= i;
            }
            return ans;
        }





        private static string ReplaceOneChar(string S, int index, string newS)
        {
            string s1 = S.Substring(0, index - 1);
            string s2 = S.Substring(index);
            return s1 + newS + s2;
        }

        private static bool PartialSum(int target, int[] Ns)
        {
            //if (target == 0) return true;
            bool[][] ps = new bool[Ns.Length + 1][];

            //i=0では、どんなWに対してもfalse
            ps[0] = new bool[target+1];
            for (int i = 0; i < ps[0].Length; i++) ps[0][i] = false;

            ps[0][0] = true;


            //考慮に入れるitemを、前から順に追加していく
            for (int i = 0; i < Ns.Length; i++)
            {
                //初期化
                ps[i + 1] = new bool[target+1];

                for (int t = 0; t <= target; t++)
                {
                    if (t >= Ns[i]) ps[i + 1][t] = ps[i][t - Ns[i]] || ps[i][t];
                    else ps[i + 1][t]=ps[i][t];
                }
            }

            return ps[Ns.Length][target];
        }

        /// <summary>
        /// 引数は、W(重さ制限)、items[重さ][価値]
        /// </summary>
        /// <param name="W"></param>
        /// <param name="items"></param>
        /// <returns></returns>
        private static int DP(int W,int[][] items)
        {
            //データ
            int[][] dp = new int[items.Length+1][];

            //i=0では、どんなWに対しても0
            dp[0] = new int[W + 1];

            //考慮に入れるitemを、前から順に追加していく
            for (int i = 0; i < items.Length; i++)
            {
                //初期化
                dp[i+1] = new int[W + 1];

                for (int w = 0; w <= W; w++)
                {
                    if (w >= items[i][0]) dp[i + 1][w] = Math.Max(dp[i][w - items[i][0]]+items[i][1], dp[i][w]);
                    else dp[i + 1][w] = dp[i][w];
                }
            }
            return dp[items.Length][W];
        }

        private static int[] Read(string input)
        {
            string[] aaa = input.Split(' ');
            int num = aaa.Length;
            int[] ints = new int[num];
            for (int i = 0; i < num; i++)
            {
                ints[i] = int.Parse(aaa[i]);
            }
            
            return ints;
        }


    }


}