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;

namespace ABC132_C
{
    class Program
    {
        static void Main(string[] args) {
            int n = int.Parse(Console.ReadLine());
            string[] input = Console.ReadLine().Split();
            int[] numbers = new int[n];
            for(int i = 0; i < n; i++) {
                numbers[i] = int.Parse(input[i]);
            }
            Array.Sort(numbers);
            int underButtomIndex = n / 2;
            int upperButtomIndex = n / 2;
            if(n % 2 == 0) {
                underButtomIndex = n / 2 - 1;
                upperButtomIndex = n / 2;
            } else {
                if(numbers[underButtomIndex] == numbers[n / 2] || numbers[upperButtomIndex] == numbers[n / 2]) {
                    Console.WriteLine(0);
                    return;
                } else {
                    Console.WriteLine(1);
                    return;
                }
                underButtomIndex = n / 2 - 1;
                upperButtomIndex = n / 2 + 1;
            }

            long sumCount = numbers[upperButtomIndex] - numbers[underButtomIndex];

            Console.WriteLine(sumCount);
        }
    }
}