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 Java8 (OpenJDK 1.8.0) (judge verdict: Time Limit Exceeded)

import java.util.Scanner;

public final class Main {
    public static void main(String[] args){
        Scanner stdIn = new Scanner(System.in);
        
        int n = stdIn.nextInt();
        int[] d = new int[n];
        for(int i=0; i<n; i++){
            d[i] = stdIn.nextInt();
        }

        for(int i=0; i<n; i++){
            int min = d[i];
            int idx = i;
            for(int j=i+1; j<n; j++){
                if(d[j]<min){
                    min = d[j];
                    idx = j;
                }
            }
            d[idx] = d[i];        
            d[i] = min;
        }

        if(n%2 == 1){
            System.out.println(0);
        }else{
            System.out.println(d[n/2] - d[n/2-1]);
        }
    }
}