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: Accepted)

import java.util.*;
import java.lang.*;

import static java.lang.StrictMath.abs;

class Main {

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        //ｎは偶数
        int n = sc.nextInt();
        int[] d = new int[n];
        int[] d2 = new int[n];
        int count =0;
        //格納
        for(int i=0; i < n; i++){
            d[i] = sc.nextInt();
            d2[i] = d[i];
        }

        //sort
        Arrays.sort(d2);
//        for(int h=0; h < n ;h++) {
//            System.out.print(d[h]);
//            System.out.println(d2[h]);
//        }
      //  真ん中の数字の判定
        int std = d2[(n/2)-1]+1;
        int std2 = d2[n/2];

            System.out.println(std2 - std+1 );

//        int ans = 0;
//        //判定
//        two: for(int l = std; l < d2[n-1]; l++) {
//            count =0;
//            for (int k = 0; k < n; k++) {
//                if (d2[k] < l) {
//                    count++;
//                }
//            }
//
//            if (count == n / 2) {
//                ans ++;
//                continue two;
//            } else if(count > n/2){
//                break two;
//
//            }
//        }
//        //出力
//        System.out.println(ans);
    }
}