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: Wrong Answer)

import java.util.*;
public class Main {
    public static int sub(int[] lst, int n, int p){
        int pivot = lst[0];
        int sm = 1;
        int eq = 1;
        for(int i = 1; i < n; i++){
            if(lst[i]==pivot){
                int sub1 = lst[eq];
                lst[eq] = lst[i];
                lst[i] = sub1;
                eq++;
                sm++;
            }else if(lst[i]<pivot){
                int sub2 = lst[sm];
                lst[sm] = lst[i];
                lst[i] = sub2;
                sm++;
            }
        }
        int[] slst = new int[sm-eq];
        int[] blst = new int[n-sm];
        for(int i = 0; i < sm-eq; i++){
            slst[i] = lst[eq+i];
        }
        for(int i = 0; i < n-sm; i++){
            blst[i] = lst[sm+i];
        }
        if(sm-eq < p+1 && p+1 <= sm){
            return pivot;
        }else if(p+1 <= sm-eq ){
            return sub(slst,sm-eq,p);
        }else{
            return sub(blst,n-sm,p-sm);
        }
    }
    public static void main(String[] args) {
        Scanner scan = new Scanner(System.in);
        int a,b;
        int n = Integer.parseInt(scan.next());
        int lst[] = new int[n];
        for(int i = 0; i < n; i++){
            lst[i] = Integer.parseInt(scan.next());
        }
        a = sub(lst,n,n/2-1);
        b = sub(lst,n,n/2);
        System.out.println(b-a);
    }
}