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.io.*;
import java.util.*;
public class Main {
	
	public static void main(String[] args) {
		InputReader in = new InputReader();
		int n = in.nextInt();
		
		int[] arr = new int[n];
		for (int i=0; i<n; i++) {
			arr[i] = in.nextInt();
		}
		Arrays.sort(arr);
		int num1 = arr[n/2 - 1];
		int num2 = arr[n/2];
		
		System.out.println(num2 - num1);
		
	}

	static class InputReader {
		public BufferedReader br;
		public StringTokenizer st;
		
		public InputReader() {
			br = new BufferedReader(new InputStreamReader(System.in));
			st = null;
		}
		
		public String next() {
			while (st == null || !st.hasMoreTokens()) {
				try {
					st = new StringTokenizer(br.readLine());
				}
				catch (IOException e) {
					throw new RuntimeException(e);
				}
			}
			return st.nextToken();
		}
		
		public int nextInt() {
			return Integer.parseInt(next());
		}
	}
}