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.ArrayList;
import java.util.Collections;
import java.util.Scanner;

public class Main {

	public static void main(String[] args) {
		// TODO 自動生成されたメソッド・スタブ
		
		Scanner sc=new Scanner(System.in);
		ArrayList<Integer>difficulty=new ArrayList<>();
		
		//問題数
		int N=sc.nextInt();
		
		//ABC問題のカウント
		int abcCount=0;
		//ARC問題のカウント
		int arcCount=0;
		//問題の数がABC、ARCともに一緒の時カウント
		int equalCount=0;
		
		//問題の難易度
		for(int i=0;i<N;i++) {
			difficulty.add(sc.nextInt());
		}
		
		int difficultyMax=Collections.max(difficulty);
		
		//難易度に応じた振り分け
		for(int i=1;i<=difficultyMax;i++) {
			for(int j=0;j<difficulty.size();j++) {
				if(difficulty.get(j)>=i) {
					arcCount++;
				}else {
					abcCount++;
				}
			}
				
				if(arcCount==abcCount) {
					equalCount++;
				}
				
	
			arcCount=0;
			abcCount=0;
		}
		System.out.println(equalCount);
		
	}

}