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.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.Arrays;

public class Main {
    private static Scanner sc;
    public static void main(String[] args) {
        Main instance = new Main();
        sc = instance.new Scanner();
        instance.solve();
    }

    private void solve() {
        try {
            int n = sc.nextInt();
            int[] d = new int[n];
            int sumd = 0;
            int maxd = 0;
            for (int i = 0; i < n; i++) {
                d[i] = sc.nextInt();
                sumd += d[i];
                if (maxd < d[i]) maxd = d[i];
            }
            Arrays.sort(d);

            int count = 0;
            for (int i = 0; i < maxd; i++) {
                int j = Arrays.binarySearch(d, i);
                if (j < 0) j = ~j;
                if (j == n/2)
                    count ++;
            }
            System.out.println(count);

        } catch (Exception e) {
            e.printStackTrace();
        }
    }

    private class Scanner {
        String[] s;
        int i;
        BufferedReader br;
        String regex = " ";

        public Scanner() {
            s = new String[0];
            i = 0;
            br = new BufferedReader(new InputStreamReader(System.in));
        }

        @Override
        protected void finalize() throws Throwable {
            try {
                super.finalize();
            } finally {
                destruction();
            }
        }

        private void destruction() throws IOException {
            if (br != null) br.close();
        }

        public String next() throws IOException {
            if (i < s.length) return s[i++];
            String st = br.readLine();
            while (st == "") st = br.readLine();
            s = st.split(regex, 0);
            i = 0;
            return s[i++];
        }

        public int nextInt() throws NumberFormatException, IOException {
            return Integer.parseInt(next());
        }

        public Long nextLong() throws NumberFormatException, IOException {
            return Long.parseLong(next());
        }

        public Double nextDouble() throws NumberFormatException, IOException {
            return Double.parseDouble(next());
        }
    }
}