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 C++14 (GCC 5.4.1) (judge verdict: Time Limit Exceeded)

#include <iostream>
#include <math.h>
#include <numeric>
#include <vector>
#include <algorithm>

using namespace std;
#define n_max 100000

int main()
{
    int N;
    vector<int> d;
    int abc_cnt = 0;
    int ans = 0;

    cin >> N;

    for(int i = 0; i < N; i++){
        int tmp;
        cin >> tmp;
        d.push_back(tmp);
    }
    
    int max = *max_element(d.begin(), d.end());
    int min = *min_element(d.begin(), d.end());
    //cout << max << endl;

    for(int k = min; k <= max; k++) {
        abc_cnt = 0;

        for(int i = 0; i < N; i++) {

            if(d[i] < k) { abc_cnt++; } 
        }
        if(abc_cnt == (N / 2)) { 
            ans++;
        }
    }

    cout << ans << endl;

    return 0;
}