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: Accepted)

#include <bits/stdc++.h>
using namespace std;

int main(int argc, char** argv){
  /*
  int N;
  cin >> N;
  priority_queue<int> d;
  
  int tmp = 0;
  int dmax = 0;
  for(int cnt0=0; cnt0<N; cnt0++){
    cin >> tmp;
    dmax = max(dmax, tmp);
    d.push(tmp);
  }
  cout << dmax;
  return 0;
  
  int ans = 0;
  for(int cnt0=1; cnt0<=dmax; cnt0++){
    int tmpnum = 0;
    priority_queue<int> dtmp = d;
    for(int cnt1=0; cnt1 < N/2 + 1; cnt1++){
      if(dtmp.top() < cnt0) tmpnum++;
      dtmp.pop();
    }
    if(tmpnum == N/2) ans++;
  }
  
  printf("%d\n", ans);
  */

  int N;
  cin >> N;
  int d[N];
  
  for(int cnt0=0; cnt0<N; cnt0++){
    cin >> d[cnt0];
  }
  sort(d, d+N);
  int* ptr_half;
//  ptr_half = (d, d+N, d[N/2 -1]);
//  int idx = (d, d+N, d[N/2 -1]);
//  printf("%d\n", *ptr_half);
  printf("%d\n", d[N/2] - d[N/2-1]);
}