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 (GCC 5.4.1) (judge verdict: Time Limit Exceeded)

#include<stdio.h>
#include<stdlib.h>

/*int cmp(const void *a, const void *b){
  if((int *)a>(int *)b)
  return 1;
  else if((int *)a<(int *)b)
  return -1;
  else
  return 0;
}*/

int main(void){
    int n;
    int ans=0;
    int i,j;
    int tmp;
    int center;
    int idxright,idxleft;
    int right,left;
    int start;

    scanf("%d",&n);
    int d[n];
    for(i=0;i<n;i++){
      scanf("%d",&d[i]);
    }
    center=n/2;
    for(i=0;i<center+1;i++){
      for(j=i+1;j<n;j++){
        if(d[i]>d[j]){
          tmp=d[i];
          d[i]=d[j];
          d[j]=tmp;
        }
      }
    }
    //qsort(d,n,sizeof(int),cmp);

    idxright=center;
    idxleft=center-1;
    right=d[idxright];
    left=d[idxleft];

    ans=right-left;
    printf("%d",ans);

    return 0;
}