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 (Clang 3.8.0) (judge verdict: Wrong Answer)

#include <stdio.h>

void sort(int d[],int n){
  int i,j;
  for( int s=0;s<n;s++){
    printf("%d\n",d[s]);
  }
  
  
   for (i = 0; i < n - 1; i++){
      for (j = n - 1; j >= i + 1; j--){ 
         if(d[j-1]<d[j]){
            int box=d[j-1];
            d[j-1]=d[j];
            d[j]=box;     
         }   
      }
   }
   for( int s=0;s<n;s++){
    printf("%d",d[s]);
  }
}
int main(){
  int n,i;
  int d[100000];
  scanf("%d",&n);
  for( i=0;i<n;i++){
    scanf("%d",&d[i]);
  }
  printf("\n");
  sort(d,n);
  printf("%d%d\n",d[i/2],d[i/2-1]);
  int ans=d[i/2-1]-d[i/2];
  printf("%d",ans);
}