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>
#include<string.h>

int main(){
  int count=0;
  char s[101];
  int p[10^5],n,h[10^5];
  
  scanf("%d",&n);
  for(int i=0;i<n;i++){
    p[i]=0;h[i]=0;
    scanf("%d",&p[i]);
  }
  
  for(int j=0;j<n;j++){
    for(int k=0;k<=j;k++){
      if(k==j){
	h[k]=p[j];
      }else if(p[j]<h[k]){
	for(int i=j-1;i>=k;i--){
	  h[i+1]=h[i];
	}
	h[k]=p[j];
	k=j+1;
      }    
    }
  }

 
    //printf("%d%d",h[n/2-1],h[n/2]);
  if(h[n/2]>h[n/2-1]){count=h[n/2]-h[n/2-1];}


  printf("%d\n",count);
  
  return 0;
}