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>
 
// useage : swap(&int1,&int2);
// 値入れ替え
int swap( int *c1, int *c2){
  int tmp;
  tmp=*c1;
  *c1=*c2;
  *c2=tmp;
  return 0;
}
 
// useage : sortAsc(list,n);
// 昇順
int sortAsc( int *list,int size){ 
  for(int i=size;i>(size/2-1);i--){
    int maxIndex = 0; 
    int maxVal   = 0;
    for(int j=0;j<i;j++){
      if(list[j] > maxVal){
        maxVal   = list[j];
        maxIndex = j;
      }
    }
    swap(&list[maxIndex],&list[i-1]);
  }
  return 0;
}
 
int main(void) {
 
  // 入力
   int n;
  scanf("%d",&n);
   int list[n];
  for(int i=0;i<n;i++){
    scanf("%d",&list[i]);
  }
 
  // 昇順sort
  sortAsc(list,n); 
  int separator=0;
  int halfN    =n/2;
  int result   =list[halfN]-list[halfN-1];

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