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: Accepted)

#include<stdio.h>
#include <stdbool.h>
void swap(long num1, long num2, long a[100000]){
   long temp = a[num1];
   a[num1] = a[num2];
   a[num2] = temp;
}
long partition(long left, long right, long pivot, long a[100000]){
   long leftPointer = left -1;
   long rightPointer = right;

   while(true){
	
      while(a[++leftPointer] < pivot){
         //khong lam gi
      }
	  while(rightPointer > 0 && a[--rightPointer] > pivot){
         //khong lam gi
      }
      if(leftPointer >= rightPointer){
         break;
      }else{
         swap(leftPointer,rightPointer,a);
      }	
   }
   swap(leftPointer,right,a);
   return leftPointer;
}
// ham sap xep
void quickSort(long left, long right, long a[100000]){        
   if(right-left <= 0){
      return;   
   }else {
      long pivot = a[right];
      long partitionPoint = partition(left, right, pivot,a);
      quickSort(left,partitionPoint-1,a);
      quickSort(partitionPoint+1,right,a);
   }        
} 
int main()
{
  long d[100000],n,i;
  scanf("%ld", &n);
  for (i=0; i<n; i++) scanf("%ld", &d[i]);
  quickSort(0, n-1, d);
  printf("%ld", d[n/2]-d[n/2-1]);
  return 0;
}