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>
#define lop(i, n) for(i=0; i<n; i++)

void bsort(int a[], int n)
{
	int i, j;

	for (i = 0; i < n - 1; i++) {
		for (j = n - 1; j > i; j--) {
			if (a[j - 1] > a[j]) {
				int temp = a[j];
				a[j] = a[j - 1];
				a[j - 1] = temp;
			}
		}
	}
}

int main()
{
	int n, k=0, l=0, cnt=0, i, j, d[100000];
	
	scanf("%d", &n);
	
	lop(i, n){ scanf("%d", &d[i]); }
	
	bsort(d, n);
	
	int m=n/2;
	
	for(i=d[m-1]; i<=d[m]; i++){
		lop(j, n){
			if(d[j]<i){ k++; }
			else if(d[j]>=i){ l++; }
			
			if(k>m || l>m) break;
		}
		if(k==l){ cnt++; }
		k=0; l=0;
	}
	
	printf("%d", cnt);
	
return 0;
}