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>
int main(void)
{
	int n,d[100001],a[100001]={0},x=0,i,j;
	scanf("%d", &n);
	for(i=0; i<n; i++)
	{
		scanf("%d", &d[i]);
		a[d[i]]++;
	}
	for(i=1; i<100001; i++)
	{
		if(a[i]>0)
		x+=a[i];
		if(x>=n/2)
		{
			if(x>n/2)
			printf("0\n");
			else
			{
				x=0;
				for(j=i+1; j<100001; j++)
				{
					x++;
					if(a[j]>0)
					break;
				}
				printf("%d\n", x);
			}
			break;
		}
	}
	
	return 0;
}