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>
#include<math.h>

int main()
{
    int d[100000];
    int n,i,j=0,k,r=0,b=0,max=0,second=0;
    scanf("%d",&n);
    for(i=0;i<n;i++)
    {
        scanf("%d",&d[i]);
        if(d[i]>max)
        {
            max=d[i];
            second=max;
        }
    }

    if(n>2)
    {
    for(k=2;k<=second;k++)
    {
        r=0;
        b=0;
        for(i=0;i<n;i++)
        {
            if(d[i]>=k)r++;
            else b++;
            if(r>n/2||b>n/2) break;
        }
        if(r==b)j++;
    }
    }
    else
    {
        j=abs(d[1]-d[0]);
    }
    
    printf("%d",j);
    return 0;
}