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>

void q_sort(int numbers[], int left, int right)
{
    int pivot, l_hold, r_hold;

    l_hold = left;
    r_hold = right;
    pivot = numbers[left];
    while (left < right)
    {
        while ((numbers[right] >= pivot) && (left < right))
            right--;
        if (left != right)
        {
            numbers[left] = numbers[right];
            left++;
        }
        while ((numbers[left] <= pivot) && (left < right))
            left++;
        if (left != right)
        {
            numbers[right] = numbers[left];
            right--;
        }
    }
    numbers[left] = pivot;
    pivot = left;
    left = l_hold;
    right = r_hold;
    if (left < pivot)
        q_sort(numbers, left, pivot-1);
    if (right > pivot)
        q_sort(numbers, pivot+1, right);
}

int main()
{
    int n;
    int d[100000];

    int i;

    scanf("%d",&n);

    for(i=0;i<n;i++)
    {
        scanf("%d",&d[i]);
    }

    q_sort(d,0,n-1);
    printf("%d\n",d[n/2]-d[(n/2)-1]);

}