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 (Clang 3.8.0) (judge verdict: Runtime Error)

#include <stdio.h>

int getAns(int, int a[]);
void sortArray(int, int a[]);

int main() {
    int n;
    int p[10000] = {0};
    
    scanf("%d", &n);
    for (int i = 0; i < n - 1; i++) {
        scanf("%d ", p + i);
    }
    scanf("%d", p + n - 1);

    printf("%d", getAns(n, p));

    return 0;
}

int getAns(int n, int p[]) {
    int count = 0;
 
    sortArray(n, p);
    return p[(n / 2)] - p[(n / 2) - 1];
}

void sortArray(int n, int p[]) {
    int tmp;
    for (int i = 0; i < n - 1; i++) {
        for (int j = 1; j < n - i; j++) {
            if (p[j - 1] > p[j]) {
                tmp = p[j];
                p[j] = p[j - 1];
                p[j - 1] = tmp; 
            }
        }
    }
}