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>
#include <stdlib.h>
int d[100000];
int asc(const void *a, const void *b);

int main(void){
    int i, n;
    scanf("%d", &n);
    for(i=0;i<n;i++){
        scanf("%d", &d[i]);
    }
    /* for(i=0;i<n-1;i++){
        for(j=i+1;j<n;j++){
            if(d[i]>d[j]){
                tmp = d[i];
                d[i] = d[j];
                d[j] = d[i];
            }
        }
    }*/
    qsort(d, n, sizeof(int), asc);
    if(d[n/2-1]==d[n/2])printf("0\n");
    else
    printf("%d\n", d[n/2]-d[n/2-1]);
    return 0;
}

int asc(const void *a, const void *b){
    return *(int *)a - *(int *)b;
}