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>
void quicksort(int a[],int left,int right){
    int i=left,j=right;
    int tmp,pivot=a[i+(j-i)/2];
    while(1){
        while(a[i]<pivot)
            i++;
        while(a[j]>pivot)
            j--;
        if(i>=j)
            break;
        tmp=a[i];
        a[i]=a[j];
        a[j]=tmp;
        i++;
        j--;
    }
    if(left<i-1)
        quicksort(a,left,i-1);
    if(j+1<right)
        quicksort(a,j+1,right);
}
int main(void){
    int N,K;
    scanf("%d",&N);
    int d[N];
    for(int i=0;i<N;i++){
        scanf("%d",&d[i]);
    }
    quicksort(d,0,N-1);
    
    K=d[N/2]-d[N/2-1];
    printf("%d\n",K);
    return 0;
}