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 <string.h>
#define max(a,b) (a>b?a:b)

void swap(int *a,int *b)
{
    int c=*a;
    *a=*b;
    *b=c;
}

void quick_sort(int a[],int l,int r)
{
    if(l>=r)return;
    int temp=a[l],i=l,j=r;
    //printf("%d  ",temp);
    while(i!=j){
        while(a[j]>=temp &&i<j)j--;
        while(a[i]<=temp &&i<j)i++;
        if(i<j)swap(&a[i],&a[j]);
    }
    swap(&a[l],&a[i]);
    quick_sort(a,l,i-1);
    quick_sort(a,i+1,r);
}

int main()
{
    int n;
    scanf("%d",&n);
    int a[n];
    for(int i=0;i<n;i++)scanf("%d",&a[i]);
    quick_sort(a,0,n-1);
//    for(int i=0;i<n;i++){
//        printf("%d ",a[i]);
//    }
    printf("%d",a[n/2]-a[n/2-1]);
}