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 swap(int a[],int i,int j){
    int temp = a[i];
    a[i] = a[j];
    a[j] = temp;
}

void sort(int a[],int n){

    int i,j;
    for(i = 0;i < n;i++){
        for(j = i;j < n;j++){
            if(a[i] > a[j]){
                swap(a,i,j);
            }
        }
    }
}

int main(){

    int i,j;
    int q;
    scanf("%d",&q);
    int arr[q];
    for(i = 0;i < q;i++){
        scanf("%d",&arr[i]);
    }

    sort(arr,q);

    int cnt = 0;
//    printf("%d %d\n",arr[q / 2 - 1],arr[q / 2]);
    if(arr[q / 2 - 1] != arr[q / 2]){
        i = arr[q / 2 - 1] + 1;
        cnt = arr[q / 2] - i + 1;
    }

    printf("%d",cnt);

    return 0;
}