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: Wrong Answer)

#include<stdio.h>

void qsort(int *arr,int q){

    if(q == 0)return;

    int i,j,flag = 0;
    int key = arr[q - 1];
    int *temp = (int*)malloc(q * sizeof(int));
    int l = 0,r = q - 1;
    for(i = 0;i < q - 1;i++){
        if(arr[i] < key){
            temp[l++] = arr[i];
        }else{
            temp[r--] = arr[i];
        }
    }
    temp[l] = key;
    flag = 0;
    for(i = 0;i < q;i++){
        if(arr[i] != temp[i])
            flag = 1;

        arr[i] = temp[i];
    }

    free(temp);

    if(flag){
        qsort(arr,l);
        qsort(arr+l+1,q - l - 1);
    }
}


int main(){

    int i,j,k;
    int q;
    scanf("%d",&q);
    int arr[q];

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

    qsort(arr,q);
    /*for(i = 0;i < q;i++)
        printf("%d ", arr[i]);

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