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 qsort(int list1[], int left, int right) {
    int l = left;
    int r = right;
    int pivot = list1[(left + right) / 2]; // 中央からピボットを選択
    int temp;

    while (1) {
        while (list1[l] < pivot) l++;
        while (list1[r] > pivot) r--;

        if (l > r) break;

        // pivotより小さい値を左、大きい値を右に入れ替え
        temp = list1[l];
        list1[l] = list1[r];
        list1[r] = temp;
        

        l++, r--;
    };

    // pivotより小さい領域と大きい領域をそれぞれ再帰的にソート
    if (left < r)  qsort(list1,left,r);
    if (l < right) qsort(list1,l,right);
}

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

    return 0;
}