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>
#define NUM 1000000
void QSort(int x[ ], int left, int right);
void Swap(int x[ ], int i, int j);

int main(){
    int n;
    int d[NUM],tmp,ad,cnt1=0,cnt2=0,num;

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

    if(n % 2 != 0){
        printf("0\n");
        return 0;
    }

    QSort(d, 0, n-1);

    // printf("sort>>");
    // for(int i = 0;i<n;i++){
    //     printf("%d ",d[i]);
    // }
    // putchar('\n');
    ad = n/2;
    tmp = d[ad];

    for(int i = 0;i<n;i++){
        if(tmp > d[i]){
            cnt1++;
        }else if(tmp <= d[i]){
            cnt2++;
        }
    }
    num = d[ad]-d[ad-1];
    // printf("d[ad]:%d - d[ad-1]:%d = num:%d\n",d[ad],d[ad-1],num);
    if(cnt1 == cnt2){
        printf("%d\n",num);
    }else{
        printf("0\n");
    }
    return 0;
    

}

void QSort(int x[ ], int left, int right)
{
    int i, j;
    int pivot;

    i = left;                      /* ソートする配列の一番小さい要素の添字 */
    j = right;                     /* ソートする配列の一番大きい要素の添字 */

    pivot = x[(left + right) / 2]; /* 基準値を配列の中央付近にとる */

    while (1) {                    /* 無限ループ */

        while (x[i] < pivot)       /* pivot より大きい値が */
            i++;                   /* 出るまで i を増加させる */

        while (pivot < x[j])       /* pivot より小さい値が */
            j--;                   /*  出るまで j を減少させる */
        if (i >= j)                /* i >= j なら */
            break;                 /* 無限ループから抜ける */

        Swap(x, i, j);             /* x[i] と x[j]を交換 */
        i++;                       /* 次のデータ */
        j--;
    }

    if (left < i - 1)              /* 基準値の左に 2 以上要素があれば */
        QSort(x, left, i - 1);     /* 左の配列を Q ソートする */
    if (j + 1 <  right)            /* 基準値の右に 2 以上要素があれば */
        QSort(x, j + 1, right);    /* 右の配列を Q ソートする */
}

void Swap(int x[ ], int i, int j)
{
    int temp;

    temp = x[i];
    x[i] = x[j];
    x[j] = temp;
}