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 (Clang 3.8.0) (judge verdict: Accepted)

#include<stdio.h>
#include<math.h>
#include<string.h>
#define FOR(N) for(int i = 0;i < N;i++)
#define PRN(N) printf("%d\n",N)
#define PRS(S) printf("%s\n",S)
void swap (int *x, int *y) {
  int temp;    // 値を一時保存する変数

  temp = *x;
  *x = *y;
  *y = temp;
}

/***
* pivotを決め、
* 全データをpivotを境目に振り分け、
* pivotの添え字を返す
***/
int partition (int array[], int left, int right) {
  int i, j, pivot;
  i = left;
  j = right + 1;
  pivot = left;   // 先頭要素をpivotとする

  do {
    do { i++; } while (array[i] < array[pivot]);
    do { j--; } while (array[pivot] < array[j]);
    // pivotより小さいものを左へ、大きいものを右へ
    if (i < j) { swap(&array[i], &array[j]); }
  } while (i < j);

  swap(&array[pivot], &array[j]);   //pivotを更新

  return j;
}

/* クイックソート */
void quick_sort (int array[], int left, int right) {
  int pivot;

  if (left < right) {
    pivot = partition(array, left, right);
    quick_sort(array, left, pivot-1);   // pivotを境に再帰的にクイックソート
    quick_sort(array, pivot+1, right);
  }
}

int main(void){
  int n;
  scanf("%d",&n);
  int a[n];
  FOR(n){
    scanf("%d",&a[i]);
  }
   quick_sort(a,0,n - 1);
  if(a[n / 2 - 1] == a[n / 2]){
    PRN(0);
  }else{
    PRN(a[n / 2] - a[n / 2 - 1]);
  }
  return 0;
}