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 <stdlib.h>
#include <string.h>

int partition(int a[],int l,int r);
int quicksort(int a[],int l,int r);

int main(void){
  int N,d[100005],K,i;
  char str[1000000],*p;
  
  fgets(str,sizeof(str),stdin);
  N=atoi(str);
  
  fgets(str,sizeof(str),stdin);
  p=strtok(str," \n");
  d[0]=atoi(p);
  
  for(i=1;i<N;i++){
    p=strtok(NULL," \n");
    d[i]=atoi(p);
  }
  
  quicksort(d,0,N-1);
  
  
  K=d[(N-1)/2+1]-d[(N-1)/2];
  
  printf("%d\n",K);
  
  
  return 0;
}

int partition(int a[],int l,int r){
    int i,j,bbuff;
    long abuff,pivot;
    
    i=l-1,j=r;
    pivot=a[r];
    for(;;){
        while(a[++i]<pivot);
        
        while(i<=--j && a[j]>=pivot);
        
        if(i>=j){
            abuff=a[r],a[r]=a[i],a[i]=abuff;
            break;
        }
        
        abuff=a[i],a[i]=a[j],a[j]=abuff;
        
    }
    
    return i;
}

int quicksort(int a[],int l,int r){
    int v;
    
    if(l>=r){
        return 0;
    }
    
    v=partition(a,l,r);
    
    quicksort(a,l,v-1);
    quicksort(a,v+1,r);
    
    return 0;
    
}