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: Runtime Error)

#include<stdio.h>
#include<stdlib.h>
int merge(int a[],int p,int q,int r);
int merge_sort(int x[],int a,int b);
int main(){
    int N;
    scanf("%d",&N);
   // int d[100001];
    int i;
    int *d;
	d=(int*)malloc((N+1)*sizeof(int));
    for(i=1;i<=N;i++){
        scanf("%d",&d[i]);
    }
    merge_sort(d,1,N);
    // for(i=1;i<=N;i++){
      // printf("%d ",d[i]);
    //}
    int K;
    if(d[N/2]!=d[N/2+1]){
        K=d[N/2+1]-d[N/2];
    }
    else {
        K=0;
    }
    printf("%d\n",K);
    free(d);
    return 0;
}

int merge_sort(int x[],int a,int b){
	int m,count=0;
 	if(a>=b){
		return 0;
	}
	m=(a+b)/2;
	count+=merge_sort(x,a,m);
	count+=merge_sort(x,m+1,b);
	count+=merge(x,a,m,b);
	return count;
}
int merge(int a[],int p,int q,int r){
	int *T;
	int i=0,j=0,k=0,count=0,s;
	int t=r-p+1;
	T=(int *)malloc(t*sizeof(int));
	for(i=0;i<t;i++){
		if(a[p+j]>a[q+1+k]){
			T[i]=a[q+1+k];
			k++;
		}
		else{
			T[i]=a[p+j];
			j++;
		}
		count++;
		if(p+j==q+1||q+1+k==r+1){
			break;
		}
		
	}
	while(i!=t){
		i++;
		if(p+j==q+1){
			T[i]=a[q+1+k];
			k++;
		}
		else{
			T[i]=a[p+j];
			j++;
		}
	}
	for(s=0;s<t;s++){
		a[p+s]=T[s];
	}
	return count;
}