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: Wrong Answer)

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


void insertnumnum(
int *i_nums,
int in_len,
int in_number
);

int main()
{
        char inputa[100000];
        char inputb[100000*10];
        char numbuff[10000];
        int numbers[10000];
        int numnum;
        int match;

        fgets( inputa, 65535, stdin );
        fgets( inputb, 100000*10-1, stdin );
        if (strchr(inputa, '\n') != NULL) inputa[strlen(inputa) - 1] = '\0';
        if (strchr(inputb, '\n') != NULL) inputb[strlen(inputb) - 1] = '\0';

        numnum = 0;
        for( int i=0; inputb[i] !=0; i++ ) {
                if( inputb[i] == ' ' ) {
                        continue;
                }

                int j=0;
                while( inputb[i] != 0 && inputb[i] != ' ' ) {
                        numbuff[j++] = inputb[i++];
                }
                numbuff[j]=0;
                insertnumnum(numbers, numnum, atoi(numbuff));
                numnum++;
                /* for( int k =0; k < numnum; k++ ) { printf("<%d>", numbers[k]); } printf("\n"); */

                if( inputb[i-1] == 0 ) {
                        i--;
                }
        }

        int N;
        N = atoi( inputa);
        if( numbers[ N/2-1 ] == numbers[ N/2 ] ) {
                printf("0\n");
        }
        printf("%d\n", numbers[N/2]-numbers[N/2-1]);

        return 0;
}

void insertnumnum(
int *i_nums,
int in_len,
int in_number
) {
        int o_nums[100000];

        int i_nums_i = 0;
        int o_nums_i = 0;

        /* printf("input=%d\n", in_number ); */
        for( i_nums_i = 0; i_nums_i < in_len; i_nums_i++ ) {
                if( i_nums[i_nums_i] > in_number ) {
                   o_nums[o_nums_i++] =in_number;
                   break;
                }
                else {
                        o_nums[o_nums_i++] = i_nums[i_nums_i];
                }
        }
        for( ; i_nums_i < in_len; i_nums_i++ ) {
                o_nums[o_nums_i++] = i_nums[i_nums_i];
        }

        if( i_nums_i == o_nums_i ) {
                o_nums[o_nums_i]  = in_number;
        }

        for( int i = 0; i < in_len+1; i++ ) {
                i_nums[i] = o_nums[i];
        }
}