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 PHP7 (7.0.15) (judge verdict: Wrong Answer)

<?php
/**
 * ABC132
 * C - Divide the Problems
 */

// $s = trim(fgets(STDIN));
// fscanf(STDIN, "%d %d", $a, $b);
// $n = array_map('intval', explode(' ', trim(fgets(STDIN))));

fscanf(STDIN, "%d", $n);
$d = array_map('intval', explode(' ', trim(fgets(STDIN))));
sort($d);
$result = 0;
$min = min($d);
$max = max($d);
for ($i = $min; $i < $max; $i++) {
    $j = binary_search($d, 0, count($d), $i, function($a, $b) {
        return ($a <=> $b);
    });
    if ($j < 0) {
        $j = abs($j) - 1;
    }
    if ($j * 2 === $n) {
        $result++;
    }
}

echo $result, PHP_EOL;

function binary_search(array $haystack, $start_index, $end_index, $needle, callable $value_compare_func)
{
    $low = $start_index;
    $hi = $end_index - 1;

    while ($low <= $hi) {
        $mid = intdiv($hi - $low, 2) + $low;
        $compare = call_user_func($value_compare_func, $haystack[$mid], $needle);

        if ($compare < 0) {
            $low = $mid + 1;
        } elseif ($compare > 0) {
            $hi = $mid - 1;
        } else {
            return $mid;
        }
    }
    return -($low + 1);
}