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 Scala (2.11.7) (judge verdict: Accepted)

import scala.math._

object Main extends App {
  val sc = new java.util.Scanner(System.in)
  val n = sc.nextInt
  val p = List.fill(n)(sc.nextLong)

  val middle = p.size / 2
  val sorted = p.sorted
  val ans = sorted(middle) - sorted(middle - 1)

  println(ans)

  ///////////////////////////////////////////

  def bigIterator(start: BigInt, end: BigInt, step: BigInt = 1) =
    Iterator.iterate(start)(_ + step).takeWhile(_ <= end)

  def lcm(a :Int, b :Int): Int =
    a / gcd(a, b) * b

  def lcm(a :Long, b :Long): Long =
    a / gcd(a, b) * b

  def gcd(a: Int, b: Int): Int =
    if (b == 0) a else gcd(b, a % b)

  def gcd(a: Long, b: Long): Long =
    if (b == 0) a else gcd(b, a % b)
}