Solve in Python:
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-----
6
9 1 4 4 6 7

-----Sample Output-----
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.
N = int(input())
d = list(map(int, input().split()))
d.sort()
print(d[N//2] - d[N//2-1])