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 Rust (1.42.0) (judge verdict: Accepted)

use proconio::input;
#[allow(unused_imports)]
use std::collections::HashSet;

fn read<T: std::str::FromStr>() -> T {
    let mut s = String::new();
    std::io::stdin().read_line(&mut s).ok();
    s.trim().parse().ok().unwrap()
}

#[allow(dead_code)]
fn read_vec<T: std::str::FromStr>() -> Vec<T> {
    let mut s = String::new();
    std::io::stdin().read_line(&mut s).ok();
    s.trim()
        .split_whitespace()
        .map(|e| e.parse().ok().unwrap())
        .collect()
}

type pair = (usize, usize);

// max = (n-1)(n-2)/2
fn main() {
    input! {
        n: usize,
        mut d: [i32; n] 
    }
    let answer = solve(n, &mut d);
    println!("{}", answer);
}

fn solve(n: usize, d: &mut [i32]) -> i32 {
    // 1. sort
    d.sort();

    // 2. threshold should middle of d
    //     possible ways = diff between <left of middle> and <right of middle>
    let mid = n / 2;
    return d[mid] - d[mid-1];

}