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::{fastout, input};
use std::io::*;
use std::str::FromStr;

use std::cmp::{max, min};

fn main() {
    input! {
        n:usize,
        d: [i64;n],
    }

    let mut d = d;
    d.sort();
    let mut cnt = 0;
    for i in 0..*d.iter().max().unwrap() {
        let idx = d.upper_bound(i);
        if idx == n / 2 {
            cnt += 1;
        }
    }
    println!("{}", cnt);
}

use std::cmp::Ordering;

pub trait BinarySearch<T> {
    fn lower_bound(&self, x: T) -> usize;
    fn upper_bound(&self, x: T) -> usize;
}

impl<T: Ord> BinarySearch<T> for [T] {
    fn lower_bound(&self, x: T) -> usize {
        let mut low = 0;
        let mut high = self.len();

        while low != high {
            let mid = (low + high) / 2;
            match self[mid].cmp(&x) {
                Ordering::Less => {
                    low = mid + 1;
                }
                Ordering::Equal | Ordering::Greater => {
                    high = mid;
                }
            }
        }
        low
    }
    fn upper_bound(&self, x: T) -> usize {
        let mut low = 0;
        let mut high = self.len();

        while low != high {
            let mid = (low + high) / 2;
            match self[mid].cmp(&x) {
                Ordering::Less | Ordering::Equal => {
                    low = mid + 1;
                }
                Ordering::Greater => {
                    high = mid;
                }
            }
        }
        low
    }
}