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 JavaScript (Node.js 12.16.1) (judge verdict: Accepted)

"use strict"
const create = () => {
    const input = require("fs").readFileSync("/dev/stdin", "utf8").trim().split("\n");
    const res = {
        "list": input, "index": 0, "max": input.length,
        "hasNext": function() {return this.index < this.max},
        "next": function() { if (!this.hasNext()) {throw "ArrayIndexOutOfBoundsException";}else{return this.list[this.index++];}}
    };
    return res;
}
const o = create();
const next = () => o.next()
const nextInt = () => parseInt(o.next())
const nextStrArray = () => o.next().split(" ")
const nextIntArray = () => o.next().split(" ").map(el => parseInt(el))
const nextCharArray = () => o.next().split("")
const hasNext = () => o.hasNext()
const myout = (...x) => console.log(...x)
const Main = () => {
    let n = nextInt()
    let d = nextIntArray()
    d.sort((a,b) => a-b)
    let v = d[n/2]
    let prev = d[n/2-1]
    console.log(v-prev)
}
Main()