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 Go (1.6) (judge verdict: Wrong Answer)

package main

import (
	"fmt"
	"sort"
	"errors"
)

func f(d []int, k int) (int, int) {
	arc, abc := 0, 0
	for _, t := range d {
		if t >= k {
			arc += 1
		} else {
			abc += 1
		}
	}
	return arc, abc
}

func lower_bound(d []int) (int, error) {
	a, b := 0, 100000

	for a + 1 < b {
		t := (a + b) / 2
		arc, abc := f(d, t)

		if arc > abc {
			a = t
		} else {
			b = t
		}
	}
	arc, abc := f(d, b)
	if arc != abc {
		return 0, errors.New("")
	} else {
		return b, nil
	}
}

func upper_bound(d []int) (int, error) {
	a, b := 0, 100000

	for a + 1 < b {
		t := (a + b) / 2
		arc, abc := f(d, t)
		if arc >= abc {
			a = t
		} else {
			b = t
		}
	}
	arc, abc := f(d, a)
	if arc != abc {
		return 0, errors.New("")
	} else {
		return a, nil
	}
}

func main() {
	var n int
	fmt.Scanf("%d", &n)

	d := make([]int, n)
	for i := 0; i < n; i++ {
		fmt.Scanf("%d", &d[i])
	}
	sort.Ints(d)

	lower, err1 := lower_bound(d)
	upper, err2 := upper_bound(d)

	if err1 != nil || err2 != nil {
		fmt.Println(0)
	} else {
		fmt.Println(upper - lower + 1)
	}
}