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 C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>

#define ld long double
#define ll long long int
#define ull unsigned long long int
#define vi vector<int>
#define vl vector<ll>
#define vvi vector< vector<int> >
#define vvl vector< vector<ll> >
#define repd(i, a, b) for (int i=(a);i<(b);i++)
#define rep(i, n) repd(i,0,n)
#define ALL(v) v.begin(), v.end()
#define INF 1e9

using namespace std;

/**
 * calculate GCD(greatest common divisor)
 * @param a
 * @param b
 * @return
 */
unsigned euclidean_gcd(unsigned a, unsigned b) {
    if (a < b) return euclidean_gcd(b, a);
    unsigned r;
    while ((r = a % b)) {
        a = b;
        b = r;
    }
    return b;
}

/**
 * change minimum
 * @tparam T
 * @param a
 * @param b
 * @return bool
 */
template<class T>
inline bool changeMinimum(T &a, T b) {
    if (a > b) {
        a = b;
        return true;
    }
    return false;
}

/**
 * change maximum
 * @tparam T
 * @param a
 * @param b
 * @return bool
 */
template<class T>
inline bool changeMaximum(T &a, T b) {
    if (a < b) {
        a = b;
        return true;
    }
    return false;
}

/**
 * output answer
 * @tparam T
 * @param answer
 * @return
 */
template<typename T>
T output(T answer) {
    cout << answer << endl;
    return 0;
}

/** ----from here ---------------------------------------------------------- */

int main() {

    int N;
    cin >> N;

    vector<int> d(100010, 0);

    for (int i = 0; i < N; ++i) {
        int t;
        cin >> t;
        d[t]++;
    }
    int ans =0;
    if(N %2 ==1 ){
        cout << ans <<endl;
        return 0;
    }
    int K = N/2;
    long long dp[100010];
    for (int i = 0; i < 100008; ++i) {
        dp[i+1] = dp[i] + d[i];
        if(dp[i+1] == K){
            ans++;
        }
    }
    cout << ans <<endl;

}