Problem:
Score : 600 points

Problem Statement

You are given an integer sequence A of length N and an integer K.
You will perform the following operation on this sequence Q times:

Choose a contiguous subsequence of length K, then remove the smallest element among the K elements contained in the chosen subsequence (if there are multiple such elements, choose one of them as you like).

Let X and Y be the values of the largest and smallest element removed in the Q operations. You would like X-Y to be as small as possible.
Find the smallest possible value of X-Y when the Q operations are performed optimally.

Constraints

1 \leq N \leq 2000

1 \leq K \leq N

1 \leq Q \leq N-K+1

1 \leq A_i \leq 10^9

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N K Q
A_1 A_2 ... A_N

Output

Print the smallest possible value of X-Y.

Sample Input 1

5 3 2
4 3 1 5 2

Sample Output 1

1

In the first operation, whichever contiguous subsequence of length 3 we choose, the minimum element in it is 1.
Thus, the first operation removes A_3=1 and now we have A=(4,3,5,2).
In the second operation, it is optimal to choose (A_2,A_3,A_4)=(3,5,2) as the contiguous subsequence of length 3 and remove A_4=2.
In this case, the largest element removed is 2, and the smallest is 1, so their difference is 2-1=1.

Sample Input 2

10 1 6
1 1 2 3 5 8 13 21 34 55

Sample Output 2

7

Sample Input 3

11 7 5
24979445 861648772 623690081 433933447 476190629 262703497 211047202 971407775 628894325 731963982 822804784

Sample Output 3

451211184

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

#include <bits/stdc++.h>
#define X first
#define Y second
#define pb push_back
#define sz(a) (int)a.size()
#define int long long

using namespace std;

int a[2010];

vector <pair <int, int>> v;

int kok[2010];
int pref[2010];

vector <int> sok;

mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());

int ans = 1e10;

signed main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    int n, k, q;
    cin >> n >> k >> q;
    for (int i = 0; i < n; i++) {
        cin >> a[i];
        v.pb({a[i], i});
    }
    sort(v.begin(), v.end());
    int kak = 0;
    for (int j = 1; j < n; j++) {
        if (v[j].first != v[j - 1].first) {
            shuffle(v.begin() + kak, v.begin() + j, rng);
            kak = j;
        }
    }
    for (int i = 0; i <= n - q; i++) {
        sok.clear();
        for (int j = 0; j < i; j++)
            sok.pb(v[j].second);
        sort(sok.begin(), sok.end());
        for (int j = 0; j <= n; j++) {
            kok[j] = 0;
            pref[j] = 0;
        }
        for (int j = i; j < i + q; j++) {
            kok[v[j].second] = 1;
        }
        for (int j = 1; j <= n; j++) {
            pref[j] = pref[j - 1] + kok[j - 1];
        }
        int nch = 0;
        bool f = true;
        for (int j = 0; j < sz(sok); j++) {
            if (pref[sok[j]] - pref[nch] > 0 && sok[j] - nch - pref[sok[j]] + pref[nch] < k - 1) {
                f = false;
                break;
            }
            nch = sok[j] + 1;
        }
        if (pref[n] - pref[nch] > 0 && n - nch - pref[n] + pref[nch] < k - 1) {
            f = false;
        }
        if (f)
            ans = min(ans, v[i + q - 1].first - v[i].first);
    }
    cout << ans << endl;
    return 0;
}