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>
using namespace std;
vector<int> clean(vector<int>& s, int k) {
    if (s.size() < k) {
        return vector<int> {};
    }
    vector<int> res;
    sort(s.begin(), s.end());
    for (int i = 0; i < int(s.size()) - k + 1; i++) {
        res.push_back(s[i]);
    }
    while (s.size()) s.pop_back();
    return res;
}
int main(){
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    int n, k, q;
    cin >> n >> k >> q;
    vector<int> a(n), e;
    for (int i = 0; i < n; i++) {
        cin >> a[i];
    }
    e = a;
    sort(a.begin(), a.end());
    int ans = INT_MAX;
    for (int i = 0; i < n; i++) {
        vector<int> s, v;
        for (int j = 0; j < n; j++) {
            if (a[j] < a[i]) {
                vector<int> p = clean(s, k);
                for (auto x: p) {
                    v.push_back(x);
                }
            } else {
                s.push_back(a[j]);
            }
        }
        vector<int> p = clean(s, k);
        for (auto x: p) {
            v.push_back(x);
        }
        if (v.size() >= q) {
            sort(v.begin(), v.end());
            ans = min(ans, v[q - 1] - v[0]);
        }
    }
    cout << ans << endl;

}