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;

int main() {
    int N, K, Q;
    cin >> N >> K >> Q;
    vector<int> A(N);
    for(int i = 0; i < N; i++) {
        cin >> A[i];
    }
    int answer = *max_element(A.begin(), A.end()) - *min_element(A.begin(), A.end());
    for(int s = 0; s < N; s++) {
        vector<pair<int, int>> ps;
        for(int i = 0; i < N; i++) {
            ps.push_back(make_pair(A[i], i));
        }
        sort(ps.begin(), ps.end());

        deque<pair<int, int>> queue;
        set<int> indexes;

        vector<int> gets;

        for(int i = 0; i < N; i++) {
            while (queue.size() > 0 && queue.front().first < ps[i].first) {
                auto p = queue.front(); queue.pop_front();
                indexes.insert(p.second);
            }

            auto it = indexes.lower_bound(ps[i].second);
            const int right = (it != indexes.end() ? *it : 2 * N);
            const int left = (it != indexes.begin() ? *(--it) : -2 * N);
            if (i >= s && (abs(ps[i].second - left) >= K || abs(ps[i].second - right) >= K)) {
                gets.push_back(ps[i].first);
            } else {
                queue.push_back(ps[i]);
            }

            if (gets.size() == Q) {
                break;
            }
        }

        if (gets.size() == Q) {
            answer = min(answer, gets[Q-1] - gets[0]);
        }
    }
    cout << answer << endl;
}