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: Accepted)

#include <bits/stdc++.h>

void optimize_cpp_stdio() {
    std::ios::sync_with_stdio(false);
    std::cout.tie(nullptr);
    std::cin.tie(nullptr);
}

int main() {
    optimize_cpp_stdio();
    int n, k, q;
    std::cin >> n >> k >> q;
    std::vector<int> a(n);
    for (int i = 0; i < n; i++)
        std::cin >> a[i];
    std::vector<int> s = a;
    std::sort(s.begin(), s.end());
    auto new_end = std::unique(s.begin(), s.end());
    s.erase(new_end, s.end());
    auto ok = [&](int x) {
        for (int low : s) {
            int high = low + x, count = 0;
            for (int i = 0, j = 0; i < n; i = j) {
                int valid = 0;
                while (j < n && low <= a[j])
                    valid += (a[j++] <= high);
                count += std::max(0, std::min(valid, j - i - k + 1));
                while (j < n && a[j] < low)
                    j++;
            }
            if (q <= count)
                return true;
        }
        return false;
    };
    int lo = -1, hi = s.back() - s.front();
    while (lo + 1 != hi) {
        int mi = (lo + hi) / 2;
        (ok(mi) ? hi : lo) = mi;
    }
    std::cout << hi << "\n";
    return 0;
}