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 <iostream>
#include <cstdio>
#include <queue>
#include <algorithm>
using namespace std;
int ans;
int n, k, q, a[2010], ls[2010];
priority_queue <int, vector <int>, greater <int> > q1, q2;
int main () {
	scanf ("%d%d%d", &n, &k, &q);
	for (int i = 1; i <= n; i++) scanf ("%d", &a[i]), ls[i] = a[i];
	sort (ls + 1, ls + n + 1);
	ans = ls[n] - ls[1];
	for (int p = 1; p <= n; p++) {
		for (int i = 1; i <= n; i++) if (a[i] >= ls[p]) {
			int j = i; while (j < n && a[j + 1] >= ls[p]) j++;
			if (j - i + 1 < k) { i = j; continue; }
			for (int x = i; x <= j; x++) q1.push (a[x]);
			for (int x = j - i + 1; x >= k; x--) q2.push (q1.top ()), q1.pop ();
			while (!q1.empty ()) q1.pop ();
			i = j;
		}
		int mx = 0; bool ok = true;
		for (int i = 1; i <= q; i++) {
			if (q2.empty ()) { ok = false; break; }
			mx = q2.top (), q2.pop ();
		}
		while (!q2.empty ()) q2.pop ();
		if (ok) ans = min (ans, mx - ls[p]);
	}
	printf ("%d\n", ans);
	return 0;
}