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>
using namespace std;

const int MAXN = 2005;

int n, k, q;

int a[MAXN];
int dx[MAXN], top = 0;

void work(int L, int R)
{
	if (L > R) return;
	static int stk[MAXN];
	int stk_top = 0, len = R-L+1;
	for (int i = L; i <= R; i++) stk[++stk_top] = a[i];
	sort(stk+1, stk+stk_top+1);
	for (int i = 1; i <= len-k+1; i++)
		dx[++top] = stk[i];
}

int main()
{
	scanf("%d%d%d", &n, &k, &q);
	for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
	int ans = 1e9;
	for (int i = 1; i <= n; i++) {
		int last = 0;
		top = 0;
		for (int j = 1; j <= n; j++) {
			if (a[j] < a[i]) {
				work(last+1, j-1);
				last = j;
			}
		}
		work(last+1, n);
		sort(dx+1, dx+top+1);
		if (top >= q) ans = min(ans, dx[q]-dx[1]);
	}
	cout << ans << endl;
	return 0;
}