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 Java8 (OpenJDK 1.8.0) (judge verdict: Wrong Answer)

import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.util.Collections;
import java.util.List;

public class Main {
	public static void main(String[] args) throws Exception {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		String[] sa = br.readLine().split(" ");
		int n = Integer.parseInt(sa[0]);
		int k = Integer.parseInt(sa[1]);
		int q = Integer.parseInt(sa[2]);
		sa = br.readLine().split(" ");
		int[] a = new int[n];
		for (int i = 0; i < n; i++) {
			a[i] = Integer.parseInt(sa[i]);
		}
		br.close();

		int ans = Integer.MAX_VALUE;
		for (int x = 0; x < n; x++) {
			int min = a[x];
			List<Integer> list = new ArrayList<>();
			List<Integer> list2 = new ArrayList<>();
			for (int i = 0; i < n; i++) {
				if (a[i] >= min) {
					list.add(a[i]);
				} else {
					int rem = Math.max(list.size() - k + 1, 0);
					if (rem > 0) {
						list2.addAll(list);
					}
					list.clear();
				}
			}
			if (list.size() - k + 1 > 0) {
				Collections.sort(list);
				for (int i = 0; i < list.size() - k + 1; i++) {
					list2.add(list.get(i));
				}
			}
			if (list2.size() >= q) {
				Collections.sort(list2);
				if (list2.get(0) == min) {
					int val = list2.get(q - 1) - min;
					ans = Math.min(ans, val);
				}
			}
		}
		System.out.println(ans);
	}
}