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: Accepted)

import java.util.ArrayList;
import java.util.List;
import java.util.Scanner;

public class Main {
  public static void main(String[] args) {
    Scanner sc = new Scanner(System.in);
    int N = sc.nextInt();
    int K = sc.nextInt();
    int Q = sc.nextInt();
    int array[] = new int[N];
    for (int i = 0; i < N; i++) {
      array[i] = sc.nextInt();
    }
    solve(N, K, Q, array);
  }

  private static void solve(int N, int K, int Q, int array[]) {
    List<Integer> list = new ArrayList<>();
    List<Integer> bigList = new ArrayList<>();
    int ans = Integer.MAX_VALUE;

    for (int i = 0; i < N; i++) {
      int selected = array[i]; // smallest
      for (int j = 0; j < N; j++) {
        if (array[j] >= selected) {
          list.add(array[j]);
        } else {
          addToBigList(list, bigList, K);
        }
      }
      addToBigList(list, bigList, K);
      if (bigList.size() >= Q) {
        bigList.sort(Integer::compare);
        int qthSmallest = bigList.get(Q-1);
        int tmp = qthSmallest - selected;
        if (tmp < ans) {
          ans = tmp;
        }
      }
      bigList.clear();
    }
    System.out.println(ans);
  }
  
  private static void addToBigList(List<Integer> list, List<Integer> bigList, int K) {
    if (list.size() >=  K) {
      list.sort(Integer::compare);
      bigList.addAll(list.subList(0, list.size()-K+1));
    }
    list.clear();
  }
}