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.*;
import java.io.*;

import static java.lang.System.in;

public class Main {
    public static void main(String[] args){
        Scanner sc = new Scanner(System.in);
        int n=sc.nextInt(), K=sc.nextInt(), Q=sc.nextInt();
        int[] a = new int[n];
        for(int i=0;i<n;i++) a[i]=sc.nextInt();
        HashSet<Integer> vis = new HashSet<>();
        PriorityQueue<Integer> small = new PriorityQueue<>(), big = new PriorityQueue<>();
        int ans = Integer.MAX_VALUE;
        for(int i=0;i<n;i++){
            if(!vis.add(a[i])) continue;
            big.clear();
            int left=0;
            while(left<n){
                if(a[left]<a[i]) {
                    left++;
                    continue;
                }
                small.clear();
                while(left<n&&a[left]>=a[i]){
                    small.add(a[left++]);
                }
                while(small.size()>=K) big.add(small.poll());
            }
            if(big.size()<Q) continue;
            for(int j=1;j<Q;j++) big.poll();
            ans = Math.min(ans,big.peek()-a[i]);
        }
        System.out.println(ans);
    }
}