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.util.*;

public class Main{
    
    
    public static void main(String[] args){
        Scanner sc = new Scanner(System.in);
        
        int inf = (int)1e9;
        
        int n = Integer.parseInt(sc.next());
        int k = Integer.parseInt(sc.next());
        int q = Integer.parseInt(sc.next());
        int[] a = new int[n];
        Val_idx[] b = new Val_idx[n];
        for(int i=0; i<n; i++){
            a[i] = Integer.parseInt(sc.next());
            b[i] = new Val_idx(a[i], i);
        }
        Arrays.sort(b);
        
        int l = b[0].idx;
        int r = b[0].idx;
        int i = 1;
        for(; i<=n-k-q+1; i++){
            //避けられるなら次
            if(simyu(i, b, n, k, q)) continue;
            //
            break;
        }
        
        //System.out.println(i);
        
        
        long ans = b[q-1].val - b[0].val;
        for(int j=0; j<i; j++){
            long tmp = b[j+q-1].val - b[j].val;
            ans = Math.min(ans, tmp);
        }
        
        System.out.println(ans);
    }
    
    public static boolean simyu(int ti, Val_idx[] b, int n, int k, int q){
        int[] c = new int[ti+2];
        for(int i=0; i<ti; i++){
            c[i+1] = b[i].idx;
        }
        c[0] = -1;
        c[ti+1] = n;
        Arrays.sort(c);
        
        int sum = 0;
        for(int i=0; i<c.length-1; i++){
            int tmp = c[i+1] - c[i] - k;
            sum += Math.max(0, tmp);
        }
        
        if(sum >= q){
            return true;
        }else{
            return false;
        }
    }
}

class Val_idx implements Comparable<Val_idx>{
    
    public int val;
    public int idx;
    
    public Val_idx(int val, int idx){
        this.val = val;
        this.idx = idx;
    }
    
    @Override
    public int compareTo(Val_idx o) {
        return this.val - o.val;
    }
}