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.*;

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[] A = new int[N];

        for(int i = 0; i < N; i++){
            A[i] = sc.nextInt();
        }
        Arrays.sort(A);

        long ans = 1000000000;

        for(int i = 0; i < N; i++){
//            System.out.println("A[" + i + "]:" + A[i]);
        }
        for(int i = 0; i < N - K - Q + 2; i++){
 //           System.out.println("A[" + (i + Q - 1) + "]-A[" + i + "]:" + (A[i + Q - 1] - A[i]));
            ans = Math.min(ans, A[i + Q - 1] - A[i]);
        }
        System.out.println(ans);
    }
}