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 C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>

using namespace std;

const int INF = 1e9;

int n, k, q, a[2005];

int main(){
	scanf("%d%d%d", &n, &k, &q);
	
	for(int i = 0; i < n; i++)
	scanf("%d", &a[i]);
	
	int res = INF;
	
	for(int i = 0; i < n; i++){
		//fix smallest = a[i]
		
		vector <int> tmp, best;
		
		for(int j = 0; j < n; j++){
			if(a[j] >= a[i])
			tmp.push_back(a[j]);
			else{
				if((int)tmp.size() >= k){
					sort(tmp.begin(), tmp.end());
					for(int ii = 0; ii < (int)tmp.size() - k + 1; ii++)
					best.push_back(tmp[ii]);
				}
				tmp.clear();
			}
		}
		
		if((int)tmp.size() >= k){
			sort(tmp.begin(), tmp.end());
			for(int ii = 0; ii < (int)tmp.size() - k + 1; ii++)
			best.push_back(tmp[ii]);
		}
		tmp.clear();
		
		sort(best.begin(), best.end());
		
		if((int)best.size() >= q)
		res = min(res, best[q - 1] - a[i]);
	}
	
	printf("%d\n", res);
}