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: Wrong Answer)

#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;

int N,K,Q;
int A[2010];
int main(){
	cin >> N >> K >> Q;
	vector<int> v(N);
	for(int i=0;i<N;i++){
		cin >> A[i];
		v[i] = A[i];
	}
	sort(v.begin(),v.end());
	int ans = v[Q-1]-v[0];
	for(int i=0;i<N;i++){
		vector<int> removed,pre_removed;
		int c = 0;
		for(int j=0;j<N;j++){
			if(A[j]>v[i]){
				pre_removed.push_back(A[j]);
				c++;
			}else{
				if(c>=K){
					sort(pre_removed.begin(),pre_removed.end());
					for(int k=0;k<=c-K;k++){
						removed.push_back(pre_removed[k]);
					}
				}
				pre_removed.clear();
				c = 0;
			}
		}
		if(removed.size()>=Q){
			sort(removed.begin(),removed.end());
			ans = min(ans,removed[Q-1]-removed[0]);
		}
		removed.clear();
	}
	cout << ans << endl;
}