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;

typedef long long ll;

const int maxn = 2000 + 10;

int n, k, Q, a[maxn], b[maxn];

bool check(int x, int y){
	for (int i = 1; i <= n; i++){
		if (a[i] < x)
			b[i] = 0;
		else if (a[i] >= x and a[i] <= y)
			b[i] = 1;
		else
			b[i] = 2;
	}
	int t = 0, cnt = 0, len = 0;
	for (int j = 1; j <= n + 1; j++){
		if (b[j] == 0){
			t += max(0, min(len - k + 1, cnt));
			cnt = 0;
			len = 0;
		}
		else{
			if (b[j] == 1)
				cnt ++;
			len ++;
		}
	}
	return t >= Q;
}

int main(){
	ios_base::sync_with_stdio (false);
	cin >> n >> k >> Q;
	vector<int> v;
	for (int i = 1; i <= n; i++){
		cin >> a[i];
		v.push_back(a[i]);
	}
	sort(v.begin(), v.end());
	int answer = 1000 * 1000 * 1000 + 1;
	for (int j = 0; j < n; j++){
		int x = v[j];
		for (int i = 1; i <= n; i++)
			if (a[i] < x)
				b[i] = 0;
		int lo = j - 1, hi = n;
		while (hi - lo > 1){
			int mid = (lo + hi) >> 1;
			if (check(v[j], v[mid]))
				hi = mid;
			else
				lo = mid;
		}
		if (hi != n)
			answer = min(answer, v[hi] - v[j]);
	}
	cout << answer << endl;
}