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 N = 2087, inf = 1e9 + 7;
int n,k,q;
int a[N],g[N],ct[N],si[N];
int solve(int bi)
{
    int mn = a[si[bi]];
    int m = 0;
    memset(ct,0,sizeof(ct));
    for (int i = 0; i < n; ++i) {
        if (a[i] < mn) {
            ++m;
            continue;
        }
        g[i] = m;
        ++ct[m];
    }
    int rm = q, mx = inf;
    for (int i = bi; i < n && rm; ++i) {
        int j = si[i];
        if (ct[g[j]] >= k) {
            --ct[g[j]];
            --rm;
            mx = a[j];
        }
    }
    return rm ? inf : mx - mn;
}
int main()
{
    ios::sync_with_stdio(0);
    cin.tie(0);
    cin >> n >> k >> q;
    for (int i = 0; i < n; ++i) {
        cin >> a[i];
        si[i] = i;
    }
    sort(si,si+n,[&](int i,int j)->bool{return a[i] < a[j];});
    int ans = inf;
    for (int i = 0; i < n; ++i) {
        int v = solve(i);
        if (v == inf)
            break;
        ans = min(ans, v);
    }
    cout << ans << endl;
}