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 <iostream>
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
ll n,m,k,a[2000],ans=1000000000;
vector<ll> u,v;
int main(void){
    cin>>n>>m>>k;
    if(k==1){cout<<0<<endl;return 0;}
    for(int i=0;i<n;i++){
        cin>>a[i];
    }
    for(int i=0;i<n;i++){
        v={};
        u={};
        for(int j=0;j<n;j++){
            if(i==j)continue;
            if(a[i]<=a[j]){
                v.push_back(a[j]);
            }
            else{
                if(v.size()>=m){
                    sort(v.begin(),v.end());
                    for(int i=0;i<min((ll)v.size()-m+1LL,k-1LL);i++)u.push_back(v[i]);
                }
                v={};
            }
        }
        if(v.size()>=m){
            sort(v.begin(),v.end());
            for(int i=0;i<min((ll)v.size()-m+1,k-1);i++)u.push_back(v[i]);
        }
        //cout<<u.size()<<" "<<i<<endl;
        if(u.size()<k-1)continue;
        sort(u.begin(),u.end());
        //cout<<u[0]<<endl;
        ans=min(ans,u[k-2]-a[i]);
        //cout<<ans<<endl;
    }
    cout<<ans<<endl;
}