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 <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define all(x) (x).begin(),(x).end()
const int mod=1000000007,MAX=2001,INF=1<<30;

int main(){
    
    int N,K,Q;cin>>N>>K>>Q;
    vector<int> A(N),B(N-K+1);//B[i]はi~i+K-1の最小値
    for(int i=0;i<N;i++){
        cin>>A[i];
    }
    deque<int> deq(N);
    int s=0,t=0;
    for(int i=0;i<N;i++){
        while(s<t&&A[deq[t-1]]>=A[i]) t--;
        deq[t]=i;
        t++;
        
        if(i-K+1>=0){
            B[i-K+1]=A[deq[s]];
            if(deq[s]==i-K+1) s++;
        }
    }
    
    int ans=INF;
    
    if(Q==1) cout<<0<<endl;
    else{
        for(int i=0;i<N;i++){
            bool flag=false;
            if(N-i<Q) continue;
            for(int j=max(0,i-K+1);j<=i;j++){
                if(B[j]==A[i]){
                    flag=true;
                    break;
                }
            }
            if(!flag) continue;
            
            vector<int> S[N],T;
            int start=0,cnt=0;
            
            for(int j=0;j<N;j++){
                if(i==j) continue;
                
                if(A[j]<A[i]){
                    start=j+1;
                }else{
                    S[start].push_back(A[j]);
                }
            }
            
            for(int j=0;j<N;j++){
                if(S[j].size()>=K) cnt+=(S[j].size()-K+1);
            }
            if(cnt<Q-1) continue;
            
            for(int j=0;j<N;j++){
                if(S[j].size()>=K){
                    sort(all(S[j]));
                    for(int k=0;k<min(int(S[j].size()-K+1),Q-1);k++){
                        T.push_back(S[j][k]);
                    }
                }
            }
            
            sort(all(T));
            
            ans=min(ans,T[Q-2]-A[i]);
        }
        
        cout<<ans<<endl;
    }
}