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;

int main()
{
    long long n,k,q;
    cin>>n>>k>>q;
    vector<long long>nms;
    for(int i=0;i<n;i++){
        long long t;
        cin>>t;
        nms.push_back(t);
    }
    long long x= LONG_LONG_MAX;
    long long y= LONG_LONG_MIN;
    int cnt=0;
    sort(nms.begin(),nms.end());
    for(int i=0;i+k-1<nms.size()&&cnt<q;i++){
            long long mn= LONG_LONG_MAX;
            int idx=-1;
            int j=i;
        for(;j<i+k;j++){
            long long lsmn=mn;
            mn=min(mn,nms[j]);
            if(mn!=lsmn)
                idx=j;
        }
        cnt++;
        x=min(x,nms[idx]);
        y=max(y,nms[idx]);
        nms.erase(nms.begin()+idx);
        if(idx==i)
            i--;
    }
    cout << y-x << endl;
    return 0;
}