Problem:
Score : 600 points

Problem Statement

We have 3N cards arranged in a row from left to right, where each card has an integer between 1 and N (inclusive) written on it. The integer written on the i-th card from the left is A_i.

You will do the following operation N-1 times:

Rearrange the five leftmost cards in any order you like, then remove the three leftmost cards. If the integers written on those three cards are all equal, you gain 1 point.

After these N-1 operations, if the integers written on the remaining three cards are all equal, you will gain 1 additional point.

Find the maximum number of points you can gain.

Constraints

1 \leq N \leq 2000

1 \leq A_i \leq N

Input

Input is given from Standard Input in the following format:

N
A_1 A_2 \cdots A_{3N}

Output

Print the maximum number of points you can gain.

Sample Input 1

2
1 2 1 2 2 1

Sample Output 1

2

Let us rearrange the five leftmost cards so that the integers written on the six cards will be 2\ 2\ 2\ 1\ 1\ 1 from left to right.

Then, remove the three leftmost cards, all of which have the same integer 2, gaining 1 point.

Now, the integers written on the remaining cards are 1\ 1\ 1.

Since these three cards have the same integer 1, we gain 1 more point.

In this way, we can gain 2 points - which is the maximum possible.

Sample Input 2

3
1 1 2 2 3 3 3 2 1

Sample Output 2

1

Sample Input 3

3
1 1 2 2 2 3 3 3 1

Sample Output 3

3

Solution in C++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;
using ll=long long;
using P = pair<ll,ll>;
const ll INF = ll(1e18)+5;

int main() {
    ll N;
    cin >> N;
    vector<ll> A(3*N);
    vector<set<ll>> s(N);
    for(int i=0;i<3*N;i++){
        ll a;
        cin >> a;
        --a;
        A[i] = a;
        s[a].insert(i);
    }
    map<ll,ll> mp;
    for(int i=0;i<2;i++){
        mp[A[i]]++;
    }
    ll ans =0;
    for(int i=0;i<N-1;i++){
        // cerr <<"i=" << i << endl;
        for(int j=0;j<3;j++){
            mp[A[(i+1)*3+j-1]]++;
        }
        bool skip = false;
        vector<P> val;
        for(auto itr = mp.begin();itr != mp.end();itr++){
            if(itr->second >=3){
                // cerr <<"found at " << itr->first << endl;
                mp[itr->first] -= 3;
                skip = true;
                ans++;
                break;
            }
            if(itr->second == 2){
               auto itr2 = s[itr->first].lower_bound((i+2)*3-1);
               if(itr2== s[itr->first].end()){
                   val.push_back(P(INF,itr->first));
                   continue;
               }
               ll v = *itr2;
               val.push_back(P(v,itr->first));
            }
            if(itr->second == 1){
                auto itr2 = s[itr->first].lower_bound((i+2)*3-1);
                 if(itr2== s[itr->first].end()){
                     val.push_back(P(INF,itr->first));
                   continue;
               }
                itr2++;
                if(itr2== s[itr->first].end()){
                    val.push_back(P(INF,itr->first));
                   continue;
               }
                ll v = *itr2;
               val.push_back(P(v,itr->first));
            }
        }
        if(skip){
            // cerr <<"skipped" << endl;
            continue;
        }
        sort(val.rbegin(),val.rend());
        ll erased = 0;
        for(int j=0;j<val.size();j++){
            if(erased<3){
                ll d = min(3-erased,mp[val[j].second]);
                mp[val[j].second]-=d;
                erased+=d;
            }
            if(erased>3){
                return -1;
            }
        }
    }
    mp[A.back()]++;
    for(auto itr = mp.begin();itr != mp.end();itr++){
        // cerr <<"itr.first=" <<itr->first <<" second =" << itr->second << endl;
            if(itr->second >=3){
                mp[itr->first] -= 3;
                ans++;
                break;
            }
    }
    cout << ans << endl;

    return 0;
}