<p><img alt="" src="https://assets.leetcode.com/uploads/2020/02/14/cinema_seats_1.png" style="width: 400px; height: 149px;" /></p>

<p>A cinema&nbsp;has <code>n</code>&nbsp;rows of seats, numbered from 1 to <code>n</code>&nbsp;and there are ten&nbsp;seats in each row, labelled from 1&nbsp;to 10&nbsp;as shown in the figure above.</p>

<p>Given the array <code>reservedSeats</code> containing the numbers of seats already reserved, for example, <code>reservedSeats[i] = [3,8]</code>&nbsp;means the seat located in row <strong>3</strong> and labelled with <b>8</b>&nbsp;is already reserved.</p>

<p><em>Return the maximum number of four-person groups&nbsp;you can assign on the cinema&nbsp;seats.</em> A four-person group&nbsp;occupies four&nbsp;adjacent seats <strong>in one single row</strong>. Seats across an aisle (such as [3,3]&nbsp;and [3,4]) are not considered to be adjacent, but there is an exceptional case&nbsp;on which an aisle split&nbsp;a four-person group, in that case, the aisle split&nbsp;a four-person group in the middle,&nbsp;which means to have two people on each side.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<p><img alt="" src="https://assets.leetcode.com/uploads/2020/02/14/cinema_seats_3.png" style="width: 400px; height: 96px;" /></p>

<pre>
<strong>Input:</strong> n = 3, reservedSeats = [[1,2],[1,3],[1,8],[2,6],[3,1],[3,10]]
<strong>Output:</strong> 4
<strong>Explanation:</strong> The figure above shows the optimal allocation for four groups, where seats mark with blue are already reserved and contiguous seats mark with orange are for one group.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> n = 2, reservedSeats = [[2,1],[1,8],[2,6]]
<strong>Output:</strong> 2
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> n = 4, reservedSeats = [[4,3],[1,4],[4,6],[1,7]]
<strong>Output:</strong> 4
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= n &lt;= 10^9</code></li>
	<li><code>1 &lt;=&nbsp;reservedSeats.length &lt;= min(10*n, 10^4)</code></li>
	<li><code>reservedSeats[i].length == 2</code></li>
	<li><code>1&nbsp;&lt;=&nbsp;reservedSeats[i][0] &lt;= n</code></li>
	<li><code>1 &lt;=&nbsp;reservedSeats[i][1] &lt;= 10</code></li>
	<li>All <code>reservedSeats[i]</code> are distinct.</li>
</ul>

class Solution {
public:

     int findCount(set<int> &seats){
        int count = 0;
        //try fitting family in 2,3,4,5
        if(!seats.contains(2) && !seats.contains(3) && !seats.contains(4) && !seats.contains(5)){
            count++;
        }
        //try fitting family in 6,7,8,9
        if(!seats.contains(6) && !seats.contains(7) && !seats.contains(8) && !seats.contains(9)){
            count++;
        }
        //this means we have already assigned the maximum possible seats, so just return
        if(count==1||count==2){
            return count;
        }
        //try fitting family in 4,5,6,7
        if(!seats.contains(4) && !seats.contains(5) && !seats.contains(6) && !seats.contains(7)){
            count++;
        }
        return count;
    }


    int maxNumberOfFamilies(int n, vector<vector<int>>& reservedSeats) {

        map<int,set<int>> mp;

        for(int i=0;i<reservedSeats.size();i++){
            set<int> s;
            if(mp.find(reservedSeats[i][0]) == mp.end()){
                s.insert(reservedSeats[i][1]);
                mp[reservedSeats[i][0]] = s;
            }else
            mp[reservedSeats[i][0]].insert(reservedSeats[i][1]);
        }
        // cout<<mp.size()<<endl;
        int ans = 2 * (n-mp.size());

        

        for(auto it = mp.begin();it!= mp.end();it++){
            set<int> seats = it->second;

            for(auto it2 = seats.begin();it2!=seats.end();it2++) cout<<*(it2)<<" , "<<endl;
            ans += findCount(seats);
            // cout<<ans<<endl;
        }

        return ans;


        
    }
};

/**

select a row

a b
b-a seats

2-5 or 4-7 or 6-9


**/


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