<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 maxNumberOfFamilies(int n, vector<vector<int>>& reservedSeats) {
        // 3 possible outcomes to seat them
        // [2,5] , [4,7] , [6,9]

        // but first, put all the people in map
        // lets divide it to 4 sections
        // [2,3] , [4,5] , [6,7] , [8,9]
        // [2,3] -> 2, [4,5] -> 4 , [6,7] -> 6 , [8,9] -> 8
        // meaning [2n , 2n+1] -> 2n ; meaning k -> (k/2)*2;

        int result = 0;
        int ROW = 0;
        int SEAT = 1;
        typedef std::map<int, int> seatMap;
        std::map<int,seatMap> rowMap;
        for(std::vector<int> seat: reservedSeats){
            int currSeat = (seat[SEAT]/2)*2;
            if (currSeat == 0 ||  currSeat == 10){
                continue;
            }
            rowMap[seat[ROW]][(seat[SEAT]/2)*2] = 1;
        }

        for (const auto& row : rowMap){
            int rowNum = row.first;
            seatMap& seats = rowMap[rowNum];
            seatMap::iterator it;
            if (seats.find(4) == seats.end()) {
                if (seats.find(2) == seats.end()) {
                    result++;
                    continue;
                }
                if (seats.find(6) == seats.end()){
                    result++;
                    continue;
                }
            }

            if (seats.find(8) == seats.end()){
                if (seats.find(6) == seats.end()) {
                    result++;
                    continue;
                }
            }

        }

        result += (n - rowMap.size())*2;

        return result;


    }

};
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