<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) {
        sort(reservedSeats.begin(),reservedSeats.end());
        int lastRow = 0;
        int ans = 0;
        vector<bool> taken = vector<bool>(10);
        for(int i = 0; i < reservedSeats.size(); i++){
            if(reservedSeats[i][0] != lastRow){
                ans += (reservedSeats[i][0] - lastRow -1)*2;
                bool first = !taken[2] & !taken[3] & !taken[4] & !taken[5];
                bool middle = !taken[4] & !taken[5] & !taken[6] & !taken[7];
                bool last = !taken[6] & !taken[7] & !taken[8] & !taken[9];

                if(first & last){
                    ans += 2;
                } else if(first | last | middle){
                    ans += 1;
                }

                lastRow = reservedSeats[i][0];
                taken = vector<bool>(10);
            }
            taken[reservedSeats[i][1]] = true;
        }

        bool first = !taken[2] & !taken[3] & !taken[4] & !taken[5];
        bool middle = !taken[4] & !taken[5] & !taken[6] & !taken[7];
        bool last = !taken[6] & !taken[7] & !taken[8] & !taken[9];

        if(first & last){
            ans += 2;
        } else if(first | last | middle){
            ans += 1;
        }

        if(n > lastRow){
            ans += (n-lastRow)*2;
        }

        return ans - 2;
    }
};
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