<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) {
        int seats=n*2;
        int j=0;
        sort(reservedSeats.begin(),reservedSeats.end());
        while(j<reservedSeats.size()){
            bool isLeftLeft=false,isLeftMid=false,isRightMid=false,isRightRight=false;
            int curr=reservedSeats[j][0];
            while(j<reservedSeats.size() && reservedSeats[j][0]==curr){
                int seat=reservedSeats[j][1];
                if(seat==2 || seat==3) isLeftLeft=true;
                else if(seat==4 || seat==5) isLeftMid=true;
                else if(seat==6 || seat==7) isRightMid=true;
                else if(seat==8 || seat==9) isRightRight=true;
                j++;
            }
            if(isLeftLeft && isRightMid || isRightRight && isLeftMid || isLeftMid && isRightMid){
                seats=seats-2;
            }
            else if(isLeftLeft || isLeftMid || isRightMid || isRightRight){
                seats=seats-1;
            }
        }
        return seats;
    }
};
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