<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 n1 = reservedSeats.size();
        unordered_map<int,vector<int>> mp;
        for(int i=0; i<n1; i++)
        {
            mp[reservedSeats[i][0]].push_back(reservedSeats[i][1]); 
        }

        int ans = (n - mp.size())*2;

        for(auto m: mp)
        {
            bool flag = false;
            vector<bool> contains(11,0);
            for(auto it: m.second)
            {
                contains[it] = 1;
            }

            if(!contains[2] && !contains[3] && !contains[4] && !contains[5])
            {
                ans++;
                flag = true;
            }

            if(!contains[6] && !contains[7] && !contains[8] && !contains[9])
            {
                ans++;
                flag = true;
            }

            if(!flag)
            {
                if(!contains[4] && !contains[5] && !contains[6] && !contains[7])
                {
                    ans++;
                }
            }
        }

        return ans;
    }
};
74721