human
I found an interesting problem on https://leetcode.com/problems/cinema-seat-allocation/:
A cinema has n rows of seats, numbered from 1 to n and there are ten seats in each row, labelled from 1 to 10 as shown in the figure above.
Given the array reservedSeats containing the numbers of seats already reserved, for example, reservedSeats[i] = [3,8] means the seat located in row 3 and labelled with 8 is already reserved.
Return the maximum number of four-person groups you can assign on the cinema seats. A four-person group occupies four adjacent seats in one single row. Seats across an aisle (such as [3,3] and [3,4]) are not considered to be adjacent, but there is an exceptional case on which an aisle split a four-person group, in that case, the aisle split a four-person group in the middle, which means to have two people on each side.
 
Example 1:

Input: n = 3, reservedSeats = [[1,2],[1,3],[1,8],[2,6],[3,1],[3,10]]
Output: 4
Explanation: 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.

Example 2:
Input: n = 2, reservedSeats = [[2,1],[1,8],[2,6]]
Output: 2

Example 3:
Input: n = 4, reservedSeats = [[4,3],[1,4],[4,6],[1,7]]
Output: 4

 
Constraints:

1 <= n <= 10^9
1 <= reservedSeats.length <= min(10*n, 10^4)
reservedSeats[i].length == 2
1 <= reservedSeats[i][0] <= n
1 <= reservedSeats[i][1] <= 10
All reservedSeats[i] are distinct.

I tried it in Python, but could not do it. Can you solve it?

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Can you provide a the Python code you requested?
assistant
#Here is the Python code you requested.

class Solution:
    def maxNumberOfFamilies(self, n: int, reservedSeats: List[List[int]]) -> int:
        rowseats = defaultdict(list)
        for s in reservedSeats:
            rowseats[s[0]].append(s[1])
            
        subtract = 0
        for _, seats in rowseats.items():
            left = True
            right = True
            middle = True
            for s in seats:
                if 2 <= s <= 5:
                    left = False
                if 4 <= s <= 7:
                    middle = False
                if 6 <= s <= 9:
                    right = False
            if left and right:
                subtract += 0
            elif left or right or middle:
                subtract += 1
            else:
                subtract += 2
                
        return n*2 - subtract