cinema_seat_allocation.py

# https://leetcode.com/problems/cinema-seat-allocation/description/
# 1386. 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.

from collections import defaultdict
from typing import List

class Solution:
    def maxNumberOfFamilies(self, n: int, reservedSeats: List[List[int]]) -> int:
        ''' Cinema Seat Allocation '''
        # l2-5 r6-9 m4-7
        output = n * 2
        occupied = defaultdict(set)
        for el in reservedSeats:
            # ВЕСА: l, c, r = 1, 2, 7
            if 2 <= el[1] <= 5:
                occupied[el[0]].add(1)
                if 4 <= el[1] <= 7:
                    occupied[el[0]].add(2)
            elif 6 <= el[1] <= 9:
                occupied[el[0]].add(7)
                if 4 <= el[1] <= 7:
                    occupied[el[0]].add(2)
        for places in occupied.values():
            # occupied[row]:
            # =1 >> -1
            # =3 >> -1
            # =9 >> -1
            # =10 >> -2
            if sum(places) == 10:
                output -= 2
            else:
                output -= 1
        return output

n, reserved = 3, [[1,2],[1,3],[1,8],[2,6],[3,1],[3,10]]
# Output: 4
n, reserved = 2, [[2,1],[1,8],[2,6]]
# Output: 2
n, reserved = 4, [[4,3],[1,4],[4,6],[1,7]]
# Output: 4
Solution().maxNumberOfFamilies(n, reserved)