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1386. Cinema Seat Allocation

Difficulty Topics

Description

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.

Solution

cinema-seat-allocation.py
class Solution:
    def maxNumberOfFamilies(self, n: int, rs: List[List[int]]) -> int:
        d = collections.defaultdict(map)

        for r,c in rs:
            if r not in d:
                d[r] = {c}
            else:
                d[r].add(c)

        res = 2 * n
        for r in d:
            ref = d[r]
            cnt = 0
            if 2 not in ref and 3 not in ref and 4 not in ref and 5 not in ref: cnt += 1

            if 6 not in ref and 7 not in ref and 8 not in ref and 9 not in ref: cnt += 1

            if 4 not in ref and 5 not in ref and 6 not in ref and 7 not in ref and cnt == 0: cnt += 1

            res += (cnt - 2)

        return res