【leetcode】1386. Cinema Seat Allocation

Topics are as follows:

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 families you can allocate on the cinema seats. A four-person family occupies fours seats in one row, that are next to each other. Seats across an aisle (such as [3,3] and [3,4]) are not considered to be next to each other, however, It is permissible for the four-person family to be separated by an aisle, but in that case, exactly two people have to sit on each side of the aisle.

 

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 families, where seats mark with blue are already reserved and 
contiguous seats mark with orange are for one family. 

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.

Problem-solving ideas: This title is actually very simple. If the seat between 2 and 9 have not been occupied, then this row of seats two persons; if 2 to 6 or 4 to 6 or 8 to 10 wherein a is not occupied, which sit one row.

code show as below:

class Solution(object):
    def maxNumberOfFamilies(self, n, reservedSeats):
        """
        :type n: int
        :type reservedSeats: List[List[int]]
        :rtype: int
        """
        res = 0
        dic = {}
        for row,seat in reservedSeats:
            dic[row] = dic.setdefault(row,[]) + [seat]
        res += (n-len(dic))*2

        def check(seats,l):
            flag = True
            for i in l:
                if i in seats:
                    flag = False
                    break
            return flag

        for seats in dic.itervalues():
            l = range(2,10)
            flag = check(seats,l)
            if flag:
                res += 2
                continue

            l = range(2,6)
            flag = check(seats, l)
            if flag:
                res += 1
                continue

            l = range(6, 10)
            flag = check(seats, l)
            if flag:
                res += 1
                continue

            l = range(4, 8)
            flag = check(seats, l)
            if flag:
                res += 1
                continue

        return res