Cinema Seat Allocation (https://leetcode.com/problems/cinema-seat-allocation/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.
the logic is perfectly good, but it takes more space.
My code :
class Solution(object): def find(self, row): str1 = "".join(map(str, row)) # Concatenate the integers in the row list ans = 0 i = 0 while i < len(str1) - 4: if str1[i:i + 4] == "0000" and i % 2 != 0: ans += 1 i += 4 else: i += 1 return ans def maxNumberOfFamilies(self, n, reservedSeats):""" :type n: int :type reservedSeats: List[List[int]] :rtype: int""" ans = 0 cinema = [[0] * 10 for _ in range(n)] for li in reservedSeats: cinema[li[0] - 1][li[1] - 1] = 1 for row in cinema: ans += self.find(row) return ansExample 1:
Seats looks like this:
Input: n = 3, reservedSeats = [[1,2],[1,3],[1,8],[2,6],[3,1],[3,10]]Output: 4Explanation: 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: 2Example 3:
Input: n = 4, reservedSeats = [[4,3],[1,4],[4,6],[1,7]]Output: 4