LeetCode | 1386. Cinema Seat Allocation安排电影院座位【Python】
LeetCode 1386. Cinema Seat Allocation安排电影院座位【Medium】【Python】【哈希表】
Problem
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 <= 10^91 <= reservedSeats.length <= min(10*n, 10^4)reservedSeats[i].length == 21 <= reservedSeats[i][0] <= n1 <= reservedSeats[i][1] <= 10- All
reservedSeats[i]are distinct.
问题
如上图所示，电影院的观影厅中有 n 行座位，行编号从 1 到 n ，且每一行内总共有 10 个座位，列编号从 1 到 10 。
给你数组 reservedSeats ，包含所有已经被预约了的座位。比如说，researvedSeats[i]=[3,8] ，它表示第 3 行第 8 个座位被预约了。
请你返回 最多能安排多少个 4 人家庭 。4 人家庭要占据 同一行内连续 的 4 个座位。隔着过道的座位（比方说 [3,3] 和 [3,4]）不是连续的座位，但是如果你可以将 4 人家庭拆成过道两边各坐 2 人，这样子是允许的。
示例 1：
输入：n = 3, reservedSeats = [[1,2],[1,3],[1,8],[2,6],[3,1],[3,10]]
输出：4
解释：上图所示是最优的安排方案，总共可以安排 4 个家庭。蓝色的叉表示被预约的座位，橙色的连续座位表示一个 4 人家庭。
示例 2：
输入：n = 2, reservedSeats = [[2,1],[1,8],[2,6]]
输出：2
示例 3：
输入：n = 4, reservedSeats = [[4,3],[1,4],[4,6],[1,7]]
输出：4
提示：
1 <= n <= 10^91 <= reservedSeats.length <= min(10*n, 10^4)reservedSeats[i].length == 21 <= reservedSeats[i][0] <= n1 <= reservedSeats[i][1] <= 10- 所有
reservedSeats[i]都是互不相同的。
思路
哈希表
正面考虑容易超时/超内存，可以从反面来考虑。
统计出被预约的位置，然后反向筛选一下可以坐的位置。
Python3代码
from typing import List
class Solution:
def maxNumberOfFamilies(self, n: int, reservedSeats: List[List[int]]) -> int:
left, right, mid = set(), set(), set()
count = 0
# 统计被预约的位置
for r, c in reservedSeats:
if r in left and r in right and r in mid:
continue
if c < 6 and c > 1:
left.add(r)
if c < 10 and c > 5:
right.add(r)
if c < 8 and c > 3:
mid.add(r)
for i in (left|right|mid):
# 预约位置在两边：1 or 10
if i not in left and i not in right:
count += 2
# 预约位置在左边/右边：2 or 3 or 8 or 9
elif i not in mid:
count += 1
# 预约位置在中间：4 or 5 or 6 or 7
elif i not in left or i not in right:
count += 1
# 反向筛选一下，没有被预约的行最多可以坐两家人
count += 2*(n - len(left|right|mid))
return count
