LeetCode 1386. Cinema Seat Allocation安排電影院座位【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.
問題
如上圖所示，電影院的觀影廳中有 n 行座位，行編號從 1 到 n ，且每一行內總共有 10 個座位，列編號從 1 到 10 。
給你數組 reservedSeats ，包含所有已經被預約了的座位。比如說，researvedSeats[i]=[3,8] ，它表示第 3 行第 8 個座位被預約了。
請你返回 最多能安排多少個 4 人家庭 。4 人家庭要佔據 同一行內連續 的 4 個座位。隔着過道的座位（比方說 [3,3] 和 [3,4]）不是連續的座位，但是如果你可以將 4 人家庭拆成過道兩邊各坐 2 人，這樣子是允許的。
示例 1：
輸入：n = 3, reservedSeats = [[1,2],[1,3],[1,8],[2,6],[3,1],[3,10]]
輸出：4
解釋：上圖所示是最優的安排方案，總共可以安排 4 個家庭。藍色的叉表示被預約的座位，橙色的連續座位表示一個 4 人家庭。
示例 2：
輸入：n = 2, reservedSeats = [[2,1],[1,8],[2,6]]
輸出：2
示例 3：
輸入：n = 4, reservedSeats = [[4,3],[1,4],[4,6],[1,7]]
輸出：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
正面考慮容易超時/超內存，可以從反面來考慮。
統計出被預約的位置，然後反向篩選一下可以坐的位置。
Python3代碼
from typing import List
class Solution:
def maxNumberOfFamilies(self, n: int, reservedSeats: List[List[int]]) -> int:
left, right, mid = set(), set(), set()
count = 0
# 統計被預約的位置
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):
# 預約位置在兩邊：1 or 10
if i not in left and i not in right:
count += 2
# 預約位置在左邊/右邊：2 or 3 or 8 or 9
elif i not in mid:
count += 1
# 預約位置在中間：4 or 5 or 6 or 7
elif i not in left or i not in right:
count += 1
# 反向篩選一下，沒有被預約的行最多可以坐兩家人
count += 2*(n - len(left|right|mid))
return count