LeetCode | 1386. Cinema Seat Allocation安排电影院座位【Python】

Myth丶恋晨 2023-07-18 03:25 24阅读 0赞

> LeetCode 1386. Cinema Seat Allocation安排电影院座位【Medium】【Python】【哈希表】

### Problem ###

[LeetCode][]

![format_png][]

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:**

![format_png 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.

### 问题 ###

[力扣][Link 1]

![format_png][]

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：**

![format_png 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 <= 10^9`
 *  `1 <= reservedSeats.length <= min(10*n, 10^4)`
 *  `reservedSeats[i].length == 2`
 *  `1 <= reservedSeats[i][0] <= n`
 *  `1 <= 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

### GitHub链接 ###

[Python][]

### 参考 ###

[Python 反向筛选][Python 1]

[LeetCode]: https://leetcode.com/problems/cinema-seat-allocation/
[format_png]: /images/20230528/44db9fb7d0044208884417b87feb158e.png
[format_png 1]: /images/20230528/f2eba510f7ef4f5b9a5f7690ea38de96.png
[Link 1]: https://leetcode-cn.com/problems/cinema-seat-allocation/
[Python]: https://github.com/Wonz5130/LeetCode-Solutions/blob/master/solutions/1386-Cinema-Seat-Allocation/1386.py
[Python 1]: https://leetcode-cn.com/problems/cinema-seat-allocation/solution/python-fan-xiang-shai-xuan-by-ch3cook/