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Java-based LeetCode algorithm problem solutions, regularly updated

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1386\. Cinema Seat Allocation

Medium

![](https://assets.leetcode.com/uploads/2020/02/14/cinema_seats_1.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 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.

**Example 1:**

![](https://assets.leetcode.com/uploads/2020/02/14/cinema_seats_3.png)

**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 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:** 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.