Class Solution


  • public class Solution
    extends Object
    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.
    • Constructor Detail

      • Solution

        public Solution()
    • Method Detail

      • maxNumberOfFamilies

        public int maxNumberOfFamilies​(int n,
                                       int[][] reservedSeats)