public class BookMyShow
extends Object
2286 - Booking Concert Tickets in Groups\.
Hard
A concert hall has `n` rows numbered from `0` to `n - 1`, each with `m` seats, numbered from `0` to `m - 1`. You need to design a ticketing system that can allocate seats in the following cases:
* If a group of `k` spectators can sit **together** in a row.
* If **every** member of a group of `k` spectators can get a seat. They may or **may not** sit together.
Note that the spectators are very picky. Hence:
* They will book seats only if each member of their group can get a seat with row number **less than or equal** to `maxRow`. `maxRow` can **vary** from group to group.
* In case there are multiple rows to choose from, the row with the **smallest** number is chosen. If there are multiple seats to choose in the same row, the seat with the **smallest** number is chosen.
Implement the `BookMyShow` class:
* `BookMyShow(int n, int m)` Initializes the object with `n` as number of rows and `m` as number of seats per row.
* `int[] gather(int k, int maxRow)` Returns an array of length `2` denoting the row and seat number (respectively) of the **first seat** being allocated to the `k` members of the group, who must sit **together**. In other words, it returns the smallest possible `r` and `c` such that all `[c, c + k - 1]` seats are valid and empty in row `r`, and `r <= maxRow`. Returns `[]` in case it is **not possible** to allocate seats to the group.
* `boolean scatter(int k, int maxRow)` Returns `true` if all `k` members of the group can be allocated seats in rows `0` to `maxRow`, who may or **may not** sit together. If the seats can be allocated, it allocates `k` seats to the group with the **smallest** row numbers, and the smallest possible seat numbers in each row. Otherwise, returns `false`.
**Example 1:**
**Input**
["BookMyShow", "gather", "gather", "scatter", "scatter"]
[[2, 5], [4, 0], [2, 0], [5, 1], [5, 1]]
**Output:** [null, [0, 0], [], true, false]
**Explanation:**
BookMyShow bms = new BookMyShow(2, 5); // There are 2 rows with 5 seats each
bms.gather(4, 0); // return [0, 0]
// The group books seats [0, 3] of row 0.
bms.gather(2, 0); // return []
// There is only 1 seat left in row 0,
// so it is not possible to book 2 consecutive seats.
bms.scatter(5, 1); // return True
// The group books seat 4 of row 0 and seats [0, 3] of row 1.
bms.scatter(5, 1); // return False
// There is only one seat left in the hall.
**Constraints:**
* 1 <= n <= 5 * 104
* 1 <= m, k <= 109
* `0 <= maxRow <= n - 1`
* At most 5 * 104 calls **in total** will be made to `gather` and `scatter`.