# bookmyshow/paytm seat allocation for groups

A cinema has `n`

rows of seats, numbered from 1 to `n`

and there are ten seats in each row, labeled 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 labeled with **8** is already reserved.

*Return the maximum number of four-person groups you can assign to 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:**

**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.

**Complexity Analysis**

- Time complexity:
**O(n)**. The complete array of size n is traversed. - Space complexity:
. An additional dictionary is used to map the row to seat information.*O*(*n*)