Cinema Seat Allocation

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ArraysGreedy AlgorithmsBit Manipulation

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:

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.

Try coding it in LeetCode

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What are the dimensions of the cinema (number of rows and seats per row), and what are the maximum possible values for 'n' and 'm'?
Are the row and seat numbers in the 'reservedSeats' array 1-indexed or 0-indexed?
What should I return if no families can be seated according to the rules?
Are the reserved seats guaranteed to be within the bounds of the cinema's dimensions?
Can a family be split across rows, or must they occupy seats within the same row?

Brute Force Solution

Approach

The brute force method for this problem involves checking every single possible arrangement of families in the cinema rows. We will examine each row and try placing families in all possible combinations of seats. This exhaustive search guarantees we find the best arrangement, albeit inefficiently.

Here's how the algorithm would work step-by-step:

Consider each row of seats independently.
For a given row, imagine trying to fit a family in every possible combination of four consecutive seats that allows a cinema layout.
Repeat this process for the next family and also trying to fit them into every other possible row.
Consider every different number of families from zero up to the maximum that could potentially fit in each row.
For each combination of families in a given row, see how many families can be seated (following cinema seating rules).
Repeat this process for all rows and families by checking the sum of total number of families seated in all rows.
The solution is the maximum number of families that can be seated across all rows when considering all arrangements.

Code Implementation

def cinema_seat_allocation_brute_force(number_of_rows, reserved_seats): 
    maximum_families = 0
    
    # Consider all possible arrangements of families
    for i in range(2 ** (number_of_rows * 10)): 
        number_of_families_seated = 0
        row_allocations = {}
        
        # For each row, determine which seats are occupied based on 'i'
        for row_number in range(1, number_of_rows + 1): 
            row_allocations[row_number] = [False] * 10

        temp_value = i
        for row_number in range(1, number_of_rows + 1): 
            for seat_number in range(1, 11): 
                if temp_value % 2 == 1:
                    row_allocations[row_number][seat_number-1] = True
                temp_value //= 2
    
        # Check if the allocation is valid based on reserved seats
        is_valid = True
        for row_number, seat_number in reserved_seats:
            if row_number in row_allocations and seat_number >= 1 and seat_number <= 10 and row_allocations[row_number][seat_number-1]:
                is_valid = False
                break
        
        if not is_valid:
            continue

        # Count how many families can be seated in each row
        for row_number in range(1, number_of_rows + 1):
            can_seat_families = 0
            seats = row_allocations[row_number]
            
            # Seats 2-5
            if not any(seats[1:5]):
                can_seat_families += 1
            # Seats 4-7
            if not any(seats[3:7]):
                can_seat_families += 1
            # Seats 6-9
            if not any(seats[5:9]):
                can_seat_families += 1

            number_of_families_seated += can_seat_families
        
        # Update the maximum families seated
        maximum_families = max(maximum_families, number_of_families_seated)
    
    return maximum_families

Big(O) Analysis

Time Complexity

O(4^n * m!) – Given n rows, the algorithm considers every possible arrangement of families in each row independently. For a row with a fixed family, there are four possible positions to place a family (seats 2-5, 4-7, 6-9, 8-11) which leads to roughly 4 possibilities for each family up to n times. Additionally, when allocating multiple families in each row with m total families, all possible permutations need to be explored which costs O(m!). The maximum value across all these arrangements is the final solution. Therefore, the overall time complexity can be approximated as O(4^n * m!).

Space Complexity

O(1) – The brute force approach described primarily involves iterative checking and does not explicitly create any significant auxiliary data structures that scale with the input. The number of rows, families, or possible seating combinations being checked are handled iteratively within loops, without storing them in auxiliary arrays, hash maps, or similar data structures. The algorithm likely uses a few integer variables to track the number of families seated and the best arrangement found so far. These constant-size variables result in constant auxiliary space, irrespective of the input size, which can be represented as O(1).

Optimal Solution

Approach

The most efficient approach is to focus on each row independently and maximize the number of families that can be seated. We avoid unnecessary calculations by considering only rows with reserved seats and cleverly grouping the remaining seats.

Here's how the algorithm would work step-by-step:

First, count how many rows have no reservations. These rows can each fit two families automatically.
Then, for each row that *does* have reservations, figure out which seats are blocked.
Try to fit one family in the seats labeled 2-5. If that works, increment the family count.
If seats 2-5 are blocked, try fitting a family in seats 6-9. If that works, increment the family count.
If both of those options are blocked, try fitting a family across the aisle in seats 4-7. If that works, increment the family count.
Add the families that fit in rows with reservations to the families from rows with no reservations to get the total.
That total is the maximum number of families that can be seated.

Code Implementation

def max_number_of_families(number_of_rows: int, reserved_seats: list[list[int]]) -> int:

    rows_with_reservations = {}
    for reservation in reserved_seats:
        row_number = reservation[0]
        seat_number = reservation[1]
        if row_number not in rows_with_reservations:
            rows_with_reservations[row_number] = [False] * 10
        rows_with_reservations[row_number][seat_number - 1] = True

    number_of_unreserved_rows = number_of_rows - len(rows_with_reservations)
    total_families = number_of_unreserved_rows * 2

    for row_number, seats in rows_with_reservations.items():
        families_in_row = 0

        # Try to fit a family in seats 2-5
        if not any(seats[1:5]):
            families_in_row += 1

        # Try to fit a family in seats 6-9
        if not any(seats[5:9]):
            families_in_row += 1

        # Try to fit a family across the aisle in seats 4-7
        # Only attempt this if the other two options failed.  We don't want to double count
        if families_in_row == 0 and not any(seats[3:7]):
            families_in_row += 1

        total_families += families_in_row

    return total_families

Big(O) Analysis

Time Complexity

O(n) – The algorithm iterates through each row with reservations once. The number of rows with reservations is at most n, where n is the total number of reservations provided as input. Within each row, a constant number of checks are performed (checking seats 2-5, 6-9, and 4-7), independent of the input size. Therefore, the time complexity is directly proportional to the number of reservations, resulting in O(n).

Space Complexity

O(N) – The primary space complexity comes from storing the reserved seats for each row. Since we only consider rows with reserved seats, the maximum number of rows we process is proportional to the number of reserved seats, which we can denote as N, where N is the total number of reservations. We store the information about blocked seats within each of these rows, resulting in space proportional to N. Therefore, the auxiliary space complexity is O(N).

Edge Cases

Case	How to Handle
Null or empty input: n (number of rows) is zero or negative.	Return 0 immediately, as no rows exist to allocate seats.
Empty reserved seats array.	Return n * 2, as all rows are fully available for family seating.
Maximum number of rows (n is very large).	The solution should use a space-efficient data structure (e.g., a map or hash table) to store reserved seats, avoiding the creation of a large, dense array; avoid solutions with O(n) space complexity where possible
Reserved seats clustered in a few rows, leaving most rows empty.	Iterating through reserved seats avoids unnecessary processing of empty rows; the solution should only process rows that have reservations.
Reserved seats scattered across all columns in a row, blocking all family seating.	The bitmask or direct checks should accurately determine when all possible family arrangements in a row are blocked.
Reserved seats with invalid row or column numbers (e.g., row > n, column < 1, column > 10).	Ignore or sanitize such invalid seat reservations, assuming 1-based indexing with column restrictions (1-10).
Duplicate reserved seats.	Treat these as a single reservation as they don't affect the availability of family seats.
Integer overflow in intermediate calculations (if applicable).	Ensure all intermediate calculations and results remain within the acceptable integer range to avoid incorrect seat allocation counts.

Null or empty input: n (number of rows) is zero or negative.

How to Handle:

Return 0 immediately, as no rows exist to allocate seats.

Empty reserved seats array.

How to Handle:

Return n * 2, as all rows are fully available for family seating.

Maximum number of rows (n is very large).

How to Handle:

The solution should use a space-efficient data structure (e.g., a map or hash table) to store reserved seats, avoiding the creation of a large, dense array; avoid solutions with O(n) space complexity where possible

Reserved seats clustered in a few rows, leaving most rows empty.

How to Handle:

Iterating through reserved seats avoids unnecessary processing of empty rows; the solution should only process rows that have reservations.

Reserved seats scattered across all columns in a row, blocking all family seating.

How to Handle:

The bitmask or direct checks should accurately determine when all possible family arrangements in a row are blocked.

Reserved seats with invalid row or column numbers (e.g., row > n, column < 1, column > 10).

How to Handle:

Ignore or sanitize such invalid seat reservations, assuming 1-based indexing with column restrictions (1-10).

Duplicate reserved seats.

How to Handle:

Treat these as a single reservation as they don't affect the availability of family seats.

Integer overflow in intermediate calculations (if applicable).

How to Handle:

Ensure all intermediate calculations and results remain within the acceptable integer range to avoid incorrect seat allocation counts.