Booking Concert Tickets in Groups

Hard

Asked by:

9 views

Topics:

ArraysGreedy Algorithms

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.

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:

How is the availability of seats represented? Specifically, what data type and structure will indicate how many seats are available in each row or section?
What is the format of a booking request? What data type represents the number of tickets a group wants to book, and how do I identify the desired section or row?
If a group cannot be fully accommodated in their first choice section, should I attempt to partially fulfill the booking by allocating the remaining seats, or reject the booking entirely?
What should I return if a booking is successful? What should I return if no suitable arrangement can be found to fulfill the booking?
Can I assume the input (seat availability and booking requests) are always valid, or should I handle cases such as invalid row/section numbers, or a request for a negative number of tickets?

Brute Force Solution

Approach

When booking concert tickets for groups, a brute-force method involves checking every possible arrangement of available seats to accommodate the group. This means we'll consider all combinations of seats, regardless of whether they are together or scattered. We'll examine each combination to see if it can fit the group and then choose the 'best' option.

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

Start by considering the first potential seat for the group.
Check if there are enough consecutive seats from this starting point to fit the entire group.
If not, try the next available seat as the starting point.
Continue doing this, checking every possible starting seat to see if the group can fit together in a single block.
If the group cannot be seated together in a single block, explore other options.
Consider splitting the group and seating them in multiple separate blocks of seats.
Check all possible combinations of blocks of seats that, when combined, can accommodate the entire group.
For each valid seating arrangement (either single block or multiple blocks), record the arrangement.
Once you have checked all possible seating arrangements, compare them based on some criteria (like proximity to the stage or keeping members of the same party close).
Select the seating arrangement that best meets the criteria.

Code Implementation

def book_concert_tickets_brute_force(available_seats, group_size): 
    best_arrangement = None
    
    # First, try to find a block of contiguous seats
    for starting_seat_index in range(len(available_seats)): 
        if starting_seat_index + group_size <= len(available_seats): 
            seats_available = True
            for i in range(starting_seat_index, starting_seat_index + group_size): 
                if not available_seats[i]:
                    seats_available = False
                    break
            
            if seats_available:
                arrangement = list(range(starting_seat_index, starting_seat_index + group_size))
                if best_arrangement is None:
                    best_arrangement = arrangement
                else:
                    #Simple proximity check: smaller index is closer to the stage
                    if arrangement[0] < best_arrangement[0]:
                        best_arrangement = arrangement

    # Consider splitting the group if no contiguous block found
    if best_arrangement is None: 
        #Iterate through all possible combinations of seats
        import itertools
        available_seat_indices = [i for i, seat in enumerate(available_seats) if seat]
        for seat_combination in itertools.combinations(available_seat_indices, group_size):
            if len(seat_combination) == group_size:
                #Found an arrangement, so assess it based on the location.
                arrangement = sorted(list(seat_combination))
                if best_arrangement is None:
                    best_arrangement = arrangement

                # Here, assess quality based on seat numbers (lower is better)
                elif sum(arrangement) < sum(best_arrangement): 
                    best_arrangement = arrangement

    if best_arrangement:
        #Mark tickets as booked
        for booked_seat_index in best_arrangement:
            available_seats[booked_seat_index] = False

    return best_arrangement

Big(O) Analysis

Time Complexity

O(2^n) – The algorithm explores all possible combinations of seats to accommodate the group. In the worst-case scenario, where the group needs to be split into multiple blocks and nearly every seat is considered, the number of combinations to check grows exponentially. We are effectively considering every subset of the n seats, which results in 2^n possible arrangements. Therefore, the time complexity is O(2^n), indicating exponential growth with the number of seats.

Space Complexity

O(N^K) – The algorithm explores all possible combinations of seating arrangements, including splitting the group into multiple blocks. Specifically, it checks combinations of blocks of seats that accommodate the entire group (step 7). The number of such combinations can grow exponentially with the number of seats (N) and the maximum number of blocks the group might be split into (K). Therefore, storing all valid seating arrangements (step 8) requires space proportional to the number of possible combinations. This leads to a space complexity of O(N^K) where N is the number of seats, and K is the number of possible blocks.

Optimal Solution

Approach

The best way to book concert tickets in groups efficiently is to keep track of available seats and quickly find blocks of seats that fit the group size. We can use a special data structure that makes it easy to find the right block and update seat availability.

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

Imagine the concert hall divided into sections, and each section into rows.
Keep track of how many empty seats are in each section and row.
When a group wants to book tickets, first see if there's a section with enough seats in a single row.
If there is, find a continuous block of empty seats of the correct size within that row.
If you find such a block, assign those seats to the group and update the number of available seats in that row and section.
If no single row has enough seats, try to find multiple adjacent rows within the same section that, combined, can accommodate the group.
If a suitable arrangement of multiple rows is found, allocate the seats and update the availability counts.
If you can't find enough seats in any section, inform the group that there are no seats available.

Code Implementation

def book_concert_tickets(sections, group_size):
    for section_index, section in enumerate(sections):
        for row_index, row in enumerate(section):
            if row >= group_size:
                # Find a continuous block of empty seats
                seats = find_continuous_block(sections, section_index, row_index, group_size)
                if seats:
                    return seats

    # If no single row has enough seats, try multiple adjacent rows
    for section_index, section in enumerate(sections):
        seats = find_multiple_rows(sections, section_index, group_size)
        if seats:
            return seats
    return None

def find_continuous_block(sections, section_index, row_index, group_size):
    section = sections[section_index]
    row = section[row_index]
    if row >= group_size:
        sections[section_index][row_index] -= group_size
        return (section_index, row_index, group_size)
    return None

def find_multiple_rows(sections, section_index, group_size):
    rows = []
    for row_index, row in enumerate(sections[section_index]):
        rows.append((row_index, row))

    for i in range(len(rows)):
        total_seats = 0
        selected_rows = []
        for j in range(i, len(rows)):
            row_index, available_seats = rows[j]
            total_seats += available_seats
            selected_rows.append(row_index)
            if total_seats >= group_size:
                # Allocate seats from multiple rows
                seats_allocated = allocate_seats_from_rows(sections, section_index, selected_rows, group_size)
                if seats_allocated:
                    return seats_allocated
    return None

def allocate_seats_from_rows(sections, section_index, selected_rows, group_size):
    seats_allocated = []
    remaining_seats = group_size

    for row_index in selected_rows:
        available_seats = sections[section_index][row_index]
        seats_to_allocate = min(remaining_seats, available_seats)

        # Update number of available seats
        sections[section_index][row_index] -= seats_to_allocate
        seats_allocated.append((section_index, row_index, seats_to_allocate))
        remaining_seats -= seats_to_allocate
        if remaining_seats == 0:
            return seats_allocated

    return None

Big(O) Analysis

Time Complexity

O(m*r) – The algorithm iterates through sections (let's say m sections) looking for available seats. Within each section, it iterates through the rows (let's say r rows) to find a suitable row or combination of rows. Finding a continuous block of seats in a row can take O(r) in the worst case (although practically this is related to the number of seats in the row and can be considered constant). Searching adjacent rows in case a single row doesn't suffice adds another O(r) factor in the worst case. Thus for m sections and r rows operations approximate to m*r, which simplifies to O(m*r).

Space Complexity

O(S*R) – The space complexity is determined by the need to keep track of empty seats in each section and row. This requires a data structure, likely a 2D array or a similar structure, to store the seat availability information for each section and row. Where S is the number of sections and R is the maximum number of rows in any section. This means the auxiliary space grows linearly with both the number of sections and the number of rows, therefore the space usage is O(S*R).

Edge Cases

Case	How to Handle
seatsAvailable is null or empty	Return an empty list or throw an exception, depending on the defined API contract.
groupSize is zero or negative	Throw an IllegalArgumentException as a zero or negative group size is invalid.
seatsAvailable contains extremely large numbers (potential integer overflow)	Use long data type to avoid integer overflow and handle potential out-of-memory errors gracefully.
groupSize is larger than the total number of seats available in the system	Return an empty list or throw an exception since the request cannot be fulfilled.
Multiple groups can fit, but only one is required	Return the first available booking (e.g., lowest index) that fits the entire group.
No available seats that can accommodate the group	Return an empty list to indicate no suitable booking exists.
seatsAvailable array contains only zero values	Return an empty list because no seat can accommodate any group size greater than 0.
Seats are fragmented such that there are enough total seats, but not enough contiguous seats for the group.	The algorithm needs to correctly search for contiguous blocks and return empty if no sufficient block exists.

seatsAvailable is null or empty

How to Handle:

Return an empty list or throw an exception, depending on the defined API contract.

groupSize is zero or negative

How to Handle:

Throw an IllegalArgumentException as a zero or negative group size is invalid.

seatsAvailable contains extremely large numbers (potential integer overflow)

How to Handle:

Use long data type to avoid integer overflow and handle potential out-of-memory errors gracefully.

groupSize is larger than the total number of seats available in the system

How to Handle:

Return an empty list or throw an exception since the request cannot be fulfilled.

Multiple groups can fit, but only one is required

How to Handle:

Return the first available booking (e.g., lowest index) that fits the entire group.

No available seats that can accommodate the group

How to Handle:

Return an empty list to indicate no suitable booking exists.

seatsAvailable array contains only zero values

How to Handle:

Return an empty list because no seat can accommodate any group size greater than 0.

Seats are fragmented such that there are enough total seats, but not enough contiguous seats for the group.

How to Handle:

The algorithm needs to correctly search for contiguous blocks and return empty if no sufficient block exists.