Introduction

In this blog post, we will explore how to solve the LeetCode problem "1386" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

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.

Explanation

Here's a breakdown of the solution:

• Focus on Rows with Reservations: The vast majority of rows will likely be entirely free. We only need to analyze rows that have at least one reserved seat.
• Efficient Row Analysis: For rows with reservations, use bit manipulation to efficiently check for available groups of four. Represent the row as a bitmask where each bit corresponds to a seat (1 indicating reserved).

• Maximize Group Count: For each row, check for the possibility of allocating groups in the following positions [1-4], [5-8], and [3-6].

• Runtime Complexity: O(k), where k is the number of reserved seats.

• Storage Complexity: O(k), where k is the number of reserved seats (for storing the row-wise reservations).

Code

    def max_number_of_families(n: int, reservedSeats: list[list[int]]) -> int:
    """
    Calculates the maximum number of four-person groups that can be assigned in a cinema.

    Args:
        n: The number of rows in the cinema.
        reservedSeats: A list of reserved seats, where each element is a list [row, seat].

    Returns:
        The maximum number of four-person groups that can be assigned.
    """

    reserved_rows = {}
    for row, seat in reservedSeats:
        if row not in reserved_rows:
            reserved_rows[row] = 0
        reserved_rows[row] |= (1 << (seat - 1))

    count = (n - len(reserved_rows)) * 2  # Each unreserved row can fit two groups

    for row, mask in reserved_rows.items():
        groups = 0
        # Check if we can fit a group in seats 2-5 (bits 1-4)
        if (mask & 0b0111100000) == 0:
            groups += 1
        # Check if we can fit a group in seats 6-9 (bits 5-8)
        if (mask & 0b0000011110) == 0:
            groups += 1
        # Check if we can fit a group in seats 4-7 (bits 3-6)
        elif (mask & 0b0001111000) == 0:
            groups += 1

        count += groups

    return count