Maximum Cinema Seat Allocation in C++

In this article, you will learn how to find maximum cinema seat allocation in C++.

Overview:

A movie theatre with many rows and a set number of seats in each row is given to you. There are N chairs in each row, and the positions are all sequential. With the limitation that a family requires 4 adjacent seats, you must arrange the chairs so that the largest number of people may be accommodated.

For ease,

• The seats in a row are numbered from 1 to N.
• Each family occupies exactly 4 seats.
• Some seats may already be blocked or occupied, which means a family cannot sit there.
• The problem is to find the maximum number of families that can be seated across all rows.

Approach:

This problem can be solved using efficient algorithms to calculate the maximum number of families. The main considerations are:

1. Identify seat blocks: Divide rows into sections where 4 consecutive seats are available.
2. Constraints: Blocked or occupied seats, which limit possible family placements.
3. Optimizing placement: Greedily assign families to valid segments.

Input Format:

1. R: Number of rows.
2. N: Number of seats per row.
3. B: Number of blocked seats.
4. A list of blocked seats, where each entry specifies the row and seat index.

Algorithm:

1. Data Representation: It represents each row as a vector or bitmask to quickly identify blocked seats and free regions.
2. Valid Seat Configurations:
   • A family can sit in four seats at indices (x, x+1, x+2, x+3) if none of these are blocked.
   • If N < 4, it is impossible to seat any family.
3. Iterate Through Rows:
   For each row, calculate:
   • The total number of valid 4-seat groups.
   • Ensure blocked seats do not overlap with these groups.
4. Optimization:
   • Use bitmasks for faster checks of consecutive free seats.
   • For each row, attempt to maximize families.
5. Output: Total families that can be seated across all rows.

Example:

Let us take an example to illustrate the maximum cinema seat allocation in C++.

Output:

Input:

Output:

Maximum number of families that can be seated: 6   

Code Explanation:

1. Data Structures:
   • unordered_map<int, set<int>>: It stores blocked seats for each row.
   • vector<int>: It represents each row’s seating arrangement to mark free and blocked seats.
2. Key functions:
   • maxFamiliesInRow: It calculates the maximum number of families that can fit in one row based on blocked seats.
   • maxCinemaFamilies: It iterates through all rows and aggregates the maximum families across rows.
3. Input/Output:
   • Input: Rows, seats per row, and blocked seat positions.
   • Output: Maximum families that can be seated.

Complexity Analysis:

1. Time Complexity:
   • Iterating through rows: O(R)O(R)O(R).
   • Processing blocked seats for a row: O(B)O(B)O(B) (for set operations).
   • Checking configurations: O(N)O(N)O(N).
   • Overall: O(R+B+N)O(R + B + N)O(R+B+N).
2. Space Complexity:
   • Storage for blocked seats: O(B)O(B)O(B).
   • Row-wise storage for seat status: O(N)O(N)O(N).
   • Overall: O(B+N)O(B + N)O(B+N).

Example Walkthrough

Input:

• Rows: R=3R = 3R=3
• Seats per row: N=10N = 10N=10
• Blocked seats: B=4B = 4B=4 (positions: (1, 4), (1, 5), (2, 2), (3, 9))

Output:

Maximum families = 444

Explanation:

1. Row 1: Blocked seats (4, 5) prevent families in the middle block. Left and right blocks are available → 222 families.
2. Row 2: Blocked seat (2) does not affect placement. Both left and right blocks are free → 222 families.
3. Row 3: Blocked seat (9) affects the right block. Only left block is free → 111 family. Total families = 2+2+1=42 + 2 + 1 = 42+2+1=4.

Optimizations:

1. Bitmasking: Instead of storing rows as vectors, use bitmasks for efficient space and time.
2. Parallel Processing: Rows are independent; computations can be parallelized.
3. Early Termination: If a row has too many blocked seats, skip unnecessary checks.

Conclusion:

In conclusion, this solution describes an efficient and scalable algorithm in C++ for the “Maximum Cinema Seat Allocation” problem. Combining modular design with careful optimizations ensures that the program is efficient in handling real-world constraints, such as large seating arrangements and hundreds of blocked seats.