How do you solve Cinema Seat Allocation on LeetCode?

Cinema Seat Allocation (LeetCode #1386) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Understanding how to maximize the number of groups by checking configurations is crucial.

What is the brute force solution for Cinema Seat Allocation?

The brute force approach for Cinema Seat Allocation is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves checking each row for possible groups of four adjacent seats. We will iterate through the reserved seats and determine how many groups can be formed based on the available seats in each row. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Cinema Seat Allocation?

Cinema Seat Allocation uses the following concepts: Array, Hash Table, Greedy, Bit Manipulation. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Cinema Seat Allocation?

Always clarify the problem constraints and edge cases with the interviewer. Explain your thought process clearly while coding, as it helps the interviewer understand your approach. Consider discussing both brute force and optimal solutions to show your understanding of trade-offs.

What are common mistakes when solving Cinema Seat Allocation?

Not considering the aisle separation when counting groups. Failing to account for the fact that two families can occupy the same row under certain conditions.

#1386

Cinema Seat Allocation

Medium

ArrayHash TableGreedyBit ManipulationHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

The optimal solution uses a greedy approach to maximize the number of four-person groups by checking the reserved seats in each row and determining the best possible allocation for two families per row.

⚙️

Algorithm

3 steps

1Step 1: Create a set to track reserved seats.
2Step 2: For each row, check the reserved seats and determine how many groups can be formed based on the available seats.
3Step 3: Use conditions to check for the three possible configurations of groups in a row and count them accordingly.

solution.py15 lines

1# Full working Python code
2class Solution:
3    def maxNumberOfFamilies(self, n: int, reservedSeats: List[List[int]]) -> int:
4        reserved = set((r, s) for r, s in reservedSeats)
5        total_groups = 0
6        for row in range(1, n + 1):
7            groups = 0
8            if (row, 2) not in reserved and (row, 3) not in reserved and (row, 4) not in reserved and (row, 5) not in reserved:
9                groups += 1
10            if (row, 6) not in reserved and (row, 7) not in reserved and (row, 8) not in reserved and (row, 9) not in reserved:
11                groups += 1
12            if (row, 4) not in reserved and (row, 5) not in reserved and (row, 6) not in reserved and (row, 7) not in reserved:
13                groups += 1
14            total_groups += groups
15        return total_groups

ℹ

Complexity note: This complexity is linear because we only iterate through the reserved seats and check each row once, making it efficient even for larger inputs.

1Understanding how to maximize the number of groups by checking configurations is crucial.
2Using a set to track reserved seats allows for O(1) lookups, making the solution efficient.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.