What is the time complexity of Cinema Seat Allocation?

The optimal solution has a time complexity of O(m).

What data structures are used to solve Cinema Seat Allocation?

This problem uses Array, Hash Table, Greedy, Bit Manipulation. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Cinema Seat Allocation?

Insight 1: The large constraint on 'n' (up to 10^9) suggests that we cannot iterate through all rows. Instead, we only need to consider rows that have reserved seats. Insight 2: We can use a hash table (or map) to store the reserved seats for each row. This allows us to quickly check if a seat is reserved in a specific row.

Which companies ask LeetCode 1386. Cinema Seat Allocation in interviews?

This problem is frequently asked at Geico, Zoho.

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Cinema Seat Allocation - LeetCode 1386 Solution

Cinema Seat Allocation - Complete Solution Guide

Cinema Seat Allocation is LeetCode problem 1386, a Medium level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

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

Detailed Explanation

The problem asks us to find the maximum number of four-person groups that can be seated in a cinema with 'n' rows of seats, where each row has 10 seats numbered from 1 to 10. We are given an array 'reservedSeats' containing the row and seat number of reserved seats. A four-person group must occupy four adjacent seats in a single row. Seats are considered adjacent unless split by an aisle (between seats 3 and 4, and between seats 7 and 8), with the exception that the aisle may split the group in the middle (2 people on each side). The goal is to maximize the number of four-person groups we can seat given the reserved seats.

Solution Approach

The solution uses a hash table to store the reserved seats for each row. It then iterates through the rows with reserved seats and checks if it's possible to seat a four-person group in any of the three possible positions (left, middle, and right). First, calculate the number of rows where it's guaranteed to have 2 four-person groups available (unreserved rows). Then, iterate through the rows containing reservations. For each reserved row, check if each of the left, middle, and right positions are available and calculate the possible groups.

Step-by-Step Algorithm

Step 1: Create a hash table (or map) to store the reserved seats for each row. The key is the row number, and the value is a set of reserved seat numbers in that row.
Step 2: Iterate through the 'reservedSeats' array and populate the hash table.
Step 3: Calculate the number of rows that are completely free (no reservations). For these rows, we can seat two four-person groups per row. This value will initialize the 'result' variable.
Step 4: Iterate through the rows that have reserved seats (the keys in the hash table).
Step 5: For each row with reserved seats, check if it's possible to seat a four-person group in the left ([2, 3, 4, 5]), middle ([4, 5, 6, 7]), and right ([6, 7, 8, 9]) positions.
Step 6: If both left and right positions are available, add 2 to the 'result'. Otherwise, if either left, middle, or right positions are available, add 1 to the 'result'.
Step 7: Return the 'result'.

Key Insights

Insight 1: The large constraint on 'n' (up to 10^9) suggests that we cannot iterate through all rows. Instead, we only need to consider rows that have reserved seats.
Insight 2: We can use a hash table (or map) to store the reserved seats for each row. This allows us to quickly check if a seat is reserved in a specific row.
Insight 3: For each row, we only need to check three possible positions for seating a four-person group: seats [2, 3, 4, 5], [4, 5, 6, 7], and [6, 7, 8, 9].

Complexity Analysis

Time Complexity: O(m)

Space Complexity: O(m)

Topics

This problem involves: Array, Hash Table, Greedy, Bit Manipulation.

Companies

Asked at: Geico, Zoho.

Cinema Seat Allocation | LeetCode 1386

Medium

Array Hash Table Greedy Bit Manipulation

Time: O(m)Space: O(m)

Solution Code

Available in:

import java.util.HashMap;
import java.util.HashSet;
import java.util.Map;
import java.util.Set;

class Solution {
    public int maxNumberOfFamilies(int n, int[][] reservedSeats) {
        Map<Integer, Set<Integer>> reservedRows = new HashMap<>();
        for (int[] seat : reservedSeats) {
            int row = seat[0];
            int col = seat[1];
            if (!reservedRows.containsKey(row)) {
                reservedRows.put(row, new HashSet<>());
            }
            reservedRows.get(row).add(col);
        }
        
        int result = (n - reservedRows.size()) * 2;
        
        for (int rowKey : reservedRows.keySet()) {
            Set<Integer> reservedCols = reservedRows.get(rowKey);
            
            boolean leftPossible = true;
            for (int col : new int[]{2, 3, 4, 5}) {
                if (reservedCols.contains(col)) {
                    leftPossible = false;
                    break;
                }
            }
            
            boolean middlePossible = true;
            for (int col : new int[]{4, 5, 6, 7}) {
                if (reservedCols.contains(col)) {
                    middlePossible = false;
                    break;
                }
            }
            
            boolean rightPossible = true;
            for (int col : new int[]{6, 7, 8, 9}) {
                if (reservedCols.contains(col)) {
                    rightPossible = false;
                    break;
                }
            }
            
            if (leftPossible && rightPossible) {
                result += 2;
            } else if (leftPossible || middlePossible || rightPossible) {
                result += 1;
            }
        }
        
        return result;
    }
}

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Problem Info

Description

Problem 1386: Cinema Seat Allocation - Detailed explanation and solution approaches.

Topics

Array Hash Table Greedy Bit Manipulation

Companies

GeicoZoho

Stats

Difficulty:Medium

Languages:4

Companies:2

Problem #:1386

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

7 steps

1
Step 1: Create a hash table (or map) to store the reserved seats for each row. The key is the row number, and the value is a set of reserved seat numbers in that row.
2
Step 2: Iterate through the 'reservedSeats' array and populate the hash table.
3
Step 3: Calculate the number of rows that are completely free (no reservations). For these rows, we can seat two four-person groups per row. This value will initialize the 'result' variable.
4
Step 4: Iterate through the rows that have reserved seats (the keys in the hash table).
5
Step 5: For each row with reserved seats, check if it's possible to seat a four-person group in the left ([2, 3, 4, 5]), middle ([4, 5, 6, 7]), and right ([6, 7, 8, 9]) positions.
6
Step 6: If both left and right positions are available, add 2 to the 'result'. Otherwise, if either left, middle, or right positions are available, add 1 to the 'result'.
7
Step 7: Return the 'result'.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(m)

💾

Space Complexity

O(m)

📝

Why?

The time complexity is dominated by the iterations through the reserved seats to construct the hash table and then iterate through the distinct rows which contains a reserved seat. The space complexity is determined by the storage required for the hash table, which in the worst case, stores all reserved seats across all the distinct rows.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Bit Manipulation

Represent each row's reserved seats as a bitmask. Then, use bitwise operations to check for available seating arrangements. This approach can be more efficient for checking seating arrangements within a row.

Time: O(m)Space: O(n) or O(m) depending on how rows are represented. O(n) is you track bitmasks for all rows. O(m) if only rows with reserved seats are tracked.

Trade-offs: More complex to implement, but can be more efficient for determining the availability of seats in a row. Might be a better approach for very large datasets.

✨

Key Insights

1Insight 1: The large constraint on 'n' (up to 10^9) suggests that we cannot iterate through all rows. Instead, we only need to consider rows that have reserved seats.
2Insight 2: We can use a hash table (or map) to store the reserved seats for each row. This allows us to quickly check if a seat is reserved in a specific row.
3Insight 3: For each row, we only need to check three possible positions for seating a four-person group: seats [2, 3, 4, 5], [4, 5, 6, 7], and [6, 7, 8, 9].

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Hash Table

A data structure that maps keys to values for quick lookup. Like a dictionary!

Greedy Algorithm

Making the best choice at each step without looking ahead.

Edge Cases

Edge case 1: No reserved seats. In this case, the algorithm correctly calculates the maximum number of families as n * 2.
Edge case 2: All seats are reserved. In this case, the algorithm correctly calculates the maximum number of families as 0.
Edge case 3: Only some seats are reserved in some rows, the algorithm needs to check each possibility in each of those rows.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Incorrectly calculating the number of available groups in a row with reserved seats. It's important to consider all three possible positions (left, middle, and right) and the overlap between them.

✗Common mistake 2: Iterating through all possible rows (1 to n) even if they don't have any reserved seats. This leads to unnecessary computations and increased time complexity. Remember to only iterate through rows that actually have reservations.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Seat allocation systems in movie theaters, airplanes, and concert halls. The algorithm can be used to optimize the seating arrangement to maximize the number of groups that can be seated together.

🏦Real-world scenario 2: Resource allocation in cloud computing. The algorithm can be used to allocate contiguous blocks of memory or processing power to different tasks, taking into account any reserved or unavailable resources.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

Medium problems are the bread and butter of technical interviews. Practice explaining your approach before coding.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: October 5, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJul 14

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJul 7

❓

Frequently Asked Questions

What is the most efficient way to solve array problems?

Most array problems can be solved efficiently using techniques like two-pointer approach, sliding window, or hash maps. The key is identifying whether the array is sorted (enabling binary search) or if tracking seen elements (using hash maps) can help achieve O(n) time complexity.

What are the trade-offs of using a hash table?

Hash tables provide O(1) average-case lookup, insertion, and deletion, making them ideal for frequency counting and duplicate detection. However, they use O(n) extra space and may degrade to O(n) time in worst-case scenarios with hash collisions. Always consider if the space trade-off is acceptable.

What techniques are commonly tested in medium-difficulty problems?

Medium problems often require combining multiple techniques: two-pointer with hash maps, BFS/DFS with state tracking, dynamic programming with optimization, or divide-and-conquer approaches. They test your ability to recognize patterns and choose the right combination of techniques.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Cinema Seat Allocation"?

For this medium-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether hash table techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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