What is the time complexity of Maximum Profit of Operating a Centennial Wheel?

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

What data structures are used to solve Maximum Profit of Operating a Centennial Wheel?

This problem uses Array, Simulation. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Maximum Profit of Operating a Centennial Wheel?

Insight 1: Simulating the wheel's operation is crucial. We need to track the number of waiting customers, boarded customers, and rotations. Insight 2: The problem can be divided into two parts: processing the given customer array and then handling the remaining waiting customers (if any) after the array is processed.

Which companies ask LeetCode 1599. Maximum Profit of Operating a Centennial Wheel in interviews?

This problem is frequently asked at peak6.

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Maximum Profit of Operating a Centennial Wheel - LeetCode 1599 Solution

Maximum Profit of Operating a Centennial Wheel - Complete Solution Guide

Maximum Profit of Operating a Centennial Wheel is LeetCode problem 1599, 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

You are the operator of a Centennial Wheel that has four gondolas , and each gondola has room for up to four people . You have the ability to rotate the gondolas counterclockwise , which costs you runningCost dollars. You are given an array customers of length n where customers[i] is the number of new customers arriving just before the i th rotation (0-indexed). This means you must rotate the wheel i times before the customers[i] customers arrive . You cannot make customers wait if there is room

Detailed Explanation

The problem describes a Centennial Wheel operation with four gondolas, each holding a maximum of four people. Customers arrive before each rotation, and the goal is to determine the minimum number of rotations needed to achieve maximum profit. Profit is calculated as (number of boarded customers * boardingCost) - (number of rotations * runningCost). The wheel can be stopped at any time, making subsequent rotations free. The function should return the minimum number of rotations to maximize profit, or -1 if no positive profit can be achieved.

Solution Approach

The solution uses a simulation approach. It iterates through the `customers` array, updating the number of waiting customers and calculating the current profit after each rotation. It keeps track of the maximum profit encountered so far and the corresponding number of rotations. After processing the `customers` array, the solution handles any remaining waiting customers. It calculates the profit from full gondola rotations and partial gondola rotations and updates the maximum profit and rotations accordingly. The final result is the number of rotations that yielded the maximum profit.

Step-by-Step Algorithm

Step 1: Initialize `waiting_customers`, `rotations`, `current_profit`, `max_profit`, and `max_profit_rotations` to 0, 0, 0, 0, and -1 respectively.
Step 2: Iterate through the `customers` array. In each iteration:
Step 3: Increment `rotations` by 1.
Step 4: Add the current `new_customers` to `waiting_customers`.
Step 5: Calculate `boarded` customers as the minimum of `waiting_customers` and 4.
Step 6: Update `waiting_customers` by subtracting `boarded`.
Step 7: Update `current_profit` by adding `boarded * boardingCost - runningCost`.
Step 8: If `current_profit` is greater than `max_profit`, update `max_profit` and `max_profit_rotations`.
Step 9: After the loop, check if there are any `waiting_customers` remaining.
Step 10: If `waiting_customers` > 0, calculate `profit_per_full_gondola` as `4 * boardingCost - runningCost`.
Step 11: If `profit_per_full_gondola` > 0, calculate `full_rotations` as `waiting_customers // 4`.
Step 12: If `full_rotations` > 0, update `current_profit` by adding `full_rotations * profit_per_full_gondola` and `rotations` by adding `full_rotations`. Update `max_profit` and `max_profit_rotations` if necessary.
Step 13: Calculate `remaining_customers` as `waiting_customers % 4`.
Step 14: If `remaining_customers` > 0, calculate `profit_per_partial_gondola` as `remaining_customers * boardingCost - runningCost`.
Step 15: If `profit_per_partial_gondola` > 0, update `current_profit` by adding `profit_per_partial_gondola` and `rotations` by incrementing it. Update `max_profit` and `max_profit_rotations` if necessary.
Step 16: Return `max_profit_rotations`.

Key Insights

Insight 1: Simulating the wheel's operation is crucial. We need to track the number of waiting customers, boarded customers, and rotations.
Insight 2: The problem can be divided into two parts: processing the given customer array and then handling the remaining waiting customers (if any) after the array is processed.
Insight 3: Optimizing the number of rotations based on the profit derived from full gondola rotations and partial gondola rotations is key to finding the optimal solution after processing the given input customer array.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(1)

Topics

This problem involves: Array, Simulation.

Companies

Asked at: peak6.

Maximum Profit of Operating a Centennial Wheel | LeetCode 1599

Medium

Array Simulation

Time: O(n)Space: O(1)

Solution Code

Available in:

class Solution {
    public int minOperationsMaxProfit(int[] customers, int boardingCost, int runningCost) {
        int waiting_customers = 0;
        int rotations = 0;
        int current_profit = 0;
        
        int max_profit = 0;
        int max_profit_rotations = -1;
        
        for (int new_customers : customers) {
            rotations++;
            waiting_customers += new_customers;
            
            int boarded = Math.min(waiting_customers, 4);
            waiting_customers -= boarded;
            
            current_profit += boarded * boardingCost - runningCost;
            
            if (current_profit > max_profit) {
                max_profit = current_profit;
                max_profit_rotations = rotations;
            }
        }
        
        if (waiting_customers > 0) {
            int profit_per_full_gondola = 4 * boardingCost - runningCost;
            
            if (profit_per_full_gondola > 0) {
                int full_rotations = waiting_customers / 4;
                if (full_rotations > 0) {
                    current_profit += full_rotations * profit_per_full_gondola;
                    rotations += full_rotations;
                    if (current_profit > max_profit) {
                        max_profit = current_profit;
                        max_profit_rotations = rotations;
                    }
                }
                
                int remaining_customers = waiting_customers % 4;
                if (remaining_customers > 0) {
                    int profit_per_partial_gondola = remaining_customers * boardingCost - runningCost;
                    if (profit_per_partial_gondola > 0) {
                        current_profit += profit_per_partial_gondola;
                        rotations++;
                        if (current_profit > max_profit) {
                            max_profit = current_profit;
                            max_profit_rotations = rotations;
                        }
                    }
                }
            }
        }
        
        return max_profit_rotations;
    }
}

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Problem Info

Description

Problem 1599: Maximum Profit of Operating a Centennial Wheel - Detailed explanation and solution approaches.

Topics

Array Simulation

Companies

peak6

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:1599

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

16 steps

1
Step 1: Initialize `waiting_customers`, `rotations`, `current_profit`, `max_profit`, and `max_profit_rotations` to 0, 0, 0, 0, and -1 respectively.
2
Step 2: Iterate through the `customers` array. In each iteration:
3
Step 3: Increment `rotations` by 1.
4
Step 4: Add the current `new_customers` to `waiting_customers`.
5
Step 5: Calculate `boarded` customers as the minimum of `waiting_customers` and 4.
6
Step 6: Update `waiting_customers` by subtracting `boarded`.
7
Step 7: Update `current_profit` by adding `boarded * boardingCost - runningCost`.
8
Step 8: If `current_profit` is greater than `max_profit`, update `max_profit` and `max_profit_rotations`.
9
Step 9: After the loop, check if there are any `waiting_customers` remaining.
10
Step 10: If `waiting_customers` > 0, calculate `profit_per_full_gondola` as `4 * boardingCost - runningCost`.
11
Step 11: If `profit_per_full_gondola` > 0, calculate `full_rotations` as `waiting_customers // 4`.
12
Step 12: If `full_rotations` > 0, update `current_profit` by adding `full_rotations * profit_per_full_gondola` and `rotations` by adding `full_rotations`. Update `max_profit` and `max_profit_rotations` if necessary.
13
Step 13: Calculate `remaining_customers` as `waiting_customers % 4`.
14
Step 14: If `remaining_customers` > 0, calculate `profit_per_partial_gondola` as `remaining_customers * boardingCost - runningCost`.
15
Step 15: If `profit_per_partial_gondola` > 0, update `current_profit` by adding `profit_per_partial_gondola` and `rotations` by incrementing it. Update `max_profit` and `max_profit_rotations` if necessary.
16
Step 16: Return `max_profit_rotations`.

💡

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(n)

👍 Linear - Good! Visits each element once.

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The time complexity is determined by the single loop that iterates through the customer array. The space complexity is constant because the solution only uses a few integer variables to store intermediate results, regardless of the input array's size.

📖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

Dynamic Programming

While not very efficient for this problem due to the constraints, you could theoretically use dynamic programming to store the maximum profit for each number of rotations. This would involve a state like dp[i] = max profit after i rotations.

Time: O(n), where n is the total number of possible rotations (which could be much larger than the size of the input array).Space: O(n), to store the DP table.

Trade-offs: Dynamic Programming is overkill for this problem as the simulation approach provides sufficient performance with a simpler implementation. It is not suitable because the space complexity is higher.

✨

Key Insights

1Insight 1: Simulating the wheel's operation is crucial. We need to track the number of waiting customers, boarded customers, and rotations.
2Insight 2: The problem can be divided into two parts: processing the given customer array and then handling the remaining waiting customers (if any) after the array is processed.
3Insight 3: Optimizing the number of rotations based on the profit derived from full gondola rotations and partial gondola rotations is key to finding the optimal solution after processing the given input customer array.

📚

Key Terms Explained

For Students

Array

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

Edge Cases

Edge case 1: If no positive profit is possible at any point. The solution initializes `max_profit_rotations` to -1 and only updates it if a positive profit is encountered. Thus, if `max_profit` remains 0, the function correctly returns -1.
Edge case 2: If there are many remaining customers after processing the input array. The solution handles this by calculating and performing additional rotations based on full and partial gondola loads, optimizing for maximum profit.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not considering the remaining waiting customers after processing the `customers` array. The solution needs to handle these customers optimally.

✗Common mistake 2: Incorrectly calculating the profit by not subtracting the `runningCost` for each rotation. The code must accurately reflect the profit formula.

✗Common mistake 3: Not checking if a profitable rotation with the remaining customers can occur, which will result in not finding the optimal number of rotations.

✅

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: Optimizing resource allocation in queuing systems, where resources (gondolas) have limited capacity and customers arrive at different rates.

🏦Real-world scenario 2: Maximizing revenue in amusement parks or theme parks by strategically managing ride operations based on customer flow and operational costs.

💡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: November 7, 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

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJul 13

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jul 6

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

@AlgoMasterJun 29

❓

Frequently Asked Questions

When should I use in-place modification vs creating a new array?

Use in-place modification when space complexity matters (O(1) space requirement) and the original array can be modified. Create a new array when you need to preserve the original data or when the problem involves significant restructuring that would complicate in-place logic.

How much time should I spend on a medium problem in interviews?

In a 45-minute interview, aim for: 5-10 minutes understanding and clarifying, 5-10 minutes discussing approach, 20-25 minutes coding, and 5-10 minutes testing. If stuck after 15 minutes of thinking, verbalize your thought process and ask for hints.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Maximum Profit of Operating a Centennial Wheel"?

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 simulation techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

🤖 Need help with this problem?