Maximum Profit of Operating a Centennial Wheel

Maximum Profit of Operating a Centennial Wheel

1. What is this problem about?

The Maximum Profit of Operating a Centennial Wheel interview question is a simulation problem. You operate a Ferris wheel with 4 gondolas. Each rotation costs a certain amount, and each passenger pays a certain amount. You are given a sequence of groups of people arriving at each rotation.

The rules:

1. At each rotation, you can load at most 4 people from the waiting line into one gondola.
2. You only gain profit when people board.
3. You must decide when to stop the wheel to maximize your total profit.

The goal is to find the minimum number of rotations needed to reach the maximum profit. If no profit is possible, return -1.

2. Why is this asked in interviews?

This Maximum Profit of Operating a Centennial Wheel coding problem tests your ability to carefully follow rules and implement a step-by-step simulation. It evaluates how you manage state (the number of people waiting, the total profit, and the rotation count) over time. While not algorithmically complex, it requires attention to detail and correct handling of "leftover" people who arrive after the initial input sequence ends.

3. Algorithmic pattern used

The Array, Simulation interview pattern is applied.

1. Maintain variables for waiting_customers, total_profit, max_profit, and best_rotation.
2. Iterate through the arrival array, adding arriving people to waiting_customers.
3. In each step (rotation):
   • Take min(4, waiting_customers) people.
   • Update waiting_customers.
   • Update total_profit += boarded * boarding_cost - running_cost.
   • Update max_profit and best_rotation if the current profit is strictly greater than the previous max.
4. Continue the simulation as long as there are people in the arrival array OR people still waiting in line.

4. Example explanation

Boarding cost: 10, Running cost: 20. Arrivals: [10, 0, 0].

1. Rotation 1: 10 arrive. Board 4. Waiting 6. Profit: (4*10)-20 = 20. Max=20, Best=1.
2. Rotation 2: 0 arrive. Board 4. Waiting 2. Profit: 20 + (4*10)-20 = 40. Max=40, Best=2.
3. Rotation 3: 0 arrive. Board 2. Waiting 0. Profit: 40 + (2*10)-20 = 40. (Not strictly greater, so Best stays 2).

Max Profit: 40 at 2 rotations.

5. Common mistakes candidates make

In the Maximum Profit of Operating a Centennial Wheel coding problem, candidates often forget to continue the simulation after the arrival array is exhausted but people are still waiting in line. Another error is not correctly handling the "strictly greater" condition for the best rotation. Some might also struggle with the loop condition, potentially leading to an infinite loop if they don't decrement the waiting count correctly.

6. Interview preparation tip

For simulation problems, write down the state variables you need before you start coding. Clear variable names like boarded_this_turn or current_profit make the logic much easier to follow and debug. Always consider the "draining" phase of a simulation where the input stops but the process continues.