Watch 9 video solutions for Maximum Profit of Operating a Centennial Wheel, a medium level problem involving Array, Simulation. This walkthrough by Fraz has 2,503 views views. Want to try solving it yourself? Practice on FleetCode or read the detailed text solution.
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 ith 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 in the gondola. Each customer pays boardingCost dollars when they board on the gondola closest to the ground and will exit once that gondola reaches the ground again.
You can stop the wheel at any time, including before serving all customers. If you decide to stop serving customers, all subsequent rotations are free in order to get all the customers down safely. Note that if there are currently more than four customers waiting at the wheel, only four will board the gondola, and the rest will wait for the next rotation.
Return the minimum number of rotations you need to perform to maximize your profit. If there is no scenario where the profit is positive, return -1.
Example 1:
Input: customers = [8,3], boardingCost = 5, runningCost = 6 Output: 3 Explanation: The numbers written on the gondolas are the number of people currently there. 1. 8 customers arrive, 4 board and 4 wait for the next gondola, the wheel rotates. Current profit is 4 * $5 - 1 * $6 = $14. 2. 3 customers arrive, the 4 waiting board the wheel and the other 3 wait, the wheel rotates. Current profit is 8 * $5 - 2 * $6 = $28. 3. The final 3 customers board the gondola, the wheel rotates. Current profit is 11 * $5 - 3 * $6 = $37. The highest profit was $37 after rotating the wheel 3 times.
Example 2:
Input: customers = [10,9,6], boardingCost = 6, runningCost = 4 Output: 7 Explanation: 1. 10 customers arrive, 4 board and 6 wait for the next gondola, the wheel rotates. Current profit is 4 * $6 - 1 * $4 = $20. 2. 9 customers arrive, 4 board and 11 wait (2 originally waiting, 9 newly waiting), the wheel rotates. Current profit is 8 * $6 - 2 * $4 = $40. 3. The final 6 customers arrive, 4 board and 13 wait, the wheel rotates. Current profit is 12 * $6 - 3 * $4 = $60. 4. 4 board and 9 wait, the wheel rotates. Current profit is 16 * $6 - 4 * $4 = $80. 5. 4 board and 5 wait, the wheel rotates. Current profit is 20 * $6 - 5 * $4 = $100. 6. 4 board and 1 waits, the wheel rotates. Current profit is 24 * $6 - 6 * $4 = $120. 7. 1 boards, the wheel rotates. Current profit is 25 * $6 - 7 * $4 = $122. The highest profit was $122 after rotating the wheel 7 times.
Example 3:
Input: customers = [3,4,0,5,1], boardingCost = 1, runningCost = 92 Output: -1 Explanation: 1. 3 customers arrive, 3 board and 0 wait, the wheel rotates. Current profit is 3 * $1 - 1 * $92 = -$89. 2. 4 customers arrive, 4 board and 0 wait, the wheel rotates. Current profit is 7 * $1 - 2 * $92 = -$177. 3. 0 customers arrive, 0 board and 0 wait, the wheel rotates. Current profit is 7 * $1 - 3 * $92 = -$269. 4. 5 customers arrive, 4 board and 1 waits, the wheel rotates. Current profit is 11 * $1 - 4 * $92 = -$357. 5. 1 customer arrives, 2 board and 0 wait, the wheel rotates. Current profit is 13 * $1 - 5 * $92 = -$447. The profit was never positive, so return -1.
Constraints:
n == customers.length1 <= n <= 1050 <= customers[i] <= 501 <= boardingCost, runningCost <= 100Problem Overview: You operate a Centennial Wheel where each rotation boards up to 4 customers. Every boarded customer pays a boarding cost, while each rotation incurs a running cost. Given an array representing arriving customers per rotation, determine the rotation number that produces the maximum profit. If the wheel never becomes profitable, return -1.
Approach 1: Greedy Approach for Immediate Boarding (O(n) time, O(1) space)
This approach simulates the wheel operation rotation by rotation. Track the number of waiting customers, board up to four each rotation, and update profit using profit += boarded * boardingCost - runningCost. Maintain variables for current profit, maximum profit seen so far, and the rotation where that maximum occurs. Continue processing while customers are arriving or still waiting in the queue.
The greedy insight: boarding as many customers as possible immediately always maximizes short-term profit because revenue is proportional to boarded riders. The simulation uses simple counters and an iteration over the arrival array, which makes it a classic simulation problem combined with basic array traversal.
Approach 2: Mathematical Computation and Optimization (O(n) time, O(1) space)
This version still iterates through arrivals but focuses on profit deltas and avoids unnecessary rotations once continuing would only reduce profit. Track cumulative customers and determine whether additional rotations after processing the array remain profitable. If remaining waiting customers can fill profitable rotations (where 4 * boardingCost > runningCost), compute those rotations mathematically rather than simulating one-by-one.
The optimization reduces redundant simulation when many customers remain after the input array ends. Instead of repeatedly boarding four customers until the queue empties, compute how many full profitable rotations exist and update profit directly. This keeps the implementation concise while maintaining constant memory usage.
Both solutions rely on simple counters and profit tracking, making them natural examples of greedy decision making combined with simulation.
Recommended for interviews: The greedy simulation approach is the expected solution. It clearly models the wheel operation and demonstrates control over queue management, state tracking, and incremental profit computation. The mathematical optimization is a nice improvement once the simulation logic is correct, but interviewers mainly want to see the O(n) simulation with careful handling of remaining customers.
| Approach | Time | Space | When to Use |
|---|---|---|---|
| Greedy Approach for Immediate Boarding | O(n) | O(1) | General case. Best for interviews and easiest to implement using direct simulation. |
| Mathematical Computation and Optimization | O(n) | O(1) | When many customers remain after processing the input array and you want to compute profitable rotations directly. |