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Maximum Profit of Operating a Centennial Wheel

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.length
  • 1 <= n <= 105
  • 0 <= customers[i] <= 50
  • 1 <= boardingCost, runningCost <= 100

Solution Explanation: Simulating the Centennial Wheel

This problem involves simulating the operation of a centennial wheel to maximize profit. The solution uses a simulation approach to track the number of customers, the profit, and the number of rotations.

Approach:

  1. Initialization: We initialize variables to track the profit (t), maximum profit (mx), waiting customers (wait), rotation count (i), and the result (ans, initially -1).

  2. Simulation Loop: The while loop continues as long as there are waiting customers or more customers to arrive. Inside the loop:

    • Customer Arrival: New customers are added to the wait variable.
    • Boarding: A maximum of 4 customers board the gondola (up = min(4, wait)).
    • Update Wait: The number of boarded customers is subtracted from the waiting list.
    • Profit Update: The profit t is updated by adding the boarding cost for the boarded customers and subtracting the running cost for the rotation.
    • Maximum Profit Check: If the current profit t exceeds the maximum profit mx, we update mx and store the current rotation count i in ans.
    • Increment Rotation: The rotation counter i is incremented.

  3. Return Value: After the loop completes, the function returns ans, representing the minimum number of rotations needed to achieve the maximum profit. If the maximum profit never becomes positive, -1 is returned.

Time Complexity: O(n), where n is the length of the customers array. The loop iterates through the array at most once.

Space Complexity: O(1). The algorithm uses a constant amount of extra space to store variables.

Code Examples (Python):

class Solution:
    def minOperationsMaxProfit(self, customers: List[int], boardingCost: int, runningCost: int) -> int:
        ans = -1
        mx = t = 0
        wait = 0
        i = 0
        while wait or i < len(customers):
            wait += customers[i] if i < len(customers) else 0
            up = min(4, wait)
            wait -= up
            t += up * boardingCost - runningCost
            i += 1
            if t > mx:
                mx = t
                ans = i
        return ans

The code in other languages (Java, C++, Go, TypeScript, Rust) follows a very similar structure, reflecting the same simulation logic. The only differences are in the syntax and specific library functions used (e.g., min function).