Introduction

In this blog post, we will explore how to solve the LeetCode problem "1599" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

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 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

Explanation

Here's the breakdown of the problem and the optimal solution:

• High-Level Approach:

  • Iterate through the customer array, simulating rotations and calculating the profit at each step.
  • Keep track of the maximum profit seen so far and the corresponding number of rotations.
  • After processing all given customer values, continue rotations until all waiting customers have boarded.

• Complexity:

  • Runtime Complexity: O(n + m), where n is the length of the customers array and m is the total number of additional rotations needed to board all waiting customers. In the worst case m could be large if many customers are left waiting after processing the initial array.
  • Storage Complexity: O(1) - We use only a few constant extra variables.

Code

    def min_operations_max_profit(customers, boardingCost, runningCost):
    """
    Calculates the minimum rotations to maximize profit for a Centennial Wheel.

    Args:
        customers (list[int]): Number of new customers arriving before each rotation.
        boardingCost (int): Cost per customer boarding the gondola.
        runningCost (int): Cost per rotation of the wheel.

    Returns:
        int: The minimum number of rotations needed to maximize profit, or -1 if no positive profit is possible.
    """

    waiting = 0
    boarded = 0
    rotations = 0
    max_profit = -1
    best_rotations = -1

    for i in range(len(customers)):
        waiting += customers[i]
        board = min(4, waiting)
        boarded += board
        waiting -= board
        rotations += 1
        profit = boarded * boardingCost - rotations * runningCost

        if profit > max_profit:
            max_profit = profit
            best_rotations = rotations

    # Handle remaining waiting customers
    while waiting > 0:
        board = min(4, waiting)
        boarded += board
        waiting -= board
        rotations += 1
        profit = boarded * boardingCost - rotations * runningCost

        if profit > max_profit:
            max_profit = profit
            best_rotations = rotations

    if max_profit <= 0:
        return -1
    else:
        return best_rotations