How do you solve Maximum Profit of Operating a Centennial Wheel on LeetCode?

Maximum Profit of Operating a Centennial Wheel (LeetCode #1599) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: The maximum number of customers that can board in one rotation is 4.

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

The optimal solution for Maximum Profit of Operating a Centennial Wheel runs in O(n) time and uses O(1) space. The time complexity is O(n) because we iterate through the customers array only once, maintaining a running total of customers and profit.

What is the brute force solution for Maximum Profit of Operating a Centennial Wheel?

The brute force approach for Maximum Profit of Operating a Centennial Wheel is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach simulates every possible number of rotations and calculates the profit for each scenario. This method is straightforward but inefficient, as it checks every combination of rotations and customers. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Maximum Profit of Operating a Centennial Wheel?

Maximum Profit of Operating a Centennial Wheel uses the following concepts: Array, Simulation. The recommended patterns to study are: Sliding Window, Greedy.

What are the interview tips for Maximum Profit of Operating a Centennial Wheel?

Always clarify the problem constraints and edge cases. Think about how to optimize your solution from the start. Practice simulating the problem with small examples before coding.

What are common mistakes when solving Maximum Profit of Operating a Centennial Wheel?

Not accounting for the maximum capacity of gondolas. Failing to break early when profit becomes negative.

#1599

Maximum Profit of Operating a Centennial Wheel

Medium

ArraySimulationSliding WindowGreedy

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(1)

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Intuition

Time O(n)Space O(1)

The optimal approach uses a sliding window technique to efficiently calculate profits while keeping track of customers waiting and the number of rotations. This reduces the need for nested loops and allows us to find the maximum profit in linear time.

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Algorithm

4 steps

1Step 1: Initialize variables for total customers, current profit, maximum profit, and the number of rotations.
2Step 2: Use a sliding window to iterate through the customers array, updating the total customers and profit for each rotation.
3Step 3: If profit exceeds maximum profit, update the maximum profit and rotations.
4Step 4: If at any point the profit becomes negative, break out of the loop as further rotations will not help.

solution.py17 lines

1def maxProfit(customers, boardingCost, runningCost):
2    n = len(customers)
3    total_customers = 0
4    current_profit = 0
5    max_profit = float('-inf')
6    rotations = 0
7    for i in range(n):
8        total_customers += customers[i]
9        served = min(total_customers, 4)
10        total_customers -= served
11        current_profit = served * boardingCost - (i + 1) * runningCost
12        if current_profit > max_profit:
13            max_profit = current_profit
14            rotations = i + 1
15        if current_profit < 0:
16            break
17    return rotations if max_profit > 0 else -1

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Complexity note: The time complexity is O(n) because we iterate through the customers array only once, maintaining a running total of customers and profit.

1The maximum number of customers that can board in one rotation is 4.
2Profit must be calculated carefully considering both boarding and running costs.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.