Introduction

In this blog post, we will explore how to solve the LeetCode problem "502" 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

Suppose LeetCode will start its IPO soon. In order to sell a good price of its shares to Venture Capital, LeetCode would like to work on some projects to increase its capital before the IPO. Since it has limited resources, it can only finish at most k distinct projects before the IPO. Help LeetCode design the best way to maximize its total capital after finishing at most k distinct projects. You are given n projects where the ith project has a pure profit profits[i] and a minimum capital of capital[i] is needed to start it. Initially, you have w capital. When you finish a project, you will obtain its pure profit and the profit will be added to your total capital. Pick a list of at most k distinct projects from given projects to maximize your final capital, and return the final maximized capital. The answer is guaranteed to fit in a 32-bit signed integer. Example 1: Input: k = 2, w = 0, profits = [1,2,3], capital = [0,1,1] Output: 4 Explanation: Since your initial capital is 0, you can only start the project indexed 0. After finishing it you will obtain profit 1 and your capital becomes 1. With capital 1, you can either start the project indexed 1 or the project indexed 2. Since you can choose at most 2 projects, you need to finish the project indexed 2 to get the maximum capital. Therefore, output the final maximized capital, which is 0 + 1 + 3 = 4. Example 2: Input: k = 3, w = 0, profits = [1,2,3], capital = [0,1,2] Output: 6 Constraints: 1 <= k <= 105 0 <= w <= 109 n == profits.length n == capital.length 1 <= n <= 105 0 <= profits[i] <= 104 0 <= capital[i] <= 109

Explanation

Here's a breakdown of the solution:

• Greedy Approach: At each step, select the most profitable project that can be started with the current capital.
• Priority Queue (Max Heap): Use a max heap to efficiently find the most profitable project among the available ones.

• Two Priority Queues: Use one priority queue to store the projects that can be started currently based on capital, and another one to store the projects that haven't been started yet.

• Runtime Complexity: O(n log n + k log n), where n is the number of projects and k is the maximum number of projects to be selected.

• Storage Complexity: O(n)

Code

    import heapq

def findMaximizedCapital(k: int, w: int, profits: list[int], capital: list[int]) -> int:
    """
    Maximizes the total capital after finishing at most k distinct projects.

    Args:
        k: The maximum number of distinct projects that can be finished.
        w: The initial capital.
        profits: A list of pure profits for each project.
        capital: A list of minimum capital required for each project.

    Returns:
        The final maximized capital.
    """

    n = len(profits)
    projects = sorted(zip(capital, profits))  # Sort projects based on required capital

    available_projects = []  # Max heap to store available projects (profit, capital)
    project_index = 0

    for _ in range(k):
        # Add projects that can be started with current capital to the heap
        while project_index < n and projects[project_index][0] <= w:
            heapq.heappush(available_projects, -projects[project_index][1])  # Negate profit for max heap
            project_index += 1

        # If no project can be started, break
        if not available_projects:
            break

        # Select the most profitable project and update capital
        w += -heapq.heappop(available_projects)

    return w