IPO Leetcode Problem 502 [Python Solution]

IPO LeetCode problem is a Heap / Priority Queue category and classified “Hard” challenge. We’ll provide a detailed explanation of the problem

Also, a Python solution, and a thorough analysis of time and space complexity.

So, if you’re ready to dive into the world of coding and algorithmic problem-solving, let’s get started!

Problem Overview

The Problem Statement

Suppose LeetCode is preparing for an Initial Public Offering (IPO).

To maximize its capital before the IPO, LeetCode needs to work on various projects.

However, there is a limit to the number of projects (k) that can be completed before the IPO.

Each project has associated pure profits and a minimum capital requirement.

LeetCode starts with an initial capital, denoted as W.

After completing a project, its pure profit is added to the total capital.

The goal is to select a list of at most k distinct projects from the given set of projects to maximize the final total capital.

The answer should be the maximized capital after choosing these projects.

The key constraint is that a project can only be selected if the available capital (W) is greater than or equal to the project’s minimum capital requirement.

The final maximized capital should fit within a 32-bit signed integer.

Example

Let’s take an example to illustrate the problem:

Input:

• k = 2
• w = 0
• profits = [1, 2, 3]
• capital = [0, 1, 1]

Output: 4

Explanation:

• Initially, your capital is 0, so you can only start the project indexed 0.
• After finishing project 0, you will obtain a profit of 1, and your capital becomes 1.
• With a capital of 1, you can either start project 1 or project 2.
• Since you can choose at most 2 projects, you need to finish project 2 to maximize your capital.
• Therefore, the final maximized capital is 0 + 1 + 3 = 4.

Constraints

Here are the constraints for this problem:

• 1 <= k <= 105
• 0 <= w <= 109
• The lengths of profits and capital arrays are the same, denoted as n.
• 1 <= n <= 105
• 0 <= profits[i] <= 104
• 0 <= capital[i] <= 109

Now that we understand the problem statement and its constraints, let’s proceed to explore potential solutions.

Understanding the Constraints

Before we dive into the solution, let’s discuss the constraints of this problem.

Understanding these constraints is essential when designing an efficient algorithm.

1. The value of k can be as large as 105. This means that we may need to select a considerable number of projects.

Our solution should be optimized for this scenario.

2. The initial capital w can be as large as 109. This indicates that we may start with a substantial amount of capital.

It’s crucial to make the most of this capital while considering the project constraints.

3. The lengths of the profits and capital arrays can be as large as 105. This implies that we need to handle a significant number of projects.

The solution should not have a time complexity that grows exponentially with the number of projects.

4. The profit values profits[i] can be as high as 104. These are the potential gains we can achieve by completing a project.

We should maximize these gains.

5. The capital requirements capital[i] can be as high as 109. These requirements determine whether we can undertake a project.

Efficiently selecting projects that fit within our capital is essential.

Now that we have a clear understanding of the problem and its constraints, let’s explore the solution.

IPO LeetCode Problem Solution

To solve this problem efficiently, we will use two priority queues: one for maximizing profits (a Max Heap) and another for minimizing capital requirements (a Min Heap).

This approach allows us to select projects with the highest profits while ensuring that we meet the capital requirements.

Here’s the Python solution:

def findMaximizedCapital(self, k: int, w: int, profits: List[int], capital: List[int]) -> int:
        # Create a Max Heap for maximizing profits
        maxProfit = []

        # Create a Min Heap for minimizing capital requirements
        minCapital = [(c, p) for c, p in zip(capital, profits)]
        heapq.heapify(minCapital)

        for i in range(k):
            # Select projects with capital requirements we can afford
            while minCapital and minCapital[0][0] <= w:
                c, p = heapq.heappop(minCapital)
                # Add the negative of profit to Max Heap for maximizing profits
                heapq.heappush(maxProfit, -p)

            if not maxProfit:
                break

            # Take the project with the highest profit
            w += -1 * heapq.heappop(maxProfit)

        return w

Let’s break down this solution step by step:

1. We initialize two heaps: maxProfit as a Max Heap (for maximizing profits) and minCapital as a Min Heap (for minimizing capital requirements).
2. We transform the project data into a list of tuples (capital, profit) and heapify it.

The tuples in minCapital will be used to track the projects with their respective capital requirements and profits.

3. We iterate k times, which represents the number of projects we can select.
4. Within each iteration, we check the projects in minCapital.

If the capital requirement of the project at the top of the Min Heap is less than or equal to our current capital w, we can take that project.

We remove it from minCapital and add its profit to maxProfit.

Note that we add the negative value of profit to maxProfit to turn it into a Max Heap.

5. If there are projects available in maxProfit (i.e., projects we can afford with our current capital), we pop the project with the highest profit (negative value), add it to our capital, and continue.
6. If maxProfit is empty, we break the loop since we’ve selected all available profitable projects.
7. Finally, we return the updated capital, which represents the maximized capital after selecting the projects.

This solution is efficient because it uses two heaps to keep track of available projects based on both profit and capital requirements.

It ensures that we always select the most profitable projects that meet our capital constraints.

Now, let’s analyze the time and space complexity of this solution.

Time and Space Complexity

Time Complexity

The time complexity of this solution is primarily determined by the heapq operations, which have a time complexity of O(log n) for both insertion and removal of elements.

We perform these operations for both maxProfit and minCapital.

Let’s break down the time complexity:

1. Converting the capital and profits arrays into a list of tuples and heapifying minCapital takes O(n * log n) time.
2. The main loop runs for k iterations, and in each iteration, we perform heap operations (insert and pop) on both `maxProfit

andminCapital`.

Each of these operations takes O(log n) time.

3. Since k can be at most 105 and the total number of projects (n) can also be at most 105, the worst-case time complexity of this solution is O(k * log n).

Space Complexity

The space complexity of this solution is determined by the space used to store the two heaps (maxProfit and minCapital) and the list of tuples.

1. The space used for maxProfit and minCapital is O(n) since both contain all the projects.
2. The space used for the list of tuples (minCapital) is O(n) as well.

Overall, the space complexity of this solution is O(n).

Reasoning Behind Our Approach

This approach is based on the idea of using two heaps to efficiently select projects while considering both profit and capital requirements.

By maintaining a Max Heap for profits and a Min Heap for capital, we can iteratively select the most profitable projects that we can afford.

This approach optimizes both time and space complexity and ensures we maximize our capital before the IPO.

In summary, the key insights of our approach are as follows:

1. Use a Max Heap to prioritize projects with higher profits.
2. Use a Min Heap to prioritize projects with lower capital requirements.
3. Iterate through projects, selecting the most profitable ones within our capital constraints.
4. Continue this process until we have selected the desired number of projects or there are no more profitable projects we can afford.

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Conclusion

In this guide, we’ve explored the IPO problem from LeetCode, a challenging problem that involves selecting projects to maximize capital before an IPO.

We’ve provided a Python solution that efficiently solves the problem by using two heaps (Max Heap and Min Heap) to prioritize projects based on profit and capital requirements.

This approach ensures that we select the most profitable projects while staying within our budget.

We’ve also analyzed the time and space complexity of our solution, taking into account the constraints of the problem.

This solution is well-suited for large datasets and allows us to handle a substantial number of projects efficiently.

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Happy coding!

Question Link: IPO LeetCode Problem

Now, it’s your turn to explore, practice, and apply this solution to more coding challenges.

Keep coding and stay curious!