IPO

Hard

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ArraysGreedy Algorithms

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 i^th 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

Try coding it in LeetCode

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What are the possible ranges for the initial capital, capital requirements, and profit values? Can they be negative?
Can the input arrays `capital` and `profits` be empty? If so, what should I return?
If I am unable to complete the maximum number of projects (`k`) due to insufficient capital, what should I return? Should I return the maximum capital I could achieve?
Are there any constraints on the length of the `capital` and `profits` arrays? Are they guaranteed to be the same length?
If multiple projects are available with the same capital requirement and would yield the same profit, is there a preferred way to select which one to prioritize?

Brute Force Solution

Approach

The brute force method for the IPO problem is like trying every single possible combination of projects to maximize profit. We explore every sequence of project selections within the available capital limit. It's a very thorough, but inefficient, way to find the best outcome.

Here's how the algorithm would work step-by-step:

Start with an empty list of chosen projects and your initial capital.
Consider each available project one at a time and see if you can afford it with your current capital.
If you can afford it, 'try' selecting that project. This means hypothetically add its profit to your capital.
After 'trying' a project, consider the remaining available projects and repeat the process of checking affordability and 'trying' projects until you've 'tried' all affordable projects up to the maximum allowed number of projects.
Go back and 'try' selecting a different initial project to see if a better combination exists.
Keep track of the total capital you would have after 'trying' all possible combinations of affordable projects within the allowed number of projects.
Compare the final capital from all the different combinations, and choose the combination that results in the highest final capital. This is your best possible outcome.

Code Implementation

def ipo_brute_force(maximum_number_of_projects, initial_capital, profits, capital_requirements):
    number_of_projects = len(profits)
    maximum_final_capital = initial_capital

    for i in range(1 << number_of_projects):
        # Generate all possible subsets of projects
        selected_projects = []
        for j in range(number_of_projects):
            if (i >> j) & 1:
                selected_projects.append(j)

        if len(selected_projects) > maximum_number_of_projects:
            continue

        current_capital = initial_capital
        number_of_projects_completed = 0

        # Simulate the project selection process
        for project_index in selected_projects:
            if current_capital >= capital_requirements[project_index]:
                current_capital += profits[project_index]
                number_of_projects_completed += 1
            else:
                current_capital = initial_capital -1
                break

        if current_capital > maximum_final_capital:
            # Update the maximum capital obtained so far
            maximum_final_capital = current_capital

    return maximum_final_capital

Big(O) Analysis

Time Complexity

O(n^k) – The brute force approach explores all possible combinations of projects. We have 'n' projects and can choose up to 'k' of them. The algorithm essentially generates all subsets of size up to 'k' from the set of 'n' projects. The number of such subsets can be approximated as n choose k (nCk), which expands to n! / (k! * (n-k)!). In the worst case, where k is close to n, the number of combinations grows rapidly, leading to a time complexity of O(n^k).

Space Complexity

O(K) – The brute force solution outlined involves exploring every possible combination of projects. In the worst-case, the algorithm might 'try' up to K projects, where K is the maximum allowed number of projects. Each 'try' might require storing information about the selected projects and their profits, leading to a potential recursion depth of K. Therefore, the auxiliary space used for the call stack could grow linearly with K, resulting in O(K) space complexity.

Optimal Solution

Approach

The core idea is to prioritize the most promising projects at each stage based on available capital. We want to maximize profits by intelligently selecting projects. It's a greedy approach, selecting the best option available at each step to reach the overall best result.

Here's how the algorithm would work step-by-step:

First, gather all potential projects and sort them into two groups: projects we can afford and projects we can't afford, based on our starting capital.
From the group of affordable projects, select the one that offers the highest profit. We'll call this the 'best' project.
Add the profit earned from the 'best' project to our existing capital. Now we have more money to invest.
Repeat the process. Update the groups of affordable and unaffordable projects based on our new capital.
Continue selecting the most profitable affordable projects and updating our capital until we have completed the desired number of projects.

Code Implementation

def maximize_capital(number_of_projects, initial_capital, project_capitals, project_profits):
    available_projects = []
    unavailable_projects = []

    for i in range(len(project_capitals)):
        unavailable_projects.append((project_capitals[i], project_profits[i]))

    current_capital = initial_capital

    for _ in range(number_of_projects):
        # Divide projects into affordable and unaffordable groups.
        while unavailable_projects and unavailable_projects[0][0] <= current_capital:
            available_projects.append(unavailable_projects.pop(0))

        unavailable_projects.sort()

        if not available_projects:
            break

        # Select the most profitable project from affordable projects.
        best_project_profit = 0
        best_project_index = -1
        for i in range(len(available_projects)):
            if available_projects[i][1] > best_project_profit:
                best_project_profit = available_projects[i][1]
                best_project_index = i

        # Add the profit to the current capital.
        current_capital += best_project_profit

        del available_projects[best_project_index]

    return current_capital

Big(O) Analysis

Time Complexity

O(n log n) – The algorithm involves sorting the projects initially, which takes O(n log n) time, where n is the number of projects. The main loop iterates k times (at most n). Inside the loop, finding affordable projects may take O(n) in the worst case. Selecting the most profitable project from affordable projects could take O(n) if we don't use a heap. However, using a max heap to maintain affordable project profits allows us to extract the maximum profit in O(log n). Rebuilding the affordable projects each step still will amount to O(n log n) in the worst case. Therefore, the overall time complexity is dominated by the sorting and heap operations, resulting in O(n log n).

Space Complexity

O(N) – The algorithm maintains two groups of projects: affordable and unaffordable. In the worst case, all N projects could initially be stored in either the affordable or unaffordable group before being processed. Therefore, the space required to store these groups scales linearly with the number of projects N. Consequently, the auxiliary space complexity is O(N).

Edge Cases

Case	How to Handle
Empty capital or profits array	Return 0 as no projects are possible with empty inputs.
k is 0, meaning no projects can be executed	Return initial capital W, as no investments are possible.
Capital array size does not match profits array size	Throw an IllegalArgumentException or return an error code to indicate invalid input.
All capital requirements are higher than the initial capital	Return the initial capital as no projects can be executed.
k is larger than the number of available projects	Execute all possible projects by iterating through all project options instead of just k times.
Profits contain negative numbers	The algorithm should still work correctly by adding the negative profits to the current capital.
Capital values are extremely large leading to potential overflow	Use long data types for calculations involving capital and profits to prevent integer overflow.
Capital and Profits arrays contain duplicate entries that could change order	The priority queue handles duplicate entries correctly and the algorithm will still produce optimal result.

Empty capital or profits array

How to Handle:

Return 0 as no projects are possible with empty inputs.

k is 0, meaning no projects can be executed

How to Handle:

Return initial capital W, as no investments are possible.

Capital array size does not match profits array size

How to Handle:

Throw an IllegalArgumentException or return an error code to indicate invalid input.

All capital requirements are higher than the initial capital

How to Handle:

Return the initial capital as no projects can be executed.

k is larger than the number of available projects

How to Handle:

Execute all possible projects by iterating through all project options instead of just k times.

Profits contain negative numbers

How to Handle:

The algorithm should still work correctly by adding the negative profits to the current capital.

Capital values are extremely large leading to potential overflow

How to Handle:

Use long data types for calculations involving capital and profits to prevent integer overflow.

Capital and Profits arrays contain duplicate entries that could change order

How to Handle:

The priority queue handles duplicate entries correctly and the algorithm will still produce optimal result.