Ipo - Leetcode Solution

Note: This solution assumes understanding of IPO problem statement and is for educational purposes only.

The IPO problem on LeetCode is a greedy algorithm problem that can be solved using a priority queue. The problem statement is as follows:

You are given a set of investment projects with their respective profits, and a way to split them into smaller projects. You are also given an initial capital starting from 0.

At each step, you can choose to invest in one of the available projects, with the goal of maximizing your total capital. However, you cannot invest in a project if your current capital is less than the project's minimum capital required.

You are allowed to undertake multiple projects simultaneously, but the total initial cost of all the projects cannot exceed your available capital.

Write a function to find the maximum capital you can obtain.

The function definition is: def findMaximizedCapital(k: int, w: int, profits: List[int], capital: List[int]) -> int:

The inputs are:

• k: the maximum number of projects you can undertake simultaneously.
• w: your initial capital.
• profits: a list of integers representing the profits of the investment projects.
• capital: a list of integers representing the minimum capital required to invest in each project.

The output should be the maximum capital you can obtain after investing in k projects.

The first step in solving this problem is to create a list of tuples for each project, where each tuple contains the project's minimum capital required and its corresponding profit.

projects = [(c, p) for c, p in zip(capital, profits)]

Next, we sort the projects in order of increasing minimum capital required.

projects.sort()

To keep track of the available projects, we use two priority queues:

• a max heap (profits_heap) to keep track of the profits of available projects, and
• a min heap (capital_heap) to keep track of the minimum capital required for each project.

We start by adding all the projects with a minimum capital less than or equal to w to the profits_heap.

profits_heap = [] capital_heap = [] for i in range(len(projects)): if projects[i][0] <= w: heappush(profits_heap, -projects[i][1]) else: heappush(capital_heap, projects[i])

Next, we iterate k times and at each iteration we choose the project with the highest profit from the profits_heap and add its profit to w. We then add all the projects that require capital less than or equal to w to the profits_heap.

for i in range(k): if len(profits_heap) == 0: break p = -heappop(profits_heap) w += p while len(capital_heap) > 0 and capital_heap[0][0] <= w: c, p = heappop(capital_heap) heappush(profits_heap, -p)

Finally, we return w, which is the maximum capital we can obtain after investing in k projects.

return w

The time complexity of this solution is O(nlogn + klogn), where n is the number of projects and k is the number of projects we can undertake simultaneously. The space complexity is O(n).

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