---
# System prepended metadata

title: Leetcode 502. IPO
tags: [Leetcode]

---

[link](https://leetcode.com/problems/ipo)

---

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

---

The findMaximizedCapital function aims to maximize the capital by selecting projects based on capital and profit. It takes the following parameters:

k: The maximum number of projects that can be selected.
w: The initial capital.
profits: A list containing the profit value for each project.
capital: A list containing the required capital to start each project.
The function first creates two heaps, minCapital and maxProfit.

minCapital: This heap is used to store pairs (c, p) representing the capital c and profit p for each project. The heap is created from the capital and profits lists using the heapq.heapify method.

maxProfit: This is a max heap that stores the negative values of profits for projects that are feasible with the current available capital. It will store the profits in a way that the project with the highest profit has the highest priority (top of the heap).

The function then iterates k times to select the k best projects.

In each iteration:

While there are projects in minCapital and the top project has a capital less than or equal to the available capital w, pop the project (c, p) from minCapital and push the negative profit -p into maxProfit. This means that the project is feasible with the current available capital, and we add its profit to maxProfit.

If there are no feasible projects in maxProfit (i.e., no project can be started with the current available capital), break the loop since we cannot select any more projects.

Otherwise, pop the top element from maxProfit (the project with the highest profit), negate it to get the actual profit, and add it to the available capital w.

Repeat the next iteration to select the next best project.

#### Solution 1
```python=
class Solution:
    def findMaximizedCapital(self, k: int, w: int, profits: List[int], capital: List[int]) -> int:
        maxProfit = []
        minCapital = [(c, p) for c, p in zip(capital, profits)]
        heapq.heapify(minCapital)

        for _ in range(k):
            while minCapital and minCapital[0][0] <= w:
                c, p = heapq.heappop(minCapital)
                heapq.heappush(maxProfit, -p)
            if not maxProfit:
                break
            w += -(heapq.heappop(maxProfit))

        return w
```
O(T): O(k log n)
O(S): O(n)
