502.IPO(Hard).md

502. IPO (Hard)

Date and Time: Aug 31, 2024, 0:33 (EST)

Link: https://leetcode.com/problems/ipo/

Question:

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 <= 10^5

• 0 <= w <= 10^9

• n == profits.length

• n == capital.length

• 1 <= n <= 10^5

• 0 <= profits[i] <= 10^4

• 0 <= capital[i] <= 10^9

Walk-through:

1. maxHeap to store all profits that we can take (all capital[i] <= w). minHeap to store the pairs of [capital[i], profits[i]], when minHeap[0][0] <= w (capital[i] <= w), we can append these profits to maxHeap.

2. We can compare the value of required capital of minHeap, then repeatedly add all satisfied profits into maxHeap. Then, we check if maxHeap is empty or not to break the loop early. Otherwise, we just add negate of the top of maxHeap to w, and return this after range(k).

Python Solution:

class Solution:
    def findMaximizedCapital(self, k: int, w: int, profits: List[int], capital: List[int]) -> int:
        minHeap, maxHeap = [(c, p) for c, p in zip(capital, profits)], []
        heapq.heapify(minHeap)
        
        for _ in range(k):
            while minHeap and minHeap[0][0] <= w:
                c, p = heapq.heappop(minHeap)
                heapq.heappush(maxHeap, -p)
            if not maxHeap:
                break
            w += -heapq.heappop(maxHeap)
        return w

Time Complexity: $O(k*log\ n)$, k is the number of times we run the heappop() and heappush(), each takes $O(log\ n)$ because we maintain a size of n heap. n is total projects.
Space Complexity: $O(n)$