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IPO

Difficulty: Hard

Problem Description

LeetCode aims to maximize its total capital before an IPO by completing at most k distinct projects. Given the profits and required capital for each project, determine the maximum capital achievable.

Key Insights

  • You can only start projects if your current capital meets their minimum requirement.
  • Each completed project increases your capital by its profit amount.
  • The goal is to select up to k projects that yield the highest profits while satisfying the capital constraints.
  • A greedy approach using a max-heap (priority queue) can efficiently select projects based on the highest profits.

Space and Time Complexity

Time Complexity: O(n log n)
Space Complexity: O(n)

Solution

To solve the problem, we can follow these steps:

  1. Pair each project’s profit with its required capital and store them in a list.
  2. Sort the projects based on their required capital to efficiently find which projects can be started with the current capital.
  3. Use a max-heap (priority queue) to keep track of the profits of the projects that can be completed with the available capital.
  4. Iteratively select the project with the highest profit from the heap up to k times, updating the capital after each selected project.

Code Solutions

import heapq

def findMaximizedCapital(k, w, profits, capital):
    # Pair profits with their corresponding capitals
    projects = sorted(zip(capital, profits))
    
    max_heap = []
    index = 0
    n = len(profits)
    
    # Iterate k times to select up to k projects
    for _ in range(k):
        # Add all projects that can be started with current capital w to the max heap
        while index < n and projects[index][0] <= w:
            heapq.heappush(max_heap, -projects[index][1])  # Push profits as negative for max heap
            index += 1
        
        # If there are projects to choose from, select the one with the highest profit
        if max_heap:
            w -= heapq.heappop(max_heap)  # Pop the maximum profit (note: negative)
    
    return w
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