IPO
Problem Description
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
Solve it on LeetCode
Approaches
1. Brute Force Approach
Intuition:
The brute force method involves examining all possible combinations of projects and selecting those which maximize the profit within the given constraint of initial capital. This approach is straightforward but computationally expensive, especially when the number of projects is large or the constraints are tight.
Steps:
- Iterate through all the projects.
- For each project, check if it can be started with the available capital.
- Choose the project giving the maximum profit that can be accommodated by your current capital.
- Update the capital with the profit obtained and repeat the process for
kselections.
Code:
- Time Complexity: O(k * n), where (n) is the number of projects and (k) is the number of possible selections.
- Space Complexity: O(1). Only a few extra variables are used.
2. Greedy Approach using Max Heap
Intuition:
To optimize the profit selection, we use a max heap to always extract the project that offers the maximum profit which can be started given the current capital. We keep track of projects whose capital requirement is within our current capital using a min-heap.
Steps:
- Pair up profits and capital into a list of jobs.
- Sort this list based on the capital required.
- Use a max-heap to manage the projects that can be started with the available capital.
- For
kselections: - Push all projects into the max-heap that can be started with the current capital.
- If the max-heap is empty, break out as no further projects can be started.
- Extract project with the maximum profit, and update the capital with the profit obtained.
Code:
- Time Complexity: O(n * log n + k * log n). Sorting the projects takes O(n * log n), and managing the max heap for up to (k) operations takes O(k *log n).
- Space Complexity: O(n) due to storing projects and the space used by the heap.