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 several projects. For each project i, it has a pure profit Pi and a minimum capital of Ci is needed to start the corresponding project. 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.
To sum up, pick a list of at most k distinct projects from given projects to maximize your final capital, and output your final maximized capital.
Example 1:
Input: k=2, W=0, Profits=[1,2,3], Capital=[0,1,1].Output: 4Explanation: 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.
Note:
- You may assume all numbers in the input are non-negative integers.
- The length of Profits array and Capital array will not exceed 50,000.
- The answer is guaranteed to fit in a 32-bit signed integer.
假设LeetCode即将开始其首次公开募股。为了向Venture Capital出售其股票的优惠价格，LeetCode希望在IPO之前开展一些项目以增加其资本。由于资源有限，它只能在IPO之前完成最多k个不同的项目。帮助LeetCode设计在完成最多k个不同项目后使得资本最多的最佳方法。
你有几个项目。 对于每个项目i，它具有纯利润Pi，并且需要最小资本Ci来启动相应的项目。 最初你有W资本，完成项目后，您将获得纯利润，利润将被添加到您的总资本中。总之，从给定项目列表中选择最多k个不同项目，最大化最终资本，并输出您的最终最大化的资本。
解法：Priority Queue，保持当前可能的最大利润。 一旦达到所需资金就插入可能的利润。
Java:
public class Solution { public int findMaximizedCapital(int k, int W, int[] Profits, int[] Capital) { PriorityQueue pqCap = new PriorityQueue<>((a, b) -> (a[0] - b[0])); PriorityQueue pqPro = new PriorityQueue<>((a, b) -> (b[1] - a[1])); for (int i = 0; i < Profits.length; i++) { pqCap.add(new int[] {Capital[i], Profits[i]}); } for (int i = 0; i < k; i++) { while (!pqCap.isEmpty() && pqCap.peek()[0] <= W) { pqPro.add(pqCap.poll()); } if (pqPro.isEmpty()) break; W += pqPro.poll()[1]; } return W; }} Python:
def findMaximizedCapital(self, k, W, Profits, Capital): heap = [] projects = sorted(zip(Profits, Capital), key=lambda l: l[1]) i = 0 for _ in range(k): while i < len(projects) and projects[i][1] <= W: heapq.heappush(heap, -projects[i][0]) i += 1 if heap: W -= heapq.heappop(heap) return W
Python:
def findMaximizedCapital(self, k, W, Profits, Capital): current = [] future = sorted(zip(Capital, Profits))[::-1] for _ in range(k): while future and future[-1][0] <= W: heapq.heappush(current, -future.pop()[1]) if current: W -= heapq.heappop(current) return W
C++:
class Solution {public: int findMaximizedCapital(int k, int W, vector & Profits, vector & Capital) { priority_queue > maxH; priority_queue , vector >, greater >> minH; for (int i = 0; i < Capital.size(); ++i) { minH.push({Capital[i], Profits[i]}); } for (int i = 0; i < k; ++i) { while (!minH.empty() && minH.top().first <= W) { auto t = minH.top(); minH.pop(); maxH.push({t.second, t.first}); } if (maxH.empty()) break; W += maxH.top().first; maxH.pop(); } return W; }}; C++:
class Solution {public: int findMaximizedCapital(int k, int W, vector & Profits, vector & Capital) { priority_queue q; multiset > s; for (int i = 0; i < Capital.size(); ++i) { s.insert({Capital[i], Profits[i]}); } for (int i = 0; i < k; ++i) { for (auto it = s.begin(); it != s.end(); ++it) { if (it->first > W) break; q.push(it->second); s.erase(it); } if (q.empty()) break; W += q.top(); q.pop(); } return W; }};