LeetCode 502 - IPO

Difficulty: hard

Problem Description

English (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 $i^{th}$ 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

Chinese (IPO)

假设 力扣（LeetCode）即将开始 IPO 。为了以更高的价格将股票卖给风险投资公司，力扣 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 力扣 设计完成最多 k 个不同项目后得到最大总资本的方式。

给你 n 个项目。对于每个项目 i ，它都有一个纯利润 profits[i] ，和启动该项目需要的最小资本 capital[i] 。

最初，你的资本为 w 。当你完成一个项目时，你将获得纯利润，且利润将被添加到你的总资本中。

总而言之，从给定项目中选择 最多 k 个不同项目的列表，以 最大化最终资本 ，并输出最终可获得的最多资本。

答案保证在 32 位有符号整数范围内。

示例 1：

输入：k = 2, w = 0, profits = [1,2,3], capital = [0,1,1]
输出：4
解释：
由于你的初始资本为 0，你仅可以从 0 号项目开始。
在完成后，你将获得 1 的利润，你的总资本将变为 1。
此时你可以选择开始 1 号或 2 号项目。
由于你最多可以选择两个项目，所以你需要完成 2 号项目以获得最大的资本。
因此，输出最后最大化的资本，为 0 + 1 + 3 = 4。

示例 2：

输入：k = 3, w = 0, profits = [1,2,3], capital = [0,1,2]
输出：6

提示：

• 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

Solution

C++

class Solution {
public:
    int findMaximizedCapital(int k, int w, vector<int>& profits, vector<int>& capitals) {
        int n = profits.size();
        vector<ProjectInfo> project_infos;
        project_infos.reserve(n);

        for (int i = 0; i < n; ++i) {
            project_infos.emplace_back(profits[i], capitals[i]);
        }
        sort(project_infos.begin(), project_infos.end());

        priority_queue<int> max_profits;
        int has_capital_index = 0;
        while (k-- > 0) {
            while (has_capital_index < n && project_infos[has_capital_index].capital <= w) {
                max_profits.push(project_infos[has_capital_index].profit);
                ++has_capital_index;
            }
            if (max_profits.empty()) {
                break;
            }

            w += max_profits.top();
            max_profits.pop();
        }

        return w;
    }

private:
    struct ProjectInfo {
        ProjectInfo(int p, int c) : profit(p), capital(c) {}

        int profit{};
        int capital{};

        bool operator<(const ProjectInfo& other) const {
            return capital < other.capital;
        }
    };
};