502.IPO

tags: Hard,Array,Greedy,Sorting,Heap

502. IPO

題目描述

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.

範例

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

解答

C++

思路:

1. 有k次機會做project, 初始資金為w, 每個project有各自的資金要求, 超過才能做, 做完利潤加入資金內, 找出最大化的最終資金
2. 每個project若在符合的資金條件下, 選最大利潤的做即可, 用priority queue存
3. 為了找符合當前資金的所有project, 把project照資金要求做sort後遍歷即可

class Solution {
public:
    int findMaximizedCapital(int k, int w, vector<int>& profits, vector<int>& capital) {        
        vector<pair<int,int>> projects;
        for(int i = 0; i < capital.size(); i++)
            projects.push_back({capital[i], profits[i]});
        sort(begin(projects), end(projects), [](pair<int,int>& p1, pair<int,int>& p2){
            return p1.first < p2.first;
        });
        priority_queue<int> q;
        int index = 0, n = profits.size();
        while(k > 0)
        {
            while(index < n && projects[index].first <= w)
                q.push(projects[index++].second);
            if(q.empty())
                break;
            w += q.top();
            q.pop();
            k--;
        }
        return w;
    }
};

Time:

$O(n\ast log(n))$ sort project的時間
Extra Space:

$O(n)$

XD Feb 23, 2023

Reference

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