LeetCode - 502. IPO

Hard

雖然這題是 Hard，但解題思路其實很簡單

題目

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

解題思路

用 priority queue 直接 Greedy 取

把所有 project 的 capital 和 profits 都放到一個 struct 名為 Project 所開的 array projects 中

透過 operator overloading 來對 projects 排序 (根據 capital 由小到大排序)

將 w 另存到 ansCapital 來做動態更新求解
把所有小於等於我們初始擁有的 ansCapital 的 project 的 profit 都 push 進 pq 裡

priority queue 預設會根據權重（在這裡是 profit）由大到小排序，越大的越 top、越先被 pop 出

讓 ansCapital += 從 pq 裡 pop 的最大值 (最大的 profits)
接下來跑下一輪看有哪些 project 小於等於 ansCapital

讓這樣的循環跑 k 次，最後回傳更新到最後的 ansCapital

Solution Code

class Solution {
    public:
        struct Project{
            int profit;
            int capital;

            // operator overloading
            // 由小到大排序
            bool const operator < (Project &item) const{
                return capital < item.capital;
            }
        };
        int findMaximizedCapital(int k, int w, vector<int>& profits, vector<int>& capital) {
            int n = profits.size();
            Project projects[n];

            for(int i = 0; i < n; ++i){
                projects[i].profit = profits[i];
                projects[i].capital = capital[i];
            }
            sort(projects, projects + n);

            int cnt = 0;
            int ansCapital = w;
            priority_queue<int> pq;
            for(int i = 0; i < k; ++i){
                while(cnt < n && projects[cnt].capital <= ansCapital){
                    pq.push(projects[cnt].profit);
                    cnt++;
                }

                if(pq.empty()) break;

                ansCapital += pq.top();
                pq.pop();
            }

            return ansCapital;
        }
};