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502. 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 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

Solution:

思路是：

  • 我们要在当前资本 w 可启动的项目中，挑选利润最大的一个来做。
  • 随着项目完成，资本 w 会增加，从而解锁更多可启动的项目。 所以我们需要两个堆来高效管理：
  1. minCapitalHeap：按 capital[i] 升序排列的最小堆，用来找出当前资本可启动的项目。
  2. maxProfitHeap：按 profit[i] 降序排列的最大堆，用来从可启动的项目中挑选利润最大的一个。
class Solution {
    public int findMaximizedCapital(int k, int w, int[] profits, int[] capital) {
       int n = profits.length;

       // minCapitalHeap: 按启动所需资本从小到大排序
       PriorityQueue<int[]> minCapitalHeap = new PriorityQueue<>((a, b) -> a[0] - b[0]);

       // mapProfitHeap: 按利润从大到小排序
       PriorityQueue<int[]> maxProfitHeap = new PriorityQueue<>((a, b) -> b[1] - a[1]);

       for (int i = 0; i < n; i++){
        minCapitalHeap.offer(new int[]{capital[i], profits[i]});
       }

       // 最多左k个项目
       for (int i = 0; i < k; i++){
        // 把所有当前资本可负担的项目放入maxProfitHeap
        while(!minCapitalHeap.isEmpty() && minCapitalHeap.peek()[0] <= w){
            int[] project = minCapitalHeap.poll();
            maxProfitHeap.offer(project);
        }

        // 如果没有可做的项目, 提前结果
        if (maxProfitHeap.isEmpty()){
            break;
        }

        // 选择利润最高的项目
        w = w + maxProfitHeap.poll()[1];
       }

       return w;

    }
}

// TC: O(nlogn)
// SC: O(n)