LeetCode每日一题(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

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用两个 BinaryHeap, 一个保存当前无法投资的项目，按项目的 capital 来排序, 一个保存当前可以投资的项目， 按项目的 profit倒序排序， 这样我们每投资一个项目， 就从当前不能投资的项目里取出 capital 要求比当前 w 小的项目， 这些项目就是新增的可以投资的项目， 然后把这些项目放到可投资项目的 heap 里， 这一步是为了保证我们每次都投收益最高的项目。最后从可投资项目里 pop 出收益最高的项目。把项目的 profit 添加到 w 中。如此循环， 直到投资了 k 个项目或者没有可投资的项目了

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use std::cmp::Reverse;
use std::collections::BinaryHeap;

impl Solution {
    pub fn find_maximized_capital(
        mut k: i32,
        mut w: i32,
        profits: Vec<i32>,
        capital: Vec<i32>,
    ) -> i32 {
        let mut available = BinaryHeap::new();
        let mut unavailable: BinaryHeap<Reverse<(i32, i32)>> = profits
            .into_iter()
            .zip(capital)
            .map(|(p, c)| Reverse((c, p)))
            .collect();
        while k > 0 {
            while let Some(Reverse((c, p))) = unavailable.pop() {
                if c > w {
                    unavailable.push(Reverse((c, p)));
                    break;
                }
                available.push((p, c));
            }
            if let Some((p, _)) = available.pop() {
                w += p;
            } else {
                return w;
            }
            k -= 1;
        }
        w
    }
}