<p>Suppose LeetCode will start its <strong>IPO</strong> 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 <strong>IPO</strong>. Since it has limited resources, it can only finish at most <code>k</code> distinct projects before the <strong>IPO</strong>. Help LeetCode design the best way to maximize its total capital after finishing at most <code>k</code> distinct projects.</p>

<p>You are given <code>n</code> projects where the <code>i<sup>th</sup></code> project has a pure profit <code>profits[i]</code> and a minimum capital of <code>capital[i]</code> is needed to start it.</p>

<p>Initially, you have <code>w</code> capital. When you finish a project, you will obtain its pure profit and the profit will be added to your total capital.</p>

<p>Pick a list of <strong>at most</strong> <code>k</code> distinct projects from given projects to <strong>maximize your final capital</strong>, and return <em>the final maximized capital</em>.</p>

<p>The answer is guaranteed to fit in a 32-bit signed integer.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> k = 2, w = 0, profits = [1,2,3], capital = [0,1,1]
<strong>Output:</strong> 4
<strong>Explanation:</strong> 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.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> k = 3, w = 0, profits = [1,2,3], capital = [0,1,2]
<strong>Output:</strong> 6
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= k &lt;= 10<sup>5</sup></code></li>
	<li><code>0 &lt;= w &lt;= 10<sup>9</sup></code></li>
	<li><code>n == profits.length</code></li>
	<li><code>n == capital.length</code></li>
	<li><code>1 &lt;= n &lt;= 10<sup>5</sup></code></li>
	<li><code>0 &lt;= profits[i] &lt;= 10<sup>4</sup></code></li>
	<li><code>0 &lt;= capital[i] &lt;= 10<sup>9</sup></code></li>
</ul>

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

        Project(int id, int profit, int capital) : id(id), profit(profit), capital(capital) { }

        bool operator==(const Project& rhs) const
        {
            return id == rhs.id;
        }
    };

    struct SortByProfit
    {
        bool operator()(const Project& lhs, const Project& rhs)
        {
            return lhs.profit < rhs.profit;
        }
    };

    struct SortByCapital
    {
        bool operator()(const Project& lhs, const Project& rhs)
        {
            return lhs.capital > rhs.capital;
        }
    };

    using profit_queue = std::priority_queue<Project, std::vector<Project>, SortByProfit>;
    using capital_queue = std::priority_queue<Project, std::vector<Project>, SortByCapital>;

    int findMaximizedCapital(int k, int w, vector<int>& profits, vector<int>& capital) {
        // Always want to perform the most profitable option that costs less than w.
        auto capitalQueue = capital_queue();
        for (auto i = 0; i < profits.size(); ++i)
        {
            capitalQueue.emplace(i, profits[i], capital[i]);
        }

        auto profitQueue = profit_queue();

        return findMaximizedCapital(k, w, capitalQueue, profitQueue);
    }

    int findMaximizedCapital(int k, int w, capital_queue& capitalQueue, profit_queue& profitQueue)
    {
        //std::cout << "iteration k=" << k << " w=" << w << std::endl;
        if (k == 0)
            return w;

        // Sort the projects with captial < w by profit.
        while (! capitalQueue.empty())
        {
            const auto top = capitalQueue.top();
            //std::cout << "available project " << top.id << " capital=" << top.capital << " profit=" << top.profit << std::endl;
            if (top.capital <= w)
            {
                capitalQueue.pop();
                profitQueue.push(top);
            }
            else
                break;
        }

        if (! profitQueue.empty())
        {
            const auto project = profitQueue.top();
            //std::cout << "next project " << project.id << " capital=" << project.capital << " profit=" << project.profit << std::endl;
            profitQueue.pop();
            w += project.profit;
            return findMaximizedCapital(k - 1, w, capitalQueue, profitQueue);
        }
        else
        {
            return w;
        }
    }
};
118112