<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:
    int findMaximizedCapital(int k, int w, vector<int>& profits,
                             vector<int>& capital) {
        const auto n = profits.size();

        const auto ids_sorted = invoke([&] {
            vector<int> ids(n);
            iota(ids.begin(), ids.end(), 0);
            sort(ids.begin(), ids.end(),
                 [&](const auto& i, const auto& j) { return capital[i] < capital[j]; });
            return ids;
        });

        vector<int> avail_profits;
        int i = 0;
        do {
            for (; i < n && capital[ids_sorted[i]] <= w; i++) {
                avail_profits.push_back(profits[ids_sorted[i]]);
                push_heap(avail_profits.begin(), avail_profits.end());
            }

            if (avail_profits.empty())
                break;

            w += avail_profits.front();
            pop_heap(avail_profits.begin(), avail_profits.end());
            avail_profits.pop_back();
        } while (--k > 0);

        return w;
    }
};
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