<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>

struct Project {
    int profit;
    int capital;

    bool operator< (const Project& b) const {
        return this->profit < b.profit;
    }
};

class Solution {
    int lastLEQ(const vector<Project>& projects, int cap) {
        const int len = projects.size();
        int l = 0, r = len - 1;
        int result = -1;

        while (l <= r) {
            int mid = (l + r) >> 1;
            if (projects[mid].capital <= cap) {
                result = mid;
                l = mid + 1;
            } else {
                r = mid - 1;
            }
        }

        return result;
    }
public:
    int findMaximizedCapital(int k, int w, vector<int>& profits, vector<int>& capital) {
        const int len = profits.size();
        vector<Project> projects(len);
        for (int i = 0; i < len; i++) {
            projects[i] = Project {
                .profit = profits[i],
                .capital = capital[i],
            };
        }

        sort(projects.begin(), projects.end(), [&] (auto& a, auto& b) {
            return a.capital < b.capital;
        });

        // for (int i = 0; i < len; i++) {
        //     cout << "capital=" << projects[i].capital
        //          << ", profits=" << projects[i].profit << endl;
        // }

        if (projects[0].capital > w) return w;

        priority_queue<Project> pq;
        int last_left = -1;
        
        for (int i = 0; i < k; i++) {
            // add all project whose capital <= w into pq
            int left = lastLEQ(projects, w);
            for (int j = last_left + 1; j <= left; j++)
                pq.push(projects[j]);
            last_left = left;
            if (pq.empty()) break;
            auto top = pq.top();
            pq.pop();
            w += top.profit;
        }

        return w;
    }
};
129487