<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;
    int visited;
    project(int a, int b):profit(a), capital(b), visited(false){}
};

bool cmp(project a, project b){
    return a.capital < b.capital;
}

class Solution {

public:
    int findMaximizedCapital(int k, int w, vector<int>& profits, vector<int>& capital) {
        int n = profits.size();
        vector<project> projects;
        vector<project*> heap;

        //1. create a vector of projects
        for (int i = 0; i < n; i++){ 
            projects.push_back(project(profits[i], capital[i]));
        }

        //2. sort the projects based on required capital
        sort(projects.begin(), projects.end(), cmp);

        int p = 0;
        int wealth = w;
        for (int i = 0; i < k; i++){
            //3.1. build heap for maximum profits (from projects with available capital)
            while (p < n && projects[p].capital <= wealth){
                //printf ("project available: %d, capital: %d, wealth: %d\n", p, projects[p].capital, wealth);
                heapPush(heap, &projects[p]);
                p++;
            }
            //3.2. add do-able projects based on new wealth to heap (from capital list)
            if (heap.size() > 0){
                //printf ("do project with profit %d\n", heap[0]->profit);
                wealth += heap[0]->profit;
                heapPop(heap);
            }else{
                break;
            }
        }

        return wealth;

    }

    void swap(vector<project*>& vec, int i, int j){
        project* temp = vec[i];
        vec[i] = vec[j];
        vec[j] = temp;
    }

    void siftDown(vector<project*>& vec, int start){
        int p = start;
        while (p < vec.size()){
            int l = p*2+1;
            int r = p*2+2;
            int smallest = p;
            if (l < vec.size() && vec[l]->profit > vec[smallest]->profit){
                smallest = l;
            }
            if (r < vec.size() && vec[r]->profit > vec[smallest]->profit){
                smallest = r;
            }
            if (smallest != p){
                swap(vec, smallest, p);
                p = smallest;
            }else{
                break;
            }
        }
    }

    void siftUp(vector<project*>& vec, int start){
        int p = start;
        while (p >= 0){
            int l = (p-1)/2;
            int smallest = p;
            if (l < vec.size() && vec[l]->profit < vec[smallest]->profit){
                smallest = l;
            }
            if (smallest != p){
                swap(vec, smallest, p);
                p = smallest;
            }else{
                break;
            }
        }
    }

    void heapPush(vector<project*>& vec, project* proj){
        vec.push_back(proj);
        siftUp(vec, vec.size()-1);
    }


    void heapPop(vector<project*>& vec){
        swap(vec, 0, vec.size()-1);
        vec.pop_back();
        siftDown(vec, 0);
    }
};
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