<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 Projects{
    public:
    int profit; 
    int capital;
    Projects(int p , int c){
        profit = p ; 
        capital = c;

    }
};
class compareProfits{
    public:
    bool operator()(Projects a , Projects b){
        return  a.profit <  b.profit;
    }
};
class compareCapital{
    public:
    bool operator()(Projects a , Projects 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<Projects> ans;
        for(int i = 0 ; i<profits.size();i++){
            ans.push_back({profits[i] , capital[i]});

        }
        priority_queue<Projects,vector<Projects> , compareCapital>minCapitalHeap;
        priority_queue<Projects ,vector<Projects>, compareProfits>maxProfitHeap;
        for(int i = 0 ; i<n;i++ ){
            minCapitalHeap.push(ans[i]);

        }
        int currentCapital = w;
        for(int i = 0 ; i<k ;i++){
            while(!minCapitalHeap.empty() && minCapitalHeap.top().capital <= currentCapital){
                maxProfitHeap.push(minCapitalHeap.top());
                minCapitalHeap.pop();
                
            }

            if(maxProfitHeap.empty()){
                break;
            }
            currentCapital += maxProfitHeap.top().profit;
            maxProfitHeap.pop();

        }
        return currentCapital;
    }
    
};
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