<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) {
        int n = profits.size();

        vector<int> idxArr;
        for(int i=0; i<n; i++)
            idxArr.push_back(i);

        sort(idxArr.begin(), idxArr.end(), [&](int idx1, int idx2){
            return capital[idx1] < capital[idx2];
        });

        if(w < capital[idxArr[0]])
            return w;
        
        priority_queue<int> pq;
        int totalCapital = w;
        int j = 0;

        while(k--){
            // Add all the possible projects
            while(j < n && totalCapital >= capital[idxArr[j]]){
                pq.push(profits[idxArr[j]]);
                j++;
            }

            if(pq.empty())
                return totalCapital;

            // Fetch the one with max profit
            int top = pq.top();
            pq.pop();

            totalCapital += top;
        }

        return totalCapital;
    }
};

// class Solution {
// public:
//     int findMaximizedCapital(int k, int w, vector<int>& profits, vector<int>& capital) {
//         ios_base::sync_with_stdio(false);
//         cin.tie(NULL);
//         cout.tie(NULL);
        
//         int n = profits.size();

//         vector<pair<int, int>> project;
//         for(int i=0; i<n; i++){
//             project.push_back({capital[i], profits[i]});
//         }

//         sort(project.begin(), project.end());
//         if(w < project[0].first)
//             return w;
        
//         priority_queue<int> pq;
//         int totalCapital = w;
//         int j = 0;

//         while(k--){
//             // Add all the possible projects
//             while(j < n && totalCapital >= project[j].first){
//                 pq.push(project[j].second);
//                 j++;
//             }

//             if(pq.empty())
//                 return totalCapital;

//             // Fetch the one with max profit
//             int top = pq.top();
//             pq.pop();

//             totalCapital += top;
//         }

//         return totalCapital;
//     }
// };
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