<p>You are given two integers <code>n</code> and <code>k</code> and two integer arrays <code>speed</code> and <code>efficiency</code> both of length <code>n</code>. There are <code>n</code> engineers numbered from <code>1</code> to <code>n</code>. <code>speed[i]</code> and <code>efficiency[i]</code> represent the speed and efficiency of the <code>i<sup>th</sup></code> engineer respectively.</p>

<p>Choose <strong>at most</strong> <code>k</code> different engineers out of the <code>n</code> engineers to form a team with the maximum <strong>performance</strong>.</p>

<p>The performance of a team is the sum of its engineers&#39; speeds multiplied by the minimum efficiency among its engineers.</p>

<p>Return <em>the maximum performance of this team</em>. Since the answer can be a huge number, return it <strong>modulo</strong> <code>10<sup>9</sup> + 7</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> n = 6, speed = [2,10,3,1,5,8], efficiency = [5,4,3,9,7,2], k = 2
<strong>Output:</strong> 60
<strong>Explanation:</strong> 
We have the maximum performance of the team by selecting engineer 2 (with speed=10 and efficiency=4) and engineer 5 (with speed=5 and efficiency=7). That is, performance = (10 + 5) * min(4, 7) = 60.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> n = 6, speed = [2,10,3,1,5,8], efficiency = [5,4,3,9,7,2], k = 3
<strong>Output:</strong> 68
<strong>Explanation:
</strong>This is the same example as the first but k = 3. We can select engineer 1, engineer 2 and engineer 5 to get the maximum performance of the team. That is, performance = (2 + 10 + 5) * min(5, 4, 7) = 68.
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> n = 6, speed = [2,10,3,1,5,8], efficiency = [5,4,3,9,7,2], k = 4
<strong>Output:</strong> 72
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= k &lt;= n &lt;= 10<sup>5</sup></code></li>
	<li><code>speed.length == n</code></li>
	<li><code>efficiency.length == n</code></li>
	<li><code>1 &lt;= speed[i] &lt;= 10<sup>5</sup></code></li>
	<li><code>1 &lt;= efficiency[i] &lt;= 10<sup>8</sup></code></li>
</ul>

class Solution {
public:
    
    // sort the engineeres on the basis of efficiency in decreasing order
    
    static bool compare(vector<int>& a, vector<int>& b)
    {
        return a[0] > b[0];
    }
    
    int maxPerformance(int n, vector<int>& speed, vector<int>& efficiency, int k) {
        
        long long mod = 1e9 + 7;
        
        vector<vector<int>> engineers(n);
        
        for(int i = 0; i < n; i++)
        {
            engineers[i] = {efficiency[i], speed[i]};
        }
        
        // sort the engineeres on the basis of efficiency in decreasing order
        
        sort(engineers.begin(), engineers.end(), compare);
        
        // declare a min. heap, which will store the speeds
        
        priority_queue<long long, vector<long long>, greater<long long>> pq;
        
        long long total_speed = 0;
        
        long long max_performance = INT_MIN;
        
        for(int i = 0; i < n; i++)
        {
            if(i < k)
            {
                long long curr_speed = engineers[i][1];
                
                total_speed += curr_speed;
                
                long long min_efficiency = engineers[i][0];
                
                long long curr_performance = total_speed * min_efficiency;
                
                max_performance = max(max_performance, curr_performance);
                
                pq.push(curr_speed);
            }
            else
            {
                long long curr_speed = engineers[i][1];
                
                if(curr_speed > pq.top())
                {
                    total_speed -= pq.top();
                    
                    total_speed += curr_speed;
                    
                    pq.pop();
                    
                    pq.push(curr_speed);
                    
                    long long min_efficiency = engineers[i][0];
                    
                    long long curr_performance = total_speed * min_efficiency;
                    
                    max_performance = max(max_performance, curr_performance);
                }
            }
        }
        
        return (max_performance % mod);
    }
};
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