<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>

// Solution:
// sort the efficiency, select a efficiency E, then select top K max speed engineer who's efficiency >= E, use a min heap with k-1 size to achieve this
// Time: O(N*logN), Space: O(N)

class Solution {
public:
    using LL = long long;

    struct Staff {
        int s, e;
        bool operator < (const Staff& rhs) const {
            return s > rhs.s;
        }
    };

    int maxPerformance(int n, vector<int>& speed, vector<int>& efficiency, int k) {
        vector<Staff> v;
        priority_queue<Staff> q;
        for (int i = 0; i < n; ++i) {
            v.push_back({speed[i], efficiency[i]});
        }
        // descreasing order, O(N*logN)
        sort(v.begin(), v.end(), [] (const Staff& u, const Staff& v) { return u.e > v.e; });
        LL ans = 0, sum = 0;
        // so triky and elegent here, O(N) to get the answer
        // sort efficiency in descreasing order, then add each element to the priority queue
        // one by one, at first, the number of valid elements will less than k
        for (int i = 0; i < n; ++i) {
            // so interesting here
            LL minE = v[i].e;
            LL sumS = sum + v[i].s;
            ans = max(ans, sumS * minE);
            q.push(v[i]); 
            sum += v[i].s;
            // always leave k-1 item in the heap because the next item will has the 
            // min efficiency
            if (q.size() == k) {
                sum -= q.top().s;
                q.pop();
            }
        }
        return ans % (int(1E9) + 7);
    }
};
41796