maximum-performance-of-a-team

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Algorithms

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>ith</code> engineer respectively.

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

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

Return the maximum performance of this team. Since the answer can be a huge number, return it modulo <code>109 + 7</code>.

&nbsp;
Example 1:

<pre>
Input: n = 6, speed = [2,10,3,1,5,8], efficiency = [5,4,3,9,7,2], k = 2
Output: 60
Explanation: 
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>

Example 2:

<pre>
Input: n = 6, speed = [2,10,3,1,5,8], efficiency = [5,4,3,9,7,2], k = 3
Output: 68
Explanation:
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>

Example 3:

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

&nbsp;
Constraints:

<ul>
	<li><code>1 &lt;= k &lt;= n &lt;= 105</code></li>
	<li><code>speed.length == n</code></li>
	<li><code>efficiency.length == n</code></li>
	<li><code>1 &lt;= speed[i] &lt;= 105</code></li>
	<li><code>1 &lt;= efficiency[i] &lt;= 108</code></li>
</ul>

第 180 场周赛

weekly-contest-180

maxPerformance

高级专家

Maximum Performance of a Team

Maximum Fruits Harvested After at Most K Steps

maximum-fruits-harvested-after-at-most-k-steps

摘水果

Greedy

Array

Sorting

Heap (Priority Queue)

给定两个整数 <code>n</code> 和 <code>k</code>，以及两个长度为 <code>n</code> 的整数数组 <code>speed</code> 和<code> efficiency</code>。现有 <code>n</code> 名工程师，编号从 <code>1</code> 到 <code>n</code>。其中 <code>speed[i]</code>&nbsp;和 <code>efficiency[i]</code>&nbsp;分别代表第 <code>i</code>&nbsp;位工程师的速度和效率。

从这 <code>n</code> 名工程师中最多选择 <code>k</code> 名不同的工程师，使其组成的团队具有最大的团队表现值。

团队表现值&nbsp;的定义为：一个团队中「所有工程师速度的和」乘以他们「效率值中的最小值」。

请你返回该团队的​​​​​​最大团队表现值，由于答案可能很大，请你返回结果对 <code>10^9 + 7</code> 取余后的结果。

&nbsp;

示例 1：

<pre>
输入：n = 6, speed = [2,10,3,1,5,8], efficiency = [5,4,3,9,7,2], k = 2
输出：60
解释：
我们选择工程师 2（speed=10 且 efficiency=4）和工程师 5（speed=5 且 efficiency=7）。他们的团队表现值为 performance = (10 + 5) * min(4, 7) = 60 。
</pre>

示例 2：

<pre>
输入：n = 6, speed = [2,10,3,1,5,8], efficiency = [5,4,3,9,7,2], k = 3
输出：68
解释：
此示例与第一个示例相同，除了 k = 3 。我们可以选择工程师 1 ，工程师 2 和工程师 5 得到最大的团队表现值。表现值为 performance = (2 + 10 + 5) * min(5, 4, 7) = 68 。
</pre>

示例 3：

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

&nbsp;

提示：

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