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105 lines
3.6 KiB
Markdown
105 lines
3.6 KiB
Markdown
# [1383. Maximum Performance of a Team](https://leetcode.com/problems/maximum-performance-of-a-team/)
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## 题目
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You are given two integers `n` and `k` and two integer arrays `speed` and `efficiency` both of length `n`. There are `n` engineers numbered from `1` to `n`. `speed[i]` and `efficiency[i]` represent the speed and efficiency of the `ith` engineer respectively.
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Choose **at most** `k` different engineers out of the `n` engineers to form a team with the maximum **performance**.
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The performance of a team is the sum of their engineers' speeds multiplied by the minimum efficiency among their engineers.
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Return *the maximum performance of this team*. Since the answer can be a huge number, return it **modulo** `109 + 7`.
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**Example 1:**
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```
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Input: n = 6, speed = [2,10,3,1,5,8], efficiency = [5,4,3,9,7,2], k = 2
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Output: 60
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Explanation:
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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.
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```
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**Example 2:**
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```
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Input: n = 6, speed = [2,10,3,1,5,8], efficiency = [5,4,3,9,7,2], k = 3
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Output: 68
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Explanation:
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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.
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```
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**Example 3:**
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```
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Input: n = 6, speed = [2,10,3,1,5,8], efficiency = [5,4,3,9,7,2], k = 4
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Output: 72
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```
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**Constraints:**
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- `1 <= <= k <= n <= 105`
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- `speed.length == n`
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- `efficiency.length == n`
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- `1 <= speed[i] <= 105`
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- `1 <= efficiency[i] <= 108`
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## 题目大意
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公司有编号为 1 到 n 的 n 个工程师，给你两个数组 speed 和 efficiency ，其中 speed[i] 和 efficiency[i] 分别代表第 i 位工程师的速度和效率。请你返回由最多 k 个工程师组成的 最大团队表现值 ，由于答案可能很大，请你返回结果对 10^9 + 7 取余后的结果。团队表现值 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」。
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## 解题思路
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- 题目要求返回最大团队表现值，表现值需要考虑速度的累加和，和效率的最小值。即使速度快，效率的最小值很小，总的表现值还是很小。先将效率从大到小排序。从效率高的工程师开始选起，遍历过程中维护一个大小为 k 的速度最小堆。每次遍历都计算一次团队最大表现值。扫描完成，最大团队表现值也筛选出来了。具体实现见下面的代码。
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## 代码
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```go
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package leetcode
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import (
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"container/heap"
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"sort"
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)
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func maxPerformance(n int, speed []int, efficiency []int, k int) int {
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indexes := make([]int, n)
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for i := range indexes {
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indexes[i] = i
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}
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sort.Slice(indexes, func(i, j int) bool {
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return efficiency[indexes[i]] > efficiency[indexes[j]]
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})
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ph := speedHeap{}
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heap.Init(&ph)
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speedSum := 0
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var max int64
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for _, index := range indexes {
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if ph.Len() == k {
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speedSum -= heap.Pop(&ph).(int)
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}
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speedSum += speed[index]
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heap.Push(&ph, speed[index])
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max = Max(max, int64(speedSum)*int64(efficiency[index]))
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}
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return int(max % (1e9 + 7))
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}
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type speedHeap []int
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func (h speedHeap) Less(i, j int) bool { return h[i] < h[j] }
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func (h speedHeap) Swap(i, j int) { h[i], h[j] = h[j], h[i] }
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func (h speedHeap) Len() int { return len(h) }
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func (h *speedHeap) Push(x interface{}) { *h = append(*h, x.(int)) }
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func (h *speedHeap) Pop() interface{} {
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res := (*h)[len(*h)-1]
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*h = (*h)[:h.Len()-1]
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return res
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}
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func Max(a, b int64) int64 {
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if a > b {
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return a
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}
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return b
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}
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``` |