Add solution 1383

leetcode/1383.Maximum-Performance-of-a-Team/1383. Maximum Performance of a Team.go

@@ -0,0 +1,48 @@
+package leetcode
+
+import (
+	"container/heap"
+	"sort"
+)
+
+func maxPerformance(n int, speed []int, efficiency []int, k int) int {
+	indexes := make([]int, n)
+	for i := range indexes {
+		indexes[i] = i
+	}
+	sort.Slice(indexes, func(i, j int) bool {
+		return efficiency[indexes[i]] > efficiency[indexes[j]]
+	})
+	ph := speedHeap{}
+	heap.Init(&ph)
+	speedSum := 0
+	var max int64
+	for _, index := range indexes {
+		if ph.Len() == k {
+			speedSum -= heap.Pop(&ph).(int)
+		}
+		speedSum += speed[index]
+		heap.Push(&ph, speed[index])
+		max = Max(max, int64(speedSum)*int64(efficiency[index]))
+	}
+	return int(max % (1e9 + 7))
+}
+
+type speedHeap []int
+
+func (h speedHeap) Less(i, j int) bool  { return h[i] < h[j] }
+func (h speedHeap) Swap(i, j int)       { h[i], h[j] = h[j], h[i] }
+func (h speedHeap) Len() int            { return len(h) }
+func (h *speedHeap) Push(x interface{}) { *h = append(*h, x.(int)) }
+func (h *speedHeap) Pop() interface{} {
+	res := (*h)[len(*h)-1]
+	*h = (*h)[:h.Len()-1]
+	return res
+}
+
+func Max(a, b int64) int64 {
+	if a > b {
+		return a
+	}
+	return b
+}

leetcode/1383.Maximum-Performance-of-a-Team/1383. Maximum Performance of a Team_test.go

@@ -0,0 +1,55 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question1383 struct {
+	para1383
+	ans1383
+}
+
+// para 是参数
+// one 代表第一个参数
+type para1383 struct {
+	n          int
+	speed      []int
+	efficiency []int
+	k          int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans1383 struct {
+	one int
+}
+
+func Test_Problem1383(t *testing.T) {
+
+	qs := []question1383{
+
+		{
+			para1383{6, []int{2, 10, 3, 1, 5, 8}, []int{5, 4, 3, 9, 7, 2}, 2},
+			ans1383{60},
+		},
+
+		{
+			para1383{6, []int{2, 10, 3, 1, 5, 8}, []int{5, 4, 3, 9, 7, 2}, 3},
+			ans1383{68},
+		},
+
+		{
+			para1383{6, []int{2, 10, 3, 1, 5, 8}, []int{5, 4, 3, 9, 7, 2}, 4},
+			ans1383{72},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 1383------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans1383, q.para1383
+		fmt.Printf("【input】:%v       【output】:%v\n", p, maxPerformance(p.n, p.speed, p.efficiency, p.k))
+	}
+	fmt.Printf("\n\n\n")
+}

leetcode/1383.Maximum-Performance-of-a-Team/README.md

@@ -0,0 +1,105 @@
+# [1383. Maximum Performance of a Team](https://leetcode.com/problems/maximum-performance-of-a-team/)
+
+## 题目
+
+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.
+
+Choose **at most** `k` different engineers out of the `n` engineers to form a team with the maximum **performance**.
+
+The performance of a team is the sum of their engineers' speeds multiplied by the minimum efficiency among their engineers.
+
+Return *the maximum performance of this team*. Since the answer can be a huge number, return it **modulo** `109 + 7`.
+
+**Example 1:**
+
+```
+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.
+```
+
+**Example 2:**
+
+```
+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.
+```
+
+**Example 3:**
+
+```
+Input: n = 6, speed = [2,10,3,1,5,8], efficiency = [5,4,3,9,7,2], k = 4
+Output: 72
+```
+
+**Constraints:**
+
+- `1 <= <= k <= n <= 105`
+- `speed.length == n`
+- `efficiency.length == n`
+- `1 <= speed[i] <= 105`
+- `1 <= efficiency[i] <= 108`
+
+## 题目大意
+
+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」。
+
+## 解题思路
+
+- 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。
+
+## 代码
+
+```go
+package leetcode
+
+import (
+	"container/heap"
+	"sort"
+)
+
+func maxPerformance(n int, speed []int, efficiency []int, k int) int {
+	indexes := make([]int, n)
+	for i := range indexes {
+		indexes[i] = i
+	}
+	sort.Slice(indexes, func(i, j int) bool {
+		return efficiency[indexes[i]] > efficiency[indexes[j]]
+	})
+	ph := speedHeap{}
+	heap.Init(&ph)
+	speedSum := 0
+	var max int64
+	for _, index := range indexes {
+		if ph.Len() == k {
+			speedSum -= heap.Pop(&ph).(int)
+		}
+		speedSum += speed[index]
+		heap.Push(&ph, speed[index])
+		max = Max(max, int64(speedSum)*int64(efficiency[index]))
+	}
+	return int(max % (1e9 + 7))
+}
+
+type speedHeap []int
+
+func (h speedHeap) Less(i, j int) bool  { return h[i] < h[j] }
+func (h speedHeap) Swap(i, j int)       { h[i], h[j] = h[j], h[i] }
+func (h speedHeap) Len() int            { return len(h) }
+func (h *speedHeap) Push(x interface{}) { *h = append(*h, x.(int)) }
+func (h *speedHeap) Pop() interface{} {
+	res := (*h)[len(*h)-1]
+	*h = (*h)[:h.Len()-1]
+	return res
+}
+
+func Max(a, b int64) int64 {
+	if a > b {
+		return a
+	}
+	return b
+}
+```