feat: leetcode 857

Author: deweyou

README.md

@@ -2,7 +2,7 @@
 
 <p align="center">
 <!-- TOPICS COUNT START -->
-<img src="https://img.shields.io/badge/-进度:490-green" alt="进度:490">
+<img src="https://img.shields.io/badge/-进度:491-green" alt="进度:491">
 <!-- TOPICS COUNT END -->
 <a href="./assets/docs/TOPICS.md"><img src="https://img.shields.io/badge/-题库目录-blue" alt="题库目录"></a>
 <a href="./assets/docs/CATEGORIES.md"><img src="https://img.shields.io/badge/-题库分类-red" alt="题库分类"></a>

assets/data/categories.json

@@ -3502,6 +3502,12 @@
         "path": "./problemset/set-intersection-size-at-least-two/README.md",
         "difficulty": "困难"
       },
+      {
+        "id": "857",
+        "title": "雇佣 K 名工人的最低成本",
+        "path": "./problemset/minimum-cost-to-hire-k-workers/README.md",
+        "difficulty": "困难"
+      },
       {
         "id": "871",
         "title": "最低加油次数",
@@ -4174,6 +4180,12 @@
         "path": "./problemset/find-median-from-data-stream/README.md",
         "difficulty": "困难"
       },
+      {
+        "id": "857",
+        "title": "雇佣 K 名工人的最低成本",
+        "path": "./problemset/minimum-cost-to-hire-k-workers/README.md",
+        "difficulty": "困难"
+      },
       {
         "id": "1606",
         "title": "找到处理最多请求的服务器",

assets/data/topics.json

@@ -3849,6 +3849,16 @@
     "url": "https://leetcode-cn.com/problems/push-dominoes/",
     "path": "./problemset/push-dominoes/README.md"
   },
+  {
+    "id": "857",
+    "title": {
+      "cn": "雇佣 K 名工人的最低成本",
+      "en": "minimum-cost-to-hire-k-workers"
+    },
+    "difficulty": "困难",
+    "url": "https://leetcode.cn/problems/minimum-cost-to-hire-k-workers/",
+    "path": "./problemset/minimum-cost-to-hire-k-workers/README.md"
+  },
   {
     "id": "868",
     "title": {

assets/docs/CATEGORIES.md

@@ -654,6 +654,7 @@
 | 334. [递增的三元子序列](../../problemset/increasing-triplet-subsequence/README.md) | 中等 |
 | 646. [最长数对链](../../problemset/maximum-length-of-pair-chain/README.md) | 中等 |
 | 757. [设置交集大小至少为2](../../problemset/set-intersection-size-at-least-two/README.md) | 困难 |
+| 857. [雇佣 K 名工人的最低成本](../../problemset/minimum-cost-to-hire-k-workers/README.md) | 困难 |
 | 871. [最低加油次数](../../problemset/minimum-number-of-refueling-stops/README.md) | 困难 |
 | 942. [增减字符串匹配](../../problemset/di-string-match/README.md) | 简单 |
 | 1217. [玩筹码](../../problemset/minimum-cost-to-move-chips-to-the-same-position/README.md) | 简单 |
@@ -791,6 +792,7 @@
 | 239. [滑动窗口最大值](../../problemset/sliding-window-maximum/README.md) | 困难 |
 | 264. [丑数 II](../../problemset/ugly-number-2/README.md) | 中等 |
 | 295. [数据流的中位数](../../problemset/find-median-from-data-stream/README.md) | 困难 |
+| 857. [雇佣 K 名工人的最低成本](../../problemset/minimum-cost-to-hire-k-workers/README.md) | 困难 |
 | 1606. [找到处理最多请求的服务器](../../problemset/find-servers-that-handled-most-number-of-requests/README.md) | 困难 |
 
 ## 树状数组

assets/docs/TOPICS.md

@@ -770,6 +770,8 @@
 
 [838. 推多米诺](../../problemset/push-dominoes/README.md)
 
+[857. 雇佣 K 名工人的最低成本](../../problemset/minimum-cost-to-hire-k-workers/README.md)
+
 [868. 二进制间距](../../problemset/binary-gap/README.md)
 
 [871. 最低加油次数](../../problemset/minimum-number-of-refueling-stops/README.md)

package.json

@@ -31,5 +31,8 @@
     "typescript": "^4.6.3",
     "vite": "^2.9.5",
     "vitest": "^0.9.4"
+  },
+  "dependencies": {
+    "@datastructures-js/priority-queue": "^6.1.3"
   }
 }

pnpm-lock.yaml

@@ -0,0 +1,70 @@
+# 雇佣 K 名工人的最低成本
+
+> 难度：困难
+>
+> https://leetcode.cn/problems/minimum-cost-to-hire-k-workers/
+
+## 题目
+
+有 `n` 名工人。 给定两个数组 `quality` 和 `wage` ，其中，`quality[i]` 表示第 `i` 名工人的工作质量，其最低期望工资为 `wage[i]` 。
+
+现在我们想雇佣 `k` 名工人组成一个工资组。在雇佣 一组 `k` 名工人时，我们必须按照下述规则向他们支付工资：
+
+1. 对工资组中的每名工人，应当按其工作质量与同组其他工人的工作质量的比例来支付工资。
+2. 工资组中的每名工人至少应当得到他们的最低期望工资。
+
+给定整数 `k` ，返回 组成满足上述条件的付费群体所需的最小金额 。在实际答案的 `10^-5` 以内的答案将被接受。
+
+### 示例
+
+#### 示例 1：
+
+```
+输入： quality = [10,20,5], wage = [70,50,30], k = 2
+输出： 105.00000
+解释： 我们向 0 号工人支付 70，向 2 号工人支付 35。
+```
+
+#### 示例 2：
+
+```
+输入： quality = [3,1,10,10,1], wage = [4,8,2,2,7], k = 3
+输出： 30.66667
+解释： 我们向 0 号工人支付 4，向 2 号和 3 号分别支付 13.33333。
+```
+
+## 解题
+
+```ts 
+/**
+ * 贪心 + 优先队列
+ * @desc 时间复杂度 O(NlogN)  空间复杂度 O(N)
+ * @param quality
+ * @param wage
+ * @param k
+ * @returns
+ */
+export function mincostToHireWorkers(quality: number[], wage: number[], k: number): number {
+  const n = quality.length
+  const h = new Array(n).fill(0).map((_, i) => i)
+  h.sort((a, b) => {
+    return quality[b] * wage[a] - quality[a] * wage[b]
+  })
+  let res = 1e9
+  let totalq = 0.0
+  const pq = new MaxPriorityQueue<number>()
+  for (let i = 0; i < k - 1; i++) {
+    totalq += quality[h[i]]
+    pq.enqueue(quality[h[i]])
+  }
+  for (let i = k - 1; i < n; i++) {
+    const idx = h[i]
+    totalq += quality[idx]
+    pq.enqueue(quality[idx])
+    const totalc = (wage[idx] / quality[idx]) * totalq
+    res = Math.min(res, totalc)
+    totalq -= pq.dequeue()
+  }
+  return Number(res.toFixed(5))
+}
+```

problemset/minimum-cost-to-hire-k-workers/README.md

@@ -0,0 +1,25 @@
+import { describe, expect, it } from 'vitest'
+import { mincostToHireWorkers } from '.'
+
+describe('雇佣 K 名工人的最低成本', () => {
+  testCase(mincostToHireWorkers)
+})
+
+function testCase(fn: (quality: number[], wage: number[], k: number) => number) {
+  it.each([
+    [
+      [10, 20, 5],
+      [70, 50, 30],
+      2,
+      105,
+    ],
+    [
+      [3, 1, 10, 10, 1],
+      [4, 8, 2, 2, 7],
+      3,
+      30.66667,
+    ],
+  ])('示例%#', (quality, wage, k, expected) => {
+    expect(fn(quality, wage, k)).toBe(expected)
+  })
+}

problemset/minimum-cost-to-hire-k-workers/index.spec.ts

@@ -0,0 +1,33 @@
+import { MaxPriorityQueue } from '@datastructures-js/priority-queue'
+
+/**
+ * 贪心 + 优先队列
+ * @desc 时间复杂度 O(NlogN)  空间复杂度 O(N)
+ * @param quality
+ * @param wage
+ * @param k
+ * @returns
+ */
+export function mincostToHireWorkers(quality: number[], wage: number[], k: number): number {
+  const n = quality.length
+  const h = new Array(n).fill(0).map((_, i) => i)
+  h.sort((a, b) => {
+    return quality[b] * wage[a] - quality[a] * wage[b]
+  })
+  let res = 1e9
+  let totalq = 0.0
+  const pq = new MaxPriorityQueue<number>()
+  for (let i = 0; i < k - 1; i++) {
+    totalq += quality[h[i]]
+    pq.enqueue(quality[h[i]])
+  }
+  for (let i = k - 1; i < n; i++) {
+    const idx = h[i]
+    totalq += quality[idx]
+    pq.enqueue(quality[idx])
+    const totalc = (wage[idx] / quality[idx]) * totalq
+    res = Math.min(res, totalc)
+    totalq -= pq.dequeue()
+  }
+  return Number(res.toFixed(5))
+}