Minimum Cost to Hire K Workers

Description

There are N workers.  The i-th worker has a quality[i] and a minimum wage expectation wage[i].

Now we want to hire exactly K workers to form a paid group.  When hiring a group of K workers, we must pay them according to the following rules:

1. Every worker in the paid group should be paid in the ratio of their quality compared to other workers in the paid group.
2. Every worker in the paid group must be paid at least their minimum wage expectation.

Return the least amount of money needed to form a paid group satisfying the above conditions.

 

Example 1:

Input: quality = [10,20,5], wage = [70,50,30], K = 2
Output: 105.00000
Explanation: We pay 70 to 0-th worker and 35 to 2-th worker.

Example 2:

Input: quality = [3,1,10,10,1], wage = [4,8,2,2,7], K = 3
Output: 30.66667
Explanation: We pay 4 to 0-th worker, 13.33333 to 2-th and 3-th workers seperately. 

 

Note:

1. 1 <= K <= N <= 10000, where N = quality.length = wage.length
2. 1 <= quality[i] <= 10000
3. 1 <= wage[i] <= 10000
4. Answers within 10^-5 of the correct answer will be considered correct.

Solution(javascript)

const MaxHeap = () => {
  const list = []
  const parent = index => Math.floor((index - 1) / 2)
  const left = index => 2 * index + 1
  const right = index => 2 * index + 2

  const swap = (a, b) => {
    const temp = list[a]
    list[a] = list[b]
    list[b] = temp
  }
  const insert = (x) => {
    list.push(x)
    let currentIndex = list.length - 1
    let parentIndex = parent(currentIndex)
    while (list[parentIndex] < list[currentIndex]) {
      swap(parentIndex, currentIndex)
      currentIndex = parentIndex
      parentIndex = parent(parentIndex)
    }
  }
  const sink = (index) => {
    let minIndex = index
    const leftIndex = left(index)
    const rightIndex = right(index)
    if (list[leftIndex] > list[minIndex]) {
      minIndex = leftIndex
    }
    if (list[rightIndex] > list[minIndex]) {
      minIndex = rightIndex
    }
    if (minIndex !== index) {
      swap(minIndex, index)
      sink(minIndex)
    }
  }
  const size = () => list.length
  const extract = () => {
    swap(0, size() - 1)
    const min = list.pop()
    sink(0)
    return min
  }
  return {
    insert,
    size,
    extract,
  }
}
/**
 * @param {number[]} quality
 * @param {number[]} wage
 * @param {number} K
 * @return {number}
 */
const mincostToHireWorkers = function (quality, wage, K) {
  const works = quality.map((q, index) => [q, wage[index]])
  works.sort((a, b) => (a[1] / a[0]) - (b[1] / b[0]))
  let totalQuantity = 0
  const maxHeap = MaxHeap()
  let min = Infinity
  for (const work of works) {
    maxHeap.insert(work[0])
    totalQuantity += work[0]
    if (maxHeap.size() > K) {
      const max = maxHeap.extract()
      totalQuantity -= max
    }
    if (maxHeap.size() === K) {
      min = Math.min(
        min,
        totalQuantity * (work[1] / work[0]),
      )
    }
  }
  return min
}