<p>There are <code>n</code> workers. You are given two integer arrays <code>quality</code> and <code>wage</code> where <code>quality[i]</code> is the quality of the <code>i<sup>th</sup></code> worker and <code>wage[i]</code> is the minimum wage expectation for the <code>i<sup>th</sup></code> worker.</p>

<p>We want to hire exactly <code>k</code> workers to form a <strong>paid group</strong>. To hire a group of <code>k</code> workers, we must pay them according to the following rules:</p>

<ol>
	<li>Every worker in the paid group must be paid at least their minimum wage expectation.</li>
	<li>In the group, each worker&#39;s pay must be directly proportional to their quality. This means if a worker&rsquo;s quality is double that of another worker in the group, then they must be paid twice as much as the other worker.</li>
</ol>

<p>Given the integer <code>k</code>, return <em>the least amount of money needed to form a paid group satisfying the above conditions</em>. Answers within <code>10<sup>-5</sup></code> of the actual answer will be accepted.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> quality = [10,20,5], wage = [70,50,30], k = 2
<strong>Output:</strong> 105.00000
<strong>Explanation:</strong> We pay 70 to 0<sup>th</sup> worker and 35 to 2<sup>nd</sup> worker.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> quality = [3,1,10,10,1], wage = [4,8,2,2,7], k = 3
<strong>Output:</strong> 30.66667
<strong>Explanation:</strong> We pay 4 to 0<sup>th</sup> worker, 13.33333 to 2<sup>nd</sup> and 3<sup>rd</sup> workers separately.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>n == quality.length == wage.length</code></li>
	<li><code>1 &lt;= k &lt;= n &lt;= 10<sup>4</sup></code></li>
	<li><code>1 &lt;= quality[i], wage[i] &lt;= 10<sup>4</sup></code></li>
</ul>

import heapq

class Solution:
    def mincostToHireWorkers(self, quality: List[int], wage: List[int], k: int) -> float:
        #Step 1: calculate the ratio of wage[i]/quality[i] and sort workers by this ratio
        workers = sorted((w/q, q) for w, q in zip(wage, quality))

        #Step 2: Initialize max-heap and quality_sum
        quality_sum = 0
        max_heap = []
        min_cost = float('inf')

        #Step 3: Iterate through the sorted list of workers
        for r, q in workers:
            #Add the current quality to the max-heap (negative value for max-heap simulatenoulsy)
            heapq.heappush(max_heap, -q)
            quality_sum += q

            #If we have more than k workers in the heap, remove the one with the largest quality
            if len(max_heap) > k:
                quality_sum += heapq.heappop(max_heap)
            
            #If we have exactly k workers, calculate the cost and update the minimum cost
            if len(max_heap) == k:
                min_cost = min(min_cost, quality_sum*r)
        return min_cost
        
18600