Total Cost To Hire K Workers - Leetcode Solution

The "Total Cost To Hire K Workers" problem on LeetCode is a problem that asks us to find the minimum cost of hiring K workers to complete a project. Each worker has a certain quality, and if we want to hire a worker with a higher quality, we need to pay them more. Additionally, there is a ratio that dictates the minimum salary we must pay a worker based on their quality.

To solve this problem, we need to first understand the given problem statement. There are N workers, each worker has a quality (qi) and a wage (wi). Additionally, there is a ratio (r) which suggests that if we want to hire a worker with quality Q2 and we have already hired a worker with quality Q1, then we must pay Q2*r salary to Q2 worker to maintain a fair ratio.

From the problem, we are to find the minimum cost that would be spent hiring K workers. To do this, we must first sort the workers based on their expected salaries (s), where s = wi/qi. The intuition behind this is to find the workers that offer the best value for money.

We then perform a sliding window search to obtain the K workers with the lowest expected salaries. During the sliding window search, we keep track of the sum of the salaries of the K workers. Additionally, since we want to maintain a fair ratio, we ensure that the ratio is met for each worker within the window.

The algorithm for solving this problem is as follows:

1. Sort the workers based on their expected salaries s = wi/qi in ascending order.
2. Initialize a priority queue and a variable to track the quality of the Kth worker seen so far.
3. Initialize a variable to track the sum of salaries of the K workers seen so far. Set this to infinity as the initial value.
4. Loop through each worker in the sorted workers:
   1. Add the current worker's quality to the priority queue.
   2. If the length of the priority queue is greater than K, remove the worker with the max quality from the queue. (since we want to keep only the K workers with the lowest expected salaries)
   3. If the length of the priority queue is K, calculate the sum of salaries of the K workers using the ratio r.
   4. If the sum of salaries is less than the current minimum sum seen so far, update the minimum sum.
5. Return the minimum sum of salaries.

The time complexity of this algorithm is O(Nlog(N)) since we first sort the workers based on their expected salaries and then perform a sliding window search, where we add and remove elements from a priority queue. The space complexity is O(N) which is the size of the priority queue.

Here is the Python code that implements the algorithm:

import heapq

class Solution:
    def mincostToHireWorkers(self, quality: List[int], wage: List[int], k: int) -> float:
        workers = sorted([(w/q, q, w) for q, w in zip(quality, wage)]) # sort workers by expected salaries
        
        heap = [] # initialize a priority queue
        total_quality = 0 # initialize a variable for the quality of the Kth worker
        min_cost = float('inf') # initialize the minimum cost variable
        
        for s, q, w in workers:
            heapq.heappush(heap, -q) # add current worker's quality to the priority queue
            total_quality += q # update total quality
            
            if len(heap) > k: # if priority queue contains more than K workers
                total_quality += heapq.heappop(heap) # remove worker with the maximum quality
                                        
            if len(heap) == k: # if priority queue contains K workers
                cost = total_quality * s # calculate cost
                min_cost = min(min_cost, cost) # update minimum cost
                
        return min_cost

In summary, the "Total Cost To Hire K Workers" problem on LeetCode can be solved using a sliding window approach that involves the use of a priority queue. We first sort the workers by their expected salaries, and then we perform a sliding window search to obtain the K workers with the lowest expected salaries while ensuring that the ratio constraint is met.

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