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Suppose we have an array called quality for each different worker, and have another array called wages and a value K. The i-th worker has a quality[i] and a minimum wage expectation wage[i]. We want to hire K workers to form a paid group. When we are hiring a group of K workers, we must pay them according to the following rules:

Each worker in the paid group should be paid in the ratio of their quality by comparing with others in the paid group.

Every worker in the paid group must be paid at least their minimum wage expectation.

We have to find the least amount of money needed to form a paid group satisfying the above conditions.

So, if the input is like quality = [10,22,5], wage = [70,52,30] and K = 2, then the output will be 105.000 because we will pay 70 to the first worker and 35 to the 3rd worker.

To solve this, we will follow these steps −

qr := a new list

for each pair (q, w) from quality and wage, do

insert (w/q, q) at the end of qr

sort the list qr

cand := a new list, csum := 0

for i in range 0 to K - 1, do

insert -qr[i, 1] into heap cand

csum := csum + qr[i, 1]

ans := csum * qr[K - 1, 0]

for idx in range K to size of quality, do

insert -qr[i, 1] into heap cand

csum := csum + qr[idx, 1] + top element from heap cand and delete it from heap

ans = minimum of ans and (csum * qr[idx][0])

return ans

Let us see the following implementation to get better understanding

import heapq def solve(quality, wage, K): qr = [] for q, w in zip(quality, wage): qr.append([w/q, q]) qr.sort() cand, csum = [], 0 for i in range(K): heapq.heappush(cand, -qr[i][1]) csum += qr[i][1] ans = csum * qr[K - 1][0] for idx in range(K, len(quality)): heapq.heappush(cand, -qr[idx][1]) csum += qr[idx][1] + heapq.heappop(cand) ans = min(ans, csum * qr[idx][0]) return ans quality = [10,22,5] wage = [70,52,30] K = 2 print(solve(quality, wage, K))

[10,22,5], [70,52,30], 2

105

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