# Minimum Cost To Hire K Workers Problem

## Description

LeetCode Problem 857.

There are n workers. You are given two integer arrays quality and wage where quality[i] is the quality of the i^th worker and wage[i] is the minimum wage expectation for the i^th worker.

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

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

Given the integer k, return the least amount of money needed to form a paid group satisfying the above conditions. Answers within 10^-5 of the actual answer will be accepted.

Example 1:

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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^nd worker.
``````

Example 2:

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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^nd and 3^rd workers separately.
``````

Constraints:

• n == quality.length == wage.length
• 1 <= k <= n <= 10^4
• 1 <= quality[i], wage[i] <= 10^4

## Sample C++ Code

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class Solution {
public:
double mincostToHireWorkers(vector<int>& quality, vector<int>& wage, int k) {
vector<pair<double,int>> rq;
for (int i = 0; i < quality.size(); ++i) {
rq.push_back({double(wage[i]) / quality[i], quality[i]});
}
sort(rq.begin(), rq.end());
priority_queue<int> pq;
int sum = 0;
double ans = DBL_MAX;
for (auto i : rq) {
sum += i.second;
pq.push(i.second);
if (pq.size() > k)  {
sum -= pq.top();
pq.pop();
}
if (pq.size() == k) {
ans = min(ans, sum * i.first);
}
}
return ans;
}
};
``````