857. Minimum Cost to Hire K Workers

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.

Approach #1: C++.

class Solution {
public:
    double mincostToHireWorkers(vector<int>& quality, vector<int>& wage, int K) {
        vector<vector<double>> workers;
        for (int i = 0; i < wage.size(); ++i)
            workers.push_back({(double)wage[i]/quality[i], (double)quality[i]});

        sort(workers.begin(), workers.end());

        double res = INT_MAX, qsum = 0;

        priority_queue<int> pq;
        for (auto worker : workers) {
            qsum += worker[1];
            pq.push(worker[1]);
            if (pq.size() > K) qsum -= pq.top(), pq.pop();
            if (pq.size() == K) res = min(res, qsum*worker[0]);
        }
        return res;
    }
};

　　

Analysis:

In this solution we use a vector to store the ratio of wage/quality and the quality, then sort the vector with ratio.

We travel the vector when the priority_queue's size < k we add the quality to the qsum.

When priority_queue's size == K we calculate the total wages at this status.

Last we select the minimum total wages as the result.

Time Complexity

O(NlogN) for sort.
O(NlogK) for priority queue.

Approach #2: Java. [Greedy]

class Solution {
    public double mincostToHireWorkers(int[] quality, int[] wage, int K) {
        int N = quality.length;
        double ans = 1e9;

        for (int captain = 0; captain < N; ++captain) {
            double factor = (double)wage[captain] / quality[captain];
            double prices[] = new double[N];
            int t = 0;
            for (int worker = 0; worker < N; ++worker) {
                double price = factor * quality[worker];
                if (price < wage[worker]) continue;
                prices[t++] = price;
            }

            if (t < K) continue;
            Arrays.sort(prices, 0, t);
            double cand = 0;
            for (int i = 0; i < K; ++i)
                cand += prices[i];
            ans = Math.min(ans, cand);
        }

        return ans;
    }
}

　

Analysis:

Having the similar thinking with above code, but this solution don't use heap to maintain the ratio, so the time complex is bigger than above.

Time Complexity:

O(N^2 \log N)O(N2logN), where NN is the number of workers.