857. Minimum Cost to Hire K Workers

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.

----------------------
找合适的长度为k的subset。遍历整个array，每次找以[i]为baseline的符合条件的subset
O(NlogN)。注意double比较的写法。sort的时候要同时sort 3个array

class Solution {
    public double mincostToHireWorkers(int[] quality, int[] wage, int k) {
        int n = quality.length;
        Person[] person = new Person[n];
        for (int i = 0; i < n; i++) {
            person[i] = new Person(quality[i], wage[i]);
        }        
        Arrays.sort(person, (a, b) -> Double.compare(a.rate, b.rate));

        PriorityQueue<Integer> que = new PriorityQueue<>((a, b) -> b - a);
        int sum = 0;
        for (int i = 0; i < k; i++) {
            que.add(person[i].quality);
            sum += person[i].quality;
        }
        
        double min = sum * person[k - 1].rate;
        for (int i = k; i < n; i++) {
            int max = que.poll();
            que.add(person[i].quality);
            sum = sum - max + person[i].quality;
            min = Math.min(min, sum * person[i].rate);
        }
        
        return min;
    }
    
    class Person{
        public int quality;
        public int wage;
        public double rate;
        public Person(int quality, int wage) {
            this.quality = quality;
            this.wage = wage;
            rate = wage * 1.0 / quality;
        }
    }
}