LeetCode_2553_Total Cost to Hire K Workers (Java)

2553. Total Cost to Hire K Workers

Medium

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You are given a 0-indexed integer array costs where costs[i] is the cost of hiring the ith worker.

You are also given two integers k and candidates. We want to hire exactly k workers according to the following rules:

• You will run k sessions and hire exactly one worker in each session.
• In each hiring session, choose the worker with the lowest cost from either the first candidates workers or the last candidates workers. Break the tie by the smallest index.
  • For example, if costs = [3,2,7,7,1,2] and candidates = 2, then in the first hiring session, we will choose the 4th worker because they have the lowest cost [3,2,7,7,1,2].
  • In the second hiring session, we will choose 1st worker because they have the same lowest cost as 4th worker but they have the smallest index [3,2,7,7,2]. Please note that the indexing may be changed in the process.
• If there are fewer than candidates workers remaining, choose the worker with the lowest cost among them. Break the tie by the smallest index.
• A worker can only be chosen once.

Return the total cost to hire exactly k workers.

 

Example 1:

Input: costs = [17,12,10,2,7,2,11,20,8], k = 3, candidates = 4
Output: 11
Explanation: We hire 3 workers in total. The total cost is initially 0.
- In the first hiring round we choose the worker from [17,12,10,2,7,2,11,20,8]. The lowest cost is 2, and we break the tie by the smallest index, which is 3. The total cost = 0 + 2 = 2.
- In the second hiring round we choose the worker from [17,12,10,7,2,11,20,8]. The lowest cost is 2 (index 4). The total cost = 2 + 2 = 4.
- In the third hiring round we choose the worker from [17,12,10,7,11,20,8]. The lowest cost is 7 (index 3). The total cost = 4 + 7 = 11. Notice that the worker with index 3 was common in the first and last four workers.
The total hiring cost is 11.

Example 2:

Input: costs = [1,2,4,1], k = 3, candidates = 3
Output: 4
Explanation: We hire 3 workers in total. The total cost is initially 0.
- In the first hiring round we choose the worker from [1,2,4,1]. The lowest cost is 1, and we break the tie by the smallest index, which is 0. The total cost = 0 + 1 = 1. Notice that workers with index 1 and 2 are common in the first and last 3 workers.
- In the second hiring round we choose the worker from [2,4,1]. The lowest cost is 1 (index 2). The total cost = 1 + 1 = 2.
- In the third hiring round there are less than three candidates. We choose the worker from the remaining workers [2,4]. The lowest cost is 2 (index 0). The total cost = 2 + 2 = 4.
The total hiring cost is 4.

 

Constraints:

• 1 <= costs.length <= 105
• 1 <= costs[i] <= 105
• 1 <= k, candidates <= costs.length

문제 풀이

왼쪽K개중 최솟값, 오른쪽 K개중 최솟값을 O(1)에 찾으며 삽입시 log(N)으로 정렬하는 최소힙 2개를 사용했다. 그리고 서로의 구간을 침범하지 않게 left<=right를 유지했다.

코드

class Solution {
    public long totalCost(int[] costs, int k, int candidates) {
        int left, right;
        left = 0;
        right = costs.length - 1;
        PriorityQueue<Integer> left_pq = new PriorityQueue<>();
        PriorityQueue<Integer> right_pq = new PriorityQueue<>();
         
        for(int i=0; i<candidates; i++){
            if(left <= right) left_pq.offer(costs[left++]);
        }
        for(int i=0; i<candidates; i++){
            if(left <= right) right_pq.offer(costs[right--]);
        }
        long res=0;

        while(k-->0){
            int a, b;
            if(left_pq.isEmpty()) {
                res += right_pq.poll();
                continue;
            }
            else if(right_pq.isEmpty()){
                res += left_pq.poll();
                continue;
            }

            a = left_pq.peek();
            b = right_pq.peek();
            if(a <= b) {
                left_pq.poll();
                if(left <= right) left_pq.offer(costs[left++]);
                res+= a;
            }    
            else if (b < a){
                right_pq.poll();
                if(left <= right) right_pq.offer(costs[right--]);
                res += b;
            }
        }

        return res;
    }
}