[2462. Total Cost to Hire K Workers](https://leetcode.com/problems/total-cost-to-hire-k-workers/)

### 題目描述

You are given a **0-indexed** integer array `costs` where `costs[i]` is the cost of hiring the i^th^ 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 4^th^ worker because they have the lowest cost `[3,2,7,7,1,2]`.
* In the second hiring session, we will choose 1^st^ worker because they have the same lowest cost as 4^th^ 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.*

### 範例

**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` <= 10^5^
* 1 <= `costs[i]` <= 10^5^
* 1 <= `k`, `candidates` <= `costs.length`

### 解答
#### Python
不看 Discussion 根本看不懂這題在幹嘛= =

class Solution:
    def totalCost(self, costs: List[int], k: int, candidates: int) -> int:
        head_workers = costs[:candidates]
        tail_workers = costs[max(candidates, len(costs) - candidates):]
        heapq.heapify(head_workers)
        heapq.heapify(tail_workers)

        ans = 0
        next_head, next_tail = candidates, len(costs) - 1 - candidates

        for _ in range(k):
            if not tail_workers or head_workers and head_workers[0] <= tail_workers[0]:
                ans += heapq.heappop(head_workers)
                if next_head <= next_tail:
                    heapq.heappush(head_workers, costs[next_head])
                    next_head += 1
            else:
                ans += heapq.heappop(tail_workers)
                if next_head <= next_tail:
                    heapq.heappush(tail_workers, costs[next_tail])
                    next_tail -= 1
        
        return ans

> [name=Ron Chen][time=Mon, Jun 26, 2023]

### Reference

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