<p>You are given a <strong>0-indexed</strong> integer array <code>costs</code> where <code>costs[i]</code> is the cost of hiring the <code>i<sup>th</sup></code> worker.</p>

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

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

<p>Return <em>the total cost to hire exactly </em><code>k</code><em> workers.</em></p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> costs = [17,12,10,2,7,2,11,20,8], k = 3, candidates = 4
<strong>Output:</strong> 11
<strong>Explanation:</strong> We hire 3 workers in total. The total cost is initially 0.
- In the first hiring round we choose the worker from [<u>17,12,10,2</u>,7,<u>2,11,20,8</u>]. 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 [<u>17,12,10,7</u>,<u>2,11,20,8</u>]. The lowest cost is 2 (index 4). The total cost = 2 + 2 = 4.
- In the third hiring round we choose the worker from [<u>17,12,10,7,11,20,8</u>]. 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.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> costs = [1,2,4,1], k = 3, candidates = 3
<strong>Output:</strong> 4
<strong>Explanation:</strong> We hire 3 workers in total. The total cost is initially 0.
- In the first hiring round we choose the worker from [<u>1,2,4,1</u>]. 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 [<u>2,4,1</u>]. 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 [<u>2,4</u>]. The lowest cost is 2 (index 0). The total cost = 2 + 2 = 4.
The total hiring cost is 4.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= costs.length &lt;= 10<sup>5 </sup></code></li>
	<li><code>1 &lt;= costs[i] &lt;= 10<sup>5</sup></code></li>
	<li><code>1 &lt;= k, candidates &lt;= costs.length</code></li>
</ul>

class Solution:
    def totalCost(self, costs: List[int], k: int, candidates: int) -> int:
        # head_workers stores the first k workers.
        # tail_workers stores at most last k workers without any workers from the first k workers
        head_workers = costs[:candidates]
        tail_workers = costs[max(candidates, len(costs) - candidates):]
        heapify(head_workers)
        heapify(tail_workers)

        answer = 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]:
                answer += heappop(head_workers)

                # Only refill the queue if there are workers outside the two queues.
                if next_head <= next_tail:
                    heappush(head_workers, costs[next_head])
                    next_head += 1

            else:
                answer += heappop(tail_workers)

                # Only refill the queue if there are workers outside the two queues.
                if next_head <= next_tail:
                    heappush(tail_workers, costs[next_tail])
                    next_tail -= 1
            
        return answer
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