<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 UnstableMinHeap:
    def __init__(self, tuples=None):
        self.heap = []
        if tuples:
            for tp in tuples:
                self.push(tp)
    def push(self, tp):
        self.heap.append(tp)
        self.sift_up(len(self.heap) - 1)

    def pop(self):
        if len(self.heap) == 0:
            raise Exception('heap is empty')
        tp = self.heap[0]
        self.swap(0, len(self.heap)-1)
        self.heap = self.heap[:-1]
        self.sift_down()
        return tp
    def swap(self, j, k):
        tmp = self.heap[j]
        self.heap[j] = self.heap[k]
        self.heap[k] = tmp 
    def sift_up(self, j):
        while j >= 1:
            parent = (j-1) // 2
            if self.heap[parent][0] > self.heap[j][0]:
                self.swap(j, parent)
                j = parent
            else:
                break
    def sift_down(self):
        i, left_c_index, right_c_index = 0, 1, 2
        while left_c_index <= len(self.heap) - 1:
            if right_c_index >= len(self.heap):  # only has one left child
                if self.heap[left_c_index][0] < self.heap[i][0]:
                    self.swap(left_c_index, i)
                    i = left_c_index
                else:
                    break
            else:  # two children
                if self.heap[i][0] > self.heap[left_c_index][0] or self.heap[i][0] > self.heap[right_c_index][0]:
                    if self.heap[left_c_index][0] < self.heap[right_c_index][0]:
                        self.swap(left_c_index, i)
                        i = left_c_index
                    else:
                        self.swap(right_c_index, i)
                        i = right_c_index
                else:
                    break
            left_c_index = 2 * i + 1
            right_c_index = 2 * i + 2                


    # def __str__(self):
    #     return '[' + ','.join(self.heap) + ']'


class Solution:
    def totalCost(self, costs: List[int], k: int, candidates: int) -> int:
        left_end_index = candidates-1
        right_start_index = len(costs) - candidates

        min_heap = []
        if left_end_index >= right_start_index-1:
            [heapq.heappush(min_heap, x) for x in zip(costs, range(0, len(costs)))]
        else:
            [heapq.heappush(min_heap, x) for x in zip(costs[:left_end_index+1], range(left_end_index+1))]
            [heapq.heappush(min_heap, x) for x in zip(costs[right_start_index:], range(right_start_index, len(costs)))]   

        total_cost = 0
        for i in range(k):
            tp = heapq.heappop(min_heap)
            total_cost += tp[0]
            picked_index = tp[1]
            if left_end_index < right_start_index - 1:                
                if picked_index <= left_end_index:
                    left_end_index += 1
                    heapq.heappush(min_heap, (costs[left_end_index], left_end_index))
                else:
                    right_start_index -= 1
                    heapq.heappush(min_heap, (costs[right_start_index], right_start_index))
        return total_cost
                    

            


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