<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 {
public:
    long long totalCost(vector<int>& costs, int k, int candidates) {
        // priority_queue<int, vector<int>, greater<int>> head, tail;
        long long total_cost = 0;
        if(costs.size() <= k){
            for(auto& c : costs){
                total_cost += c;
            }
            return total_cost;
        }

        auto l_begin = costs.begin();
        auto l_end = costs.begin() + candidates;
        auto r_begin = costs.end() - candidates;
        auto r_end = costs.end();

        auto l_min_idx = min_element(l_begin, l_end);
        auto r_min_idx = min_element(r_begin, r_end);

        bool single = false;

        while(k > 0){
            k --;
            // cout << total_cost << endl;
            // cout << "--------" << endl;
            // cout << "l : " << *l_min_idx << endl;
            // cout << "r : " << *r_min_idx << endl;
            if(*l_min_idx <= *r_min_idx){
                total_cost += *l_min_idx;
                *l_min_idx = *l_begin;
                ++ l_begin;

                if(++l_end >= r_begin){
                    l_end = r_end;
                    break;
                }else{
                    l_min_idx = min_element(l_begin, l_end);
                }
            }else{
                total_cost += *r_min_idx;
                *r_min_idx = *(r_end - 1);
                -- r_end;

                if(--r_begin <= l_end){
                    l_end = r_end;
                    break;
                }else{
                    r_min_idx = min_element(r_begin, r_end);
                }

            }
        }

        sort(l_begin, l_end);

        while(k > 0){
            total_cost += *l_begin;
            ++ l_begin;
            k--;
        }
    
        return total_cost;
    }
};

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