<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<pair<int, int>, vector<pair<int, int>>, greater<pair<int, int>>> pq1, pq2;
        long long ans = 0;
        int n = costs.size();
        candidates = min(candidates, n);
        int l = 0, r = 0;

        for(int i = 0; i < candidates; i++) {
            pq1.push(make_pair(costs[i], i));
            l = i;
            pq2.push(make_pair(costs[n - i - 1], n - i - 1));
            r = n - i - 1;
        }

        // while(!pq1.empty()) {
        //     cout << pq1.top().first << " " << pq1.top().second << endl;
        //     pq1.pop();
        // }
        // cout << endl;
        // while(!pq2.empty()) {
        //     cout << pq2.top().first << " " << pq2.top().second << endl;
        //     pq2.pop();
        // }
        // cout << endl;
        // for(int i = 0; i < candidates; i++) {
        //     pq1.push(make_pair(costs[i], i));
        //     l = i;
        //     pq2.push(make_pair(costs[n - i - 1], n - i - 1));
        //     r = n - i - 1;
        // }

        for(int i = 0; i < k; i++)  {
            // cout << "Looking at " << pq1.top().first <<"," << pq1.top().second;
            // cout << " and " << pq2.top().first << "," << pq2.top().second << endl;
            

            if(pq2.size() == 0 || (pq1.size() > 0 && pq1.top().first < pq2.top().first)) {
                ans += pq1.top().first;
                // cout << pq1.top().first << " "<< pq1.top().second;
                pq1.pop();
                    // cout << endl << "Size "<< pq1.size() << endl;

                l++;
                if(l < n && l < r)
                    pq1.push(make_pair(costs[l], l));
            }
            else if(pq1.size() > 0 && pq2.size() > 0 && pq1.top().first == pq2.top().first)   {
                ans += pq1.top().first;
                // cout << pq1.top().first << " ";
                if(pq1.top().second < pq2.top().second)    {
                    // cout<< pq1.top().second;
                    
                    pq1.pop();
                    l++;
                    if(l < n && l < r)
                        pq1.push(make_pair(costs[l], l));
                }
                else if(pq1.top().second == pq2.top().second)   {
                    // cout<< pq1.top().second;
                    
                    pq2.pop();
                    pq1.pop();
                    l++;
                    r--;
                }
                else    {
                    // cout<< pq2.top().second;
                    pq2.pop();
                    r--;
                    if(r >= 0 && r > l)
                        pq2.push(make_pair(costs[r], r));
                }
            }
            else if(pq1.size() == 0 ||  (pq2.size() > 0 && pq1.top().first > pq2.top().first))   {
                ans += pq2.top().first;
                // cout << pq2.top().first << " " << pq2.top().second;
                pq2.pop();
                r--;
                if(r >= 0 && r > l)
                    pq2.push(make_pair(costs[r], r));
            }
            // cout << endl;
        }
        return ans;
    }
};
83245