<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>

// long long ans = 0;
//         priority_queue<pair<int,int>,vector<pair<int,int>>,greater<pair<int,int>>> left;
//         priority_queue<pair<int,int>,vector<pair<int,int>>,greater<pair<int,int>>> right;
        
//         int n = costs.size();
        
//         for(int i=0;i<n;i++){
//             left.push({costs[i],i});
//             right.push({costs[n-i-1],n-i-1});
//         }

//         if(left.top().first>right.top().first){

//         }
//         else if(left.top().first<right.top().first){

//         }
//         else{
            
//         }

//         cout<<q.top().second;

class Solution {
public:
    long long totalCost(vector<int>& costs, int k, int candidates) {

        long long ans = 0;
        priority_queue<int,vector<int>,greater<int>> left;
        priority_queue<int,vector<int>,greater<int>> right;
        unordered_map<int,bool> m;
        
        int n = costs.size();

        int currleft = 0;
        int currright = n-1;

        
        for(int i=0;i<candidates;i++){
            left.push(costs[currleft++]);            
        }

        for(int i=0;i<candidates;i++){
            if(currleft<=currright)
                right.push(costs[currright--]);           
        }

                


        
        

        for(int i=0;i<k;i++){
            if(left.empty()){
                left.push(INT_MAX);
            }

            if(right.empty()){
                right.push(INT_MAX);
            }



            if(left.top()>right.top()){
                ans += right.top();
                right.pop();
                if(currleft<=currright){
                    right.push(costs[currright]);
                    m[currright] = 1;
                    currright -= 1;
                }    

                cout<<"aa"<<endl;           
            }
            else{
                ans += left.top();
                left.pop();
                if(currleft<=currright){
                    left.push(costs[currleft]);
                    m[currleft] = 1;
                    currleft += 1;
                }
            }

            cout<<ans<<endl;

        }    

        return ans;
        
    }
};
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