<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>& c, int k, int ca) {
        vector<int> v(c.size(), 0);

        priority_queue<pair<int,int> , vector<pair<int,int>>, greater<pair<int,int>> > p1;
        priority_queue<pair<int,int> , vector<pair<int,int>>, greater<pair<int,int>> > p2;

        long long r=0;
        int i=0;
        int i1=0;
        int i2=c.size()-1;
        if(c.size()<2*ca) {
            sort(c.begin(), c.end());
            for(int i=0;i<c.size();i++) {
            r+=c[i];
            if(i==k-1)
            break;
            }
            return r;
        }

        while(i<ca) {
            v[i]=1;
            v[i2]=1;
            p1.push(make_pair(c[i], i));
            p2.push(make_pair(c[i2], i2));
            i++;
            i2--;
            if(p1.size()+p2.size()==c.size())
            break;
            
        }

        int l=0;
        vector<int> x;

        while(l<k) {
            if(p1.size() + p2.size()<2*ca) {
                while(!p1.empty()) {
                    x.push_back(p1.top().first);
                    p1.pop();
                } 
                while(!p2.empty()) {
                    x.push_back(p2.top().first);
                    p2.pop();
                }

                sort(x.begin(), x.end());
                break;
            }
            if(p1.top().first<=p2.top().first) {
                r+=p1.top().first;
                
                p1.pop();
                while(i<c.size() && p1.size()<ca) {
                    if(v[i]==0) {
                        v[i]=1;
                        p1.push(make_pair(c[i], i));
                       
                    }
                    i++;
                }
            } else {
                r+=p2.top().first;
                
                p2.pop();
                while(i2>=0 && p2.size()<ca) {
                    if(v[i2]==0) {
                        v[i2]=1;
                        p2.push(make_pair(c[i2], i2));
                        
                    }
                    i2--;
                }
            }
            l++;
        }
        int o=0;
        while(l<k) {
            r+=x[o++];
            l++;
        }

        return r;
    }
};
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