<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 {

    struct cmp {
        bool operator()(pair<int, int>& a, pair<int, int>& b) {
            return a.second > b.second ||
                   a.second == b.second && a.first > b.first;
        }
    };

public:
    long long totalCost(vector<int>& costs, int k, int candidates) {
        priority_queue<pair<int, int>, vector<pair<int, int>>, cmp> lpq, rpq;
        long long res = 0;
        int li = 0, ri = costs.size() - 1;
        unordered_set<int> selectedCands, rRepeatCands, lRepeatCands;
        while (selectedCands.size() < k) {
            while (lpq.size() - lRepeatCands.size() < candidates &&
                   li < costs.size()) {
                if (selectedCands.find(li) == selectedCands.end()) {
                    lpq.push({li, costs[li]});
                }
                li++;
            }
            while (rpq.size() - rRepeatCands.size() < candidates && ri >= 0) {
                if (selectedCands.find(ri) == selectedCands.end()) {
                    rpq.push({ri, costs[ri]});
                }
                ri--;
            }

            int curSelect;
            pair<int, int> rtop = rpq.top(), ltop = lpq.top();

            if (li == costs.size()) {
                curSelect = ltop.first;
                if (selectedCands.find(curSelect) == selectedCands.end()) {
                    res += ltop.second;
                    selectedCands.insert(curSelect);
                }
                lpq.pop();
                continue;
            }
            if (selectedCands.find(ltop.first) != selectedCands.end()){
                lpq.pop();
                continue;
            }
            if (selectedCands.find(rtop.first) != selectedCands.end()){
                rpq.pop();
                continue;
            }
            if (ltop.second <= rtop.second) {
                curSelect = ltop.first;
                if (curSelect > ri) {
                    rRepeatCands.insert(curSelect);
                }
                res += ltop.second;
                lpq.pop();
            } else {
                curSelect = rtop.first;
                if (curSelect < li) {
                    lRepeatCands.insert(curSelect);
                }
                res += rtop.second;
                rpq.pop();
            }
            selectedCands.insert(curSelect);
        }
        return res;
    }
};
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