### LeetCode Challenge #2462. Total Cost to Hire K Workers

You are given a 0-indexed integer array `costs` where `costs[i]` is the cost of hiring the `ith` worker.

You are also given two integers `k` and `candidates`. We want to hire exactly `k` workers according to the following rules:

• You will run `k` sessions and hire exactly one worker in each session.
• In each hiring session, choose the worker with the lowest cost from either the first `candidates` workers or the last `candidates` workers. Break the tie by the smallest index.
• For example, if `costs = [3,2,7,7,1,2]` and `candidates = 2`, then in the first hiring session, we will choose the `4th` worker because they have the lowest cost `[3,2,7,7,1,2]`.
• In the second hiring session, we will choose `1st` worker because they have the same lowest cost as `4th` worker but they have the smallest index `[3,2,7,7,2]`. Please note that the indexing may be changed in the process.
• If there are fewer than candidates workers remaining, choose the worker with the lowest cost among them. Break the tie by the smallest index.
• A worker can only be chosen once.

Return the total cost to hire exactly `k` workers.

Example 1:

```Input: costs = [17,12,10,2,7,2,11,20,8], k = 3, candidates = 4
Output: 11
Explanation: We hire 3 workers in total. The total cost is initially 0.
- In the first hiring round we choose the worker from [17,12,10,2,7,2,11,20,8]. 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 [17,12,10,7,2,11,20,8]. The lowest cost is 2 (index 4). The total cost = 2 + 2 = 4.
- In the third hiring round we choose the worker from [17,12,10,7,11,20,8]. 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.
```

Example 2:

```Input: costs = [1,2,4,1], k = 3, candidates = 3
Output: 4
Explanation: We hire 3 workers in total. The total cost is initially 0.
- In the first hiring round we choose the worker from [1,2,4,1]. 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 [2,4,1]. 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 [2,4]. The lowest cost is 2 (index 0). The total cost = 2 + 2 = 4.
The total hiring cost is 4.
```

Constraints:

• `1 <= costs.length <= 105 `
• `1 <= costs[i] <= 105`
• `1 <= k, candidates <= costs.length`
##### Video Solution
###### C++ Solution
```				```
class Solution {
public:
long long totalCost(vector<int>& costs, int k, int candidates) {
int n=costs.size();
priority_queue<int,vector<int>,greater<int>> left;
priority_queue<int,vector<int>,greater<int>> right;
vector<bool>visited(n,false);
int i=0,j=n-1;

for(i=0; i<min(n,candidates); i++)
{
if(!visited[i])
{
left.push(costs[i]);
visited[i]=true;
}
}
for(j=n-1; j>=n-candidates; j--)
{
if(!visited[j])
{
right.push(costs[j]);
visited[j]=true;
}
}

long long total=0;

while(k--)
{
if(!left.empty() and !right.empty())
{
if(left.top()<right.top() or left.top()==right.top())
{
total+=left.top();
left.pop();

if(!visited[i])
{
left.push(costs[i]);
visited[i]=true;
i++;
}
}
else if(right.top()<left.top())
{
total+=right.top();
right.pop();

if(!visited[j])
{
right.push(costs[j]);
visited[j]=true;
j--;
}
}
}
else if(right.empty())
{
total+=left.top();
left.pop();
}
else if(left.empty())
{
total+=right.top();
right.pop();
}
}