## 2462. Total Cost to Hire K Workers
https://leetcode.com/problems/total-cost-to-hire-k-workers/

## Two heaps
- Idea
- using two min heap from both side
- Complexity
- Time: O(NlogN + klogN), N is the # of candidates, k means # of hiring rounds
- Space: O(N)

class Solution {
public:
    long long totalCost(vector<int>& costs, int k, int candidates) {
        int l = 0, r = costs.size() - 1;
        priority_queue<int, vector<int>, greater<int>> lPq;
        priority_queue<int, vector<int>, greater<int>> rPq;

        long long cost = 0;
        while(k--){
            while(lPq.size() < candidates and l <= r){
                lPq.push(costs[l++]);
            }
            while(rPq.size() < candidates and l <= r){
                rPq.push(costs[r--]);
            }

            int pick1 = lPq.size() > 0 ? lPq.top() : INT_MAX;
            int pick2 = rPq.size() > 0 ? rPq.top() : INT_MAX;

            if (pick1 <= pick2){
                cost += pick1;
                lPq.pop();
            }
            else{
                cost += pick2;
                rPq.pop();
            }
        }

        return cost;
    }
};