26.6.23(2462. Total Cost to Hire K Workers(Medium))

class Solution {
public:
    long long totalCost(vector<int>& cost, int k, int take) 
    {
        typedef pair<int,int>p;

        int n = cost.size();
        //if 2*take >= n then we will simply push everything into a pq then run it for k times
        long long ans = 0;
        priority_queue<p,vector<p>,greater<p>>pq;

        if(2*take >= n)
        {
           for(auto i : cost) 
           {
               pq.push({i,0});
           }
           while(k--)
           {
               auto it = pq.top();
               ans += it.first;
               pq.pop();
           }
           return ans;
        }
        else
        {
            int i,j;
            //from front 
            int temp = take;
            for(i = 0; temp--;i++)
            {
                pq.push({cost[i],0}); // from front as indicator push 0;
            }
            i--;
            //from back
            temp = take;
            for(j = n-1;temp--;j--)
            {
                pq.push({cost[j],1}); // from back as indicator push 1;
            }
            j++;

            bool flag = false ;
            while(pq.size() && k--)
            {
               auto it = pq.top();
               pq.pop();

               int c = it.first; // c --> cost[i] 
               int ind = it.second; // ind --> indicator that it from first or last

               ans += c;

               (!flag && ind == 0)? i++ : j--;

               if(i >= j)
                flag = true;
               else
                (ind == 0)? pq.push({cost[i],0}) : pq.push({cost[j],1});
            }
            return ans;
        }
    }
};