Question 52 of 75 - LeetCode Weekly Challenge

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();
            }
        }
        return total;
    }
};
				
			

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