2462

2462. Total Cost to Hire K Workers (Medium)

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

Companies: Amazon, MathWorks

Related Topics:
Array, Two Pointers, Heap (Priority Queue), Simulation

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Solution 1. Heap + Two Pointers

// OJ: https://leetcode.com/problems/total-cost-to-hire-k-workers
// Author: github.com/lzl124631x
// Time: O((K + C) * logC)
// Space: O(C)
class Solution {
public:
    long long totalCost(vector<int>& A, int K, int C) {
        long long N = A.size(), i = 0, j = N - 1, ans = 0;
        priority_queue<pair<int, int>, vector<pair<int, int>>, greater<>> pq;
        for (int k = 0; k < C && i <= j; ++k, ++i) pq.emplace(A[i], i);
        for (int k = 0; k < C && i <= j; ++k, --j) pq.emplace(A[j], j);
        while (K--) {
            auto [c, index] = pq.top();
            pq.pop();
            ans += c;
            if (i <= j) {
                if (index < i) pq.emplace(A[i], i), ++i;
                else pq.emplace(A[j], j), --j;
            }
        }
        return ans;
    }
};