Welcome to Subscribe On Youtube

2462. Total Cost to Hire K Workers

Description

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

Solutions

Solution 1: Priority Queue (Min Heap)

We maintain a priority queue (min heap) for the current candidate workers, and use variables $i$ and $j$ to mark the minimum index of the frontmost worker and the minimum index of the rearmost worker. Initially, $i = \text{candidates} - 1$ and $j = n - \text{candidates}$.

First, we put the costs of the first $candidates$ workers into the priority queue, then we put the costs of the last $candidates$ workers into the priority queue. Before putting them in, we need to check whether they are already in the priority queue according to $i$ or $j$. If they are, we don’t need to put them in again.

We loop $k$ times, each time taking out the worker with the smallest cost from the priority queue and accumulating the cost. If the index $x$ of the current worker is in the index range $[0,..i]$ of the frontmost workers, we move $i$ one step to the right, and then check whether we need to put the cost of the worker corresponding to $i$ into the priority queue; if the index is in the index range $[j,..n-1]$ of the rearmost workers, we move $j$ one step to the left, and then check whether we need to put the cost of the worker corresponding to $j$ into the priority queue.

After the traversal ends, we return the accumulated cost as the answer.

The time complexity is $O(n \times \log n)$, where $n$ is the length of the array $costs$.

  • class Solution {
        public long totalCost(int[] costs, int k, int candidates) {
            PriorityQueue<int[]> q = new PriorityQueue<>((a, b) -> {
                if (a[0] == b[0]) {
                    return a[1] - b[1];
                }
                return a[0] - b[0];
            });
            int n = costs.length;
            int i = candidates - 1, j = n - candidates;
            for (int h = 0; h < candidates; ++h) {
                q.offer(new int[] {costs[h], h});
            }
            for (int h = n - candidates; h < n; ++h) {
                if (h > i) {
                    q.offer(new int[] {costs[h], h});
                }
            }
            long ans = 0;
            while (k-- > 0) {
                var e = q.poll();
                int c = e[0], x = e[1];
                ans += c;
                if (x <= i) {
                    if (++i < j) {
                        q.offer(new int[] {costs[i], i});
                    }
                }
                if (x >= j) {
                    if (--j > i) {
                        q.offer(new int[] {costs[j], j});
                    }
                }
            }
            return ans;
        }
    }
    
  • using pii = pair<int, int>;
    
    class Solution {
    public:
        long long totalCost(vector<int>& costs, int k, int candidates) {
            priority_queue<pii, vector<pii>, greater<pii>> q;
            int n = costs.size();
            int i = candidates - 1, j = n - candidates;
            for (int h = 0; h < candidates; ++h) q.push({costs[h], h});
            for (int h = n - candidates; h < n; ++h)
                if (h > i) q.push({costs[h], h});
            long long ans = 0;
            while (k--) {
                auto [c, x] = q.top();
                q.pop();
                ans += c;
                if (x <= i) {
                    if (++i < j) {
                        q.push({costs[i], i});
                    }
                }
                if (x >= j) {
                    if (--j > i) {
                        q.push({costs[j], j});
                    }
                }
            }
            return ans;
        }
    };
    
  • class Solution:
        def totalCost(self, costs: List[int], k: int, candidates: int) -> int:
            q = []
            n = len(costs)
            i, j = candidates - 1, n - candidates
            for h in range(candidates):
                q.append((costs[h], h))
            for h in range(n - candidates, n):
                if h > i:
                    q.append((costs[h], h))
            heapify(q)
            ans = 0
            for _ in range(k):
                c, x = heappop(q)
                ans += c
                if x <= i:
                    i += 1
                    if i < j:
                        heappush(q, (costs[i], i))
                if x >= j:
                    j -= 1
                    if i < j:
                        heappush(q, (costs[j], j))
            return ans
    
    
  • func totalCost(costs []int, k int, candidates int) int64 {
    	q := hp{}
    	n := len(costs)
    	i, j := candidates-1, n-candidates
    	for h := 0; h < candidates; h++ {
    		heap.Push(&q, pair{costs[h], h})
    	}
    	for h := n - candidates; h < n; h++ {
    		if h > i {
    			heap.Push(&q, pair{costs[h], h})
    		}
    	}
    	ans := 0
    	for k > 0 {
    		p := heap.Pop(&q).(pair)
    		c, x := p.c, p.x
    		ans += c
    		if x <= i {
    			i++
    			if i < j {
    				heap.Push(&q, pair{costs[i], i})
    			}
    		}
    		if x >= j {
    			j--
    			if i < j {
    				heap.Push(&q, pair{costs[j], j})
    			}
    		}
    		k--
    	}
    	return int64(ans)
    }
    
    type pair struct{ c, x int }
    type hp []pair
    
    func (h hp) Len() int           { return len(h) }
    func (h hp) Less(i, j int) bool { return h[i].c < h[j].c || h[i].c == h[j].c && h[i].x < h[j].x }
    func (h hp) Swap(i, j int)      { h[i], h[j] = h[j], h[i] }
    func (h *hp) Push(v any)        { *h = append(*h, v.(pair)) }
    func (h *hp) Pop() any          { a := *h; v := a[len(a)-1]; *h = a[:len(a)-1]; return v }
    

All Problems

All Solutions