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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
ksessions and hire exactly one worker in each session. - In each hiring session, choose the worker with the lowest cost from either the first
candidatesworkers or the lastcandidatesworkers. Break the tie by the smallest index.- For example, if
costs = [3,2,7,7,1,2]andcandidates = 2, then in the first hiring session, we will choose the4thworker because they have the lowest cost[3,2,7,7,1,2]. - In the second hiring session, we will choose
1stworker because they have the same lowest cost as4thworker but they have the smallest index[3,2,7,7,2]. Please note that the indexing may be changed in the process.
- For example, if
- 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 <= 1051 <= costs[i] <= 1051 <= 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 $costs$.
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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 } -
function totalCost(costs: number[], k: number, candidates: number): number { const n = costs.length; if (candidates * 2 >= n) { costs.sort((a, b) => a - b); return costs.slice(0, k).reduce((acc, x) => acc + x, 0); } const pq = new PriorityQueue({ compare: (a, b) => (a[0] === b[0] ? a[1] - b[1] : a[0] - b[0]), }); for (let i = 0; i < candidates; ++i) { pq.enqueue([costs[i], i]); pq.enqueue([costs[n - i - 1], n - i - 1]); } let [l, r] = [candidates, n - candidates - 1]; let ans = 0; while (k--) { const [cost, i] = pq.dequeue()!; ans += cost; if (l > r) { continue; } if (i < l) { pq.enqueue([costs[l], l++]); } else { pq.enqueue([costs[r], r--]); } } return ans; }