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Weekly Contest 318 Q3
Array
Two Pointers
Simulation
Heap (Priority Queue)

2462. Total Cost to Hire K Workers

中文文档

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)

First, we check if $candidates \times 2$ is greater than or equal to $n$. If it is, we directly return the sum of the costs of the first $k$ smallest workers.

Otherwise, we use a min heap $pq$ to maintain the costs of the first $candidates$ workers and the last $candidates$ workers.

We first add the costs and corresponding indices of the first $candidates$ workers to the min heap $pq$, and then add the costs and corresponding indices of the last $candidates$ workers to the min heap $pq$. We use two pointers $l$ and $r$ to point to the indices of the front and back candidates, initially $l = candidates$, $r = n - candidates - 1$.

Then we perform $k$ iterations, each time taking the worker with the smallest cost from the min heap $pq$ and adding its cost to the answer. If $l &gt; r$, it means that all the front and back candidates have been selected, and we skip directly. Otherwise, if the index of the current worker is less than $l$, it means it is a worker from the front, we add the cost and index of the $l$-th worker to the min heap $pq$, and then increment $l$; otherwise, we add the cost and index of the $r$-th worker to the min heap $pq$, and then decrement $r$.

After the loop ends, we return the answer.

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

Python3

class Solution:
    def totalCost(self, costs: List[int], k: int, candidates: int) -> int:
        n = len(costs)
        if candidates * 2 >= n:
            return sum(sorted(costs)[:k])
        pq = []
        for i, c in enumerate(costs[:candidates]):
            heappush(pq, (c, i))
        for i in range(n - candidates, n):
            heappush(pq, (costs[i], i))
        heapify(pq)
        l, r = candidates, n - candidates - 1
        ans = 0
        for _ in range(k):
            c, i = heappop(pq)
            ans += c
            if l > r:
                continue
            if i < l:
                heappush(pq, (costs[l], l))
                l += 1
            else:
                heappush(pq, (costs[r], r))
                r -= 1
        return ans

Java

class Solution {
    public long totalCost(int[] costs, int k, int candidates) {
        int n = costs.length;
        long ans = 0;
        if (candidates * 2 >= n) {
            Arrays.sort(costs);
            for (int i = 0; i < k; ++i) {
                ans += costs[i];
            }
            return ans;
        }
        PriorityQueue<int[]> pq
            = new PriorityQueue<>((a, b) -> a[0] == b[0] ? a[1] - b[1] : a[0] - b[0]);
        for (int i = 0; i < candidates; ++i) {
            pq.offer(new int[] {costs[i], i});
            pq.offer(new int[] {costs[n - i - 1], n - i - 1});
        }
        int l = candidates, r = n - candidates - 1;
        while (k-- > 0) {
            var p = pq.poll();
            ans += p[0];
            if (l > r) {
                continue;
            }
            if (p[1] < l) {
                pq.offer(new int[] {costs[l], l++});
            } else {
                pq.offer(new int[] {costs[r], r--});
            }
        }
        return ans;
    }
}

C++

class Solution {
public:
    long long totalCost(vector<int>& costs, int k, int candidates) {
        int n = costs.size();
        if (candidates * 2 > n) {
            sort(costs.begin(), costs.end());
            return accumulate(costs.begin(), costs.begin() + k, 0LL);
        }
        using pii = pair<int, int>;
        priority_queue<pii, vector<pii>, greater<pii>> pq;
        for (int i = 0; i < candidates; ++i) {
            pq.emplace(costs[i], i);
            pq.emplace(costs[n - i - 1], n - i - 1);
        }
        long long ans = 0;
        int l = candidates, r = n - candidates - 1;
        while (k--) {
            auto [cost, i] = pq.top();
            pq.pop();
            ans += cost;
            if (l > r) {
                continue;
            }
            if (i < l) {
                pq.emplace(costs[l], l++);
            } else {
                pq.emplace(costs[r], r--);
            }
        }
        return ans;
    }
};

Go

func totalCost(costs []int, k int, candidates int) (ans int64) {
	n := len(costs)
	if candidates*2 > n {
		sort.Ints(costs)
		for _, x := range costs[:k] {
			ans += int64(x)
		}
		return
	}
	pq := hp{}
	for i, x := range costs[:candidates] {
		heap.Push(&pq, pair{x, i})
		heap.Push(&pq, pair{costs[n-i-1], n - i - 1})
	}
	l, r := candidates, n-candidates-1
	for ; k > 0; k-- {
		p := heap.Pop(&pq).(pair)
		ans += int64(p.cost)
		if l > r {
			continue
		}
		if p.i < l {
			heap.Push(&pq, pair{costs[l], l})
			l++
		} else {
			heap.Push(&pq, pair{costs[r], r})
			r--
		}
	}
	return
}

type pair struct{ cost, i int }
type hp []pair

func (h hp) Len() int { return len(h) }
func (h hp) Less(i, j int) bool {
	return h[i].cost < h[j].cost || (h[i].cost == h[j].cost && h[i].i < h[j].i)
}
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 }

TypeScript

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;
}