2462. Total Cost to Hire K

  • Hardness: \(\color{orange}\textsf{Medium}\)
  • Ralated Topics: Array、Two Pointer、Heap (Priority Queue)、Simulation
  • \(\color{blue}\textsf{Weekly Contest 318}\)

一、題目

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 <= 10^5
  • 1 <= costs[i] <= 10^5
  • 1 <= k, candidates <= costs.length

二、分析

  • 將數組左右兩邊 candidates 長度的數字加入各別的 min heap 中，按順序比較兩個 min heap，將小的加入和中。
  • 注意當從 min heap 中取值之後，要移動指針，將新的值放入 min heap，除非當左右指針相撞(代表整個數組都在 candidate 的範圍內)。

三、解題

1. Heap (Priority Queue)

  • Time complexity: \(O(n)\)
  • Space complexity: \(O(n)\)
long long totalCost(vector<int>& costs, int k, int candidates) {
    priority_queue<int, vector<int>, greater<int>> front, end;  // 宣告兩個 min heap
    long long res = 0;
    int cnt = 0;
    int i = 0, j = costs.size()-1;
    front.push(INT_MAX);                                        // 當 min heap 為空時，必定傳回另一個 min heap 的值
    end.push(INT_MAX);
    while (i <= j && cnt < candidates) {
        front.push(costs[i++]);                                 // 左指針移動
        if (i <= j) end.push(costs[j--]);                       // 右指針移動，左右指針相撞，代表已經包含整個數組
        cnt++;
    }
    while (k--) {                                               // 取 k 個數字
        if (front.top() <= end.top()) {
            res += front.top();
            front.pop();
            if (i <= j) front.push(costs[i++]);                 // 左右指針相撞，不再加入新的值
        } else {
            res += end.top();
            end.pop();
            if (i <= j) end.push(costs[j--]);                   // 左右指針相撞，不再加入新的值
        }
    }
    return res;
}

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