Today I have done leetcode's June LeetCoding Challenge with cpp.

June LeetCoding Challenge 26

Description

Total Cost to Hire K Workers

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

Solution

auto speedup = [](){
  cin.tie(nullptr);
  cout.tie(nullptr);
  ios::sync_with_stdio(false);
  return 0;
}();
class Solution {
  using pi = pair<int, int>;
public:
  long long totalCost(vector<int>& costs, int k, int candidates) {
    int len = costs.size();
    priority_queue<pi, vector<pi>, greater<pi>> pq;
    for(int i = 0; i < candidates && i * 2 < len; ++i) {
      pq.push({costs[i], i});
      if(len - 1 - i != i) pq.push({costs[len - 1 - i], len - i - 1});
    }

    int frontPos = candidates;
    int backPos = len - 1 - candidates;
    long long answer = 0;
    while(k--) {
      auto [cost, pos] = pq.top();
      pq.pop();
      answer += cost;
      if(frontPos > backPos) {
        continue;
      }
      if(pos < frontPos) {
        pq.push({costs[frontPos], frontPos});
        frontPos += 1;
      } else {
        pq.push({costs[backPos], backPos});
        backPos -= 1;
      }
    }

    return answer;
  }
};

// Accepted
// 160/160 cases passed (232 ms)
// Your runtime beats 45.42 % of cpp submissions
// Your memory usage beats 46.64 % of cpp submissions (75.3 MB)