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

June LeetCoding Challenge 18

Description

Most Profit Assigning Work

You have n jobs and m workers. You are given three arrays: difficulty, profit, and worker where:

• difficulty[i] and profit[i] are the difficulty and the profit of the ith job, and
• worker[j] is the ability of jth worker (i.e., the jth worker can only complete a job with difficulty at most worker[j]).

Every worker can be assigned at most one job, but one job can be completed multiple times.

• For example, if three workers attempt the same job that pays $1, then the total profit will be $3. If a worker cannot complete any job, their profit is $0.

Return the maximum profit we can achieve after assigning the workers to the jobs.

Example 1:

Input: difficulty = [2,4,6,8,10], profit = [10,20,30,40,50], worker = [4,5,6,7]
Output: 100
Explanation: Workers are assigned jobs of difficulty [4,4,6,6] and they get a profit of [20,20,30,30] separately.

Example 2:

Input: difficulty = [85,47,57], profit = [24,66,99], worker = [40,25,25]
Output: 0

Constraints:

• n == difficulty.length
• n == profit.length
• m == worker.length
• 1 <= n, m <= 104
• 1 <= difficulty[i], profit[i], worker[i] <= 105

Solution

class Solution {
public:
  int maxProfitAssignment(vector<int>& difficulty, vector<int>& profit, vector<int>& workers) {
    map<int, int> jobs;
    for(int i = 0; i < profit.size(); ++i) {
      jobs[difficulty[i]] = max(jobs[difficulty[i]], profit[i]);
    }
    int mmax = jobs.begin()->second;
    for(auto &[d, p] : jobs) {
      p = max(mmax, p);
      mmax = p;
    }

    int answer = 0;
    for(auto worker: workers) {
      auto it = jobs.upper_bound(worker);
      if(it == jobs.begin()) continue;
      --it;
      answer += it->second;
    }
    return answer;
  }
};