Problem

We have jobs: difficulty[i] is the difficulty of the ith job, and profit[i] is the profit of the ith job.

Now we have some workers. worker[i] is the ability of the ith worker, which means that this worker can only complete a job with difficulty at most worker[i].

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

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

What is the most profit we can make?

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 profit of [20,20,30,30] seperately.

Notes:

• 1 <= difficulty.length = profit.length <= 10000
• 1 <= worker.length <= 10000
• difficulty[i], profit[i], worker[i]  are in range [1, 10^5]

Solution 1: Sorting + Two pointers

Time complexity: O(nlogn + mlogm)

Space complexity: O(n)

// Author: Huahua
// Running time: 90 ms
class Solution {
public:
  int maxProfitAssignment(vector<int>& difficulty, vector<int>& profit, vector<int>& worker) {
    const int n = difficulty.size();
    vector<pair<int, int>> jobs; // difficulty, profit
    
    for (int i = 0; i < n; ++i)
      jobs.emplace_back(difficulty[i], profit[i]);
    
    std::sort(jobs.begin(), jobs.end());
    std::sort(worker.begin(), worker.end());
    
    int ans = 0;
    int i = 0;
    int best = 0;
    for (int level : worker) {
      while (i < n && level >= jobs[i].first)
        best = max(best, jobs[i++].second);
      ans += best;
    }
    return ans;
  }
};

Solution 2: Bucket + Greedy

Key idea: for each difficulty D, find the most profit job whose requirement is <= D.

Three steps:

1. for each difficulty D, find the most profit job whose requirement is == D, best[D] = max{profit of difficulty D}.
2. if difficulty D – 1 can make more profit than difficulty D, best[D] = max(best[D], best[D – 1]).
3. The max profit each worker at skill level D can make is best[D].

Time complexity: O(n)

Space complexity: O(10000)

C++

// Author: Huahua
// Running time: 112 ms
class Solution {
public:
  int maxProfitAssignment(vector<int>& difficulty, vector<int>& profit, vector<int>& worker) {
    const int N = 100000;
    // max profit at difficulty i
    vector<int> max_profit(N + 1, 0);
    for (int i = 0; i < difficulty.size(); ++i)
      max_profit[difficulty[i]] = max(max_profit[difficulty[i]], profit[i]);
    for (int i = 2; i <= N; ++i)
      max_profit[i] = max(max_profit[i], max_profit[i - 1]);
    int ans = 0;
    for (int level : worker)
      ans += max_profit[level];
    return ans;
  }
};

C++ using map

// Author: Huahua
// Running time: 112 ms
class Solution {
public:
  int maxProfitAssignment(vector<int>& difficulty, vector<int>& profit, vector<int>& worker) {
    map<int, int> bests;
    for (int i = 0; i < difficulty.size(); ++i)
      bests[difficulty[i]] = max(bests[difficulty[i]], profit[i]);
    int best = 0;
    for (auto it = bests.begin(); it != bests.end(); ++it)
      it->second = best = max(it->second, best);    
    int ans = 0;
    for (int level : worker)
      ans += prev(bests.upper_bound(level))->second;
    return ans;
  }
};

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