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826. Most Profit Assigning Work 👍

  • Time: $O(n\log n + m\log m)$
  • Space: $O(n)$
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class Solution {
 public:
  int maxProfitAssignment(vector<int>& difficulty, vector<int>& profit,
                          vector<int>& worker) {
    int ans = 0;
    vector<pair<int, int>> jobs;

    for (int i = 0; i < difficulty.size(); ++i)
      jobs.emplace_back(difficulty[i], profit[i]);

    ranges::sort(jobs);
    ranges::sort(worker);

    int i = 0;
    int maxProfit = 0;

    for (const int w : worker) {
      for (; i < jobs.size() && w >= jobs[i].first; ++i)
        maxProfit = max(maxProfit, jobs[i].second);
      ans += maxProfit;
    }

    return ans;
  }
};

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class Solution {
  public int maxProfitAssignment(int[] difficulty, int[] profit, int[] worker) {
    int ans = 0;
    List<Pair<Integer, Integer>> jobs = new ArrayList<>();

    for (int i = 0; i < difficulty.length; ++i)
      jobs.add(new Pair<>(difficulty[i], profit[i]));

    Collections.sort(jobs, Comparator.comparing(Pair::getKey));
    Arrays.sort(worker);

    int i = 0;
    int maxProfit = 0;

    for (final int w : worker) {
      for (; i < jobs.size() && w >= jobs.get(i).getKey(); ++i)
        maxProfit = Math.max(maxProfit, jobs.get(i).getValue());
      ans += maxProfit;
    }

    return ans;
  }
}

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class Solution:
  def maxProfitAssignment(self, difficulty: List[int], profit: List[int], worker: List[int]) -> int:
    ans = 0
    jobs = sorted(zip(difficulty, profit))
    worker.sort(reverse=1)

    i = 0
    maxProfit = 0

    for w in sorted(worker):
      while i < len(jobs) and w >= jobs[i][0]:
        maxProfit = max(maxProfit, jobs[i][1])
        i += 1
      ans += maxProfit

    return ans