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826. Most Profit Assigning Work

Description

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

Solutions

• Java
• C++
• Python
• Go

• class Solution {
      public int maxProfitAssignment(int[] difficulty, int[] profit, int[] worker) {
          int n = difficulty.length;
          List<int[]> job = new ArrayList<>();
          for (int i = 0; i < n; ++i) {
              job.add(new int[] {difficulty[i], profit[i]});
          }
          job.sort(Comparator.comparing(a -> a[0]));
          Arrays.sort(worker);
          int res = 0;
          int i = 0, t = 0;
          for (int w : worker) {
              while (i < n && job.get(i)[0] <= w) {
                  t = Math.max(t, job.get(i++)[1]);
              }
              res += t;
          }
          return res;
      }
  }
  

• class Solution {
  public:
      int maxProfitAssignment(vector<int>& difficulty, vector<int>& profit, vector<int>& worker) {
          int n = difficulty.size();
          vector<pair<int, int>> job;
          for (int i = 0; i < n; ++i) {
              job.push_back({difficulty[i], profit[i]});
          }
          sort(job.begin(), job.end());
          sort(worker.begin(), worker.end());
          int i = 0, t = 0;
          int res = 0;
          for (auto w : worker) {
              while (i < n && job[i].first <= w) {
                  t = max(t, job[i++].second);
              }
              res += t;
          }
          return res;
      }
  };
  

• class Solution:
      def maxProfitAssignment(
          self, difficulty: List[int], profit: List[int], worker: List[int]
      ) -> int:
          n = len(difficulty)
          job = [(difficulty[i], profit[i]) for i in range(n)]
          job.sort(key=lambda x: x[0])
          worker.sort()
          i = t = res = 0
          for w in worker:
              while i < n and job[i][0] <= w:
                  t = max(t, job[i][1])
                  i += 1
              res += t
          return res
  
  

• func maxProfitAssignment(difficulty []int, profit []int, worker []int) int {
  	var job [][2]int
  	for i := range difficulty {
  		job = append(job, [2]int{difficulty[i], profit[i]})
  	}
  
  	sort.SliceStable(job, func(i, j int) bool { return job[i][0] <= job[j][0] })
  	sort.Ints(worker)
  	i, t, n, res := 0, 0, len(difficulty), 0
  	for _, w := range worker {
  		for i < n && job[i][0] <= w {
  			t = max(t, job[i][1])
  			i++
  		}
  		res += t
  	}
  	return res
  }
  

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