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 i^th job, and
worker[j] is the ability of j^th worker (i.e., the j^th 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

most-profit-assigning-work.py

class Solution:
    def maxProfitAssignment(self, difficulty: List[int], profit: List[int], worker: List[int]) -> int:
        A = sorted([(x, y) for x, y in zip(difficulty, profit)])
        difficulty.sort()
        mp = defaultdict(int)
        curr = res = 0

        for d, p in A:
            curr = max(curr, p)

            mp[d] = max(mp[d], curr)

        for ability in worker:
            d = bisect.bisect_right(difficulty, ability) - 1

            if ability >= difficulty[d]:
                res += mp[difficulty[d]]

        return res