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

Solutions

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
}

function maxProfitAssignment(difficulty: number[], profit: number[], worker: number[]): number {
    const n = profit.length;
    worker.sort((a, b) => a - b);
    const jobs = Array.from({ length: n }, (_, i) => [difficulty[i], profit[i]]);
    jobs.sort((a, b) => a[0] - b[0]);
    let [ans, mx, i] = [0, 0, 0];
    for (const w of worker) {
        while (i < n && jobs[i][0] <= w) {
            mx = Math.max(mx, jobs[i++][1]);
        }
        ans += mx;
    }
    return ans;
}

826 - Most Profit Assigning Work

826. Most Profit Assigning Work

Description

Solutions

All Problems

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826. Most Profit Assigning Work

Description

Solutions

All Problems

All Solutions