LeetCode-in-Java

826. Most Profit Assigning Work

Medium

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

Solution

public class Solution {
    public int maxProfitAssignment(int[] difficulty, int[] profit, int[] worker) {
        int n = 100000;
        int[] maxProfit = new int[n];

        for (int i = 0; i < difficulty.length; i++) {
            maxProfit[difficulty[i]] = Math.max(maxProfit[difficulty[i]], profit[i]);
        }

        for (int i = 1; i < n; i++) {
            maxProfit[i] = Math.max(maxProfit[i], maxProfit[i - 1]);
        }

        int sum = 0;
        for (int efficiency : worker) {
            sum += maxProfit[efficiency];
        }
        return sum;
    }
}

This site is open source. Improve this page.