LeetCode in Kotlin

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

class Solution {
    fun maxProfitAssignment(difficulty: IntArray, profit: IntArray, worker: IntArray): Int {
        val n = 100000
        val maxProfit = IntArray(n)
        for (i in difficulty.indices) {
            maxProfit[difficulty[i]] = maxProfit[difficulty[i]].coerceAtLeast(profit[i])
        }
        for (i in 1 until n) {
            maxProfit[i] = maxProfit[i].coerceAtLeast(maxProfit[i - 1])
        }
        var sum = 0
        for (efficiency in worker) {
            sum += maxProfit[efficiency]
        }
        return sum
    }
}

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