Most Profit Assigning Work

Similar Problems:

We have jobs: difficulty[i] is the difficulty of the ith job, and profit[i] is the profit of the ith job.

Now we have some workers. worker[i] is the ability of the ith worker, which means that this worker can only complete a job with difficulty at most worker[i].

Every worker can be assigned at most one job, but one job can be completed multiple times.

For example, if 3 people attempt the same job that pays $1, then the total profit will be $3. If a worker cannot complete any job, his profit is $0.

What is the most profit we can make?

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 profit of [20,20,30,30] seperately.

Notes:

- 1 <= difficulty.length = profit.length <= 10000
- 1 <= worker.length <= 10000
- difficulty[i], profit[i], worker[i] are in range [1, 10^5]

Github: code.dennyzhang.com

Credits To: leetcode.com

Leave me comments, if you have better ways to solve.

// https://code.dennyzhang.com/most-profit-assigning-work // Basic Ideas: dynamic programming + binarysearch // // Complexity: Time O(n*log(n)), Space O(n) func maxProfitAssignment(difficulty []int, profit []int, worker []int) int { m := make(map[int]int) for i, d := range difficulty { v, status := m[d] if status == false { m[d] = profit[i] } else { if v<profit[i] {m[d]=profit[i]} } } sort.Ints(difficulty[:]) for i := 0; i<len(profit); i++ { v, _ := m[difficulty[i]] profit[i] = v } // get most profit values dp := make([]int, len(difficulty)) dp[0] = profit[0] for i, _ := range difficulty { if i == 0 { continue } dp[i] = profit[i] if dp[i] < dp[i-1] { dp[i] = dp[i-1] } } res := 0 // binarysearch: find the first no smaller value for _, w := range worker{ item := 0 left, right := 0, len(difficulty) for left<right { mid := left + int((right-left)/2) if difficulty[mid] == w { item = dp[mid] break } else{ if difficulty[mid] > w { right = mid } else { left = mid + 1 } } } if item == 0 { if left != 0 { item = dp[left-1] } } res += item } return res }