Problem
You have n
jobs and m
workers. You are given three arrays: difficulty
, profit
, and worker
where:
difficulty[i]
andprofit[i]
are the difficulty and the profit of theith
job, andworker[j]
is the ability ofjth
worker (i.e., thejth
worker can only complete a job with difficulty at mostworker[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 <= 10^4
1 <= difficulty[i], profit[i], worker[i] <= 10^5
Solution (Java)
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;
}
}
Explain:
nope.
Complexity:
- Time complexity : O(n).
- Space complexity : O(n).