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2323. Find Minimum Time to Finish All Jobs II
Description
You are given two 0indexed integer arrays jobs
and workers
of equal length, where jobs[i]
is the amount of time needed to complete the i^{th}
job, and workers[j]
is the amount of time the j^{th}
worker can work each day.
Each job should be assigned to exactly one worker, such that each worker completes exactly one job.
Return the minimum number of days needed to complete all the jobs after assignment.
Example 1:
Input: jobs = [5,2,4], workers = [1,7,5] Output: 2 Explanation:  Assign the 2^{nd} worker to the 0^{th} job. It takes them 1 day to finish the job.  Assign the 0^{th} worker to the 1^{st} job. It takes them 2 days to finish the job.  Assign the 1^{st} worker to the 2^{nd} job. It takes them 1 day to finish the job. It takes 2 days for all the jobs to be completed, so return 2. It can be proven that 2 days is the minimum number of days needed.
Example 2:
Input: jobs = [3,18,15,9], workers = [6,5,1,3] Output: 3 Explanation:  Assign the 2^{nd} worker to the 0^{th} job. It takes them 3 days to finish the job.  Assign the 0^{th} worker to the 1^{st} job. It takes them 3 days to finish the job.  Assign the 1^{st} worker to the 2^{nd} job. It takes them 3 days to finish the job.  Assign the 3^{rd} worker to the 3^{rd} job. It takes them 3 days to finish the job. It takes 3 days for all the jobs to be completed, so return 3. It can be proven that 3 days is the minimum number of days needed.
Constraints:
n == jobs.length == workers.length
1 <= n <= 10^{5}
1 <= jobs[i], workers[i] <= 10^{5}
Solutions

class Solution { public int minimumTime(int[] jobs, int[] workers) { Arrays.sort(jobs); Arrays.sort(workers); int ans = 0; for (int i = 0; i < jobs.length; ++i) { ans = Math.max(ans, (jobs[i] + workers[i]  1) / workers[i]); } return ans; } }

class Solution { public: int minimumTime(vector<int>& jobs, vector<int>& workers) { sort(jobs.begin(), jobs.end()); sort(workers.begin(), workers.end()); int ans = 0; for (int i = 0; i < jobs.size(); ++i) ans = max(ans, (jobs[i] + workers[i]  1) / workers[i]); return ans; } };

class Solution: def minimumTime(self, jobs: List[int], workers: List[int]) > int: jobs.sort() workers.sort() return max((a + b  1) // b for a, b in zip(jobs, workers))

func minimumTime(jobs []int, workers []int) int { sort.Ints(jobs) sort.Ints(workers) ans := 0 for i, a := range jobs { b := workers[i] ans = max(ans, (a+b1)/b) } return ans }