2323. Find Minimum Time to Finish All Jobs II 🔒

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Description

You are given two 0-indexed integer arrays jobs and workers of equal length, where jobs[i] is the amount of time needed to complete the ith job, and workers[j] is the amount of time the jth 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 2nd worker to the 0th job. It takes them 1 day to finish the job.
- Assign the 0th worker to the 1st job. It takes them 2 days to finish the job.
- Assign the 1st worker to the 2nd 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 2nd worker to the 0th job. It takes them 3 days to finish the job.
- Assign the 0th worker to the 1st job. It takes them 3 days to finish the job.
- Assign the 1st worker to the 2nd job. It takes them 3 days to finish the job.
- Assign the 3rd worker to the 3rd 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 <= 105
• 1 <= jobs[i], workers[i] <= 105

Solutions

Solution 1: Greedy

To minimize the number of days required to complete all jobs, we can try to assign longer jobs to workers who can work longer hours.

Therefore, we can first sort $\textit{jobs}$ and $\textit{workers}$, then assign jobs to workers based on their indices. Finally, we calculate the maximum ratio of job time to worker time.

The time complexity is $O(n \log n)$, and the space complexity is $O(\log n)$. Here, $n$ is the number of jobs.

Python3

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))

Java

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;
    }
}

C++

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

Go

func minimumTime(jobs []int, workers []int) (ans int) {
	sort.Ints(jobs)
	sort.Ints(workers)
	for i, a := range jobs {
		b := workers[i]
		ans = max(ans, (a+b-1)/b)
	}
	return
}

TypeScript

function minimumTime(jobs: number[], workers: number[]): number {
    jobs.sort((a, b) => a - b);
    workers.sort((a, b) => a - b);
    let ans = 0;
    const n = jobs.length;
    for (let i = 0; i < n; ++i) {
        ans = Math.max(ans, Math.ceil(jobs[i] / workers[i]));
    }
    return ans;
}

Rust

impl Solution {
    pub fn minimum_time(mut jobs: Vec<i32>, mut workers: Vec<i32>) -> i32 {
        jobs.sort();
        workers.sort();
        jobs.iter()
            .zip(workers.iter())
            .map(|(a, b)| (a + b - 1) / b)
            .max()
            .unwrap()
    }
}

JavaScript

/**
 * @param {number[]} jobs
 * @param {number[]} workers
 * @return {number}
 */
var minimumTime = function (jobs, workers) {
    jobs.sort((a, b) => a - b);
    workers.sort((a, b) => a - b);
    let ans = 0;
    const n = jobs.length;
    for (let i = 0; i < n; ++i) {
        ans = Math.max(ans, Math.ceil(jobs[i] / workers[i]));
    }
    return ans;
};