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1723. Find Minimum Time to Finish All Jobs

Description

You are given an integer array jobs, where jobs[i] is the amount of time it takes to complete the ith job.

There are k workers that you can assign jobs to. Each job should be assigned to exactly one worker. The working time of a worker is the sum of the time it takes to complete all jobs assigned to them. Your goal is to devise an optimal assignment such that the maximum working time of any worker is minimized.

Return the minimum possible maximum working time of any assignment.

 

Example 1:

Input: jobs = [3,2,3], k = 3
Output: 3
Explanation: By assigning each person one job, the maximum time is 3.

Example 2:

Input: jobs = [1,2,4,7,8], k = 2
Output: 11
Explanation: Assign the jobs the following way:
Worker 1: 1, 2, 8 (working time = 1 + 2 + 8 = 11)
Worker 2: 4, 7 (working time = 4 + 7 = 11)
The maximum working time is 11.

 

Constraints:

  • 1 <= k <= jobs.length <= 12
  • 1 <= jobs[i] <= 107

Solutions

Solution 1

Thinking

Assign jobs to \(k\) workers and minimize the maximum load. The job count is small enough to search, but naive \(k^n\) is too large.

Abandon a branch once the current max load is already no better than the recorded answer. Assigning longer jobs first triggers that prune sooner.

Sort \(jobs\) descending and DFS into each worker, then undo. If a worker is still empty, skip later empty workers to cut symmetric assignments.

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class Solution:
    def minimumTimeRequired(self, jobs: List[int], k: int) -> int:
        def dfs(i):
            nonlocal ans
            if i == len(jobs):
                ans = min(ans, max(cnt))
                return
            for j in range(k):
                if cnt[j] + jobs[i] >= ans:
                    continue
                cnt[j] += jobs[i]
                dfs(i + 1)
                cnt[j] -= jobs[i]
                if cnt[j] == 0:
                    break

        cnt = [0] * k
        jobs.sort(reverse=True)
        ans = inf
        dfs(0)
        return ans

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class Solution {
    private int[] cnt;
    private int ans;
    private int[] jobs;
    private int k;

    public int minimumTimeRequired(int[] jobs, int k) {
        this.k = k;
        Arrays.sort(jobs);
        for (int i = 0, j = jobs.length - 1; i < j; ++i, --j) {
            int t = jobs[i];
            jobs[i] = jobs[j];
            jobs[j] = t;
        }
        this.jobs = jobs;
        cnt = new int[k];
        ans = 0x3f3f3f3f;
        dfs(0);
        return ans;
    }

    private void dfs(int i) {
        if (i == jobs.length) {
            int mx = 0;
            for (int v : cnt) {
                mx = Math.max(mx, v);
            }
            ans = Math.min(ans, mx);
            return;
        }
        for (int j = 0; j < k; ++j) {
            if (cnt[j] + jobs[i] >= ans) {
                continue;
            }
            cnt[j] += jobs[i];
            dfs(i + 1);
            cnt[j] -= jobs[i];
            if (cnt[j] == 0) {
                break;
            }
        }
    }
}

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class Solution {
public:
    int ans;

    int minimumTimeRequired(vector<int>& jobs, int k) {
        vector<int> cnt(k);
        ans = 0x3f3f3f3f;
        sort(jobs.begin(), jobs.end(), greater<int>());
        dfs(0, k, jobs, cnt);
        return ans;
    }

    void dfs(int i, int k, vector<int>& jobs, vector<int>& cnt) {
        if (i == jobs.size()) {
            ans = min(ans, *max_element(cnt.begin(), cnt.end()));
            return;
        }
        for (int j = 0; j < k; ++j) {
            if (cnt[j] + jobs[i] >= ans) continue;
            cnt[j] += jobs[i];
            dfs(i + 1, k, jobs, cnt);
            cnt[j] -= jobs[i];
            if (cnt[j] == 0) break;
        }
    }
};

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func minimumTimeRequired(jobs []int, k int) int {
    cnt := make([]int, k)
    ans := 0x3f3f3f3f
    sort.Slice(jobs, func(i, j int) bool {
        return jobs[i] > jobs[j]
    })
    var dfs func(int)
    dfs = func(i int) {
        if i == len(jobs) {
            mx := slices.Max(cnt)
            ans = min(ans, mx)
            return
        }
        for j := 0; j < k; j++ {
            if cnt[j]+jobs[i] >= ans {
                continue
            }
            cnt[j] += jobs[i]
            dfs(i + 1)
            cnt[j] -= jobs[i]
            if cnt[j] == 0 {
                break
            }
        }
    }
    dfs(0)
    return ans
}

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