Find Minimum Time to Finish All Jobs

Problem

You are given an integer array , where is the amount of time it takes to complete the job.

There are 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

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

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

Submit your solution at here

Solution

Intuition

I did a trick with and , then manage to pass some big test. But still, my main implementation got flaw and TLE. My approach is traverse all possibility with DFS, cut off some branch that can not lead to better result. Not sure why it's TLE. Any guidance is much appreciated.

Code

class Solution {
public:
    int minimumTimeRequired(vector<int>& jobs, int k) {
        // AC but still bad implementation, TLE with this test
        // [10001,10002,10003,10004,10005,10006,10007,10008,10009,10010,10011,1000000]
        // 10
        int n = jobs.size();
        if(k == n) {
            return *max_element(jobs.begin(), jobs.end());
        }
        if(k == n-1) {
            sort(jobs.begin(), jobs.end());
            int min_pair = jobs[0] + jobs[1];
            int max_job = jobs[n-1];
            return max(min_pair, max_job);
        }
        typedef pair<int, vector<int>> state;
        vector<int> cost(n+1, 1e9);
        stack<state> q;
        vector<int> workers(k,0);
        q.emplace(0, workers);
        map<state, bool> vis;
        while(!q.empty()) {
            auto [u, w] = q.top();
            if(vis[q.top()]) {
                q.pop();
                continue;
            }
            vis[q.top()] = true;
            q.pop();
            int next = w[k-1];
            if(next >= cost[n]) {
                continue;
            }
            cost[u] = next;
            int v = u+1;
            if(v > n) {
                continue;
            }
            for(int i = w.size()-1;i>=0;i--) {
                auto nextw = w;
                nextw[i] += jobs[v-1];
                sort(nextw.begin(), nextw.end());
                q.emplace(v, nextw);
            }
        }
        return cost[n];
    }
};