<p>You are given an integer array <code>jobs</code>, where <code>jobs[i]</code> is the amount of time it takes to complete the <code>i<sup>th</sup></code> job.</p>

<p>There are <code>k</code> workers that you can assign jobs to. Each job should be assigned to <strong>exactly</strong> one worker. The <strong>working time</strong> 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 <strong>maximum working time</strong> of any worker is <strong>minimized</strong>.</p>

<p><em>Return the <strong>minimum</strong> possible <strong>maximum working time</strong> of any assignment. </em></p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

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

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> jobs = [1,2,4,7,8], k = 2
<strong>Output:</strong> 11
<strong>Explanation:</strong> 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.</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= k &lt;= jobs.length &lt;= 12</code></li>
	<li><code>1 &lt;= jobs[i] &lt;= 10<sup>7</sup></code></li>
</ul>

#include <vector>
#include <algorithm>
#include <climits>
#include <unordered_set>

class Solution {
public:
    int minimumTimeRequired(std::vector<int>& jobs, int k) {
        std::vector<int> workers(k, 0);
        int res = INT_MAX;
        
        std::function<void(int)> dfs = [&](int curr) {
            if (curr == jobs.size()) {
                res = std::min(res, *std::max_element(workers.begin(), workers.end()));
                return;
            }
            
            std::unordered_set<int> seen; // record searched workload of workers
            for (int i = 0; i < k; i++) {
                if (seen.count(workers[i])) continue; // if we have searched this workload, skip it.
                if (workers[i] + jobs[curr] >= res) continue; // branch cutting
                seen.insert(workers[i]);
                workers[i] += jobs[curr];
                dfs(curr + 1);
                workers[i] -= jobs[curr];
            }
        };
        
        dfs(0);
        return res;
    }
};

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