<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>

class Solution:
    def __init__(self):
        self.workers = None
        self.min_time = float('inf')

    def minimumTimeRequired(self, jobs: List[int], k: int) -> int:
        if not jobs or k < 1: return 0
        self.workers = [0] * k
        jobs.sort(reverse=True)
        self.backtrack(jobs, 0)
        return self.min_time
    
    def backtrack(self, jobs: List[int], start: int):
        if start >= len(jobs):
            # all jobs assigned, sum up time
            self.min_time = min(self.min_time, max(self.workers))
            return
        
        task = jobs[start]

        # assign task to any worker
        for j in range(len(self.workers)):
            if self.workers[j] + task > self.min_time:
                # this is not best option, skip
                continue
            self.workers[j] += task
            self.backtrack(jobs, start + 1)
            self.workers[j] -= task

            # !!! this is a trick !!!
            # if after loop, there is no task assigned to worker j at all,
            # that means we don't make best use of all workers at all.
            # this must not be the best option, break current loop of try.
            if self.workers[j] == 0:
                break
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