<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 minimumTimeRequired(self, jobs, k):
        def can_do_jobs(jobs, k, limit):
            workloads = [0] * k
            return backtrack(jobs, workloads, 0, limit)
        
        def backtrack(jobs, workloads, i, limit):
            if i == len(jobs):
                return True
            job = jobs[i]
            for j in range(k):
                if workloads[j] + job <= limit:
                    workloads[j] += job
                    if backtrack(jobs, workloads, i + 1, limit):
                        return True
                    workloads[j] -= job
                if workloads[j] == 0:
                    break
            return False
        
        left, right = max(jobs), sum(jobs)
        while left < right:
            mid = (left + right) // 2
            if can_do_jobs(jobs, k, mid):
                right = mid
            else:
                left = mid + 1
        return left

# Example usage:
solution = Solution()
jobs = [3, 2, 3]
k = 3
print(solution.minimumTimeRequired(jobs, k))  # Output: 3

jobs = [1, 2, 4, 7, 8]
k = 2
print(solution.minimumTimeRequired(jobs, k))  # Output: 11
108