<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>

from typing import List
from math import inf

class Solution:
    def minimumTimeRequired(self, jobs: List[int], k: int) -> int:
        def fn(i):
            """Assign jobs to workers and find the minimum maximum time."""
            nonlocal ans  # Access the outer scope variable `ans`
            if i == len(jobs):  # All jobs are assigned
                ans = min(ans, max(time))  # Update the minimum possible maximum working time
            else:
                for kk in range(k):
                    # Avoid duplicate assignments to workers with the same current time
                    if not kk or time[kk - 1] != time[kk]:
                        time[kk] += jobs[i]  # Assign job `i` to worker `kk`
                        if max(time) < ans:  # Only proceed if current assignment is potentially better
                            fn(i + 1)  # Recursively assign next job
                        time[kk] -= jobs[i]  # Backtrack: Remove job from worker `kk`
        
        ans = inf  # Initialize the minimum maximum working time as infinity
        time = [0] * k  # Array to keep track of the working time of each worker
        fn(0)  # Start the job assignment process
        return ans  # Return the minimum possible maximum working time

# Time Complexity: O(k^N) in the worst case, where N is the number of jobs
# Space Complexity: O(k) for the recursion stack and time array

# Create an instance of Solution
sol = Solution()

# Test with examples
print(sol.minimumTimeRequired([3, 2, 3], 3))  # Output: 3
print(sol.minimumTimeRequired([1, 2, 4, 7, 8], 2))  # Output: 11

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