<p>There are <code>n</code> tasks assigned to you. The task times are represented as an integer array <code>tasks</code> of length <code>n</code>, where the <code>i<sup>th</sup></code> task takes <code>tasks[i]</code> hours to finish. A <strong>work session</strong> is when you work for <strong>at most</strong> <code>sessionTime</code> consecutive hours and then take a break.</p>

<p>You should finish the given tasks in a way that satisfies the following conditions:</p>

<ul>
	<li>If you start a task in a work session, you must complete it in the <strong>same</strong> work session.</li>
	<li>You can start a new task <strong>immediately</strong> after finishing the previous one.</li>
	<li>You may complete the tasks in <strong>any order</strong>.</li>
</ul>

<p>Given <code>tasks</code> and <code>sessionTime</code>, return <em>the <strong>minimum</strong> number of <strong>work sessions</strong> needed to finish all the tasks following the conditions above.</em></p>

<p>The tests are generated such that <code>sessionTime</code> is <strong>greater</strong> than or <strong>equal</strong> to the <strong>maximum</strong> element in <code>tasks[i]</code>.</p>

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

<pre>
<strong>Input:</strong> tasks = [1,2,3], sessionTime = 3
<strong>Output:</strong> 2
<strong>Explanation:</strong> You can finish the tasks in two work sessions.
- First work session: finish the first and the second tasks in 1 + 2 = 3 hours.
- Second work session: finish the third task in 3 hours.
</pre>

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

<pre>
<strong>Input:</strong> tasks = [3,1,3,1,1], sessionTime = 8
<strong>Output:</strong> 2
<strong>Explanation:</strong> You can finish the tasks in two work sessions.
- First work session: finish all the tasks except the last one in 3 + 1 + 3 + 1 = 8 hours.
- Second work session: finish the last task in 1 hour.
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> tasks = [1,2,3,4,5], sessionTime = 15
<strong>Output:</strong> 1
<strong>Explanation:</strong> You can finish all the tasks in one work session.
</pre>

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

<ul>
	<li><code>n == tasks.length</code></li>
	<li><code>1 &lt;= n &lt;= 14</code></li>
	<li><code>1 &lt;= tasks[i] &lt;= 10</code></li>
	<li><code>max(tasks[i]) &lt;= sessionTime &lt;= 15</code></li>
</ul>

class Solution:
    def minSessions(self, jobs: List[int], workHours: int) -> int:
        # Sort jobs in descending order to optimize the packing process
        jobs.sort(reverse=True)

        # Function to check if it's possible to assign jobs to workers with a given limit on workers
        def canFinish(jobs, workers, workHours):
            sessions = [0] * workers  # Track the work hours left for each worker

            # Try to assign each job to a worker
            def dfs(jobIndex):
                if jobIndex == len(jobs):
                    return True

                for i in range(workers):
                    # Check if the current worker can take this job
                    if sessions[i] + jobs[jobIndex] <= workHours:
                        sessions[i] += jobs[jobIndex]
                        if dfs(jobIndex + 1):
                            return True
                        sessions[i] -= jobs[jobIndex]

                    # If this worker hasn't been assigned any job and can't take this one, no point in continuing
                    if sessions[i] == 0:
                        break

                return False

            return dfs(0)

        # Use binary search to find the minimum number of workers required
        left, right = 1, len(jobs)

        while left < right:
            mid = (left + right) // 2
            if canFinish(jobs, mid, workHours):
                right = mid
            else:
                left = mid + 1

        return left


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