<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>

from functools import lru_cache
class Solution:
    def minSessions(self, tasks: List[int], sessionTime: int) -> int:
        n = len(tasks)
        tasks.sort(reverse=True)
        res = n
        breakout = set()
        
        # @lru_cache(maxsize=None)
        def backtrack(start, curr_work, sessions, bitmask, last):
            nonlocal res
            if bitmask == (1 << n) - 1:
                res = min(res, sessions + 1)
                return

            if (start, curr_work, sessions, bitmask) in breakout:
                return

            breakout.add((start, curr_work, sessions, bitmask))

            if start == n or curr_work == sessionTime:
                backtrack(0, 0, sessions + 1, bitmask, -1)
                return

            if bitmask & (1 << start) == 0 and curr_work + tasks[start] <= sessionTime and tasks[start] != last:
                bitmask |= (1 << start) 
                backtrack(start + 1, curr_work + tasks[start], sessions, bitmask, last)
                bitmask ^= (1 << start)
                last = tasks[start]
                if curr_work == 0:
                    return 
            backtrack(start + 1, curr_work, sessions, bitmask, last)
            return 

        backtrack(0, 0, 0, 0, -1)
        # print(len(breakout))
        # print(backtrack.cache_info())  # 顯示統計信息
        return res

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