<p>You are given a <strong>0-indexed</strong> integer array <code>nums</code> representing the score of students in an exam. The teacher would like to form one <strong>non-empty</strong> group of students with maximal <strong>strength</strong>, where the strength of a group of students of indices <code>i<sub>0</sub></code>, <code>i<sub>1</sub></code>, <code>i<sub>2</sub></code>, ... , <code>i<sub>k</sub></code> is defined as <code>nums[i<sub>0</sub>] * nums[i<sub>1</sub>] * nums[i<sub>2</sub>] * ... * nums[i<sub>k</sub>​]</code>.</p>

<p>Return <em>the maximum strength of a group the teacher can create</em>.</p>

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

<pre>
<strong>Input:</strong> nums = [3,-1,-5,2,5,-9]
<strong>Output:</strong> 1350
<strong>Explanation:</strong> One way to form a group of maximal strength is to group the students at indices [0,2,3,4,5]. Their strength is 3 * (-5) * 2 * 5 * (-9) = 1350, which we can show is optimal.
</pre>

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

<pre>
<strong>Input:</strong> nums = [-4,-5,-4]
<strong>Output:</strong> 20
<strong>Explanation:</strong> Group the students at indices [0, 1] . Then, we&rsquo;ll have a resulting strength of 20. We cannot achieve greater strength.
</pre>

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

<ul>
	<li><code>1 &lt;= nums.length &lt;= 13</code></li>
	<li><code>-9 &lt;= nums[i] &lt;= 9</code></li>
</ul>

class Solution:
    def maxStrength(self, nums: List[int]) -> int:
        if len(nums) == 1:
            return nums[0]

        neg_c, zero_c = 0, 0
        for n in nums:
            if n < 0:
                neg_c += 1
            elif n == 0:
                zero_c += 1

        if (neg_c == 1 and neg_c + zero_c == len(nums)) or zero_c == len(nums):
            return 0

        cache = {}
        def dfs(i, cur):
            if i >= len(nums):
                return cur
            if (i, cur) in cache:
                return cache[(i, cur)]

            cache[(i, cur)] = max(dfs(i + 1, cur), dfs(i + 1, cur * nums[i]))
            return cache[(i, cur)]

        return dfs(0, 1)
209