<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:
        nums.sort()
        i = 0
        N = len(nums)
        hasZero = False
        while i < N:
            if nums[i] == 0:
                hasZero = True
                nums[i], nums[N-1] = nums[N-1], nums[i]
                N -= 1
            i += 1

        while N > 0 and nums[N-1] == 0:
            N -= 1
        if N == 0:
            return 0
        nums = nums[:N]

        negativeCnt = N
        for i in range(N-1, -1, -1):
            if nums[i] > 0:
                negativeCnt -= 1
            else:
                break

        if negativeCnt == 1 and N == 1:
            return 0 if hasZero else nums[0]

        left = 1
        for i in range(negativeCnt, N):
            left *= nums[i]

        res = 0
        for i in range(1 << negativeCnt):
            if i.bit_count() % 2 != 0:
                continue
            cur = 1
            for j in range(negativeCnt):
                if i & (1 << j):
                    cur *= nums[j]
            res = max(res, cur)

        return res * left

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