<p>Given an integer array <code>nums</code>, find a <span data-keyword="subarray-nonempty">subarray</span> that has the largest product, and return <em>the product</em>.</p>

<p>The test cases are generated so that the answer will fit in a <strong>32-bit</strong> integer.</p>

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

<pre>
<strong>Input:</strong> nums = [2,3,-2,4]
<strong>Output:</strong> 6
<strong>Explanation:</strong> [2,3] has the largest product 6.
</pre>

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

<pre>
<strong>Input:</strong> nums = [-2,0,-1]
<strong>Output:</strong> 0
<strong>Explanation:</strong> The result cannot be 2, because [-2,-1] is not a subarray.
</pre>

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

<ul>
	<li><code>1 &lt;= nums.length &lt;= 2 * 10<sup>4</sup></code></li>
	<li><code>-10 &lt;= nums[i] &lt;= 10</code></li>
	<li>The product of any subarray of <code>nums</code> is <strong>guaranteed</strong> to fit in a <strong>32-bit</strong> integer.</li>
</ul>

class Solution:
    def maxProduct(self, nums: List[int]) -> int:
        def sub(nums):
            l = []
            r = [0] * len(nums)
            curr = 1
            for i in nums:
                curr *= i
                l.append(curr)
            curr = 1
            for i in range(len(nums)-1,-1,-1):
                curr *= nums[i]
                r[i] = curr
            if r[0] > 0:
                return r[0]
            return max(max(l),max(r), max(nums))
        
        s = None
        out = 0
        for i in range(len(nums)):
            if nums[i] == 0:
                if s is not None:
                    out = max(out, sub(nums[s:i]))
                s = None
                continue
            if s is None:
                s = i
        if s is not None:
            out = max(out, sub(nums[s:]))  
        if out==0:
            return max(nums) 
        return out


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