<p>Given an array of integers <code>nums</code>, find the maximum length of a subarray where the product of all its elements is positive.</p>

<p>A subarray of an array is a consecutive sequence of zero or more values taken out of that array.</p>

<p>Return <em>the maximum length of a subarray with positive product</em>.</p>

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

<pre>
<strong>Input:</strong> nums = [1,-2,-3,4]
<strong>Output:</strong> 4
<strong>Explanation:</strong> The array nums already has a positive product of 24.
</pre>

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

<pre>
<strong>Input:</strong> nums = [0,1,-2,-3,-4]
<strong>Output:</strong> 3
<strong>Explanation:</strong> The longest subarray with positive product is [1,-2,-3] which has a product of 6.
Notice that we cannot include 0 in the subarray since that&#39;ll make the product 0 which is not positive.</pre>

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

<pre>
<strong>Input:</strong> nums = [-1,-2,-3,0,1]
<strong>Output:</strong> 2
<strong>Explanation:</strong> The longest subarray with positive product is [-1,-2] or [-2,-3].
</pre>

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

<ul>
	<li><code>1 &lt;= nums.length &lt;= 10<sup>5</sup></code></li>
	<li><code>-10<sup>9</sup> &lt;= nums[i] &lt;= 10<sup>9</sup></code></li>
</ul>

class Solution:
    def getMaxLen(self, nums: List[int]) -> int:
        memo = {}

        def dp(idx, oddNegCount):
            if nums[idx] == 0:
                return -inf
            if idx == 0:
                if nums[idx] > 0:
                    if oddNegCount:
                        return -inf
                    else:
                        return 1
                else:
                    if oddNegCount:
                        return 1
                    else:
                        return -inf
            
            # oddNegCount will auto be False
            if (idx, oddNegCount) not in memo:
                if oddNegCount and nums[idx] < 0:
                    memo[(idx, oddNegCount)] = 1
                elif not oddNegCount and nums[idx] > 0:
                    memo[(idx, oddNegCount)] = 1
                else:
                    memo[(idx, oddNegCount)] = 0

                if nums[idx] > 0:
                    tempAns = dp(idx-1, oddNegCount)
                    if tempAns > 0:
                        memo[(idx, oddNegCount)] = max(memo[(idx, oddNegCount)], 1 + tempAns)
                else:
                    tempAns = dp(idx-1, not(oddNegCount))
                    if tempAns > 0:
                        memo[(idx, oddNegCount)] = max(memo[(idx, oddNegCount)], 1 + tempAns)

            return memo[(idx, oddNegCount)]

        final = 0
        for i in range(len(nums)):
            final = max(final, dp(i, False))
        
        # print(memo)
        return final

                
                    



        
640