<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 {
public:
    int getMaxLen(vector<int>& nums) {
        int n = nums.size();
        vector<vector<int>> zeros_pos;
        int first = 0;
        for (int i = 0; i < n; i++) {
            if (nums[i] == 0) {
                zeros_pos.push_back({first, i});
                first = i + 1;
            }
        }     
        zeros_pos.push_back({first, n});
        int ans_ = 0;
        for (auto i : zeros_pos) {
            vector<int> pos_neg;
            for (int j = i[0]; j < i[1]; j++) {
                if (nums[j] < 0) pos_neg.push_back(j);
               
            }
            if (pos_neg.size() % 2 == 0) {
                ans_ = max(ans_, i[1] - i[0]);
            } 
            else {
                ans_ = max(ans_, max(i[1] - pos_neg[0] - 1, pos_neg[pos_neg.size() - 1] - i[0]));
            }
        }
        return ans_;      
    }
};
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