<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 res = 0;

        vector<int> zeros;
        zeros.push_back(-1);
        for(int i = 0; i < nums.size(); ++i){
            if(nums[i] == 0) zeros.push_back(i);
        }
        zeros.push_back(nums.size());

        for(int i = 1; i < zeros.size(); ++i){
            int negativeCount = 0;
            int first = zeros[i-1];
            int last = zeros[i];

            for(int j = zeros[i-1] + 1; j < zeros[i]; j++){
                if(nums[j] < 0) negativeCount++;
                if(negativeCount == 1 && nums[j] < 0){
                    first = j;
                    last = j;
                }
                else if(negativeCount > 1 && nums[j] < 0){
                    last = j;
                }
            }
            if(negativeCount % 2 == 0){
                res = max(res, zeros[i] - zeros[i-1] - 1);
            }
            else{
                res = max(res, zeros[i] - first - 1);
                res = max(res, last - zeros[i-1] - 1);
            }
        }

        return res;
    }
};
61000