<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:

    // if positive
        // start with positive => 1 + next, positive
        // contiune with negative => 1 + next, negative
    // if negative
        // ignore, and check next with positive
        // start with negative => 1 + next, negative
    // if 0
        // ignore, and check next with positive
    int maximum = 0;
    pair<int, int> recursive(vector<int>& nums, int pos){
        int n = nums.size();

        // first = positive
        // second = negative

        if(pos == n){
            return {0, -INT_MAX};
        }
        auto res = recursive(nums, pos+1);
        
        if(!nums[pos]){
            maximum = max(maximum, res.first);
            return {0, -INT_MAX};
        }else if(nums[pos] > 0){
            return {res.first + 1, res.second + 1};
        }else{
            maximum = max(maximum, res.first);
            return {max(res.second + 1, 0), res.first + 1};
        }
    }

    int getMaxLen(vector<int>& nums) {
        auto res = recursive(nums, 0);
        return max(res.first, maximum);
    }
};
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