<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>

#define ll long long
class Solution {
public:
    int getMaxLen(vector<int>& nums) {
        vector<int> zs;
        zs.push_back(-1);
        for (int i = 0; i < nums.size(); i++) {
            if (nums[i] == 0)
                zs.push_back(i);
        }
        zs.push_back(nums.size());
        int ans = 0;
        for (int zi = 0; zi + 1 < zs.size(); zi++) {
            int l = zs[zi] + 1;
            int r = zs[zi + 1] - 1;
            int z = 0;
            int li = -1, ri = -1;
            for (int i = l; i <= r; i++) {
                if (nums[i] < 0) {
                    li = i;
                    break;
                }
            }

            for(int i = r; i >= l ; i--){
                if (nums[i] < 0){
                    ri = i ;
                    break;
                }
            }

            for (int i = l; i <= r; i++) {
                if (nums[i] < 0) z++;
                if (z%2==0) {
                    ans = max(ans, i - l + 1);
                }
            }

            if (z&1) {
                ans = max ({ans , r-li , ri - l}) ;
            }

        }
        return ans;
    }
};
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