<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) {
        ios::sync_with_stdio(false);
        cin.tie(nullptr);
        cout.tie(nullptr);
        int n = nums.size();
        vector<int> neg(n);
        neg[n - 1] = nums.back() < 0? 1 : 0;
        for (int i = nums.size() - 2; i >= 0; i--) {
            if (nums[i] < 0) {
                neg[i] = neg[i + 1] + 1;
            }
            else {
                neg[i] = neg[i + 1];
            }
        }  
        int ans = 0;
        int cur = 0;
        int l = 0;
        for (int i = 0; i < nums.size(); i++) {
            if (nums[i] < 0) {
                cur++;
            }
            else if (nums[i] == 0) {
                if (cur % 2 == 1) {
                    while (nums[l] > 0) {
                        l++;
                    }
                    l++;
                    cur--;
                }
                ans = max(ans, i - l);
                l = i + 1;
                cur = 0;
            }
            if (cur % 2 == 1 && i < nums.size() - 1 && neg[i + 1] == 0) {
                while (nums[l] > 0) {
                    l++;
                }
                cur--;
                l++;
            }
            if (cur % 2 == 0) {
                ans = max(ans, i - l + 1);
            }
        }
        return ans;
    }
};
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