Maximum Length of Subarray With Positive Product
<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>
Greedy; Array; Dynamic Programming
class Solution {
public:
    int getMaxLen(vector<int>& nums) {
        int n = nums.size();
        vector<int> f(n, 0), g(n, 0);
        f[0] = nums[0] > 0 ? 1 : 0;
        g[0] = nums[0] < 0 ? 1 : 0;
        int ans = f[0];

        for (int i = 1; i < n; ++i) {
            if (nums[i] > 0) {
                f[i] = f[i - 1] + 1;
                g[i] = g[i - 1] > 0 ? g[i - 1] + 1 : 0;
            } else if (nums[i] < 0) {
                f[i] = g[i - 1] > 0 ? g[i - 1] + 1 : 0;
                g[i] = f[i - 1] + 1;
            }
            ans = max(ans, f[i]);
        }
        return ans;
    }
};
