<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 n=nums.size(),curr_leng=0,curr_neg=0,result;
        vector<int> dp(n);
        vector<int>dp2(n);
        for (int i=0;i<n;i++) if (nums[i]>0) dp[i]=1;
        for (int i=0;i<n;i++) if (nums[i]>0) dp2[i]=1;
        if (nums[0]!=0)curr_leng=1;
        if (nums[0]<0)curr_neg=1;
        for (int i=1;i<n;i++)
        {
            if (nums[i]==0)
            {
                curr_leng=0;
                curr_neg=0;
                dp[i]=0;
            }
            else
            {
                curr_leng++;
                if (nums[i]<0) curr_neg++;
                if (curr_neg%2==0) dp[i]=max(dp[i],curr_leng);
                else dp[i]=max(dp[i],dp[i-1]);
            }
        }
        curr_leng=0;curr_neg=0;
         if (nums[n-1]!=0)curr_leng=1;
        if (nums[n-1]<0)curr_neg=1;
          for (int i=n-2;i>=0;i--)
        {
            if (nums[i]==0)
            {
                curr_leng=0;
                curr_neg=0;
                dp2[i]=0;
            }
            else
            {
                curr_leng++;
                if (nums[i]<0) curr_neg++;
                if (curr_neg%2==0) dp2[i]=max(dp2[i],curr_leng);
                else dp2[i]=max(dp2[i],dp2[i+1]);
            }
        }
        result=max(*max_element(dp.begin(),dp.end()),*max_element(dp2.begin(),dp2.end()));
        return result;
    }
};
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