<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>& arr) 
    {
        int n=arr.size();
        arr.push_back(0);
        map<int,int> pos;
        
        for(int i=0;i<n+1;i++)
        {
            if(arr[i]>0)
            arr[i]=1;

            if(arr[i]<0)
            arr[i]=-1;
            
            if(arr[i]==0)
            pos[i]=1;
        }
        
        int prod=1;int ans=0;
        int ansf=0;int ln=-1,fn=-1,lz=-1;
        for(int i=0;i<=n;i++)
        {
            if(pos.find(i)==pos.end())
            {
                if(arr[i]<0)
                ln=i;

                if(arr[i]<0 && fn==-1)
                fn=i;

                prod*=arr[i];
                ans++;
                // cout<<ans<<" ";
            }
            
            else{
                if(prod<0)
                {
                    ans=ans-min(i-ln,fn-lz);
                    // cout<<ans<<" "<<ln<<" "<<fn;
                }
                
                ansf=max(ans,ansf);
                ans=0;ln=-1;fn=-1;
                prod=1;lz=i;
            }
        }
        
        return ansf;
        
    }
};


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