<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() ;
        int currLen = 0 ;
        int currCnt = 0 ;
        vector<int> preCnt(n) ;
        vector<int> length(n) ;
        vector<int> firstN(n) ;
        int fn = -1 ;
        
        for( int i=0 ; i<n ; i++ ){
            if( nums[i] == 0 ){
                currLen = 0 ;
                currCnt = 0 ;
                fn = -1 ;
            }
            else if( nums[i] < 0 ){
                length[i] = ++currLen ;
                preCnt[i] = ++currCnt ;
                if( fn == -1 ) fn = i ;
            }
            else{
                length[i] = ++currLen ;
                preCnt[i] = currCnt ;
            }
            firstN[i] = fn ;
        }
        
        // for( auto i : preCnt ) cout << i << " " ;
        // cout << endl ;
        
        int ans = 0 ;
        for( int i=0 ; i<n ; i++ ){
            if( preCnt[i]%2 == 0 ){
                ans = max( ans , length[i] ) ;
            }
            else if(firstN[i] != -1){
                ans = max( ans , i-firstN[i]  ) ;
            }
        }
        return ans ;
        
    }
};
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