<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>

static int __ = []{
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
return 0;
}();
class Solution {
public:
    int helper(vector<int>&nums,int s,int e){
        int cntNeg=0;
        int firstNeg=-1,lastNeg=-1;
        for(int i=s;i<=e;i++){
            if(nums[i]<0){
               cntNeg++;
               lastNeg=i;
               if(firstNeg<0)
                 firstNeg=i;
            }
        }
        if(cntNeg%2==0)
          return (e-s)+1;
    return max(lastNeg-s,e-firstNeg);
    }
    int getMaxLen(vector<int>& nums) {
       vector<int>zero;
       zero.push_back(-1);
       int n=nums.size();
       for(int i=0;i<n;i++){
        if(nums[i]==0)
          zero.push_back(i);
       }
       zero.push_back(n);
       int ans=0;
       for(int i=1;i<zero.size();i++){
          int u=zero[i-1]+1,v=zero[i]-1;
          ans=max(ans,helper(nums,u,v));
       }
    return ans;
    }
};
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