<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) {
        vector<pair<int,int>>zero;
        int n=nums.size();
        int ct=0;
        for(int i=0;i<n;i++){
            if(nums[i]<0)ct++;
            if(nums[i]==0){ zero.push_back({ct,i});
            ct=0;
            }
    }
    zero.push_back({ct,n});
    int mx=0;
    int i=0;
    int p=zero.size();
    for(int k=0;k<p;k++){
       if(zero[k].first%2==0){
        mx=max(mx,zero[k].second-i);
        i=zero[k].second+1;
       }else{
        int start=0;
        if(k-1>=0)start=zero[k-1].second+1;
        while(i<zero[k].second){

           if(nums[i]<0){
            int mp=max(i-start,zero[k].second-(i+1));
            mx=max(mx,mp);
           }
           i++;
        }
        while(i<n&&nums[i]==0)i++;
       }
    }
    return mx;
    }
};
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