<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>

#define ll long long
class Solution {
   
public:
    int getMaxLen(vector<int>& nums) {
        ll int n = nums.size();
        for(int i=0;i<n;i++) {
            if(nums[i]!=0) nums[i]/=abs(nums[i]);
        }
        map<string,ll int> mp;

        long long int prod=1;
        ll int mx=0;
        ll int ans=0;
        ll int s=0;

        for(int i=0;i<n;i++){
            prod*=nums[i];

            if(prod==0){
                prod=1;
                s=i+1;
                mp.clear();
                continue;
            }

            if(prod>0) {
                mx=i-s+1;
                ans=max(ans,mx);
            }
    

            if(prod<0){
                if(mp.find("neg")!=mp.end()){
                    mx=i-mp["neg"];
                    ans=max(ans,mx);
                }

                else mp["neg"] = i;
                
            }


        }

        return ans;

        
    }
};
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