<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 fun(vector<int>& nums, int low, int high){

    if(low == high){
        if(nums[low] == 0){
            return 0;
        }
        if(nums[low] < 0){
            return 0;
        }
        return 1;
    }


    for(int j=low;j<=high;j++){
        if(nums[j] == 0){
            return max(fun(nums, low, j-1), fun(nums, j+1, high));
        }
    }

    int count=0;

    for(int k=low;k<=high;k++){
        if(nums[k] < 0){
            count++;
        }
    }
    int st = -1, end = -1;

    if(count % 2 == 0){
        return high-low+1;
    }

    else{
        for(int i =low;i<=high;i++){
            if(nums[i] < 0 ){
                st=i;
                break;
            }
        }

        for(int i = high;i>=low;i--){
            if(nums[i] < 0 ){
                end=i;
                break;
            }
        }


        if(st - low > high - end){
            return end-low;
        }

        return high-st;
    }

return 1;

}

int getMaxLen(vector<int>& nums) {
    int n=nums.size();

    return fun(nums, 0, n-1);
    }
};
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