A subarray of an array is a consecutive sequence of zero or more values taken out of that array.

Return the maximum length of a subarray with positive product.

Example 1:

Input: nums = [1,-2,-3,4]
Output: 4
Explanation: The array nums already has a positive product of 24.

Example 2:

Input: nums = [0,1,-2,-3,-4]
Output: 3
Explanation: 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'll make the product 0 which is not positive.

Example 3:

Input: nums = [-1,-2,-3,0,1]
Output: 2
Explanation: The longest subarray with positive product is [-1,-2] or [-2,-3].

Example 4:

Input: nums = [-1,2]
Output: 1

Example 5:

Input: nums = [1,2,3,5,-6,4,0,10]
Output: 4

Constraints:

• 1 <= nums.length <= 10^5
• -10^9 <= nums[i] <= 10^9

Solution: DP

p[i] := max length of positive products ends with arr[i]
n[i] := max length of negtive products ends with arr[i]

if arr[i] > 0: p[i] = p[i – 1] + 1, n[i] = n[i] + 1 if n[i] else 0
if arr[i] < 0: p[i] = n[i – 1] + 1 if n[i – 1] else 0, n[i] = p[i – 1] + 1
if arr[i] == 0: p[i] = n[i] = 0
ans = max(p[i])

Time complexity: O(n)
Space complexity: O(n) -> O(1)

class Solution {
public:
  int getMaxLen(vector<int>& nums) {
    int p = 0;
    int n = 0;
    int ans = 0;
    for (int x : nums) {      
      if (x > 0) {
        ++p;
        if (n) ++n;
      } else if (x < 0) {
        int tp = p;
        p = n ? n + 1 : 0;
        n = tp + 1;
      } else {
        p = n = 0;        
      }      
      ans = max(ans, p);
    }    
    return ans;
  }
};

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