# 1567 Maximum Length of Subarray With Positive Product

Given an array of integers `nums`, find the maximum length of a subarray where the product of all its elements is positive.

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].
``````

Constraints:

• `1 <= nums.length <= 105`
• `-109 <= nums[i] <= 109`
 `````` 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 `````` ``````class Solution: def getMaxLen(self, nums: List[int]) -> int: n = len(nums) # dp[0]: len of subarray ending with cur num with pos prod # dp[1]: len of subarray ending with cur num with neg prod dp = [0] * 2 if nums[0] > 0: dp[0] = 1 if nums[0] < 0: dp[1] = 1 res = dp[0] for i in range(1, n): cur = nums[i] tmp = [0] * 2 if cur > 0: tmp[0] = dp[0] + 1 if dp[1] > 0: tmp[1] = dp[1] + 1 elif cur < 0: tmp[1] = dp[0] + 1 if dp[1] > 0: tmp[0] = max(tmp[0], dp[1] + 1) dp = tmp res = max(res, dp[0]) return res``````