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].
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)
C++
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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; } }; |