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:
class Solution {
public int getMaxLen(int[] nums) {
int n = nums.length;
int count = 0, max = 0;
int l = 0, r = 0;
while (r < n) {
int temp = r;
int curr = nums[r];
if (curr < 0) {
count ++;
}
if (count % 2 == 1 && (curr == 0 || r == n - 1)) {
while (count % 2 == 1) {
int out = nums[l++];
if (out < 0) {
count --;
}
}
}
while (l < n && nums[l] == 0) l ++;
if (count % 2 == 0) {
if (curr == 0) r --;
// System.out.println("l: " + l + ", r: " + r);
max = Math.max(max, r - l + 1);
}
if (curr == 0) {
l = temp + 1;
}
r = temp + 1;
}
return max;
}
}