# 1567. Maximum Length of Subarray With Positive Product

## Description

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

## Solutions

• class Solution {
public int getMaxLen(int[] nums) {
int f1 = nums[0] > 0 ? 1 : 0;
int f2 = nums[0] < 0 ? 1 : 0;
int res = f1;
for (int i = 1; i < nums.length; ++i) {
if (nums[i] > 0) {
++f1;
f2 = f2 > 0 ? f2 + 1 : 0;
} else if (nums[i] < 0) {
int pf1 = f1, pf2 = f2;
f2 = pf1 + 1;
f1 = pf2 > 0 ? pf2 + 1 : 0;
} else {
f1 = 0;
f2 = 0;
}
res = Math.max(res, f1);
}
return res;
}
}

• class Solution {
public:
int getMaxLen(vector<int>& nums) {
int f1 = nums[0] > 0 ? 1 : 0;
int f2 = nums[0] < 0 ? 1 : 0;
int res = f1;
for (int i = 1; i < nums.size(); ++i) {
if (nums[i] > 0) {
++f1;
f2 = f2 > 0 ? f2 + 1 : 0;
} else if (nums[i] < 0) {
int pf1 = f1, pf2 = f2;
f2 = pf1 + 1;
f1 = pf2 > 0 ? pf2 + 1 : 0;
} else {
f1 = 0;
f2 = 0;
}
res = max(res, f1);
}
return res;
}
};

• class Solution:
def getMaxLen(self, nums: List[int]) -> int:
f1 = 1 if nums[0] > 0 else 0
f2 = 1 if nums[0] < 0 else 0
res = f1
for num in nums[1:]:
pf1, pf2 = f1, f2
if num > 0:
f1 += 1
if f2 > 0:
f2 += 1
else:
f2 = 0
elif num < 0:
pf1, pf2 = f1, f2
f2 = pf1 + 1
if pf2 > 0:
f1 = pf2 + 1
else:
f1 = 0
else:
f1 = 0
f2 = 0
res = max(res, f1)
return res

• func getMaxLen(nums []int) int {
f1, f2 := 0, 0
if nums[0] > 0 {
f1 = 1
}
if nums[0] < 0 {
f2 = 1
}
res := f1
for i := 1; i < len(nums); i++ {
if nums[i] > 0 {
f1++
if f2 > 0 {
f2++
} else {
f2 = 0
}
} else if nums[i] < 0 {
pf1, pf2 := f1, f2
f2 = pf1 + 1
if pf2 > 0 {
f1 = pf2 + 1
} else {
f1 = 0
}
} else {
f1, f2 = 0, 0
}
res = max(res, f1)
}
return res
}

• function getMaxLen(nums: number[]): number {
// 连续正数计数n1, 连续负数计数n2
let n1 = nums[0] > 0 ? 1 : 0,
n2 = nums[0] < 0 ? 1 : 0;
let ans = n1;
for (let i = 1; i < nums.length; ++i) {
let cur = nums[i];
if (cur == 0) {
(n1 = 0), (n2 = 0);
} else if (cur > 0) {
++n1;
n2 = n2 > 0 ? n2 + 1 : 0;
} else {
let t1 = n1,
t2 = n2;
n1 = t2 > 0 ? t2 + 1 : 0;
n2 = t1 + 1;
}
ans = Math.max(ans, n1);
}
return ans;
}