LeetCode #1567 — MEDIUM

Maximum Length of Subarray With Positive Product

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

Problem Statement

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

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Dynamic Programming · Greedy

Example 1

[1,-2,-3,4]

Example 2

[0,1,-2,-3,-4]

Example 3

[-1,-2,-3,0,1]

Step 02

Core Insight

What unlocks the optimal approach

Split the whole array into subarrays by zeroes since a subarray with positive product cannot contain any zero.
If the subarray has even number of negative numbers, the whole subarray has positive product.
Otherwise, we have two choices, either - remove the prefix till the first negative element in this subarray, or remove the suffix starting from the last negative element in this subarray.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #1567: Maximum Length of Subarray With Positive Product
class Solution {
    public int getMaxLen(int[] nums) {
        int n = nums.length;
        int[] f = new int[n];
        int[] g = new int[n];
        f[0] = nums[0] > 0 ? 1 : 0;
        g[0] = nums[0] < 0 ? 1 : 0;
        int ans = f[0];
        for (int i = 1; i < n; ++i) {
            if (nums[i] > 0) {
                f[i] = f[i - 1] + 1;
                g[i] = g[i - 1] > 0 ? g[i - 1] + 1 : 0;
            } else if (nums[i] < 0) {
                f[i] = g[i - 1] > 0 ? g[i - 1] + 1 : 0;
                g[i] = f[i - 1] + 1;
            }
            ans = Math.max(ans, f[i]);
        }
        return ans;
    }
}

// Accepted solution for LeetCode #1567: Maximum Length of Subarray With Positive Product
func getMaxLen(nums []int) int {
	n := len(nums)
	f := make([]int, n)
	g := make([]int, n)
	if nums[0] > 0 {
		f[0] = 1
	}
	if nums[0] < 0 {
		g[0] = 1
	}
	ans := f[0]

	for i := 1; i < n; i++ {
		if nums[i] > 0 {
			f[i] = f[i-1] + 1
			if g[i-1] > 0 {
				g[i] = g[i-1] + 1
			} else {
				g[i] = 0
			}
		} else if nums[i] < 0 {
			if g[i-1] > 0 {
				f[i] = g[i-1] + 1
			} else {
				f[i] = 0
			}
			g[i] = f[i-1] + 1
		}
		ans = max(ans, f[i])
	}
	return ans
}

# Accepted solution for LeetCode #1567: Maximum Length of Subarray With Positive Product
class Solution:
    def getMaxLen(self, nums: List[int]) -> int:
        n = len(nums)
        f = [0] * n
        g = [0] * n
        f[0] = int(nums[0] > 0)
        g[0] = int(nums[0] < 0)
        ans = f[0]
        for i in range(1, n):
            if nums[i] > 0:
                f[i] = f[i - 1] + 1
                g[i] = 0 if g[i - 1] == 0 else g[i - 1] + 1
            elif nums[i] < 0:
                f[i] = 0 if g[i - 1] == 0 else g[i - 1] + 1
                g[i] = f[i - 1] + 1
            ans = max(ans, f[i])
        return ans

// Accepted solution for LeetCode #1567: Maximum Length of Subarray With Positive Product
struct Solution;

impl Solution {
    fn get_max_len(nums: Vec<i32>) -> i32 {
        nums.split(|&x| x == 0)
            .map(Self::max_length)
            .max()
            .unwrap_or(0)
    }
    fn max_length(v: &[i32]) -> i32 {
        let mut neg = 0;
        let mut res = 0;
        let n = v.len();
        let mut first_neg: Option<usize> = None;
        for i in 0..n {
            if v[i] < 0 {
                neg += 1;
                if first_neg.is_none() {
                    first_neg = Some(i);
                }
            }
            if neg % 2 == 0 {
                res = res.max((i + 1) as i32);
            } else {
                if let Some(j) = first_neg {
                    res = res.max((i - j) as i32);
                }
            }
        }
        res
    }
}

#[test]
fn test() {
    let nums = vec![1, -2, -3, 4];
    let res = 4;
    assert_eq!(Solution::get_max_len(nums), res);
    let nums = vec![0, 1, -2, -3, -4];
    let res = 3;
    assert_eq!(Solution::get_max_len(nums), res);
    let nums = vec![-1, -2, -3, 0, 1];
    let res = 2;
    assert_eq!(Solution::get_max_len(nums), res);
    let nums = vec![-1, 2];
    let res = 1;
    assert_eq!(Solution::get_max_len(nums), res);
    let nums = vec![1, 2, 3, 5, -6, 4, 0, 10];
    let res = 4;
    assert_eq!(Solution::get_max_len(nums), res);
}

// Accepted solution for LeetCode #1567: Maximum Length of Subarray With Positive Product
function getMaxLen(nums: number[]): number {
    const n = nums.length;
    const f: number[] = Array(n).fill(0);
    const g: number[] = Array(n).fill(0);

    if (nums[0] > 0) {
        f[0] = 1;
    }
    if (nums[0] < 0) {
        g[0] = 1;
    }

    let ans = f[0];
    for (let i = 1; i < n; i++) {
        if (nums[i] > 0) {
            f[i] = f[i - 1] + 1;
            g[i] = g[i - 1] > 0 ? g[i - 1] + 1 : 0;
        } else if (nums[i] < 0) {
            f[i] = g[i - 1] > 0 ? g[i - 1] + 1 : 0;
            g[i] = f[i - 1] + 1;
        }

        ans = Math.max(ans, f[i]);
    }

    return ans;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(n)

Approach Breakdown

RECURSIVE

O(2ⁿ) time

O(n) space

Pure recursion explores every possible choice at each step. With two choices per state (take or skip), the decision tree has 2ⁿ leaves. The recursion stack uses O(n) space. Many subproblems are recomputed exponentially many times.

DYNAMIC PROGRAMMING

O(n × m) time

O(n × m) space

Each cell in the DP table is computed exactly once from previously solved subproblems. The table dimensions determine both time and space. Look for the state variables — each unique combination of state values is one cell. Often a rolling array can reduce space by one dimension.

Shortcut: Count your DP state dimensions → that’s your time. Can you drop one? That’s your space optimization.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

State misses one required dimension

Wrong move: An incomplete state merges distinct subproblems and caches incorrect answers.

Usually fails on: Correctness breaks on cases that differ only in hidden state.

Fix: Define state so each unique subproblem maps to one DP cell.

Using greedy without proof

Wrong move: Locally optimal choices may fail globally.

Usually fails on: Counterexamples appear on crafted input orderings.

Fix: Verify with exchange argument or monotonic objective before committing.