Maximum Length of Subarray With Positive Product

Maximum Length of Subarray With Positive Product - Problem

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.

Input & Output

Example 1 — Mixed Signs

$ Input: nums = [1,-2,-3,4]

› Output: 4

💡 Note: The entire array has positive product: 1 × (-2) × (-3) × 4 = 24 > 0, so maximum length is 4

Example 2 — With Zero

$ Input: nums = [0,1,-2,-3,-4]

› Output: 3

💡 Note: Zero breaks the array. Subarray [1,-2,-3] has product 6 > 0 with length 3, which is maximum

Example 3 — All Negative

$ Input: nums = [-1,-2,-3]

› Output: 2

💡 Note: Any two negatives give positive product. [-1,-2] or [-2,-3] both have length 2

Constraints

1 ≤ nums.length ≤ 10⁵
-10⁹ ≤ nums[i] ≤ 10⁹

Visualization

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Asked in

a Amazon 25 M Microsoft 18

The key insight is that a positive product requires an even number of negative numbers. Best approach uses dynamic programming to track lengths of subarrays ending at each position with positive/negative products. Time: O(n), Space: O(1)

Common Approaches

✓ Brute Force

⏱️ Time: O(n²) Space: O(1)

For each starting position, extend the subarray one element at a time, calculating the product and tracking the maximum length when product is positive.

Dynamic Programming

⏱️ Time: O(n) Space: O(1)

Use two variables to track the maximum length ending at current position with positive and negative products. Key insight: negative × negative = positive.

Brute Force — Algorithm Steps

For each starting index i
Extend subarray from i to j
Calculate product and check if positive
Update maximum length

Visualization

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Step-by-Step Walkthrough

Start Position

Fix starting index i

Extend Right

Extend to index j, calculate product

Check Product

If positive, update max length

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

int solution(int* nums, int n) {
    int maxLength = 0;
    
    for (int i = 0; i < n; i++) {
        long long product = 1;
        for (int j = i; j < n; j++) {
            product *= nums[j];
            if (product > 0) {
                int length = j - i + 1;
                if (length > maxLength) {
                    maxLength = length;
                }
            }
        }
    }
    
    return maxLength;
}

int main() {
    char line[10000];
    fgets(line, sizeof(line), stdin);
    
    int nums[1000];
    int n = 0;
    char* ptr = line + 1;
    
    while (*ptr && *ptr != ']') {
        if (*ptr == ',' || *ptr == ' ') {
            ptr++;
            continue;
        }
        nums[n++] = atoi(ptr);
        while (*ptr && *ptr != ',' && *ptr != ']') {
            ptr++;
        }
    }
    
    int result = solution(nums, n);
    printf("%d\n", result);
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n²)

Nested loops: outer loop n times, inner loop up to n times

⚠ Quadratic Growth

Space Complexity

O(1)

Only using variables to track product and max length

✓ Linear Space

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Medium Frequency

~25 min Avg. Time

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Smart Actions

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Input & Output

Constraints

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Related Problems

Common Approaches

Brute Force — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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