Sign of the Product of an Array - Problem

Array Math Easy

Implement a function signFunc(x) that returns:

1 if x is positive.
-1 if x is negative.
0 if x is equal to 0.

You are given an integer array nums. Let product be the product of all values in the array nums.

Return signFunc(product).

Input & Output

Example 1 — Mixed Signs

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

› Output: 1

💡 Note: Product = (-1) × 2 × (-3) × 4 = 24. Since 24 > 0, return 1.

Example 2 — Contains Zero

$ Input: nums = [1,5,0,2,-3]

› Output: 0

💡 Note: Product = 1 × 5 × 0 × 2 × (-3) = 0. Since product equals 0, return 0.

Example 3 — Odd Negatives

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

› Output: -1

💡 Note: Product = (-1) × (-2) × (-3) = -6. Since -6 < 0, return -1.

Constraints

1 ≤ nums.length ≤ 1000
-100 ≤ nums[i] ≤ 100

Visualization

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Asked in

a Amazon 15 F Facebook 10

The key insight is that we only need the sign of the product, not the actual value. Count negative numbers: if odd count, result is negative; if even count, result is positive; if any zero exists, result is zero. Best approach avoids overflow with Time: O(n), Space: O(1).

Common Approaches

✓ Calculate Full Product

⏱️ Time: O(n) Space: O(1)

Calculate the actual product by multiplying all elements, then determine if it's positive, negative, or zero. This is straightforward but can cause overflow for large arrays.

Count Negatives and Zeros

⏱️ Time: O(n) Space: O(1)

Instead of calculating the actual product, count negative numbers and check for zeros. If there's any zero, return 0. If the count of negatives is odd, return -1, otherwise return 1.

Calculate Full Product — Algorithm Steps

Step 1: Initialize product = 1
Step 2: Multiply each number in the array
Step 3: Return 1 if product > 0, -1 if product < 0, 0 if product = 0

Visualization

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Step-by-Step Walkthrough

Initialize

Start with product = 1

Multiply

Multiply each element with product

Check Sign

Return sign of final product

Code -

solution.c — C

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

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

int main() {
    char line[10000];
    fgets(line, sizeof(line), stdin);
    
    // Simple parser for JSON array
    int nums[1000];
    int count = 0;
    char* token = strtok(line + 1, ",]");
    while (token != NULL) {
        nums[count++] = atoi(token);
        token = strtok(NULL, ",]");
    }
    
    int result = solution(nums, count);
    printf("%d\n", result);
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n)

Single loop through all n elements

✓ Linear Growth

Space Complexity

O(1)

Only using one variable to store product

✓ Linear Space

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Sign of the Product of an Array - Problem

Input & Output

Constraints

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Related Problems

Common Approaches

Calculate Full Product — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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