Sign of the Product of an Array - Practice Coding Problems

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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Understanding the Visualization

Input

Array of integers with mixed signs

Analysis

Count negatives and check for zeros

Output

Return 1, -1, or 0 based on product sign

Key Takeaway

🎯 Key Insight: We only need the sign, not the actual product - count negatives to avoid overflow

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

Approach	Time	Space	Notes
✓ Count Negatives and Zeros	O(n)	O(1)	Count negative numbers and check for zeros without calculating the full product
Calculate Full Product	O(n)	O(1)	Multiply all numbers and check the sign of the result

Count Negatives and Zeros — Algorithm Steps

Step 1: Check if any element is zero
Step 2: Count negative numbers
Step 3: Return 0 if zero exists, -1 if odd negatives, 1 if even negatives

Visualization

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Step-by-Step Walkthrough

Scan Array

Check each element for zero or negative

Count Negatives

Track count of negative numbers

Determine Sign

Return based on zero existence and negative count

Code -

solution.c — C

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

int solution(int* nums, int numsSize) {
    int negativeCount = 0;
    
    for (int i = 0; i < numsSize; i++) {
        if (nums[i] == 0) {
            return 0;
        } else if (nums[i] < 0) {
            negativeCount++;
        }
    }
    
    return negativeCount % 2 == 1 ? -1 : 1;
}

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 to count negatives and check for zeros

✓ Linear Growth

Space Complexity

O(1)

Only using variables to count negatives and track zero existence

✓ Linear Space

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Count Negatives and Zeros — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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