Problem Description

You need to implement a function that determines the sign of the product of all numbers in an array.

The function signFunc(x) should return:

• 1 if x is positive
• -1 if x is negative
• 0 if x equals 0

Given an integer array nums, you need to:

1. Calculate the product of all values in the array
2. Return the sign of that product using the signFunc logic

For example:

• If nums = [1, 2, 3], the product is 6 (positive), so return 1
• If nums = [-1, 2, 3], the product is -6 (negative), so return -1
• If nums = [1, 0, 3], the product is 0, so return 0

The key insight is that you don't actually need to calculate the full product (which could cause overflow). Instead, you only need to track:

• Whether any element is 0 (which makes the entire product 0)
• The count of negative numbers (odd count means negative product, even count means positive product)

The solution traverses the array once, immediately returning 0 if any element is 0, and flipping the sign for each negative number encountered. This efficiently determines the sign without computing the actual product value.

Intuition

When we multiply numbers together, what really determines the sign of the final product? Let's think about the properties of multiplication:

1. Any number multiplied by 0 gives 0
2. Positive × Positive = Positive
3. Negative × Negative = Positive
4. Positive × Negative = Negative

This means if there's even a single 0 in the array, the entire product becomes 0 regardless of other values.

For non-zero numbers, we notice that the actual magnitude doesn't matter - only whether each number is positive or negative. A negative number flips the sign of our running product, while a positive number keeps it the same.

Consider these examples:

• [2, 3, -4]: Start positive → stays positive after 2 → stays positive after 3 → flips to negative after -4
• [-1, -2, 3]: Start positive → flips to negative after -1 → flips back to positive after -2 → stays positive after 3

We can track this sign-flipping behavior without calculating the actual product. Start with ans = 1 (positive), and for each negative number we encounter, multiply ans by -1 to flip its sign. Positive numbers don't change the sign, so we can skip them.

This approach avoids overflow issues that could occur if we actually multiplied large numbers together. We only care about the sign, not the magnitude, so we can work with just 1 and -1 values throughout the traversal.

Learn more about Math patterns.

Solution Approach

The implementation uses a direct traversal approach with a sign-tracking variable.

We initialize an answer variable ans = 1 to represent a positive sign initially. This variable will hold either 1 (positive) or -1 (negative) throughout the traversal.

The algorithm works as follows:

1. Traverse each element v in the array nums

2. Check for zero: If v == 0, immediately return 0 since any product containing zero equals zero

3. Handle negative numbers: If v < 0, multiply ans by -1 to flip its sign

   • If ans was 1, it becomes -1
   • If ans was -1, it becomes 1

4. Skip positive numbers: Positive numbers don't affect the sign, so no action needed

5. Return the final sign: After traversing all elements, ans contains the sign of the product

Here's how the code executes for nums = [-1, -2, -3, 4, 5]:

• Start: ans = 1
• Process -1: negative, so ans = 1 * (-1) = -1
• Process -2: negative, so ans = -1 * (-1) = 1
• Process -3: negative, so ans = 1 * (-1) = -1
• Process 4: positive, skip
• Process 5: positive, skip
• Return: -1

The time complexity is O(n) where n is the length of the array, as we traverse it once. The space complexity is O(1) since we only use a constant amount of extra space for the ans variable.

Example Walkthrough

Let's walk through the solution with nums = [2, -3, 0, -4].

Step 1: Initialize ans = 1 (representing positive sign)

Step 2: Process first element 2

• 2 > 0, so it's positive
• No action needed, ans remains 1

Step 3: Process second element -3

• -3 < 0, so it's negative
• Flip the sign: ans = 1 * (-1) = -1

Step 4: Process third element 0

• 0 == 0, so we found a zero
• Immediately return 0 (the product will be zero)

The function returns 0 without needing to check the remaining element -4.

Let's try another example without zeros: nums = [-2, 3, -5]

Step 1: Initialize ans = 1

Step 2: Process -2

• -2 < 0, it's negative
• Flip sign: ans = 1 * (-1) = -1

Step 3: Process 3

• 3 > 0, it's positive
• No action needed, ans remains -1

Step 4: Process -5

• -5 < 0, it's negative
• Flip sign: ans = -1 * (-1) = 1

Step 5: Return ans = 1

The product would be (-2) * 3 * (-5) = 30 which is positive, confirming our result of 1.

Solution Implementation

Time and Space Complexity

The time complexity is O(n), where n is the length of the input array nums. This is because the algorithm iterates through each element of the array exactly once in a single loop. Each operation inside the loop (checking if a value equals 0, checking if a value is negative, and multiplying by -1) takes constant time O(1).

The space complexity is O(1). The algorithm only uses a single variable ans to track the sign of the product, regardless of the input size. No additional data structures are created that grow with the input size, making the space usage constant.

Learn more about how to find time and space complexity quickly.

Common Pitfalls

1. Attempting to Calculate the Actual Product

The most common mistake is trying to compute the actual product of all elements and then determining its sign:

# WRONG APPROACH - prone to overflow
def arraySign(self, nums: List[int]) -> int:
    product = 1
    for num in nums:
        product *= num  # This can overflow!
  
    if product > 0:
        return 1
    elif product < 0:
        return -1
    else:
        return 0

Why it fails: For large arrays with big numbers, the product quickly exceeds integer limits, causing overflow errors. For example, [100, 100, 100, 100, 100] would result in 10,000,000,000 which may cause issues in some systems.

Solution: Only track the sign changes, never calculate the actual product value.

2. Forgetting to Handle Zero Early

Another pitfall is checking for zeros after processing all elements:

# INEFFICIENT APPROACH
def arraySign(self, nums: List[int]) -> int:
    sign = 1
    has_zero = False
  
    for num in nums:
        if num == 0:
            has_zero = True  # Still continues processing!
        elif num < 0:
            sign *= -1
  
    return 0 if has_zero else sign

Why it's inefficient: Once we encounter a zero, we know the answer is 0. Continuing to process remaining elements wastes time.

Solution: Return immediately upon finding a zero with an early exit strategy.

3. Using Complex Counting Logic

Some solutions unnecessarily count negative numbers and then determine the sign based on odd/even count:

# OVERCOMPLICATED APPROACH
def arraySign(self, nums: List[int]) -> int:
    negative_count = 0
  
    for num in nums:
        if num == 0:
            return 0
        if num < 0:
            negative_count += 1
  
    return -1 if negative_count % 2 == 1 else 1

Why it's overcomplicated: While this works, it adds unnecessary complexity with modulo operations and counting logic when simple sign flipping achieves the same result more intuitively.

Solution: Use direct sign multiplication (sign *= -1) which is cleaner and more readable.