Algorithms and Data Structures in Python: Basic Array Problem - Sign of the Product of an Array

Introduction

Hello folks! Welcome to a guide on cracking a cool array problem from LeetCode, which is perfect for getting your feet wet with algorithms and data structures. Whether you’re dreaming of joining big tech giants, diving into machine learning, or just leveling up your coding game, you’re in the right spot.

While these puzzles are cool and definitely sharpen your skills, if you’re all about building awesome apps or slick websites, you might want to learn more about getting real-world experience. Sure, it’s great to know your algorithms and data structures, but don’t spend all your time on them. Mixing it up with actual projects can be a more direct route to landing that dream tech job. After all, creating stuff that people can interact with is a whole different kind of satisfaction.

So, while we’re here to boost those problem-solving skills with a coding challenge, remember it’s all about balance. Ready to tackle the problem and learn something new along the way? Let’s dive in!

Problem

https://leetcode.com/problems/sign-of-the-product-of-an-array/

There is 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).

Example 1:

Input: nums = [-1,-2,-3,-4,3,2,1]
Output: 1
Explanation: The product of all values in the array is 144, and signFunc(144) = 1

Example 2:

Input: nums = [1,5,0,2,-3]
Output: 0
Explanation: The product of all values in the array is 0, and signFunc(0) = 0

Example 3:

Input: nums = [-1,1,-1,1,-1]
Output: -1
Explanation: The product of all values in the array is -1, and signFunc(-1) = -1

Solution

class Solution:
    def arraySign(self, nums: List[int]) -> int:
        sign = 1  # 1. Initialization

        # 2. Iterating Through nums
        for num in nums:
          if num == 0: # 2.1 Zero Check
            return 0  
          elif num < 0: # 2.2 Negative Check
            sign *= -1

        # 3. Result
        return sign

Step-by-Step Explanation

1. Initialization: We initiate our solution with a variable named sign set to 1. This choice is strategic, streamlining our process by presuming the product of the array of elements is positive from the start. This presumption simplifies our logic in two key ways:

• Eliminates Redundancy: By assuming a positive product, we remove the need to explicitly check for positive numbers (num > 0) in our array. Positive numbers do not alter the sign of the product. Hence, checking for them does not contribute to our final determination of the product's sign.
• Focuses on Sign-Changing Conditions: With sign initialized to 1, our attention shifts solely to conditions that would change this assumption—specifically, encountering a 0 or a negative number. A 0 immediately determines the product's sign as 0, while each negative number encountered toggles the sign variable, effectively keeping track of whether we've encountered an even or odd number of negative numbers.

2. Iterating Through nums: We loop through each number in the given array nums. For each number, we have two checks:

2.1. Zero Check: If any number is 0, the product of the entire array is 0. Thus, we immediately return 0 without further computation.

2.2. Negative Check: If a number is negative (num < 0), we invert the sign by multiplying it by -1. This step accounts for the mathematical rule that the product of two negative numbers is positive, but the product of a positive and a negative number is negative.

3. Result: After iterating through the entire array, the sign variable holds the sign of the product. We return sign, which will be either 1 for a positive product, -1 for a negative product, or 0 if any element was 0.

Additional Considerations

• Alternative Sign Checking: Instead of sign *= -1, you could use this conditional: sign = -sign if num < 0 else sign.
• Efficiency: This solution avoids an unnecessary and potentially large product calculation. Its focus on finding the sign makes it computationally efficient.

Time Complexity Analysis

The solution we provided iterates through each element of the array exactly once. The operations performed within each iteration (checking if an element is zero or negative and then possibly changing the sign of the sign variable) are constant time operations, meaning they take the same amount of time regardless of the size of the input array.

Therefore, the time complexity of the solution is O(n), where n is the number of elements in the input array nums. This linear time complexity indicates that the time it takes to execute our solution scales linearly with the input size. In other words, if the input array size doubles, the execution time of our algorithm also roughly doubles, which is an efficient scaling behavior for an algorithm.

Wrapping Up

This problem presents an excellent opportunity to practice thinking about optimization, edge cases (like zeroes), and how simple conditional logic can solve problems. Engaging with challenges like this sharpens critical programming fundamentals, empowering you to navigate beyond the realm of abstract puzzles into the practical world of coding with confidence and creativity.

Although today's problem is considered basic, it's important to recognize that mastering medium-level problems and beyond can take time and persistent effort. The journey from understanding basic concepts to tackling more complex challenges is a gradual one, filled with learning opportunities at every step. Don't lose hope if progress seems slow at times; every problem solved is a step forward in honing your skills. Consistency and patience are your allies in this journey of growth.

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