Introduction

In this blog post, we will explore how to solve the LeetCode problem "1822" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

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). 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 Constraints: 1 <= nums.length <= 1000 -100 <= nums[i] <= 100

Explanation

Here's an efficient solution to the problem:

• Core Idea: Instead of calculating the actual product, we only need to track the sign of the product. We can do this by counting the number of negative numbers.
• Zero Check: If any number is zero, the product is zero, so we can return 0 immediately.

• Sign Determination: If the number of negative numbers is even, the product is positive (sign = 1). If the number of negative numbers is odd, the product is negative (sign = -1).

• Runtime Complexity: O(n), where n is the length of the input array. Storage Complexity: O(1).

Code

    def arraySign(nums):
    """
    Calculates the sign of the product of an array of integers.

    Args:
        nums: A list of integers.

    Returns:
        1 if the product is positive, -1 if negative, and 0 if zero.
    """
    negative_count = 0
    for num in nums:
        if num == 0:
            return 0
        if num < 0:
            negative_count += 1

    if negative_count % 2 == 0:
        return 1
    else:
        return -1