Introduction

In this blog post, we will explore how to solve the LeetCode problem "2427" 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

Given two positive integers a and b, return the number of common factors of a and b. An integer x is a common factor of a and b if x divides both a and b. Example 1: Input: a = 12, b = 6 Output: 4 Explanation: The common factors of 12 and 6 are 1, 2, 3, 6. Example 2: Input: a = 25, b = 30 Output: 2 Explanation: The common factors of 25 and 30 are 1, 5. Constraints: 1 <= a, b <= 1000

Explanation

Here's the approach, followed by the Python code:

• Find the Greatest Common Divisor (GCD): The common factors of a and b are precisely the factors of their GCD. This drastically reduces the search space.

• Iterate up to the Square Root of the GCD: We only need to iterate up to the square root of the GCD to find all factors. If i is a factor of gcd, then gcd / i is also a factor. This optimization significantly improves efficiency.

• Runtime Complexity: O(sqrt(min(a, b))), Storage Complexity: O(1)

Code

    def commonFactors(a: int, b: int) -> int:
    """
    Given two positive integers a and b, return the number of common factors of a and b.
    An integer x is a common factor of a and b if x divides both a and b.

    Example 1:
    Input: a = 12, b = 6
    Output: 4
    Explanation: The common factors of 12 and 6 are 1, 2, 3, 6.

    Example 2:
    Input: a = 25, b = 30
    Output: 2
    Explanation: The common factors of 25 and 30 are 1, 5.

    Constraints:
    1 <= a, b <= 1000
    """

    def gcd(a, b):
        while(b):
            a, b = b, a % b
        return a

    g = gcd(a, b)
    count = 0
    for i in range(1, int(g**0.5) + 1):
        if g % i == 0:
            if i * i == g:
                count += 1
            else:
                count += 2
    return count