Introduction

In this blog post, we will explore how to solve the LeetCode problem "1362" 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 an integer num, find the closest two integers in absolute difference whose product equals num + 1 or num + 2. Return the two integers in any order. Example 1: Input: num = 8 Output: [3,3] Explanation: For num + 1 = 9, the closest divisors are 3 & 3, for num + 2 = 10, the closest divisors are 2 & 5, hence 3 & 3 is chosen. Example 2: Input: num = 123 Output: [5,25] Example 3: Input: num = 999 Output: [40,25] Constraints: 1 <= num <= 10^9

Explanation

Here's a solution to find the closest factors whose product is num + 1 or num + 2:

• Core Idea: Iterate downwards from the square root of num + 2 (the larger possible product) to find a factor. Once a factor is found for either num + 1 or num + 2, calculate the corresponding pair, and return. Starting from the square root ensures that we find the closest factors quickly.

• Prioritization: Check for factors of num + 1 before num + 2. If both have equally close factors, the problem statement implies choosing factors of num+1

• Optimization: Iterating downwards avoids calculating square roots repeatedly and ensures the first factor found is from the closest factor pair.

• Complexity: O(sqrt(n)) runtime and O(1) space complexity, where n is the input number num.

Code

    def closestDivisors(num: int) -> list[int]:
    """
    Finds the closest two integers in absolute difference whose product equals num + 1 or num + 2.

    Args:
        num: The input integer.

    Returns:
        A list of two integers representing the closest divisors.
    """

    best_a = 1
    best_b = num + 2
    min_diff = num + 1  # Initialize with a large difference

    for i in range(int((num + 2)**0.5), 0, -1):
        if (num + 1) % i == 0:
            a = i
            b = (num + 1) // i
            diff = abs(a - b)
            if diff < min_diff:
                min_diff = diff
                best_a = a
                best_b = b
            break  # Found factors for num + 1, no need to check num + 2 further if they're equally close

        if (num + 2) % i == 0:
            a = i
            b = (num + 2) // i
            diff = abs(a - b)
            if diff <= min_diff:  # Include equal because num+1 case is preferred
                min_diff = diff
                best_a = a
                best_b = b

    return [best_a, best_b]