Introduction

In this blog post, we will explore how to solve the LeetCode problem "1492" 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

You are given two positive integers n and k. A factor of an integer n is defined as an integer i where n % i == 0. Consider a list of all factors of n sorted in ascending order, return the kth factor in this list or return -1 if n has less than k factors. Example 1: Input: n = 12, k = 3 Output: 3 Explanation: Factors list is [1, 2, 3, 4, 6, 12], the 3rd factor is 3. Example 2: Input: n = 7, k = 2 Output: 7 Explanation: Factors list is [1, 7], the 2nd factor is 7. Example 3: Input: n = 4, k = 4 Output: -1 Explanation: Factors list is [1, 2, 4], there is only 3 factors. We should return -1. Constraints: 1 <= k <= n <= 1000 Follow up: Could you solve this problem in less than O(n) complexity?

Explanation

Here's an efficient solution to find the k-th factor of a number, addressing the time complexity constraint:

• Square Root Optimization: Iterate only up to the square root of n. If i is a factor, then n // i is also a factor. This significantly reduces the number of iterations.
• Separate Lists: Store factors less than or equal to the square root in one list, and factors greater than the square root in another. This allows us to reconstruct the sorted list of factors efficiently.

• Handle Perfect Squares: If n is a perfect square, avoid adding the square root twice.

• Time Complexity: O(sqrt(n))

• Space Complexity: O(sqrt(n))

Code

    def kthFactor(n: int, k: int) -> int:
    factors1 = []
    factors2 = []
    sqrt_n = int(n**0.5)

    for i in range(1, sqrt_n + 1):
        if n % i == 0:
            factors1.append(i)
            if i * i != n:
                factors2.append(n // i)

    factors = factors1 + factors2[::-1]

    if k > len(factors):
        return -1
    else:
        return factors[k - 1]