Introduction

In this blog post, we will explore how to solve the LeetCode problem "2269" 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

The k-beauty of an integer num is defined as the number of substrings of num when it is read as a string that meet the following conditions: It has a length of k. It is a divisor of num. Given integers num and k, return the k-beauty of num. Note: Leading zeros are allowed. 0 is not a divisor of any value. A substring is a contiguous sequence of characters in a string. Example 1: Input: num = 240, k = 2 Output: 2 Explanation: The following are the substrings of num of length k: - "24" from "240": 24 is a divisor of 240. - "40" from "240": 40 is a divisor of 240. Therefore, the k-beauty is 2. Example 2: Input: num = 430043, k = 2 Output: 2 Explanation: The following are the substrings of num of length k: - "43" from "430043": 43 is a divisor of 430043. - "30" from "430043": 30 is not a divisor of 430043. - "00" from "430043": 0 is not a divisor of 430043. - "04" from "430043": 4 is not a divisor of 430043. - "43" from "430043": 43 is a divisor of 430043. Therefore, the k-beauty is 2. Constraints: 1 <= num <= 109 1 <= k <= num.length (taking num as a string)

Explanation

• Convert the number to a string. Iterate through the string using a sliding window of size k.
  • In each iteration, extract the substring, convert it to an integer, and check if it's a divisor of the original number (and is not zero).
  • Increment a counter if the substring meets the conditions.

• Time Complexity: O(n), where n is the number of digits in num. Space Complexity: O(1)

Code

    def divisorSubstrings(num: int, k: int) -> int:
    """
    Calculates the k-beauty of an integer.

    Args:
        num: The integer to analyze.
        k: The length of the substrings.

    Returns:
        The k-beauty of the integer.
    """
    num_str = str(num)
    n = len(num_str)
    count = 0

    for i in range(n - k + 1):
        sub_str = num_str[i:i + k]
        sub_num = int(sub_str)

        if sub_num != 0 and num % sub_num == 0:
            count += 1

    return count