# Solving Leetcode Interviews in Seconds with AI: Count the Digits That Divide a Number


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "2520" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like [Chatmagic](https://www.chatmagic.app), 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, return the number of digits in num that divide num. An integer val divides nums if nums % val == 0.   Example 1:  Input: num = 7 Output: 1 Explanation: 7 divides itself, hence the answer is 1.  Example 2:  Input: num = 121 Output: 2 Explanation: 121 is divisible by 1, but not 2. Since 1 occurs twice as a digit, we return 2.  Example 3:  Input: num = 1248 Output: 4 Explanation: 1248 is divisible by all of its digits, hence the answer is 4.    Constraints:  1 <= num <= 109 num does not contain 0 as one of its digits.  

	# Explanation
	Here's the breakdown of the solution:

*   **Iterate through digits:** Convert the number to a string and iterate through each digit.
*   **Check divisibility:** For each digit, check if it divides the original number without any remainder using the modulo operator (%).
*   **Count divisible digits:** Increment a counter for each digit that divides the original number.

*   **Runtime Complexity:** O(log<sub>10</sub>(num)), **Storage Complexity:** O(1)

	
	# Code
	```python
	def count_digits(num: int) -> int:
    """
    Given an integer num, return the number of digits in num that divide num.
    An integer val divides nums if nums % val == 0.

    Example 1:
    Input: num = 7
    Output: 1
    Explanation: 7 divides itself, hence the answer is 1.

    Example 2:
    Input: num = 121
    Output: 2
    Explanation: 121 is divisible by 1, but not 2. Since 1 occurs twice as a digit, we return 2.

    Example 3:
    Input: num = 1248
    Output: 4
    Explanation: 1248 is divisible by all of its digits, hence the answer is 4.

    Constraints:
    1 <= num <= 109
    num does not contain 0 as one of its digits.
    """
    count = 0
    num_str = str(num)
    for digit in num_str:
        digit_int = int(digit)
        if num % digit_int == 0:
            count += 1
    return count
	```