Count the Digits That Divide a Number

Difficulty: Easy | Acceptance: 85.90% | Paid: No Topics: Math

Given an integer num, return the number of digits in num that divide num.

An integer val divides nums if nums % val == 0.

• Examples
• Constraints
• Iterative Approach
• String Conversion Approach

Examples

Example 1:

Input: num = 7
Output: 1
Explanation: 7 divides 7, so 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, so the answer is 4.

Constraints

1 <= num <= 10⁹

Iterative Approach

Intuition Extract each digit from the number using modulo and division operations, then check if each digit divides the original number.

Steps

• Store the original number in a temporary variable
• Iterate through each digit by taking modulo 10
• Check if the digit divides the original number
• Divide the temporary number by 10 to move to the next digit
• Count the digits that divide the number

class Solution:
    def countDigits(self, num: int) -> int:
        original = num
        count = 0
        while num > 0:
            digit = num % 10
            if original % digit == 0:
                count += 1
            num //= 10
        return count

Complexity

• Time: O(log n) where n is the number of digits in num
• Space: O(1)
• Notes: This is the most efficient approach with constant space complexity.

String Conversion Approach

Intuition Convert the number to a string and iterate through each character, converting it back to an integer to check divisibility.

Steps

• Convert the number to a string
• Iterate through each character in the string
• Convert each character back to an integer
• Check if the digit divides the original number
• Count the digits that divide the number

class Solution:
    def countDigits(self, num: int) -> int:
        count = 0
        for digit in str(num):
            if num % int(digit) == 0:
                count += 1
        return count

Complexity

• Time: O(log n) where n is the number of digits in num
• Space: O(log n) for storing the string representation
• Notes: Slightly less space-efficient than the iterative approach but more readable.