How do you solve Count the Digits That Divide a Number on LeetCode?

Count the Digits That Divide a Number (LeetCode #2520) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: Digits should be checked individually for divisibility.

What is the brute force solution for Count the Digits That Divide a Number?

The brute force approach for Count the Digits That Divide a Number is Brute Force, with O(n) time complexity and O(1) space complexity. The brute-force approach involves checking each digit of the number to see if it divides the number evenly. This is straightforward but can be inefficient for larger numbers. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Count the Digits That Divide a Number?

Count the Digits That Divide a Number uses the following concepts: Math. The recommended patterns to study are: Mathematical operations, Digit extraction.

What are the interview tips for Count the Digits That Divide a Number?

Clarify the problem requirements and constraints before coding. Think aloud while solving to demonstrate your thought process. Test your solution with edge cases to ensure robustness.

What are common mistakes when solving Count the Digits That Divide a Number?

Forgetting to check if the digit is zero before division. Not handling the number as a whole when checking divisibility.

#2520

Count the Digits That Divide a Number

Easy

MathMathematical operationsDigit extraction

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n)	O(n)
Space	O(1)	O(1)

💡

Intuition

Time O(n)Space O(1)

The optimal solution still checks each digit but avoids unnecessary conversions and uses a more efficient loop. This approach is similar to the brute force but is streamlined for performance.

⚙️

Algorithm

5 steps

1Step 1: Initialize a counter to zero.
2Step 2: Use a while loop to extract digits from the number until it becomes zero.
3Step 3: For each extracted digit, check if it divides the original number.
4Step 4: If it does, increment the counter.
5Step 5: Return the counter as the result.

solution.py9 lines

1def countDigits(num):
2    count = 0
3    original = num
4    while num > 0:
5        digit = num % 10
6        if digit != 0 and original % digit == 0:
7            count += 1
8        num //= 10
9    return count

ℹ

Complexity note: The time complexity remains O(n) as we still process each digit, but we do it more efficiently without converting to strings. The space complexity is O(1) since we only use a fixed amount of extra space.

1Digits should be checked individually for divisibility.
2Avoid division by zero by ensuring digits are non-zero.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.