How do you solve Smallest Divisible Digit Product I on LeetCode?

Smallest Divisible Digit Product I (LeetCode #3345) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(1) time complexity and O(1) space complexity. Key insight: The product of digits can be quickly computed, and checking divisibility is efficient.

What is the time complexity of Smallest Divisible Digit Product I?

The optimal solution for Smallest Divisible Digit Product I runs in O(1) time and uses O(1) space. The time complexity is O(1) since we only check a maximum of 10 numbers, making it constant time regardless of the input size.

What is the brute force solution for Smallest Divisible Digit Product I?

The brute force approach for Smallest Divisible Digit Product I is Brute Force, with O(n²) time complexity and O(1) space complexity. We will start from the given number n and check each subsequent number to see if the product of its digits is divisible by t. This is straightforward but can be inefficient if n is large. The optimal Optimal Solution approach improves this to O(1).

What algorithm or data structure is used to solve Smallest Divisible Digit Product I?

Smallest Divisible Digit Product I uses the following concepts: Math, Enumeration. The recommended patterns to study are: Enumeration, Brute Force.

What are the interview tips for Smallest Divisible Digit Product I?

Always start with a brute-force approach to understand the problem better. Make sure to explain your thought process clearly while coding. Test your code with edge cases to ensure it works under all conditions.

What are common mistakes when solving Smallest Divisible Digit Product I?

Not considering edge cases where n is already divisible by t. Overcomplicating the solution by trying to optimize prematurely.

#3345

Smallest Divisible Digit Product I

Easy

MathEnumerationEnumerationBrute Force

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(1)
Space	O(1)	O(1)

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Intuition

Time O(1)Space O(1)

Instead of checking every number sequentially, we can limit our checks to a maximum of 10 numbers, as the product of digits will cycle through combinations quickly.

⚙️

Algorithm

3 steps

1Step 1: Start from n and check if the product of its digits is divisible by t.
2Step 2: If not, increment n and check again, but only do this for up to 10 iterations.
3Step 3: Return the first number found that meets the criteria.

solution.py14 lines

1# Full working Python code
2
3def digit_product(x):
4    product = 1
5    while x > 0:
6        product *= x % 10
7        x //= 10
8    return product
9
10
11def smallest_divisible_digit_product(n, t):
12    for i in range(10):
13        if digit_product(n + i) % t == 0:
14            return n + i

ℹ

Complexity note: The time complexity is O(1) since we only check a maximum of 10 numbers, making it constant time regardless of the input size.

1The product of digits can be quickly computed, and checking divisibility is efficient.
2The constraints allow us to limit our checks to a small range, making the problem manageable.

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