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

Smallest Divisible Digit Product II (LeetCode #3348) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Understanding the properties of digit products and their divisibility can help optimize the search.

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

The brute force approach for Smallest Divisible Digit Product II is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves generating numbers starting from the given string 'num' and checking each one to see if it meets the criteria of being zero-free and having a digit product divisible by 't'. This is straightforward but inefficient. The optimal Optimal Solution approach improves this to O(n).

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

Smallest Divisible Digit Product II uses the following concepts: Math, String, Backtracking, Greedy, Number Theory. The recommended patterns to study are: Greedy Approach, Backtracking.

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

Always clarify the constraints and edge cases before diving into coding. Think about how to minimize changes to the number while achieving the goal. Practice breaking down the problem into smaller parts to simplify the solution.

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

Failing to check for zero-free condition after modifying digits. Not considering edge cases where no valid number exists.

#3348

Smallest Divisible Digit Product II

Hard

MathStringBacktrackingGreedyNumber TheoryGreedy ApproachBacktracking

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n)Space O(n)

The optimal approach involves modifying the digits of 'num' directly to find the smallest zero-free number that meets the product condition. This is more efficient as it avoids unnecessary checks and focuses on the digits that need to be changed.

⚙️

Algorithm

5 steps

1Step 1: Convert 'num' to a list of digits.
2Step 2: Calculate the product of the digits and check if it is divisible by 't'.
3Step 3: If not, iterate from the last digit to find the first digit that can be increased to make the product divisible by 't'.
4Step 4: Increment this digit and set all subsequent digits to '1' to minimize the number.
5Step 5: If the resulting number is valid (zero-free), return it; otherwise, return '-1'.

solution.py26 lines

1# Full working Python code
2
3def smallest_divisible_digit_product(num, t):
4    digits = list(map(int, num))
5    product = 1
6    for d in digits:
7        product *= d
8    if product % t == 0:
9        return num
10
11    for i in range(len(digits) - 1, -1, -1):
12        if digits[i] < 9:
13            digits[i] += 1
14            for j in range(i + 1, len(digits)):
15                digits[j] = 1
16            new_num = ''.join(map(str, digits))
17            if '0' not in new_num and product_of_digits(new_num) % t == 0:
18                return new_num
19            break
20    return '-1'
21
22def product_of_digits(num):
23    product = 1
24    for digit in num:
25        product *= int(digit)
26    return product

ℹ

Complexity note: The time complexity is O(n) because we only iterate through the digits a limited number of times, and the space complexity is O(n) for storing the digits.

1Understanding the properties of digit products and their divisibility can help optimize the search.
2Incrementing digits from the least significant to the most allows for minimal changes to achieve the desired product.

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