How do you solve Self Dividing Numbers on LeetCode?

Self Dividing Numbers (LeetCode #728) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n * d) time complexity and O(n) space complexity. Key insight: Self-dividing numbers cannot contain the digit zero.

What is the brute force solution for Self Dividing Numbers?

The brute force approach for Self Dividing Numbers is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach checks each number in the given range and verifies if it is self-dividing by examining each digit. This is straightforward but can be slow for larger ranges. The optimal Optimal Solution approach improves this to O(n * d).

What algorithm or data structure is used to solve Self Dividing Numbers?

Self Dividing Numbers uses the following concepts: Math. The recommended patterns to study are: Math, Iteration.

What are the interview tips for Self Dividing Numbers?

Always clarify edge cases, such as the presence of zero in digits. Discuss your thought process and approach before jumping into coding. Consider the efficiency of your solution and be prepared to optimize if needed.

What are common mistakes when solving Self Dividing Numbers?

Forgetting to check for the digit zero, which leads to incorrect results. Not handling the conversion of numbers to digits properly, leading to runtime errors.

#728

Self Dividing Numbers

Easy

MathMathIteration

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n * d)Space O(n)

The optimal solution still checks each number in the range, but it uses a more efficient method to validate self-dividing status by directly manipulating digits without converting to strings, which can save time.

⚙️

Algorithm

6 steps

1Step 1: Initialize an empty list to store self-dividing numbers.
2Step 2: Loop through each number from left to right.
3Step 3: For each number, use a while loop to extract each digit.
4Step 4: Check if the digit is zero or does not divide the number evenly.
5Step 5: If all digits pass the checks, add the number to the list.
6Step 6: Return the list of self-dividing numbers.

solution.py19 lines

1# Full working Python code
2
3def selfDividingNumbers(left, right):
4    result = []
5    for num in range(left, right + 1):
6        n = num
7        is_self_dividing = True
8        while n > 0:
9            digit = n % 10
10            if digit == 0 or num % digit != 0:
11                is_self_dividing = False
12                break
13            n //= 10
14        if is_self_dividing:
15            result.append(num)
16    return result
17
18# Example usage
19print(selfDividingNumbers(1, 22))

ℹ

Complexity note: The time complexity is O(n * d), where n is the number of integers in the range and d is the number of digits in the largest number (at most 5 for numbers up to 10,000). The space complexity is O(n) due to the storage of results.

1Self-dividing numbers cannot contain the digit zero.
2Each digit must divide the number evenly for it to be considered self-dividing.

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