How do you solve Smallest All-Ones Multiple on LeetCode?

Smallest All-Ones Multiple (LeetCode #3790) 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: Only numbers with odd digits can be multiples of even k.

What is the time complexity of Smallest All-Ones Multiple?

The optimal solution for Smallest All-Ones Multiple runs in O(n) time and uses O(1) space. This approach only tracks remainders, making it efficient in both time and space.

What is the brute force solution for Smallest All-Ones Multiple?

The brute force approach for Smallest All-Ones Multiple is Brute Force, with O(n²) time complexity and O(1) space complexity. Generate all numbers consisting of only the digit '1' and check their divisibility by k. This is inefficient as it checks each number sequentially. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Smallest All-Ones Multiple?

Smallest All-Ones Multiple uses the following concepts: Hash Table, Math. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Smallest All-Ones Multiple?

Explain your thought process clearly. Discuss edge cases upfront. Practice modular arithmetic problems.

What are common mistakes when solving Smallest All-Ones Multiple?

Not considering the case where k is even. Forgetting to check if count exceeds k.

#3790

Smallest All-Ones Multiple

Medium

Hash TableMathHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n)Space O(1)

Use modular arithmetic to build the number incrementally without generating large integers. This reduces the problem to tracking remainders.

⚙️

Algorithm

3 steps

1Step 1: Initialize rem as 1 % k and count as 1.
2Step 2: While rem is not 0, update rem = (rem * 10 + 1) % k and increment count.
3Step 3: If rem becomes 0, return count; otherwise, if count exceeds k, return -1.

solution.py9 lines

1def smallestAllOnesMultiple(k):
2    rem = 1 % k
3    count = 1
4    while rem != 0:
5        rem = (rem * 10 + 1) % k
6        count += 1
7        if count > k:
8            return -1
9    return count

ℹ

Complexity note: This approach only tracks remainders, making it efficient in both time and space.

1Only numbers with odd digits can be multiples of even k.
2Using modular arithmetic prevents overflow.

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