How do you solve Smallest Integer Divisible by K on LeetCode?

Smallest Integer Divisible by K (LeetCode #1015) 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: The problem can be solved using breadth-first search to avoid overflow and efficiently find the solution.

What is the time complexity of Smallest Integer Divisible by K?

The optimal solution for Smallest Integer Divisible by K runs in O(n) time and uses O(n) space. The time complexity is O(n) because we explore each remainder at most once. The space complexity is O(n) due to the queue and visited set.

What is the brute force solution for Smallest Integer Divisible by K?

The brute force approach for Smallest Integer Divisible by K is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves generating numbers made up of the digit '1' and checking each one to see if it is divisible by k. This method is straightforward but inefficient, especially for larger values of k. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Smallest Integer Divisible by K?

Smallest Integer Divisible by K uses the following concepts: Hash Table, Math. The recommended patterns to study are: Breadth-First Search, Hash Set.

What are the interview tips for Smallest Integer Divisible by K?

Always consider edge cases, such as when k is even. Explain your thought process clearly; it shows your understanding. Practice implementing both brute-force and optimal solutions to solidify your understanding.

What are common mistakes when solving Smallest Integer Divisible by K?

Students often try to generate all numbers directly, leading to overflow issues. Failing to track visited remainders can cause infinite loops.

#1015

Smallest Integer Divisible by K

Medium

Hash TableMathBreadth-First SearchHash Set

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)

Instead of generating large numbers, we can track remainders when we build the number. This way, we avoid overflow and can find the answer using a breadth-first search approach.

⚙️

Algorithm

4 steps

1Step 1: Use a queue to perform a breadth-first search starting with the number '1'.
2Step 2: Maintain a set to track visited remainders to avoid cycles.
3Step 3: For each number, calculate the new number by appending '1' and compute the new remainder.
4Step 4: If the remainder is 0, return the current length. If not, enqueue the new number if its remainder hasn't been seen before.

solution.py14 lines

1from collections import deque
2
3def smallestIntegerDivisibleByK(k):
4    queue = deque([(1, 1)])  # (current number, length)
5    visited = set()
6    while queue:
7        n, length = queue.popleft()
8        if n % k == 0:
9            return length
10        remainder = n % k
11        if remainder not in visited:
12            visited.add(remainder)
13            queue.append((remainder * 10 + 1, length + 1))
14    return -1

ℹ

Complexity note: The time complexity is O(n) because we explore each remainder at most once. The space complexity is O(n) due to the queue and visited set.

1The problem can be solved using breadth-first search to avoid overflow and efficiently find the solution.
2Tracking remainders helps in determining divisibility without generating large numbers.

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