How do you solve Smallest Missing Multiple of K on LeetCode?

Smallest Missing Multiple of K (LeetCode #3718) 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: Using a hash set allows for O(1) lookups.

What is the time complexity of Smallest Missing Multiple of K?

The optimal solution for Smallest Missing Multiple of K runs in O(n) time and uses O(n) space. The algorithm runs in O(n) due to the single pass through nums to create the set and the linear check for multiples.

What is the brute force solution for Smallest Missing Multiple of K?

The brute force approach for Smallest Missing Multiple of K is Brute Force, with O(n²) time complexity and O(n) space complexity. Check each multiple of k starting from k until we find one that is not in nums. 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 Missing Multiple of K?

Smallest Missing Multiple of K uses the following concepts: Array, Hash Table. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Smallest Missing Multiple of K?

Explain your thought process clearly. Consider edge cases, like when all multiples are present. Discuss time and space complexities.

What are common mistakes when solving Smallest Missing Multiple of K?

Forgetting to check for positive multiples only. Not using a set for efficient lookups.

#3718

Smallest Missing Multiple of K

Easy

ArrayHash TableHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n)Space O(n)

Utilize a hash set for quick lookups and iterate through multiples of k until we find the missing one.

⚙️

Algorithm

3 steps

1Step 1: Create a set from nums for O(1) lookups.
2Step 2: Initialize a variable multiple to k.
3Step 3: Check if multiple is in the set; if not, return it. If yes, increment by k and repeat.

solution.py6 lines

1def smallest_missing_multiple(nums, k):
2    num_set = set(nums)
3    multiple = k
4    while multiple not in num_set:
5        return multiple
6        multiple += k

ℹ

Complexity note: The algorithm runs in O(n) due to the single pass through nums to create the set and the linear check for multiples.

1Using a hash set allows for O(1) lookups.
2The smallest missing multiple can be found by iterating through multiples of k.

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