3718 - Smallest Missing Multiple of K (Easy)Copy Anchor Link

Problem LinkCopy Anchor Link

https://leetcode.com/problems/smallest-missing-multiple-of-k/

Problem StatementCopy Anchor Link

Given an integer array nums and an integer k, return the smallest positive multiple of k that is missing from nums.

A multiple of k is any positive integer divisible by k.

Example 1:

Input: nums = [8,2,3,4,6], k = 2

Output: 10

Explanation:

The multiples of k = 2 are 2, 4, 6, 8, 10, 12... and the smallest multiple missing from nums is 10.

Example 2:

Input: nums = [1,4,7,10,15], k = 5

Output: 5

Explanation:

The multiples of k = 5 are 5, 10, 15, 20... and the smallest multiple missing from nums is 5.

Constraints:

• 1 <= nums.length <= 100
• 1 <= nums[i] <= 100
• 1 <= k <= 100

Approach 1: EnumerationCopy Anchor Link

Convert $nums$ into a set to allow fast membership checks. Starting from k, check each positive multiple of $k$ in increasing order. If the current multiple exists in the set, add k and continue, otherwise return it. Because the multiples are checked from smallest to largest, the first missing one is guaranteed to be the answer.

Building the set takes $O(n)$ time and space. At most $n+1$ multiples need to be checked, so the overall time complexity is $O(n)$, and the space complexity is $O(n)$.

class Solution:
    def missingMultiple(self, nums: List[int], k: int) -> int:
        nums = {*nums}
        res = k
        while True:
            if res not in nums: break
            res += k
        return res