3718. Smallest Missing Multiple of K

3718. Smallest Missing Multiple of K

Difficulty: Easy

Topics: Mid Level, Array, Hash Table, Weekly Contest 472

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.

Example 3:

• Input: nums = [2,4,6,8,10], k = 2
• Output: 12

Example 4:

• Input: nums = [5,10,15,20], k = 5
• Output: 25

Example 5:

• Input: nums = [3,6,9,12,15], k = 3
• Output: 18

Example 6:

• Input: nums = [1,2,3,4,5,6,7,8,9,10], k = 1
• Output: 11

Example 7:

• Input: nums = [100], k = 1
• Output: 2

Example 8:

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

Constraints:

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

Hint:

1. Add the values in nums to a hash set
2. Iterate through the positive multiples of k and return the first one not in the hash set

Solution:

We solve the problem by first storing all array elements in a hash set for O(1) lookups. We then iterate upward through positive multiples of k starting from k itself, and return the first multiple that is not present in the set. This ensures the smallest missing multiple is found efficiently.

Approach

• Use a hash set – Convert the array into a hash set (via array_flip) to quickly check membership.
• Start from the first positive multiple – Begin checking from k (the smallest positive multiple of k).
• Increment by k – Move to the next multiple (+= k) until we find one not in the set.
• Return the missing multiple – The first missing multiple encountered is the answer.

Let's implement this solution in PHP: 3718. Smallest Missing Multiple of K

<?php
/**
 * @param Integer[] $nums
 * @param Integer $k
 * @return Integer
 */
function missingMultiple(array $nums, int $k): int
{
    ...
    ...
    ...
    /**
     * go to ./solution.php
     */
}

// Test cases
echo missingMultiple([8,2,3,4,6], 2) .  "\n";               // Output: 10
echo missingMultiple([1,4,7,10,15], 5) .  "\n";             // Output: 5
echo missingMultiple([2,4,6,8,10], 2) .  "\n";              // Output: 12
echo missingMultiple([5,10,15,20], 5) .  "\n";              // Output: 25
echo missingMultiple([3,6,9,12,15], 3) .  "\n";             // Output: 18
echo missingMultiple([1,2,3,4,5,6,7,8,9,10], 1) .  "\n";    // Output: 11
echo missingMultiple([100], 1) .  "\n";                     // Output: 2
echo missingMultiple([1,2,3], 4) .  "\n";                   // Output: 4
?>

Explanation:

• Step 1: $set = array_flip($nums); – This creates a hash map where keys are the array values, enabling constant-time existence checks.
• Step 2: $multiple = $k; – We start with the smallest positive multiple of k.
• Step 3: while (isset($set[$multiple])) { $multiple += $k; } – Keep advancing to the next multiple while the current one exists in the set.
• Step 4: Once the loop exits, $multiple is the smallest multiple of k not present in nums, so we return it.

Complexity Analysis

• Time Complexity: O(n + m) where n is the number of elements in nums (to build the set) and m is the number of multiples we check until we find a missing one. In the worst case, m could be up to the largest multiple present plus k, but since constraints are small (≤100), this is effectively constant.
• Space Complexity: O(n) – for the hash set storing up to n unique values.

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