3533. Concatenated Divisibility
Description
You are given an array of positive integers nums and a positive integer k.
A permutation of nums is said to form a divisible concatenation if, when you concatenate the decimal representations of the numbers in the order specified by the permutation, the resulting number is divisible by k.
Return the lexicographically smallest permutation (when considered as a list of integers) that forms a divisible concatenation. If no such permutation exists, return an empty list.
Example 1:
Input: nums = [3,12,45], k = 5
Output: [3,12,45]
Explanation:
| Permutation | Concatenated Value | Divisible by 5 |
|---|---|---|
| [3, 12, 45] | 31245 | Yes |
| [3, 45, 12] | 34512 | No |
| [12, 3, 45] | 12345 | Yes |
| [12, 45, 3] | 12453 | No |
| [45, 3, 12] | 45312 | No |
| [45, 12, 3] | 45123 | No |
The lexicographically smallest permutation that forms a divisible concatenation is [3,12,45].
Example 2:
Input: nums = [10,5], k = 10
Output: [5,10]
Explanation:
| Permutation | Concatenated Value | Divisible by 10 |
|---|---|---|
| [5, 10] | 510 | Yes |
| [10, 5] | 105 | No |
The lexicographically smallest permutation that forms a divisible concatenation is [5,10].
Example 3:
Input: nums = [1,2,3], k = 5
Output: []
Explanation:
Since no permutation of nums forms a valid divisible concatenation, return an empty list.
Constraints:
1 <= nums.length <= 131 <= nums[i] <= 1051 <= k <= 100
Solutions
Solution 1
Thinking
\(n \le 13\) allows a search over permutations, but rebuilding the concatenated integer from scratch repeats work. We also need the lexicographically smallest permutation.
Precompute each value’s power of ten. A subset DP stores the used set and the running remainder modulo \(k\), and reconstructs along the lexicographically smaller path. If no state works, return an empty list.
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