LeetCode-in-Java
3533. Concatenated Divisibility
Hard
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
Solution
import java.util.Arrays;
@SuppressWarnings("java:S107")
public class Solution {
public int[] concatenatedDivisibility(int[] nums, int k) {
Arrays.sort(nums);
int digits = 0;
int n = nums.length;
int[] digCnt = new int[n];
for (int i = 0; i < n; i++) {
int num = nums[i];
digits++;
digCnt[i]++;
while (num >= 10) {
digits++;
digCnt[i]++;
num /= 10;
}
}
int[] pow10 = new int[digits + 1];
pow10[0] = 1;
for (int i = 1; i <= digits; i++) {
pow10[i] = (pow10[i - 1] * 10) % k;
}
int[] res = new int[n];
return dfs(0, 0, k, digCnt, nums, pow10, new boolean[1 << n][k], 0, res, n)
? res
: new int[0];
}
private boolean dfs(
int mask,
int residue,
int k,
int[] digCnt,
int[] nums,
int[] pow10,
boolean[][] visited,
int ansIdx,
int[] ans,
int n) {
if (ansIdx == n) {
return residue == 0;
}
if (visited[mask][residue]) {
return false;
}
for (int i = 0, bit = 1; i < n; i++, bit <<= 1) {
if ((mask & bit) == bit) {
continue;
}
int newResidue = (residue * pow10[digCnt[i]] + nums[i]) % k;
ans[ansIdx] = nums[i];
if (dfs(mask | bit, newResidue, k, digCnt, nums, pow10, visited, ansIdx + 1, ans, n)) {
return true;
}
}
visited[mask][residue] = true;
return false;
}
}