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 <= 13
• 1 <= nums[i] <= 105
• 1 <= 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;
    }
}

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