LeetCode #3533 — HARD

Concatenated Divisibility

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

Solve on LeetCode
The Problem

Problem Statement

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
Patterns Used
📊Dynamic Programming→⚡Bit Manipulation→

Roadmap

  1. Brute Force Baseline
  2. Core Insight
  3. Algorithm Walkthrough
  4. Edge Cases
  5. Full Annotated Code
  6. Interactive Study Demo
  7. Complexity Analysis
Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Dynamic Programming · Bit Manipulation

Example 1

[3,12,45]
5

Example 2

[10,5]
10

Example 3

[1,2,3]
5
Step 02

Core Insight

What unlocks the optimal approach

  • Can we write a recursive solution for this?
  • Can we use bitmasks with dynamic programming to optimize the above recursion?
  • Use the idea of bitmask-based dynamic programming.
  • Use the idea to reconstruct the answer from the dynamic programming table using the state variables, such as <code>mask</code> and <code>remainder</code>.
Interview move: turn each hint into an invariant you can check after every iteration/recursion step.
Step 03

Algorithm Walkthrough

Iteration Checklist

  1. Define state (indices, window, stack, map, DP cell, or recursion frame).
  2. Apply one transition step and update the invariant.
  3. Record answer candidate when condition is met.
  4. Continue until all input is consumed.
Use the first example testcase as your mental trace to verify each transition.
Step 04

Edge Cases

Minimum Input
Single element / shortest valid input
Validate boundary behavior before entering the main loop or recursion.
Duplicates & Repeats
Repeated values / repeated states
Decide whether duplicates should be merged, skipped, or counted explicitly.
Extreme Constraints
Largest constraint values
Re-check complexity target against constraints to avoid time-limit issues.
Invalid / Corner Shape
Empty collections, zeros, or disconnected structures
Handle special-case structure before the core algorithm path.
Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #3533: Concatenated Divisibility
class Solution {
  public int[] concatenatedDivisibility(int[] nums, int k) {
    final int n = nums.length;
    int[] lengths = new int[n];
    int[] pows = new int[n];

    Arrays.sort(nums);

    for (int i = 0; i < n; ++i) {
      lengths[i] = String.valueOf(nums[i]).length();
      pows[i] = (int) Math.pow(10, lengths[i]) % k;
    }

    Integer[][] mem = new Integer[1 << n][k];
    return dp(nums, pows, mem, k, 0, 0) ? reconstruct(nums, pows, mem, k, 0, 0) : new int[0];
  }

  // Returns true if there is a way to form a number divisible by `k` using the
  // numbers in `nums`, where nums[i] is used iff `mask & (1 << i)`.
  private boolean dp(int[] nums, int[] pows, Integer[][] mem, int k, int mask, int mod) {
    if (mem[mask][mod] != null)
      return mem[mask][mod] == 1;
    if (mask == (1 << nums.length) - 1)
      return (mem[mask][mod] = mod == 0 ? 1 : 0) == 1;
    for (int i = 0; i < nums.length; ++i)
      if ((mask >> i & 1) == 0) {
        final int newMod = (mod * pows[i] + nums[i]) % k;
        if (dp(nums, pows, mem, k, mask | 1 << i, newMod))
          return (mem[mask][mod] = 1) == 1;
      }
    return (mem[mask][mod] = 0) == 1;
  }

  // Reconstructs the numbers that form a number divisible by `k` using the
  // numbers in `nums`, where nums[i] is used iff `mask & (1 << i)`.
  private int[] reconstruct(int[] nums, int[] pows, Integer[][] mem, int k, int mask, int mod) {
    for (int i = 0; i < nums.length; ++i)
      if ((mask >> i & 1) == 0) {
        final int newMod = (mod * pows[i] + nums[i]) % k;
        if (dp(nums, pows, mem, k, mask | 1 << i, newMod)) {
          int[] first = new int[] {nums[i]};
          int[] rest = reconstruct(nums, pows, mem, k, mask | 1 << i, newMod);
          int[] res = new int[first.length + rest.length];
          System.arraycopy(first, 0, res, 0, first.length);
          System.arraycopy(rest, 0, res, first.length, rest.length);
          return res;
        }
      }
    return new int[0];
  }
}
Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Press Step or Run All to begin.
Step 07

Complexity Analysis

Time
O(n × m)
Space
O(n × m)

Approach Breakdown

RECURSIVE
O(2ⁿ) time
O(n) space

Pure recursion explores every possible choice at each step. With two choices per state (take or skip), the decision tree has 2ⁿ leaves. The recursion stack uses O(n) space. Many subproblems are recomputed exponentially many times.

DYNAMIC PROGRAMMING
O(n × m) time
O(n × m) space

Each cell in the DP table is computed exactly once from previously solved subproblems. The table dimensions determine both time and space. Look for the state variables — each unique combination of state values is one cell. Often a rolling array can reduce space by one dimension.

Shortcut: Count your DP state dimensions → that’s your time. Can you drop one? That’s your space optimization.
Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

State misses one required dimension

Wrong move: An incomplete state merges distinct subproblems and caches incorrect answers.

Usually fails on: Correctness breaks on cases that differ only in hidden state.

Fix: Define state so each unique subproblem maps to one DP cell.

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