Concatenated Divisibility
Time O(nlogr + k * n * 2^n) · Space O(logr + k * 2^n) · Official statement on LeetCode
Solutions
// Time: O(nlogr + k * n * 2^n)
// Space: O(logr + k * 2^n)
// dp, backtracing
class Solution {
public:
vector<int> concatenatedDivisibility(vector<int>& nums, int k) {
const auto& length = [](int x) {
int l = 0;
for (; x > 0; x /= 10) {
++l;
}
return max(l, 1);
};
vector<int> lookup(size(nums));
for (int i = 0; i < size(nums); ++i) {
lookup[i] = length(nums[i]);
}
const int mx = ranges::max(lookup);
vector<int> pow10(mx + 1);
pow10[0] = 1 % k;
for (int i = 0; i + 1 < size(pow10); ++i) {
pow10[i + 1] = (pow10[i] * 10) % k;
}
vector<vector<bool>> dp(1 << size(nums), vector<bool>(k));
dp.back()[0] = true;
for (int mask = (size(dp) - 1) - 1; mask >= 0; --mask) {
for (int r = 0; r < k; ++r) {
for (int i = 0; i < size(nums); ++i) {
if (mask & (1 << i)) {
continue;
}
if (dp[mask | (1 << i)][(r * pow10[lookup[i]] + nums[i]) % k]) {
dp[mask][r] = true;
break;
}
}
}
}
vector<int> result;
if (!dp[0][0]) {
return result;
}
vector<pair<int,int>> order;
for (int i = 0; i < size(nums); ++i) {
order.emplace_back(nums[i], i);
}
sort(begin(order), end(order));
for (int _ = 0, mask = 0, r = 0; _ < size(nums); ++_) {
for (const auto& [_, i] : order) {
if (mask & (1 << i)) {
continue;
}
if (dp[mask | (1 << i)][(r * pow10[lookup[i]] + nums[i]) % k]) {
result.emplace_back(nums[i]);
mask |= (1 << i);
r = (r * pow10[lookup[i]] + nums[i]) % k;
break;
}
}
}
return result;
}
};
Beginner Explanation
What is Concatenated Divisibility?
Concatenated Divisibility (LeetCode #3533) is a Hard problem that primarily trains dynamic programming.
How to think about it
- Restate the goal in your own words before coding.
- Work a tiny example by hand so the invariant becomes obvious.
- Identify the pattern — this problem aligns with dynamic programming and bit manipulation.
- Only then translate the idea into code.
Why this problem matters
Hard problems force you to combine patterns and prove complexity carefully — interview gold. Official solution notes mention: DP, Bitmasks, Backtracing.
AlgoForge explanations are original teaching notes. Always open the official problem statement on LeetCode for constraints and examples.
Interview Walkthrough
Interview approach for Concatenated Divisibility
Opening (30–60 seconds)
- Clarify inputs/outputs and edge cases (empty input, single element, duplicates, overflow).
- State a brute force so the interviewer knows you can solve it naively.
- Propose the optimal direction tied to dynamic programming and bit manipulation.
Core solution narrative
- Define the state you track (pointers, DP cell, set membership, stack top, etc.).
- Explain the transition when you process the next element.
- Call out time (O(nlogr + k * n * 2^n)) and space (O(logr + k * 2^n)) before coding.
- Code cleanly; narrate variable names.
What interviewers listen for
- Correctness on edge cases
- Complexity honesty
- Ability to discuss trade-offs (e.g., hash map space vs. sort + two pointers)
Follow-up questions they may ask
- Can you solve it with less memory?
- What if the input stream is infinite / doesn't fit in RAM?
- How would tests look for adversarial inputs?
Optimized Approach
Optimized solution notes
The reference solutions on AlgoForge target O(nlogr + k * n * 2^n) time and O(logr + k * 2^n) space.
Pattern focus: dynamic programming and bit manipulation
Use the pattern as a checklist:
- dynamic programming — confirm the invariant holds after each step
- bit manipulation — confirm the invariant holds after each step
Multiple methods appear in the source solutions — compare them and explain when each is preferable.
Implementation tips
- Prefer readable names over micro-optimizations in interviews.
- Extract helpers only when they clarify (e.g., expand-around-center, DFS visit).
- After AC-level logic, re-scan for off-by-one and null checks.
Complexity Analysis
Complexity
| Measure | Bound |
|---|---|
| Time | O(nlogr + k * n * 2^n) |
| Space | O(logr + k * 2^n) |
How to justify this in an interview
- Time: count loops, map/set operations, and recursive branching; state average vs worst case if relevant.
- Space: include hash maps, recursion stack, and output allocation when the problem asks for it.
If your implementation differs from the reference, re-derive big-O from your code — never memorize a complexity you cannot defend.
Common Mistakes
Common mistakes on Concatenated Divisibility
- Skipping edge cases — empty collections, single-element inputs, max constraints.
- Wrong invariant for dynamic programming and bit manipulation — updating state too early or too late.
- Mutating input unexpectedly when the problem forbids it.
- Off-by-one in windows, ranges, or binary search bounds.
- Ignoring overflow / precision for integer arithmetic problems.
- Overengineering — jumping to an advanced structure when a simpler approach works.
Alternative Approaches
Alternatives
The source file includes more than one method. Compare:
- Primary optimized path — best complexity for typical interviews.
- Secondary approach — often brute force, sorting-based, or space-optimized variant.
Practice articulating when you would pick each (constraints, readability, follow-ups).
Edge Cases
Edge cases checklist
- Minimum input size
- Maximum input size / time limits
- Duplicates and already-sorted input
- Negative numbers / zeros (if applicable)
- Disconnected structures (graphs/trees)
- Single path vs branching recursion depth
Pattern Recognition
Spotting this pattern
Signal phrases that point to dynamic programming and bit manipulation:
- Sorted input or ability to sort without changing the answer class
- Need for contiguous subarray / substring → consider sliding window
- Need for O(1) membership → hash set/map
- Optimal substructure + overlapping subproblems → DP
- Connectivity / components → graph DFS/BFS or Union-Find
Primary topics: dynamic programming.
Follow-up Interview Questions
Follow-ups
- How does the solution change if the input is a stream?
- Can you solve it in-place?
- What if duplicates must be handled differently?
- How would you parallelize the approach?
- Design tests that would break a buggy implementation.
Practice Recommendations
What to practice next
- Re-solve Concatenated Divisibility in a second language (cpp, python).
- Drill 3–5 more problems tagged dynamic programming.
- Teach the solution out loud in under 5 minutes.
- Add this problem to your revision calendar in 3 days and 14 days.
Visualization
Study checklist
- Read the official problem statement on LeetCode
- Solve on paper / whiteboard first
- Implement the dynamic programming and bit manipulation approach
- Verify edge cases from the checklist
- State time and space complexity aloud
- Compare with the AlgoForge reference solution
- Schedule a revision session
Revision notes
Concatenated Divisibility (#3533) — Hard. Pattern: dynamic programming and bit manipulation. Complexity: O(nlogr + k * n * 2^n) time / O(logr + k * 2^n) space. Re-derive the invariant before coding.
FAQs
What is the time complexity of Concatenated Divisibility?+
The reference solutions aim for O(nlogr + k * n * 2^n) time and O(logr + k * 2^n) space. Always re-derive complexity from the code you write in the interview.
What pattern does Concatenated Divisibility use?+
It primarily maps to dynamic programming and bit manipulation, within the broader topic of dynamic programming.
Is Concatenated Divisibility good for interviews?+
Yes — as a Hard problem it is a solid practice target. Pair it with related problems in the same pattern family for spaced repetition.
Where can I read the official statement?+
Open the official LeetCode page for constraints and examples: https://leetcode.com/problems/concatenated-divisibility/