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Problem solution · C++

Concatenated Divisibility

Concatenated Divisibility: a C++ solution using dynamic programming. Learn the idea, check the complexity, and read the full code, with credit to walkccc LeetCode Solutions.

Technique
Dynamic programming
Source
walkccc LeetCode Solutions
Length
55 lines
Start with the idea.

Try the problem first. If you get stuck, read the approach below, then write your own solution. The full code is at the bottom.

Approach

Dynamic programming

For Concatenated Divisibility, the implementation records answers for smaller states and reuses them to build the requested result without repeating work.

  1. Define precisely what one DP state represents.
  2. Establish the base cases before transitions are evaluated.
  3. Process states in dependency order and combine only already-known values.

Code notes

  • 55 lines of C++ from the credited upstream file 3533.cpp.
  • The implementation visibly relies on sequence storage, cached states.
  • 3 loop blocks detected.

Complexity

Multiply the number of reachable states by the work performed for each transition, then include the stored state table in memory usage.

Check the problem constraints before deciding whether this complexity will pass.

Source

Code and credit

This code comes from walkccc LeetCode Solutions by P.-Y. Chen (walkccc) and is used under the MIT licence.

Open problem ↗Original source ↗
Full codeConcatenated Divisibility · C++C++
Use this to learn the idea, then write your own version.
class Solution { public:  vector<int> concatenatedDivisibility(vector<int>& nums, int k) {    vector<int> lengths;    vector<int> pows;     ranges::sort(nums);     for (const int num : nums) {      lengths.push_back(to_string(num).length());      pows.push_back(static_cast<int>(pow(10, lengths.back())) % k);    }     vector<vector<int>> mem(1 << nums.size(), vector<int>(k, -1));    return dp(nums, pows, mem, k, 0, 0) ? reconstruct(nums, pows, mem, k, 0, 0)                                        : vector<int>();  }  private:  // 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)`.  bool dp(const vector<int>& nums, const vector<int>& pows,          vector<vector<int>>& mem, int k, int mask, int mod) {    if (mem[mask][mod] != -1)      return mem[mask][mod] == 1;    if (mask == (1 << nums.size()) - 1)      return mod == 0;    for (int i = 0; i < nums.size(); ++i)      if ((mask >> i & 1) == 0) {        const int newMod = (mod * pows[i] + nums[i]) % k;        if (dp(nums, pows, mem, k, mask | 1 << i, newMod))          return mem[mask][mod] = 1;      }    return mem[mask][mod] = 0;  }   // Reconstructs the numbers that form a number divisible by `k` using the  // numbers in `nums`, where nums[i] is used iff `mask & (1 << i)`.  vector<int> reconstruct(const vector<int>& nums, const vector<int>& pows,                          vector<vector<int>>& mem, int k, int mask, int mod) {    for (int i = 0; i < nums.size(); ++i)      if ((mask >> i & 1) == 0) {        const int newMod = (mod * pows[i] + nums[i]) % k;        if (dp(nums, pows, mem, k, mask | 1 << i, newMod)) {          vector<int> res{nums[i]};          vector<int> rest =              reconstruct(nums, pows, mem, k, mask | 1 << i, newMod);          ranges::copy(rest, ranges::back_inserter(res));          return res;        }      }    return {};  }};

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