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3533. Concatenated Divisibility 👍

  • Time: $O(2^n \cdot k \cdot n)$
  • Space: $O(2^n \cdot k)$
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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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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];
  }
}

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class Solution:
  def concatenatedDivisibility(self, nums: list[int], k: int) -> list[int]:
    n = len(nums)
    nums.sort()
    lengths = [len(str(num)) for num in nums]
    pows = [pow(10, length, k) for length in lengths]

    @functools.lru_cache(None)
    def dp(mask: int, mod: int) -> bool:
      """
      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)`.
      """
      if mask == (1 << n) - 1:
        return mod == 0
      for i in range(n):
        if (mask >> i & 1) == 0:
          newMod = (mod * pows[i] + nums[i]) % k
          if dp(mask | 1 << i, newMod):
            return True
      return False

    def reconstruct(mask: int, mod: int) -> list[int]:
      """
      Reconstructs the numbers that form a number divisible by `k` using the
      numbers in `nums`, where nums[i] is used iff `mask & (1 << i)`.
      """
      for i in range(n):
        if (mask >> i & 1) == 0:
          newMod = (mod * pows[i] + nums[i]) % k
          if dp(mask | 1 << i, newMod):
            return [nums[i]] + reconstruct(mask | 1 << i, newMod)
      return []

    return reconstruct(0, 0) if dp(0, 0) else []