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2741. Special Permutations 👍

  • Time: $O(n^2 \cdot 2^n)$
  • Space: $O(n \cdot 2^n)$
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class Solution {
 public:
  int specialPerm(vector<int>& nums) {
    const int n = nums.size();
    const int maxMask = 1 << n;
    int ans = 0;
    vector<vector<int>> mem(n, vector<int>(maxMask));

    for (int i = 0; i < n; ++i) {
      ans += specialPerm(nums, i, 1 << i, maxMask, mem);
      ans %= kMod;
    }

    return ans;
  }

 private:
  static constexpr int kMod = 1'000'000'007;

  // Returns the number of special permutations, where the previous number is
  // nums[i] and `mask` is the bitmask of the used numbers.
  int specialPerm(const vector<int>& nums, int prev, int mask,
                  const int& maxMask, vector<vector<int>>& mem) {
    if (mask == maxMask - 1)
      return 1;
    if (mem[prev][mask] > 0)
      return mem[prev][mask];

    int res = 0;

    for (int i = 0; i < nums.size(); ++i) {
      if (mask >> i & 1)
        continue;
      if (nums[i] % nums[prev] == 0 || nums[prev] % nums[i] == 0) {
        res += specialPerm(nums, i, mask | 1 << i, maxMask, mem);
        res %= kMod;
      }
    }

    return mem[prev][mask] = res;
  }
};

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class Solution {
  public int specialPerm(int[] nums) {
    final int n = nums.length;
    final int maxMask = 1 << n;
    int ans = 0;
    int[][] mem = new int[n][maxMask];

    for (int i = 0; i < n; ++i) {
      ans += specialPerm(nums, i, 1 << i, maxMask, mem);
      ans %= kMod;
    }

    return ans;
  }

  private static final int kMod = 1_000_000_007;

  // Returns the number of special permutations, where the previous number is
  // nums[i] and `mask` is the bitmask of the used numbers.
  private int specialPerm(int[] nums, int prev, int mask, int maxMask, int[][] mem) {
    if (mask == maxMask - 1)
      return 1;
    if (mem[prev][mask] != 0)
      return mem[prev][mask];

    int res = 0;

    for (int i = 0; i < nums.length; ++i) {
      if ((mask >> i & 1) == 1)
        continue;
      if (nums[i] % nums[prev] == 0 || nums[prev] % nums[i] == 0) {
        res += specialPerm(nums, i, mask | 1 << i, maxMask, mem);
        res %= kMod;
      }
    }

    return mem[prev][mask] = res;
  }
}

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class Solution:
  def specialPerm(self, nums: list[int]) -> int:
    kMod = 1_000_000_007
    maxMask = 1 << len(nums)

    @functools.lru_cache(None)
    def dp(prev: int, mask: int) -> int:
      """
      Returns the number of special permutations, where the previous number is
      nums[i] and `mask` is the bitmask of the used numbers.
      """
      if mask == maxMask - 1:
        return 1

      res = 0

      for i, num in enumerate(nums):
        if mask >> i & 1:
          continue
        if num % nums[prev] == 0 or nums[prev] % num == 0:
          res += dp(i, mask | 1 << i)
          res %= kMod

      return res

    return sum(dp(i, 1 << i)
               for i in range(len(nums))) % kMod