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2741. Special Permutations

Description

You are given a 0-indexed integer array nums containing n distinct positive integers. A permutation of nums is called special if:

  • For all indexes 0 <= i < n - 1, either nums[i] % nums[i+1] == 0 or nums[i+1] % nums[i] == 0.

Return the total number of special permutations. As the answer could be large, return it modulo 109 + 7.

 

Example 1:

Input: nums = [2,3,6]
Output: 2
Explanation: [3,6,2] and [2,6,3] are the two special permutations of nums.

Example 2:

Input: nums = [1,4,3]
Output: 2
Explanation: [3,1,4] and [4,1,3] are the two special permutations of nums.

 

Constraints:

  • 2 <= nums.length <= 14
  • 1 <= nums[i] <= 109

Solutions

Solution 1: State Compression Dynamic Programming

We notice that the maximum length of the array in the problem does not exceed $14$. Therefore, we can use an integer to represent the current state, where the $i$-th bit is $1$ if the $i$-th number in the array has been selected, and $0$ if it has not been selected.

We define $f[i][j]$ as the number of schemes where the current selected integer state is $i$, and the index of the last selected integer is $j$. Initially, $f[0][0]=0$, and the answer is $\sum_{j=0}{n-1}f[2n-1][j]$.

Considering $f[i][j]$, if only one number is currently selected, then $f[i][j]=1$. Otherwise, we can enumerate the index $k$ of the last selected number. If the numbers corresponding to $k$ and $j$ meet the requirements of the problem, then $f[i][j]$ can be transferred from $f[i \oplus 2^j][k]$. That is:

$$ f[i][j]= \begin{cases} 1, & i=2^j\ \sum_{k=0}^{n-1}f[i \oplus 2^j][k], & i \neq 2^j \textit{ and nums}[j] \textit{ and nums}[k] \textit{ meet the requirements of the problem}\ \end{cases} $$

The final answer is $\sum_{j=0}{n-1}f[2n-1][j]$. Note that the answer may be very large, so we need to take the modulus of $10^9+7$.

The time complexity is $O(n^2 \times 2^n)$, and the space complexity is $O(n \times 2^n)$. Here, $n$ is the length of the array.

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class Solution:
    def specialPerm(self, nums: List[int]) -> int:
        mod = 10**9 + 7
        n = len(nums)
        m = 1 << n
        f = [[0] * n for _ in range(m)]
        for i in range(1, m):
            for j, x in enumerate(nums):
                if i >> j & 1:
                    ii = i ^ (1 << j)
                    if ii == 0:
                        f[i][j] = 1
                        continue
                    for k, y in enumerate(nums):
                        if x % y == 0 or y % x == 0:
                            f[i][j] = (f[i][j] + f[ii][k]) % mod
        return sum(f[-1]) % mod

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class Solution {
    public int specialPerm(int[] nums) {
        final int mod = (int) 1e9 + 7;
        int n = nums.length;
        int m = 1 << n;
        int[][] f = new int[m][n];
        for (int i = 1; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                if ((i >> j & 1) == 1) {
                    int ii = i ^ (1 << j);
                    if (ii == 0) {
                        f[i][j] = 1;
                        continue;
                    }
                    for (int k = 0; k < n; ++k) {
                        if (nums[j] % nums[k] == 0 || nums[k] % nums[j] == 0) {
                            f[i][j] = (f[i][j] + f[ii][k]) % mod;
                        }
                    }
                }
            }
        }
        int ans = 0;
        for (int x : f[m - 1]) {
            ans = (ans + x) % mod;
        }
        return ans;
    }
}

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class Solution {
public:
    int specialPerm(vector<int>& nums) {
        const int mod = 1e9 + 7;
        int n = nums.size();
        int m = 1 << n;
        int f[m][n];
        memset(f, 0, sizeof(f));
        for (int i = 1; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                if ((i >> j & 1) == 1) {
                    int ii = i ^ (1 << j);
                    if (ii == 0) {
                        f[i][j] = 1;
                        continue;
                    }
                    for (int k = 0; k < n; ++k)