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, eithernums[i] % nums[i+1] == 0ornums[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 <= 141 <= 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 $\sum_{j=0}^{n-1}f[2^n-1][j]$.
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 $\sum_{j=0}^{n-1}f[2^n-1][j]$. 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.
Python3
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
Java
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;
}
}
C++
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) {
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;
}
};
Go
func specialPerm(nums []int) (ans int) {
const mod int = 1e9 + 7
n := len(nums)
m := 1 << n
f := make([][]int, m)
for i := range f {
f[i] = make([]int, n)
}
for i := 1; i < m; i++ {
for j, x := range nums {
if i>>j&1 == 1 {
ii := i ^ (1 << j)
if ii == 0 {
f[i][j] = 1
continue
}
for k, y := range nums {
if x%y == 0 || y%x == 0 {
f[i][j] = (f[i][j] + f[ii][k]) % mod
}
}
}
}
}
for _, x := range f[m-1] {
ans = (ans + x) % mod
}
return
}
TypeScript
function specialPerm(nums: number[]): number {
const mod = 1e9 + 7;
const n = nums.length;
const m = 1 << n;
const f = Array.from({ length: m }, () => Array(n).fill(0));
for (let i = 1; i < m; ++i) {
for (let j = 0; j < n; ++j) {
if (((i >> j) & 1) === 1) {
const ii = i ^ (1 << j);
if (ii === 0) {
f[i][j] = 1;
continue;
}
for (let 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;
}
}
}
}
}
return f[m - 1].reduce((acc, x) => (acc + x) % mod);
}
Rust
impl Solution {
pub fn special_perm(nums: Vec<i32>) -> i32 {
const MOD: i32 = 1_000_000_007;
let n = nums.len();
let m = 1 << n;
let mut f = vec![vec![0; n]; m];
for i in 1..m {
for j in 0..n {
if (i >> j) & 1 == 1 {
let ii = i ^ (1 << j);
if ii == 0 {
f[i][j] = 1;
continue;
}
for k in 0..n {
if nums[j] % nums[k] == 0 || nums[k] % nums[j] == 0 {
f[i][j] = (f[i][j] + f[ii][k]) % MOD;
}
}
}
}
}
let mut ans = 0;
for &x in &f[m - 1] {
ans = (ans + x) % MOD;
}
ans
}
}