2992. Number of Self-Divisible Permutations π ο
Descriptionο
Given an integer n, return the number of permutations of the 1-indexed array nums = [1, 2, ..., n], such that it's self-divisible.
A 1-indexed array a of length n is self-divisible if for every 1 <= i <= n, gcd(a[i], i) == 1.
A permutation of an array is a rearrangement of the elements of that array, for example here are all of the permutations of the array [1, 2, 3]:
[1, 2, 3][1, 3, 2][2, 1, 3][2, 3, 1][3, 1, 2][3, 2, 1]
Example 1:
Input: n = 1 Output: 1 Explanation: The array [1] has only 1 permutation which is self-divisible.
Example 2:
Input: n = 2 Output: 1 Explanation: The array [1,2] has 2 permutations and only one of them is self-divisible: nums = [1,2]: This is not self-divisible since gcd(nums[2], 2) != 1. nums = [2,1]: This is self-divisible since gcd(nums[1], 1) == 1 and gcd(nums[2], 2) == 1.
Example 3:
Input: n = 3 Output: 3 Explanation: The array [1,2,3] has 3 self-divisble permutations: [1,3,2], [3,1,2], [2,3,1]. It can be shown that the other 3 permutations are not self-divisible. Hence the answer is 3.
Constraints:
1 <= n <= 12
Solutionsο
Solution 1: State Compression + Memoization Searchο
We can use a binary number $mask$ to represent the current permutation state, where the $i$-th bit is $1$ indicates that the number $i$ has been used, and $0$ indicates that the number $i$ has not been used yet.
Then, we design a function $dfs(mask)$, which represents the number of permutations that can be constructed from the current permutation state $mask$ and meet the requirements of the problem. The answer is $dfs(0)$.
We can use the method of memoization search to calculate the value of $dfs(mask)$.
In the process of calculating $dfs(mask)$, we use $i$ to indicate which number is to be added to the permutation. If $i \gt n$, it means that the permutation has been constructed, and we can return $1$.
Otherwise, we enumerate the numbers $j$ that have not been used in the current permutation. If $i$ and $j$ meet the requirements of the problem, then we can add $j$ to the permutation. At this time, the state becomes $mask \mid 2^j$, where $|$ represents bitwise OR operation. Since $j$ has been used, we need to recursively calculate the value of $dfs(mask \mid 2^j)$ and add it to $dfs(mask)$.
Finally, we can get the value of $dfs(0)$, which is the answer.
The time complexity is $O(n \times 2^n)$, and the space complexity is $O(2^n)$. Where $n$ is the length of the permutation.
Python3ο
class Solution:
def selfDivisiblePermutationCount(self, n: int) -> int:
@cache
def dfs(mask: int) -> int:
i = mask.bit_count() + 1
if i > n:
return 1
ans = 0
for j in range(1, n + 1):
if (mask >> j & 1) == 0 and gcd(i, j) == 1:
ans += dfs(mask | 1 << j)
return ans
return dfs(0)
Javaο
class Solution {
private int n;
private Integer[] f;
public int selfDivisiblePermutationCount(int n) {
this.n = n;
f = new Integer[1 << (n + 1)];
return dfs(0);
}
private int dfs(int mask) {
if (f[mask] != null) {
return f[mask];
}
int i = Integer.bitCount(mask) + 1;
if (i > n) {
return 1;
}
f[mask] = 0;
for (int j = 1; j <= n; ++j) {
if ((mask >> j & 1) == 0 && (i % j == 0 || j % i == 0)) {
f[mask] += dfs(mask | 1 << j);
}
}
return f[mask];
}
}
C++ο
class Solution {
public:
int selfDivisiblePermutationCount(int n) {
int f[1 << (n + 1)];
memset(f, -1, sizeof(f));
function<int(int)> dfs = [&](int mask) {
if (f[mask] != -1) {
return f[mask];
}
int i = __builtin_popcount(mask) + 1;
if (i > n) {
return 1;
}
f[mask] = 0;
for (int j = 1; j <= n; ++j) {
if ((mask >> j & 1) == 0 && (i % j == 0 || j % i == 0)) {
f[mask] += dfs(mask | 1 << j);
}
}
return f[mask];
};
return dfs(0);
}
};
Goο
func selfDivisiblePermutationCount(n int) int {
f := make([]int, 1<<(n+1))
for i := range f {
f[i] = -1
}
var dfs func(int) int
dfs = func(mask int) int {
if f[mask] != -1 {
return f[mask]
}
i := bits.OnesCount(uint(mask)) + 1
if i > n {
return 1
}
f[mask] = 0
for j := 1; j <= n; j++ {
if mask>>j&1 == 0 && (i%j == 0 || j%i == 0) {
f[mask] += dfs(mask | 1<<j)
}
}
return f[mask]
}
return dfs(0)
}
TypeScriptο
function selfDivisiblePermutationCount(n: number): number {
const f: number[] = Array(1 << (n + 1)).fill(-1);
const dfs = (mask: number): number => {
if (f[mask] !== -1) {
return f[mask];
}
const i = bitCount(mask) + 1;
if (i > n) {
return 1;
}
f[mask] = 0;
for (let j = 1; j <= n; ++j) {
if (((mask >> j) & 1) === 0 && (i % j === 0 || j % i === 0)) {
f[mask] += dfs(mask | (1 << j));
}
}
return f[mask];
};
return dfs(0);
}
function bitCount(i: number): number {
i = i - ((i >>> 1) & 0x55555555);
i = (i & 0x33333333) + ((i >>> 2) & 0x33333333);
i = (i + (i >>> 4)) & 0x0f0f0f0f;
i = i + (i >>> 8);
i = i + (i >>> 16);
return i & 0x3f;
}
Solution 2: State Compression + Dynamic Programmingο
We can rewrite the memoization search in Solution 1 into the form of dynamic programming, define $f[mask]$ to represent the number of permutations that the current permutation state is $mask$ and meet the requirements of the problem. Initially, $f[0]=1$, and the rest are $0$.
We enumerate $mask$ in the range of $[0, 2^n)$, for each $mask$, we use $i$ to represent which number is the last one to join the permutation, then we enumerate the last number $j$ added to the current permutation. If $i$ and $j$ meet the requirements of the problem, then the state $f[mask]$ can be transferred from the state $f[mask \oplus 2^(j-1)]$, where $\oplus$ represents bitwise XOR operation. We add all the values of the transferred state $f[mask \oplus 2^(j-1)]$ to $f[mask]$, which is the value of $f[mask]$.
Finally, we can get the value of $f[2^n - 1]$, which is the answer.
The time complexity is $O(n \times 2^n)$, and the space complexity is $O(2^n)$. Where $n$ is the length of the permutation.
Python3ο
class Solution:
def selfDivisiblePermutationCount(self, n: int) -> int:
f = [0] * (1 << n)
f[0] = 1
for mask in range(1 << n):
i = mask.bit_count()
for j in range(1, n + 1):
if (mask >> (j - 1) & 1) == 1 and gcd(i, j) == 1:
f[mask] += f[mask ^ (1 << (j - 1))]
return f[-1]
Javaο
class Solution {
public int selfDivisiblePermutationCount(int n) {
int[] f = new int[1 << n];
f[0] = 1;
for (int mask = 0; mask < 1 << n; ++mask) {
int i = Integer.bitCount(mask);
for (int j = 1; j <= n; ++j) {
if (((mask >> (j - 1)) & 1) == 1 && (i % j == 0 || j % i == 0)) {
f[mask] += f[mask ^ (1 << (j - 1))];
}
}
}
return f[(1 << n) - 1];
}
}
C++ο
class Solution {
public:
int selfDivisiblePermutationCount(int n) {
int f[1 << n];
memset(f, 0, sizeof(f));
f[0] = 1;
for (int mask = 0; mask < 1 << n; ++mask) {
int i = __builtin_popcount(mask);
for (int j = 1; j <= n; ++j) {
if (((mask >> (j - 1)) & 1) == 1 && (i % j == 0 || j % i == 0)) {
f[mask] += f[mask ^ (1 << (j - 1))];
}
}
}
return f[(1 << n) - 1];
}
};
Goο
func selfDivisiblePermutationCount(n int) int {
f := make([]int, 1<<n)
f[0] = 1
for mask := 0; mask < 1<<n; mask++ {
i := bits.OnesCount(uint(mask))
for j := 1; j <= n; j++ {
if mask>>(j-1)&1 == 1 && (i%j == 0 || j%i == 0) {
f[mask] += f[mask^(1<<(j-1))]
}
}
}
return f[(1<<n)-1]
}
TypeScriptο
function selfDivisiblePermutationCount(n: number): number {
const f: number[] = Array(1 << n).fill(0);
f[0] = 1;
for (let mask = 0; mask < 1 << n; ++mask) {
const i = bitCount(mask);
for (let j = 1; j <= n; ++j) {
if ((mask >> (j - 1)) & 1 && (i % j === 0 || j % i === 0)) {
f[mask] += f[mask ^ (1 << (j - 1))];
}
}
}
return f.at(-1)!;
}
function bitCount(i: number): number {
i = i - ((i >>> 1) & 0x55555555);
i = (i & 0x33333333) + ((i >>> 2) & 0x33333333);
i = (i + (i >>> 4)) & 0x0f0f0f0f;
i = i + (i >>> 8);
i = i + (i >>> 16);
return i & 0x3f;
}