# 1434. Number of Ways to Wear Different Hats to Each Other

## Description

There are n people and 40 types of hats labeled from 1 to 40.

Given a 2D integer array hats, where hats[i] is a list of all hats preferred by the ith person.

Return the number of ways that the n people wear different hats to each other.

Since the answer may be too large, return it modulo 109 + 7.

Example 1:

Input: hats = [[3,4],[4,5],[5]]
Output: 1
Explanation: There is only one way to choose hats given the conditions.
First person choose hat 3, Second person choose hat 4 and last one hat 5.


Example 2:

Input: hats = [[3,5,1],[3,5]]
Output: 4
Explanation: There are 4 ways to choose hats:
(3,5), (5,3), (1,3) and (1,5)


Example 3:

Input: hats = [[1,2,3,4],[1,2,3,4],[1,2,3,4],[1,2,3,4]]
Output: 24
Explanation: Each person can choose hats labeled from 1 to 4.
Number of Permutations of (1,2,3,4) = 24.


Constraints:

• n == hats.length
• 1 <= n <= 10
• 1 <= hats[i].length <= 40
• 1 <= hats[i][j] <= 40
• hats[i] contains a list of unique integers.

## Solutions

Solution 1: Dynamic Programming

We notice that $n$ is not greater than $10$, so we consider using DP with state compression to solve this problem.

We define $f[i][j]$ as the number of ways to assign the first $i$ hats to the people whose state is $j$. Here $j$ is a binary number, which represents a set of people. We have $f[0][0]=1$ at the beginning, and the answer is $f[m][2^n - 1]$, where $m$ is the maximum number of hats and $n$ is the number of people.

Consider $f[i][j]$. If we don’t assign the $i$-th hat to anyone, then $f[i][j]=f[i-1][j]$; if we assign the $i$-th hat to the person $k$ who likes it, then $f[i][j]=f[i-1][j \oplus 2^k]$. Here $\oplus$ denotes the XOR operation. Therefore, we can get the state transition equation:

$f[i][j]=f[i-1][j]+ \sum_{k \in like[i]} f[i-1][j \oplus 2^k]$

where $like[i]$ denotes the set of people who like the $i$-th hat.

The final answer is $f[m][2^n - 1]$, and the answer may be very large, so we need to take it modulo $10^9 + 7$.

Time complexity $O(m \times 2^n \times n)$, space complexity $O(m \times 2^n)$. Here $m$ is the maximum number of hats, which is no more than $40$ in this problem; and $n$ is the number of people, which is no more than $10$ in this problem.

• class Solution {
public int numberWays(List<List<Integer>> hats) {
int n = hats.size();
int m = 0;
for (var h : hats) {
for (int v : h) {
m = Math.max(m, v);
}
}
List<Integer>[] g = new List[m + 1];
Arrays.setAll(g, k -> new ArrayList<>());
for (int i = 0; i < n; ++i) {
for (int v : hats.get(i)) {
}
}
final int mod = (int) 1e9 + 7;
int[][] f = new int[m + 1][1 << n];
f[0][0] = 1;
for (int i = 1; i <= m; ++i) {
for (int j = 0; j < 1 << n; ++j) {
f[i][j] = f[i - 1][j];
for (int k : g[i]) {
if ((j >> k & 1) == 1) {
f[i][j] = (f[i][j] + f[i - 1][j ^ (1 << k)]) % mod;
}
}
}
}
return f[m][(1 << n) - 1];
}
}

• class Solution {
public:
int numberWays(vector<vector<int>>& hats) {
int n = hats.size();
int m = 0;
for (auto& h : hats) {
m = max(m, *max_element(h.begin(), h.end()));
}
vector<vector<int>> g(m + 1);
for (int i = 0; i < n; ++i) {
for (int& v : hats[i]) {
g[v].push_back(i);
}
}
const int mod = 1e9 + 7;
int f[m + 1][1 << n];
memset(f, 0, sizeof(f));
f[0][0] = 1;
for (int i = 1; i <= m; ++i) {
for (int j = 0; j < 1 << n; ++j) {
f[i][j] = f[i - 1][j];
for (int k : g[i]) {
if (j >> k & 1) {
f[i][j] = (f[i][j] + f[i - 1][j ^ (1 << k)]) % mod;
}
}
}
}
return f[m][(1 << n) - 1];
}
};

• class Solution:
def numberWays(self, hats: List[List[int]]) -> int:
g = defaultdict(list)
for i, h in enumerate(hats):
for v in h:
g[v].append(i)
mod = 10**9 + 7
n = len(hats)
m = max(max(h) for h in hats)
f = [[0] * (1 << n) for _ in range(m + 1)]
f[0][0] = 1
for i in range(1, m + 1):
for j in range(1 << n):
f[i][j] = f[i - 1][j]
for k in g[i]:
if j >> k & 1:
f[i][j] = (f[i][j] + f[i - 1][j ^ (1 << k)]) % mod
return f[m][-1]


• func numberWays(hats [][]int) int {
n := len(hats)
m := 0
for _, h := range hats {
m = max(m, slices.Max(h))
}
g := make([][]int, m+1)
for i, h := range hats {
for _, v := range h {
g[v] = append(g[v], i)
}
}
const mod = 1e9 + 7
f := make([][]int, m+1)
for i := range f {
f[i] = make([]int, 1<<n)
}
f[0][0] = 1
for i := 1; i <= m; i++ {
for j := 0; j < 1<<n; j++ {
f[i][j] = f[i-1][j]
for _, k := range g[i] {
if j>>k&1 == 1 {
f[i][j] = (f[i][j] + f[i-1][j^(1<<k)]) % mod
}
}
}
}
return f[m][(1<<n)-1]
}

• function numberWays(hats: number[][]): number {
const n = hats.length;
const m = Math.max(...hats.flat());
const g: number[][] = Array.from({ length: m + 1 }, () => []);
for (let i = 0; i < n; ++i) {
for (const v of hats[i]) {
g[v].push(i);
}
}
const f: number[][] = Array.from({ length: m + 1 }, () =>
Array.from({ length: 1 << n }, () => 0),
);
f[0][0] = 1;
const mod = 1e9 + 7;
for (let i = 1; i <= m; ++i) {
for (let j = 0; j < 1 << n; ++j) {
f[i][j] = f[i - 1][j];
for (const k of g[i]) {
if (((j >> k) & 1) === 1) {
f[i][j] = (f[i][j] + f[i - 1][j ^ (1 << k)]) % mod;
}
}
}
}
return f[m][(1 << n) - 1];
}