Formatted question description: https://leetcode.ca/all/1434.html

1434. Number of Ways to Wear Different Hats to Each Other (Hard)

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

Given a list of list of integers hats, where hats[i] is a list of all hats preferred by the i-th 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 10^9 + 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.


Example 4:

Input: hats = [[1,2,3],[2,3,5,6],[1,3,7,9],[1,8,9],[2,5,7]]
Output: 111


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.

Related Topics:
Dynamic Programming, Bit Manipulation

Solution 1.

• class Solution {
public int numberWays(List<List<Integer>> hats) {
final int MODULO = 1000000007;
int peopleCount = hats.size();
int hatsCount = 40;
Map<Integer, List<Integer>> map = new HashMap<Integer, List<Integer>>();
for (int i = 0; i < peopleCount; i++) {
List<Integer> curHats = hats.get(i);
for (int hat : curHats) {
List<Integer> list = map.getOrDefault(hat, new ArrayList<Integer>());
map.put(hat, list);
}
}
int[][] dp = new int[hatsCount + 1][1 << peopleCount];
dp[0][0] = 1;
for (int i = 1; i <= hatsCount; i++) {
for (int j = 0; j < 1 << peopleCount; j++)
dp[i][j] = dp[i - 1][j];
if (map.containsKey(i)) {
List<Integer> list = map.get(i);
for (int j = 0; j < 1 << peopleCount; j++) {
for (int k : list) {
if ((j & (1 << k)) == 1 << k)
dp[i][j] = (dp[i][j] + dp[i - 1][j - (1 << k)]) % MODULO;
}
}
}
}
return dp[hatsCount][(1 << peopleCount) - 1];
}
}

############

class Solution {
private static final int MOD = (int) 1e9 + 7;

public int numberWays(List<List<Integer>> hats) {
List<Integer>[] d = new List[41];
Arrays.setAll(d, k -> new ArrayList<>());
int n = hats.size();
int mx = 0;
for (int i = 0; i < n; ++i) {
for (int h : hats.get(i)) {
mx = Math.max(mx, h);
}
}
long[][] dp = new long[mx + 1][1 << n];
dp[0][0] = 1;
for (int i = 1; i < mx + 1; ++i) {
for (int j : d[i]) {
if (((mask >> j) & 1) == 1) {
}
}
}
}
return (int) dp[mx][(1 << n) - 1];
}
}

• // OJ: https://leetcode.com/problems/number-of-ways-to-wear-different-hats-to-each-other/
// Time: O(2^N * M)
// Space: O(2^N)
class Solution {
public:
int numberWays(vector<vector<int>>& hats) {
vector<vector<int>> persons(40);
int N = hats.size(), mod = 1e9+7;
for (int i = 0; i < N; ++i) {
for (int h : hats[i]) persons[h - 1].push_back(i);
}
for (int i = 0; i < 40; ++i) {
for (int j = (1 << N) - 1; j >= 0; --j) {
for (int p : persons[i]) {
if (j & (1 << p)) continue;
masks[j | (1 << p)] %= mod;
}
}
}
return masks[(1 << N) - 1];
}
};

• class Solution:
def numberWays(self, hats: List[List[int]]) -> int:
d = defaultdict(list)
for i, h in enumerate(hats):
for v in h:
d[v].append(i)
n = len(hats)
mx = max(max(h) for h in hats)
dp = [[0] * (1 << n) for _ in range(mx + 1)]
dp[0][0] = 1
mod = int(1e9) + 7
for i in range(1, mx + 1):
for mask in range(1 << n):
for j in d[i]:
if (mask >> j) & 1:
return dp[mx][(1 << n) - 1]


• func numberWays(hats [][]int) int {
d := make([][]int, 41)
mx := 0
for i, h := range hats {
for _, v := range h {
d[v] = append(d[v], i)
mx = max(mx, v)
}
}
dp := make([][]int, mx+1)
n := len(hats)
for i := range dp {
dp[i] = make([]int, 1<<n)
}
dp[0][0] = 1
mod := int(1e9) + 7
for i := 1; i <= mx; i++ {
for _, j := range d[i] {
if ((mask >> j) & 1) == 1 {
}
}
}
}
return dp[mx][(1<<n)-1]
}

func max(a, b int) int {
if a > b {
return a
}
return b
}

• 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];
}