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.

Solution

number-of-ways-to-wear-different-hats-to-each-other.py

class Solution:
    def numberWays(self, hats: List[List[int]]) -> int:
        M = 10 ** 9 + 7
        n, maxN = len(hats), 41
        completed_mask = (1 << n) - 1
        s = collections.defaultdict(list)

        for i, hatList in enumerate(hats):
            for hat in hatList:
                s[hat].append(i)

        @cache
        def dfs(hat, mask):
            if mask == completed_mask: return 1
            if hat >= maxN: return 0

            res = dfs(hat + 1, mask) # skip current round

            for index in s[hat]:
                if mask & (1 << index): continue

                res = (res + dfs(hat + 1, mask | (1 << index))) % M

            return res

        return dfs(0, 0)