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

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.

// OJ: https://leetcode.com/problems/number-of-ways-to-wear-different-hats-to-each-other/
// Author: github.com/lzl124631x
// 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;
        vector<int> masks(1 << N);
        masks[0] = 1;
        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)] += masks[j];
                    masks[j | (1 << p)] %= mod;
                }
            }
        }
        return masks[(1 << N) - 1];
    }
};