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.

// 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;
        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];
    }
};

Java

  • 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>());
                    list.add(i);
                    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];
        }
    }
    
  • // 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;
            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];
        }
    };
    
  • print("Todo!")
    

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