Number of Ways to Wear Different Hats to Each Other

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.

/**
 * The first idea come to our mind is that we will do straightforward like the
 * problem said: Return the number of ways that the n people wear different hats
 * to each other. We will assign n people with different hats and use dp to
 * calculate number of ways.
 */
// But that will result in TLE
class Solution {
    int mod = 1_000_000_007;

    public int numberWays(List<List<Integer>> hats) {
        int n = hats.size();
        // h2p[i] indicates the list of people who can wear i_th hat
        List<Integer>[] h2p = new List[41];
        for (int i = 1; i <= 40; i++)
            h2p[i] = new ArrayList<>();
        for (int i = 0; i < n; i++)
            for (int hat : hats.get(i))
                h2p[hat].add(i);
        // max value of full mask can be 2^n -> 1024
        Integer[][] dp = new Integer[41][1024];
        // Using a bitmask where each bit shows whether the ith perseon has been
        // assigned or not
        return dfs(h2p, (1 << n) - 1, 1, 0, dp);
        // we are using 2 masks here
        // 1st-> (1<<n)-1 -> this shows the final mask configuration we want to reach
        // 2nd-> 0 -> this shows the current mask configuration
    }

    // dp[i][j] -> ways to assign ith hat to mask value of j
    // Remember that (i+1)th hat will be assigned after the ith hat
    int dfs(List<Integer>[] h2p, int allMask, int hat, int assignedPeople, Integer[][] dp) {
        if (assignedPeople == allMask)
            return 1; // Check if assigned different hats to all people
        if (hat > 40)
            return 0; // no more hats to process
        if (dp[hat][assignedPeople] != null)
            return dp[hat][assignedPeople];
        int ans = dfs(h2p, allMask, hat + 1, assignedPeople, dp); // Don't wear this hat
        for (int p : h2p[hat]) {
            // Skip if person `p` was assigned hat
            if (((assignedPeople >> p) & 1) == 1)
                continue;
            // Wear this hat for p_th person
            ans += dfs(h2p, allMask, hat + 1, assignedPeople | (1 << p), dp);
            ans %= mod;
        }
        dp[hat][assignedPeople] = ans;
        return ans;
    }
}