<p>There are <code>n</code> people and <code>40</code> types of hats labeled from <code>1</code> to <code>40</code>.</p>

<p>Given a 2D integer array <code>hats</code>, where <code>hats[i]</code> is a list of all hats preferred by the <code>i<sup>th</sup></code> person.</p>

<p>Return <em>the number of ways that the <code>n</code> people wear different hats to each other</em>.</p>

<p>Since the answer may be too large, return it modulo <code>10<sup>9</sup> + 7</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> hats = [[3,4],[4,5],[5]]
<strong>Output:</strong> 1
<strong>Explanation:</strong> 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.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> hats = [[3,5,1],[3,5]]
<strong>Output:</strong> 4
<strong>Explanation:</strong> There are 4 ways to choose hats:
(3,5), (5,3), (1,3) and (1,5)
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> hats = [[1,2,3,4],[1,2,3,4],[1,2,3,4],[1,2,3,4]]
<strong>Output:</strong> 24
<strong>Explanation:</strong> Each person can choose hats labeled from 1 to 4.
Number of Permutations of (1,2,3,4) = 24.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>n == hats.length</code></li>
	<li><code>1 &lt;= n &lt;= 10</code></li>
	<li><code>1 &lt;= hats[i].length &lt;= 40</code></li>
	<li><code>1 &lt;= hats[i][j] &lt;= 40</code></li>
	<li><code>hats[i]</code> contains a list of <strong>unique</strong> integers.</li>
</ul>

class Solution {
public:
    int n, M = 1e9 + 7;
    
    int dfs(vector<vector<int>>& hatsToPeople, vector<vector<int>>& dp, int hat, int mask) {
        // if the mask is able to reache the end state, 
        // i.e. all people can wear some hats, then return 1
        if (mask == (1 << n) - 1) return 1;
        // if the current hat is out of 40, return 0
        if (hat > 40) return 0;
        // if we calculated the result before, return it directly
        if (dp[mask][hat] != -1) return dp[mask][hat];
        // get number of ways if we skip this hat
        long res = dfs(hatsToPeople, dp, hat + 1, mask);
        // since the current hat can be assigned to different people
        // we try all of them
        for (auto& person: hatsToPeople[hat]) {
            // if this person has been assigned with a hat, then skip it
            if ((1 << person) & mask) continue;
            // then we can assign this hat to this person
            (res += dfs(hatsToPeople, dp, hat + 1, mask | (1 << person))) %= M;
        }
        // memoize the result
        return dp[mask][hat] = res;
    }
    
    int numberWays(vector<vector<int>>& hats) {
        // number of people
        n = hats.size();
        // for each hat, we map to different people
        vector<vector<int>> hatsToPeople(41);
        // iterate each person
        for (int i = 0; i < n; i++) {
            // iterate i-th person's preferred hats
            for (auto &h : hats[i]) {
                // assign hat h to person i
                hatsToPeople[h].push_back(i);
            }
        }
        // init dp
        vector<vector<int>> dp(1 << n, vector<int>(41, -1));
        // dfs
        return dfs(hatsToPeople, dp, 1, 0);
    }
};
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