<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 numberWays(vector<vector<int>>& hats) {
        // there are 40 hats
        // and at most 10 persons.
        // to state compression, we should take 10 persons as state instead of 40 hats.
        // memo[human_state][hat_idx] = 
        //     given the human state, meaning who has hat and who does not.
        //     at current hat_idx, how many ways to waer different hats to each other?
        vector<unordered_set<int>> preference(hats.size());
        for(int i = 0; i<hats.size(); i++) {
            for(int h : hats.at(i)) preference.at(i).insert(h-1);
        }
        vector<vector<int>> memo(1<<hats.size(), vector<int>(40, -1));
        return dfsMemo(hats.size(), preference, memo, 0, 0);
    }
    
    int dfsMemo(int n,
                const vector<unordered_set<int>>& preference,
                vector<vector<int>>& memo,
                int state, int hat_idx) {
        if(state==((1<<n) - 1)) return 1;
        if(hat_idx>=40) return 0; // out of boundary
        if(memo[state][hat_idx]!=-1) return memo[state][hat_idx];
        
        long long res = 0;
        for(int i = 0; i<preference.size(); i++) {
            if(state & (1<<i)) continue; // this guy has hat already
            if(preference.at(i).count(hat_idx)==0) continue; // this guy dose not like it.
            int new_state = state | (1<<i);    
            res = (res + dfsMemo(n, preference, memo, new_state, hat_idx+1))%MOD;
        }
        // ignore this hat.
        res = (res + dfsMemo(n, preference, memo, state, hat_idx+1))%MOD;
        return memo[state][hat_idx] = res;
    }
private:
    static const int MOD=1e9+7;
};
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