<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:
    static const int MOD = 1e9 + 7;
    unordered_map<int/*hat index*/, vector<int>/*people to satisfy*/> g_graph;
    unordered_map<int/*hat index*/, unordered_map<int/*state*/, int/*cnt*/>> dp;

    using GraphItr = decltype(g_graph.begin());

    int numberWays(vector<vector<int>>& hats) {
        for (int i = 0; i < hats.size(); i++) {
            for (auto& eachHat : hats[i]) {
                g_graph[eachHat].push_back(i);
            }
        }

        int numPeople = hats.size();
        int remainigPeople = (0x01 << numPeople) - 1;
        return func(g_graph.begin(), remainigPeople);
    }

    int func(GraphItr curItr, int peopleToSatisfy) {
        if (peopleToSatisfy == 0) {
            return 1;
        }

        if (curItr == g_graph.end()) {
            return 0;
        }

        if (dp[curItr->first].count(peopleToSatisfy)) {
            return dp[curItr->first][peopleToSatisfy];
        }

        int cnt = 0;
        // first we can choose not to use current one
        cnt += func(next(curItr), peopleToSatisfy);
        cnt = cnt % MOD;
        for (auto& eachPeople : curItr->second) {
            if ((0x01 << eachPeople) & peopleToSatisfy) {
                // can satisfy
                cnt += func(next(curItr), (0x01 << eachPeople) ^ peopleToSatisfy);
                cnt = cnt % MOD;
            }
        }

        dp[curItr->first][peopleToSatisfy] = cnt;
        return cnt;
    }
};
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