<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:
    //3 -> 0
    //4 -> 0, 1
    //5 -> 1, 2
    static constexpr int MOD = 1e9+7;

    //ending case is where we have everyone with a hat, so the mask has all n people with hats, or the 1 is turned on
    int numberWays(vector<vector<int>>& hats) {
        int n = hats.size();
        //for every hat which person can wear this particular hat
        //dp[current_hat][mask] = the number of ways for the n people to wear the hats given we are at this hat and we have this mask (these people have already found a hat)
        map<int, vector<int>> hat_mapping;
        for(int i = 0; i < n; ++i)
        {
            for(auto hat : hats[i])
            {
                hat_mapping[hat].push_back(i);
            }
        }
        map<pair<int,int>, int> cache;
        return countNumberWays(hat_mapping.begin(), 0, hat_mapping, cache, n);
    }
private:
    int countNumberWays(auto it, int mask, map<int, vector<int>>& hat_mapping, map<pair<int,int>, int>& cache, const int n)
    {
        if(__builtin_popcount(mask) == n)
        {
            return 1;
        }
        if(it == hat_mapping.end())
        {
            return 0;
        }
        auto curr_pos = make_pair(it -> first, mask);
        if(cache.count(curr_pos))
        {
            return cache[curr_pos];
        }
        int curr_value = 0;
        for(auto person : it -> second)
        {
            if((mask >> person) & 1) continue;
            int new_mask = mask | (1 << person);
            auto new_it = it;
            ++new_it;
            curr_value = (curr_value % MOD) + (countNumberWays(new_it, new_mask, hat_mapping, cache, n) % MOD);
            curr_value = curr_value % MOD;
        }
        auto new_it = it;
        ++new_it;
        curr_value = (curr_value % MOD) + (countNumberWays(new_it, mask, hat_mapping, cache, n) % MOD);
        curr_value = curr_value % MOD;
        cache[curr_pos] = curr_value;
        return curr_value;
    }
};
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