<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:
    long t[41][1 << 10]; //memo table
    int mod = 1e9 + 7;
    int numberWays(vector<vector<int>>& hats) {
        
        vector <vector<int>> interim;
        int maxId = 0;
        
        for(int i = 0; i < hats.size(); i++)
        {
            vector <int> temp(41, 0);
            for(int j = 0; j < hats[i].size(); j++)
            {
                temp[hats[i][j]] = 1;
                maxId = max(maxId, hats[i][j]);
            }
            
            interim.push_back(temp);
        }
        
        hats.clear();
        hats = interim; /*changing hat vector such that for every person i there is a vector of 
		size 41 that denotes whether the particular person holds the hat[a] where a is [0, 41], where 0 is not useful.*/
        
        memset(t, -1, sizeof(t));
        
        long ans = solve(hats, 1, (1 << hats.size()) - 1, maxId); /*maxId is for the max number 
		of hats in the total treasure*/
        
        return (int)ans;
    }
    
    long solve(vector<vector<int>>& hats, int i, int mask, int m) //here i is the hat number
    {
        if(!mask) //when every person has a hat
            return 1;
        
        if(i > m) // if the hat number crosses the maximum number for hats
            return 0;
        
        if(t[i][mask] != -1)
            return t[i][mask]; //memo
        
        long ans = 0;
        for(int person = 0; person < hats.size(); person++) /*for every hat i, this piece of 
		code checks every person, if he/she holds the hat, the code progresses*/
            if(hats[person][i] && (mask & (1 << person))) /* progress if the person holds the 
			hat and if the person is not already asigned a hat*/
                ans += solve(hats, i + 1, mask ^ (1 << person), m); /*mask change changes the
				status of the person person to 0, hence he cannot be used further to hold any hat.*/
        
        ans += solve(hats, i + 1, mask, m); /* this takes care of the situation when we 
		have to skip the particular ith hat.*/
        
        return t[i][mask] = ans % mod;
    }
};
10600