<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:
    #define ll long long 
    const int MOD = 1e9 + 7 ;
    ll dp[41][1 << 10] ;
    int n ;
    unordered_map<int,vector<int>> mpp ;
    
    ll solve(int hat , int mask){
        int all = (1 << n) - 1; //mask for all persons visited
        
        if(mask == all) return 1 ;
        if(hat > 40) return 0 ;
        
        if(dp[hat][mask] != -1) return dp[hat][mask] ;
		
		//two options 
        ll op1 = solve(hat + 1 , mask) ; //skip this hat
        ll op2 = 0 ; //assign this hat to valid person
        
        for(auto &i : mpp[hat]){
            //check if ith person has already hat assigned 
            if(mask & (1 << i)) continue ;
            op2 += solve(hat + 1 , mask | (1 << i)) ;
            op2 %= MOD ;
        }
        
        return dp[hat][mask] = (op1 + op2) % MOD ;
    }
    
    int numberWays(vector<vector<int>>& hats) {
        n = hats.size() ;
        memset(dp,-1,sizeof(dp)) ;
        
        for(int i = 0 ; i < n ; ++i )
            for(auto &x : hats[i]) mpp[x].push_back(i);
        
        return solve(1,0) % MOD ;
    }
};
12000