<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 dp[41][100000];
    int mod = 1000000007;
    int getNumberOfWays(int start, int mask, vector<vector<int>> &assign, int n) {

        if(mask == (1 << n) - 1) {
            return 1;
        }
        if(start > 40) {
            return 0;
        }

        if(dp[start][mask] != -1) {
            return dp[start][mask];
        }
        
        // Skip the hat
        long long ans = 0;
        // Take the hat
        for(int i = 1; i <= n; i++) {
            if((mask & (1 << (i - 1))) == 0 && assign[i][start] == 1) {
               ans = ans + (long long)getNumberOfWays(start + 1, (mask | (1 << (i - 1))), assign, n);
               ans = ans % mod;
            }
        }
        ans = ans + (long long)getNumberOfWays(start + 1, mask, assign, n);
        ans = ans % mod;
        return dp[start][mask] = ans;
        
    }
    int numberWays(vector<vector<int>>& hats) {
        // dp[i][mask] = Number of ways to distribute hats from i to 40 such that the ith hat is given to that person which is not present in the mask
        int n = hats.size();
        memset(dp, -1, sizeof(dp));
        vector<vector<int>> assign(n+1, vector<int>(41, 0));
        for(int i = 0; i < n; i++) {
            for(int j = 0; j < hats[i].size(); j++) {
                assign[i+1][hats[i][j]] = 1;
            }
        }
        return getNumberOfWays(1, 0, assign, n);
    }
};
28100