<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>

using LL = long long;

class Solution {
public:
  int numberWays(vector<vector<int>>& A) {
    int n = A.size();
    int MOD = 1e9+7;

    map<int, vector<int>> mp;
    for (int i = 0; i < n; i++) {
      for (int j = 0; j < A[i].size(); j++) {
        int hat = A[i][j];
        mp[hat].push_back(i);
      }
    }

    vector<vector<LL>> dp(41, vector<LL>(1<<n));
    dp[0][0] = 1;
    for (int i = 1; i <= 40; i++) {
      dp[i] = dp[i-1];
      for (int state = 0; state < (1<<n); state++) {
        for (auto p : mp[i]) {
          if ((state & (1 << p))) { continue; } // if p has taken ANY hat, skip

          dp[i][ ( state | (1<<p) ) ] += dp[i-1][state];
          dp[i][ ( state | (1<<p) ) ] %= MOD;

          //printf("%d %d = %d\n", i, state | (1<<p), dp[i][ ( state | (1<<p) ) ]);
        }
      }
    }
    return dp[40][ (1<<n)-1 ];
  }
};
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