<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 mod = 1000000007;
    vector<vector<int>> dp;  // Memoization table for DP

    // DP function to calculate the number of ways
    int solve(int peopleMask, int idHat, const vector<vector<int>>& hatToPeople, int numPeople) {
        if (peopleMask == (1 << numPeople) - 1) return 1; // All people have hats
        if (idHat > 40) return 0; // No more hats available

        if (dp[peopleMask][idHat] != -1) return dp[peopleMask][idHat]; // Memoized result

        // Option 1: Skip this hat
        long long ans = solve(peopleMask, idHat + 1, hatToPeople, numPeople);

        // Option 2: Try to assign this hat to any person who can wear it
        for (int person : hatToPeople[idHat]) {
            if (peopleMask & (1 << person)) continue; // Person already has a hat

            ans = (ans + solve(peopleMask | (1 << person), idHat + 1, hatToPeople, numPeople)) % mod;
        }

        return dp[peopleMask][idHat] = ans; // Memoize the result
    }

    int numberWays(vector<vector<int>>& hats) {
        int numPeople = hats.size();
        vector<vector<int>> hatToPeople(41); // hats[i] gives the list of people who can wear hat i

        // Populate hat to people mapping
        for (int person = 0; person < numPeople; ++person) {
            for (int hat : hats[person]) {
                hatToPeople[hat].push_back(person);
            }
        }

        // Initialize DP table with -1
        dp = vector<vector<int>>(1 << numPeople, vector<int>(41, -1));

        // Start the DP from hat 1 with no people having hats
        return solve(0, 1, hatToPeople, numPeople);
    }
};

13400