Number of Ways to Wear Different Hats to Each Other
<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 the number of ways that <code>n</code> people can wear <strong>different</strong> hats from each other.</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>
Bit Manipulation; Array; Dynamic Programming; Bitmask
func numberWays(hats [][]int) int {
	n := len(hats)
	m := 0
	for _, h := range hats {
		m = max(m, slices.Max(h))
	}
	g := make([][]int, m+1)
	for i, h := range hats {
		for _, v := range h {
			g[v] = append(g[v], i)
		}
	}
	const mod = 1e9 + 7
	f := make([][]int, m+1)
	for i := range f {
		f[i] = make([]int, 1<<n)
	}
	f[0][0] = 1
	for i := 1; i <= m; i++ {
		for j := 0; j < 1<<n; j++ {
			f[i][j] = f[i-1][j]
			for _, k := range g[i] {
				if j>>k&1 == 1 {
					f[i][j] = (f[i][j] + f[i-1][j^(1<<k)]) % mod
				}
			}
		}
	}
	return f[m][(1<<n)-1]
}