Profitable Schemes
<p>There is a group of <code>n</code> members, and a list of various crimes they could commit. The <code>i<sup>th</sup></code> crime generates a <code>profit[i]</code> and requires <code>group[i]</code> members to participate in it. If a member participates in one crime, that member can&#39;t participate in another crime.</p>

<p>Let&#39;s call a <strong>profitable scheme</strong> any subset of these crimes that generates at least <code>minProfit</code> profit, and the total number of members participating in that subset of crimes is at most <code>n</code>.</p>

<p>Return the number of schemes that can be chosen. Since the answer may be very large, <strong>return it modulo</strong> <code>10<sup>9</sup> + 7</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> n = 5, minProfit = 3, group = [2,2], profit = [2,3]
<strong>Output:</strong> 2
<strong>Explanation:</strong> To make a profit of at least 3, the group could either commit crimes 0 and 1, or just crime 1.
In total, there are 2 schemes.</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> n = 10, minProfit = 5, group = [2,3,5], profit = [6,7,8]
<strong>Output:</strong> 7
<strong>Explanation:</strong> To make a profit of at least 5, the group could commit any crimes, as long as they commit one.
There are 7 possible schemes: (0), (1), (2), (0,1), (0,2), (1,2), and (0,1,2).</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= n &lt;= 100</code></li>
	<li><code>0 &lt;= minProfit &lt;= 100</code></li>
	<li><code>1 &lt;= group.length &lt;= 100</code></li>
	<li><code>1 &lt;= group[i] &lt;= 100</code></li>
	<li><code>profit.length == group.length</code></li>
	<li><code>0 &lt;= profit[i] &lt;= 100</code></li>
</ul>
Array; Dynamic Programming
func profitableSchemes(n int, minProfit int, group []int, profit []int) int {
	m := len(group)
	f := make([][][]int, m)
	for i := range f {
		f[i] = make([][]int, n+1)
		for j := range f[i] {
			f[i][j] = make([]int, minProfit+1)
			for k := range f[i][j] {
				f[i][j][k] = -1
			}
		}
	}
	const mod = 1e9 + 7
	var dfs func(i, j, k int) int
	dfs = func(i, j, k int) int {
		if i >= m {
			if k >= minProfit {
				return 1
			}
			return 0
		}
		if f[i][j][k] != -1 {
			return f[i][j][k]
		}
		ans := dfs(i+1, j, k)
		if j+group[i] <= n {
			ans += dfs(i+1, j+group[i], min(k+profit[i], minProfit))
		}
		ans %= mod
		f[i][j][k] = ans
		return ans
	}
	return dfs(0, 0, 0)
}