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
class Solution {
public:
    int profitableSchemes(int n, int minProfit, vector<int>& group, vector<int>& profit) {
        int m = group.size();
        int f[m][n + 1][minProfit + 1];
        memset(f, -1, sizeof(f));
        const int mod = 1e9 + 7;
        function<int(int, int, int)> dfs = [&](int i, int j, int k) -> int {
            if (i >= m) {
                return k == minProfit ? 1 : 0;
            }
            if (f[i][j][k] != -1) {
                return f[i][j][k];
            }
            int 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;
            return f[i][j][k] = ans;
        };
        return dfs(0, 0, 0);
    }
};