<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>

from functools import lru_cache

def pr(n, groups, profits, min_profits):
    size = len(groups)
    
    @lru_cache(maxsize=None)
    def helper(start, left, mp):
        if start == size - 1:
            if left < groups[start]:
                # We can not do this
                return 0 if mp > 0 else 1
            # we can do this job but cannot reach our goal
            if profits[start] < mp:
                return 0
            # we can do this job and reach our goal
            return 1 if mp > 0 else 2 # We can exclude and reach our goal
        
        # incl or excl
        incl = excl = 0
        excl = helper(start + 1, left, mp)
        if left >= groups[start]:
            incl = helper(start + 1, left - groups[start], max(mp - profits[start], 0))
        return incl + excl
        
    return helper(0, n, min_profits) % (10 ** 9 + 7)

class Solution:
    def profitableSchemes(self, n: int, minProfit: int, group: List[int], profit: List[int]) -> int:
        return pr(n, group, profit, minProfit)
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