<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>

class Solution:
    def profitableSchemes(self, n: int, minProfit: int, group: List[int], profit: List[int]) -> int:
        # Define the modulo value
        MOD = 10**9 + 7
        
        # Initialize the DP table
        # dp[i][j][k] represents the number of ways to achieve profit j with i members for the first k crimes
        dp = [[[0] * (minProfit + 1) for _ in range(n + 1)] for _ in range(len(group) + 1)]
        
        # Base case: There's one way to achieve 0 profit with 0 members for 0 crimes
        dp[0][0][0] = 1
        
        for k in range(1, len(group) + 1):
            members_required = group[k - 1]
            profit_generated = profit[k - 1]
            
            for i in range(n + 1):
                for j in range(minProfit + 1):
                    # If we don't commit the current crime
                    dp[k][i][j] = dp[k - 1][i][j]
                    
                    # If we commit the current crime
                    if i >= members_required:
                        dp[k][i][j] = (dp[k][i][j] + dp[k - 1][i - members_required][max(0, j - profit_generated)]) % MOD
        
        # Sum up all ways to achieve at least minProfit with any number of members
        result = 0
        for i in range(n + 1):
            result = (result + dp[len(group)][i][minProfit]) % MOD
        
        return result

# Example usage:
solution = Solution()
n = 5
minProfit = 3
group = [2, 2]
profit = [2, 3]
print(solution.profitableSchemes(n, minProfit, group, profit))  # Output: 2
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