<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 {
public:
    const static int MOD = 1000000007;
    int dp[101][101][2001];
    int solve(int n, int i, int profit, int minProfit, int size, vector<int>& group, vector<int>& profits){
        if(n<0) return 0;
        if(i >= size) return profit >= minProfit;

        if(dp[i][n][profit]!=-1) return dp[i][n][profit];

        long long count = 0;
        count += solve(n-group[i], i+1, profit+profits[i], minProfit, size, group, profits);
        count += solve(n, i+1, profit, minProfit, size, group, profits);

        return dp[i][n][profit] = count % MOD;
    }
    int profitableSchemes(int n, int minProfit, vector<int>& group, vector<int>& profit) {
        memset(dp, -1, sizeof(dp));
        int size = profit.size();
        int ans = solve(n, 0, 0, minProfit, size, group, profit);
        return ans;
    }
};
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