<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:
    int profitableSchemes(int n, int minProfit, vector<int>& group, vector<int>& profit) {
        int mod = 1e9+7;
        int size=profit.size();
        // int ppl =accumulate(group.begin(),group.end(),0);
        // int prt =accumulate(begin(profit),end(profit),0);
        vector<vector<vector<int>>>dp(size+1,vector<vector<int>>(n+2,vector<int>(minProfit+1,-1)));
        function<int(int,int,int)>solve = [&](int ind, int ppl, int prt)
        {
            if(ppl>n) return 0;
            if(ind==size)
            {
                if(ppl<=n && prt>=minProfit) return 1;
                else return 0;
            }
            if(dp[ind][ppl][prt]!=-1) return dp[ind][ppl][prt];
            int not_take = solve(ind+1,ppl,prt);
            int take = solve(ind+1,min(n+1,ppl+group[ind]),min(minProfit,prt+profit[ind]));
            return dp[ind][ppl][prt]= (take+not_take)%mod;
        };
        return solve(0,0,0);
    }
};
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