<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 mod = 1e9+7;
long long func(int n, int mini,vector<int>& group,vector<int>& profit,int start, vector<vector<vector<int>>> &dp){
    int sz = group.size();
   if(start>=sz){
    if(mini<=0 && n>=0) return 1;
    return 0;
   }
    if(dp[start][n][mini]!=-1) return dp[start][n][mini];
    long long take=0,nottake=0;
    if(group[start]<=n){
 take =  (func(n-group[start],max(mini-profit[start],0), group,profit,start+1,dp))%mod;
    };
    nottake = (func(n,mini,group, profit,start+1,dp))%mod;
    return dp[start][n][mini]= (take+nottake)%mod;
}
    int profitableSchemes(int n, int minProfit, vector<int>& group, vector<int>& profit) {
        int sz = group.size();
       vector<vector<vector<int>>> dp(sz,vector<vector<int>>(n+4,vector<int>(minProfit+7,-1)));
       return func(n,minProfit,group,profit,0,dp);
    }
};
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