<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 m=group.size(),mod =1e9+7;
        vector<vector<vector<long long>>>dp(m,vector<vector<long long>>(n+1,vector<long long>(minProfit+1,0)));
        dp[0][0][0]=1;
        for(int i=0;i<m;i++){
            int p = profit[i];
            int num = group[i];
            if(i==0){
                if(group[i]<=n){
                    if(profit[i]>minProfit){
                        dp[i][group[i]][minProfit]++;
                    }
                    else dp[i][group[i]][profit[i]]++;
                }
                continue;
            }
            for(int j=0;j<=n;j++){
                for(int k=0;k<=minProfit;k++){
                    dp[i][j][k]+=dp[i-1][j][k];
                    dp[i][j][k]%=mod;
                    if(group[i]+j<=n){
                        dp[i][j+group[i]][min(minProfit,k+profit[i])]+=dp[i-1][j][k];
                        dp[i][j+group[i]][min(minProfit,k+profit[i])]%=mod;
                    }
                }
            }
        }
        // for(auto x:dp){
        //     for(auto y:x){
        //         for(auto z:y){
        //             cout<<z<<" ";
        //         }
        //         cout<<"\n";
        //     }
        //     cout<<"\n";
        // }
        long long ans=0;
        for(int i=0;i<=n;i++){
                ans+=dp[m-1][i][minProfit];
                ans%=mod;
        }
       return ans;
    }
};
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