<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 int MOD = 1e9+7;
    int solve(int i, int groupSize, int minProfit, vector<int>& group, vector<int>& profit, int n, vector<vector<vector<int>>>& dp){
        if(groupSize>n && minProfit>0)  return 0;
        if(i==group.size() && minProfit<=0)   return 1;
        if(i==group.size() && minProfit>0)   return 0;
        if(dp[i][groupSize][minProfit]!=-1) return dp[i][groupSize][minProfit];
        
        // if(minProfit <= 0){
        //     int take = 0;
        //     if(group[i]+groupSize<=n){
        //         take = solve(i+1, groupSize+group[i], minProfit-profit[i], group, profit, n);
        //     }
        //     int notTake = solve(i+1, groupSize, minProfit, group, profit, n);
        //     return 1+take+notTake;
        // }
        int take = 0;
        if(group[i]+groupSize<=n){
            take = solve(i+1, groupSize+group[i], max(0, minProfit - profit[i]), group, profit, n, dp);
        }
        int notTake = solve(i+1, groupSize, minProfit, group, profit, n, dp);
        return dp[i][groupSize][minProfit] = (take+notTake)%MOD;
    }

    int profitableSchemes(int n, int minProfit, vector<int>& group, vector<int>& profit) {
        vector<vector<vector<int>>> dp(group.size(), vector<vector<int>>(n + 100, vector<int>(minProfit + 1, -1)));
        return solve(0, 0, minProfit, group, profit, n, dp);
    }
};
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