<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;
    using ll = long long int;
    using pii = pair<int, int>;
    vector<vector<vector<int>>> dp;
    
    ll rec(int i, int n, int p, int minProfit, vector<pii>& gp){
        if(i >= gp.size()){
            if(p>=minProfit) return 1;
            return 0;
        }

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

        ll take=0, nottake=0;
        if(gp[i].first <= n){
            int newprofit = p+gp[i].second;
            if(p >= minProfit) newprofit = p;

            take = rec(i+1, n-gp[i].first, newprofit, minProfit, gp) %mod;
        }
        nottake = rec(i+1, n, p, minProfit, gp) %mod;
        
        return dp[i][n][p] = ((take+nottake) %mod);
    }

    int profitableSchemes(int n, int minProfit, vector<int>& group, vector<int>& profit) {
        vector<pii> gp;
        for(int i=0; i<group.size(); i++){
            gp.push_back({group[i], profit[i]});
        }
        sort(gp.begin(), gp.end());

        dp.resize(gp.size()+1, vector<vector<int>>(n+1, vector<int>(300, -1)));
        ll ans = rec(0, n, 0, minProfit, gp) %mod;
        return (int)ans;
    }
};
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