<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:
    //dp[i][j][k] means ending with current crime, j profit, k weight. 
    //dp[i][j][k] = dp[i-1][j][k] + dp[i-1][j-profit[i]][k-group[i]];
    const int modVal = pow(10, 9) + 7;
    int profitableSchemes(int n, int minProfit, vector<int>& group, vector<int>& profit) {
        int sumProfit = accumulate(profit.begin(), profit.end(), 0);
        int res = 0;
        vector<vector<int>> dp(sumProfit + 1, vector<int>(n + 1, 0));
        dp[0][0] = 1;
        for(int i = 0; i < group.size(); i++) {
            for(int j = sumProfit; j >= profit[i]; j--) {
                for(int k = n; k >= group[i]; k--) {
                    dp[j][k] = (dp[j][k] + dp[j-profit[i]][k-group[i]]) % modVal;
                }
            }
        }
        
        for(int j = minProfit; j <= sumProfit; j++) {
            for(int k = 0; k <= n; k++) {
                res = (res + dp[j][k]) % modVal;
            }
        }
        return res;
    }
};
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