The Question
Coding

Profitable Schemes

You are given $n$ members and a set of $m$ potential crimes. Each crime $i$ requires $group[i]$ members and produces $profit[i]$ profit. A member can participate in at most one crime. A 'profitable scheme' is a subset of crimes such that the total number of members required does not exceed $n$, and the total profit generated is at least $minProfit$. Return the total number of distinct profitable schemes possible. As the result can be large, return it modulo $10^9 + 7$. Constraints: - $1 \le n, minProfit \le 100$ - $1 \le group.length = profit.length \le 100$ - $0 \le group[i], profit[i] \le 100$
Python
Dynamic Programming
Knapsack Problem
Questions & Insights

Clarifying Questions

What are the maximum constraints for $n$, $minProfit$, and the number of crimes?
Assumption: Based on typical competitive programming constraints for this problem, $n \le 100$, $minProfit \le 100$, and the number of crimes $\le 100$.
Can a crime require zero members or result in zero profit?
Assumption: Yes, though usually $group[i] \ge 1$. If $group[i] = 0$, it simply adds profit without consuming member capacity. If $profit[i] = 0$, it consumes members without contributing to the profit goal.
Should we count empty subsets?
Assumption: An empty subset (zero members, zero profit) is valid if $minProfit = 0$.
How do we handle total profit that exceeds $minProfit$?
Assumption: Any profit greater than or equal to $minProfit$ satisfies the condition. We can cap the profit state at $minProfit$ to keep the DP table bounded.

Thinking Process

Identify the Pattern: This is a variation of the 0/1 Knapsack Problem. Instead of maximizing value within a weight limit, we need to count subsets that satisfy two constraints: a maximum capacity constraint (members \le n) and a minimum threshold constraint (profit \ge minProfit).
Define the DP State: We need to track the number of crimes considered, the number of members used, and the profit earned. Let dp[i][j][k] be the number of ways to choose a subset from the first i crimes using j members to achieve at leastk profit.
State Transition: For each crime i with requirement g = group[i] and profit p = profit[i]:
If we don't pick crime i, the number of ways is dp[i-1][j][k].
If we pick crime i (and j \ge g), the number of ways is dp[i-1][j-g][\max(0, k-p)]. We use \max(0, k-p) because if the current crime provides more profit than needed to reach k, any state that previously reached "at least 0 profit" now contributes to "at least k profit".
Space Optimization: Since the state for crime i only depends on i-1, we can reduce the 3D DP to 2D (dp[j][k]) by iterating through members (j) and profit (k) in reverse order to avoid using the same crime multiple times for the same state.
Implementation Breakdown

Problem Set

Functional Requirements:
Calculate the number of subsets of crimes where \sum group \le n and \sum profit \ge minProfit.
Return the count modulo 10^9 + 7.
Constraints:
1 \le n, minProfit \le 100
1 \le group.length = profit.length \le 100
0 \le group[i], profit[i] \le 100

Approach

Algorithm: Dynamic Programming (2D Space-Optimized 0/1 Knapsack).
Data Structure: 2D Array (Matrix).
Complexity:
Time: O(C \cdot N \cdot P), where C is the number of crimes, N is the number of members, and P is minProfit.
Space: O(N \cdot P).

Implementation

Wrap Up

Advanced Topics

Readability vs. Performance: In Python, using numpy could significantly speed up these matrix operations, but it's not standard in competitive programming environments. For pure Python, the reverse iteration is crucial for space efficiency.
Alternative DP State: One could define dp[k] as a 1D array where each element is a list or another array representing member counts. This might be slightly more intuitive for some but harder to implement efficiently in Python.
Scaling to Large $n$ or $minProfit: If n and minProfit were much larger (e.g., 10^5), this O(N \cdot P) approach would fail. If the number of crimes was very small (< 20), we could use bitmask DP or meet-in-the-middle.