Introduction

In this blog post, we will explore how to solve the LeetCode problem "879" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

There is a group of n members, and a list of various crimes they could commit. The ith crime generates a profit[i] and requires group[i] members to participate in it. If a member participates in one crime, that member can't participate in another crime. Let's call a profitable scheme any subset of these crimes that generates at least minProfit profit, and the total number of members participating in that subset of crimes is at most n. Return the number of schemes that can be chosen. Since the answer may be very large, return it modulo 109 + 7. Example 1: Input: n = 5, minProfit = 3, group = [2,2], profit = [2,3] Output: 2 Explanation: 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. Example 2: Input: n = 10, minProfit = 5, group = [2,3,5], profit = [6,7,8] Output: 7 Explanation: 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). Constraints: 1 <= n <= 100 0 <= minProfit <= 100 1 <= group.length <= 100 1 <= group[i] <= 100 profit.length == group.length 0 <= profit[i] <= 100

Explanation

Here's the solution:

• Dynamic Programming: Use a 3D DP table dp[i][j][k] to store the number of schemes using crimes up to index i with at most j members and at least k profit.
• State Transitions: Iterate through the crimes. For each crime, either include it or exclude it. Update the DP table based on these two choices. Carefully handle boundary conditions and constraints.

• Modulo Arithmetic: Apply modulo operation at each step to prevent overflow.

• Time Complexity: O(n minProfit m), where n is the number of members, minProfit is the minimum profit, and m is the number of crimes.

• Space Complexity: O(n minProfit m)

Code

    def profitableSchemes(n: int, minProfit: int, group: list[int], profit: list[int]) -> int:
    """
    Calculates the number of profitable schemes.

    Args:
        n: The number of members.
        minProfit: The minimum profit required.
        group: A list of group sizes for each crime.
        profit: A list of profits for each crime.

    Returns:
        The number of profitable schemes modulo 10^9 + 7.
    """
    MOD = 10**9 + 7
    m = len(group)

    # dp[i][j][k] represents the number of schemes using crimes up to index i
    # with at most j members and at least k profit
    dp = [[[0] * (minProfit + 1) for _ in range(n + 1)] for _ in range(m + 1)]

    # Initialize the base case: no crimes, 0 members, 0 profit
    for j in range(n + 1):
        dp[0][j][0] = 1

    for i in range(1, m + 1):
        g = group[i - 1]
        p = profit[i - 1]
        for j in range(n + 1):
            for k in range(minProfit + 1):
                # Exclude the current crime
                dp[i][j][k] = dp[i - 1][j][k]

                # Include the current crime if possible
                if j >= g:
                    dp[i][j][k] = (dp[i][j][k] + dp[i - 1][j - g][max(0, k - p)]) % MOD

    # Sum up all schemes with at least minProfit profit and at most n members
    result = 0
    for j in range(n + 1):
        result = (result + dp[m][j][minProfit]) % MOD

    return result