Introduction

In this blog post, we will explore how to solve the LeetCode problem "1815" 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 donuts shop that bakes donuts in batches of batchSize. They have a rule where they must serve all of the donuts of a batch before serving any donuts of the next batch. You are given an integer batchSize and an integer array groups, where groups[i] denotes that there is a group of groups[i] customers that will visit the shop. Each customer will get exactly one donut. When a group visits the shop, all customers of the group must be served before serving any of the following groups. A group will be happy if they all get fresh donuts. That is, the first customer of the group does not receive a donut that was left over from the previous group. You can freely rearrange the ordering of the groups. Return the maximum possible number of happy groups after rearranging the groups. Example 1: Input: batchSize = 3, groups = [1,2,3,4,5,6] Output: 4 Explanation: You can arrange the groups as [6,2,4,5,1,3]. Then the 1st, 2nd, 4th, and 6th groups will be happy. Example 2: Input: batchSize = 4, groups = [1,3,2,5,2,2,1,6] Output: 4 Constraints: 1 <= batchSize <= 9 1 <= groups.length <= 30 1 <= groups[i] <= 109

Explanation

Here's the breakdown of the solution:

• Core Idea: The problem can be efficiently solved using dynamic programming (DP) with memoization. The state of the DP represents the counts of groups with specific remainders when divided by batchSize. We iterate through all possible group arrangements by trying each remainder group at each step.
• State Representation: The DP state is a tuple representing the counts of groups for each possible remainder (0 to batchSize - 1). This allows us to track the number of groups of each remainder type that are still available.

• Optimization: Memoization is critical to avoid recomputing the same subproblems, significantly reducing the time complexity.

• Runtime Complexity: O(batchSize^{groups.length}). However, since groups.length <= 30, and batchSize <= 9 in the worst case the DP memoization will greatly improve performance, making it acceptable. Storage Complexity: O(batchSize^{groups.length}).

Code

    from functools import lru_cache

def maxHappyGroups(batchSize, groups):
    """
    Calculates the maximum number of happy groups after rearranging the groups.

    Args:
        batchSize: The batch size of donuts.
        groups: A list of integers representing the number of customers in each group.

    Returns:
        The maximum possible number of happy groups.
    """

    counts = [0] * batchSize
    for group in groups:
        counts[group % batchSize] += 1

    @lru_cache(maxsize=None)
    def solve(state, remainder):
        """
        Recursive function with memoization to find the maximum happy groups.

        Args:
            state: A tuple representing the counts of groups for each remainder.
            remainder: The current remainder from the previous groups.

        Returns:
            The maximum number of happy groups for the given state and remainder.
        """

        if sum(state) == 0:
            return 0

        max_happy = 0
        for i in range(batchSize):
            if state[i] > 0:
                new_state = list(state)
                new_state[i] -= 1
                new_state = tuple(new_state)

                if (remainder == 0):
                    max_happy = max(max_happy, 1 + solve(new_state, (remainder + i) % batchSize))
                else:
                    max_happy = max(max_happy, solve(new_state, (remainder + i) % batchSize))

        return max_happy

    counts = tuple(counts)
    ans = solve(counts, 0)
    return ans