Solution Explanation for Maximum Number of Groups Getting Fresh Donuts

This problem involves finding the maximum number of groups that can receive fresh donuts, given a batch size and the number of customers in each group. The key insight is to use dynamic programming with bit manipulation to efficiently explore the possible group orderings.

Core Idea:

1. Remainders: The crucial aspect is the remainder when the number of customers in a group is divided by the batchSize. Groups with a remainder of 0 always get fresh donuts. Groups with non-zero remainders need careful ordering to maximize the number of happy groups.

2. State Representation: We use a bitmask (state) to represent the current arrangement of groups with non-zero remainders. Each bit in the mask corresponds to a group; a set bit indicates the group has been used in the arrangement.

3. Dynamic Programming: The dfs (depth-first search) function recursively explores possible arrangements. The function's arguments are:

   • state: The bitmask representing the used groups.
   • mod: The current remainder of customers served (modulo batchSize).

4. Base Case: If all groups with non-zero remainders are used (state == mask), the function returns 0 or 1 depending on whether the final remainder mod is 0.

5. Recursive Step: The function iterates through unused groups (groups with non-zero remainders). For each group, it recursively calls itself with an updated state (including the current group) and an updated mod. It selects the arrangement that maximizes the number of happy groups.

6. Memoization: To avoid redundant calculations, memoization (caching results) is used to store the results of previous dfs calls for each state and remainder.

Time Complexity Analysis:

The time complexity is dominated by the dfs function. In the worst case, it explores all possible combinations of arranging the groups with non-zero remainders. The number of such groups can be up to 30. The number of possible states is 230. For each state, we iterate through at most 30 groups. Thus the time complexity is O(2N * N), where N is the number of groups with non-zero remainders. However, the use of memoization significantly reduces the effective time complexity in practice. It becomes closer to O(B * 2N), where B is the batchSize. This is because the memoization avoids recalculating the same states.

Space Complexity Analysis:

The space complexity is dominated by the memoization table. The size of the memoization table is O(2N * B), where N is the number of groups with non-zero remainders and B is the batchSize. The auxiliary space used by the recursive calls is also proportional to N in the worst case, which is relatively small compared to the memoization table.

Code Examples (Python):

from functools import cache
 
class Solution:
    def maxHappyGroups(self, batchSize: int, groups: List[int]) -> int:
        @cache
        def dfs(state, mod):
            if state == mask:
                return mod == 0
            res = 0
            for i, v in enumerate(remainders):
                if state >> i & 1 == 0:
                    res = max(res, dfs(state | 1 << i, (mod + v) % batchSize))
            return res
 
        remainders = [v % batchSize for v in groups if v % batchSize]
        mask = (1 << len(remainders)) - 1
        happy_groups = sum(1 for v in groups if v % batchSize == 0)
        return happy_groups + dfs(0, 0)
 

This Python code efficiently implements the solution using bit manipulation, dynamic programming, and memoization. Other languages like Java, C++, and Go can implement the same algorithm with similar time and space complexity characteristics. The core logic involving bitmasks and recursive state exploration remains the same.