How do you solve Maximum Number of Groups Getting Fresh Donuts on LeetCode?

Maximum Number of Groups Getting Fresh Donuts (LeetCode #1815) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Understanding how to partition groups based on their sizes and the batch size is crucial.

What is the brute force solution for Maximum Number of Groups Getting Fresh Donuts?

The brute force approach for Maximum Number of Groups Getting Fresh Donuts is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves generating all possible permutations of the groups and checking how many happy groups can be formed for each permutation. This is straightforward but inefficient due to the factorial growth of permutations. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Maximum Number of Groups Getting Fresh Donuts?

Maximum Number of Groups Getting Fresh Donuts uses the following concepts: Array, Dynamic Programming, Bit Manipulation, Memoization, Bitmask. The recommended patterns to study are: Dynamic Programming, Bit Manipulation, Greedy Algorithms.

What are the interview tips for Maximum Number of Groups Getting Fresh Donuts?

Always clarify the problem constraints and edge cases before jumping into coding. Discuss your thought process and potential approaches with the interviewer. Practice dynamic programming problems to become comfortable with state transitions.

What are common mistakes when solving Maximum Number of Groups Getting Fresh Donuts?

Failing to consider the effect of group sizes on the remainder when divided by batchSize. Not utilizing the frequency of group sizes effectively, leading to unnecessary complexity.

#1815

Maximum Number of Groups Getting Fresh Donuts

Hard

ArrayDynamic ProgrammingBit ManipulationMemoizationBitmaskDynamic ProgrammingBit ManipulationGreedy Algorithms

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

The optimal solution utilizes dynamic programming and bit manipulation to efficiently calculate the maximum number of happy groups by tracking the remainders when groups are served. This avoids the need to check every permutation.

⚙️

Algorithm

3 steps

1Step 1: Create a frequency array to count how many groups have sizes that give the same remainder when divided by batchSize.
2Step 2: Use dynamic programming to explore combinations of these groups while keeping track of the current remainder.
3Step 3: Maximize the number of happy groups by ensuring that the sum of group sizes in each partition is 0 mod batchSize.

solution.py12 lines

1def maxHappyGroups(batchSize, groups):
2    from collections import Counter
3    freq = Counter(g % batchSize for g in groups)
4    dp = [-1] * batchSize
5    dp[0] = 0
6    for r in range(batchSize):
7        if freq[r] > 0:
8            for j in range(batchSize):
9                if dp[j] != -1:
10                    new_rem = (j + r) % batchSize
11                    dp[new_rem] = max(dp[new_rem], dp[j] + freq[r])
12    return dp[0] + (1 if freq[0] > 0 else 0)

ℹ

Complexity note: The time complexity is O(n) since we iterate through the groups to create the frequency array and then through the batchSize for dynamic programming. The space complexity is O(n) due to the frequency and dp arrays.

1Understanding how to partition groups based on their sizes and the batch size is crucial.
2Using dynamic programming can significantly reduce the complexity of the problem.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.