Problem Description

You have a donut shop that bakes donuts in batches of size batchSize. The shop has a strict rule: all donuts from one batch must be served before any donuts from the next batch can be served.

Multiple groups of customers will visit your shop, where groups[i] represents the number of customers in the i-th group. Each customer wants exactly one donut.

When a group arrives at the shop, all customers in that group must be served before any customers from the next group. A group is considered "happy" if they receive fresh donuts - meaning the first customer of their group doesn't get a donut that was left over from serving the previous group.

For example, if batchSize = 4 and a group of 3 customers is served, there will be 1 leftover donut. If the next group arrives, they would get that 1 leftover donut first, making them unhappy since they didn't get all fresh donuts.

You can rearrange the order in which groups visit the shop to maximize happiness. Your goal is to find the maximum number of groups that can be happy.

Key points:

• Groups must be served completely before the next group
• A group is happy only if they start with a fresh batch (no leftovers from previous group)
• You can rearrange the visiting order of groups
• Return the maximum possible number of happy groups

Intuition

The key insight is that a group is happy when they start with no leftover donuts. This happens when the total number of donuts served to all previous groups is a multiple of batchSize.

Think about it this way: if we've served exactly 12 donuts and batchSize = 4, we've used exactly 3 complete batches with no leftovers. The next group will be happy because they'll get fresh donuts.

This means we care about the remainder when dividing the cumulative sum of customers by batchSize. A group starting at position i is happy if the sum of all group sizes before position i modulo batchSize equals 0.

We can make several observations:

1. Groups with size divisible by batchSize: These groups are special. If a group has exactly batchSize customers (or any multiple), they consume complete batches and leave no leftovers. We can greedily place all such groups first - they'll all be happy and won't affect subsequent groups.

2. Remainder matters, not actual size: For groups whose size isn't divisible by batchSize, only the remainder matters. A group of 7 customers with batchSize = 4 has the same effect as a group of 3 customers - both leave 3 donuts for the next batch.

3. The problem becomes arrangement of remainders: After handling complete batches, we need to arrange groups based on their remainders [1, 2, ..., batchSize-1] to maximize happy groups. The first group in any arrangement is always happy (starts fresh), and subsequent groups are happy when the cumulative remainder equals 0.

4. State representation: Since we only care about how many groups have each possible remainder value, we can represent the state as a frequency count of remainders. With at most 30 groups and remainder values from 1 to batchSize-1, we can encode this state efficiently using bit manipulation where every 5 bits represent the count of one remainder value.

The solution uses dynamic programming with memoization to try different arrangements of these remainder groups, tracking the running remainder to determine when a new group would be happy.

Learn more about Memoization, Dynamic Programming and Bitmask patterns.

Solution Approach

The solution uses greedy strategy combined with state compression and memoized search (dynamic programming).

Step 1: Handle Groups Divisible by BatchSize

First, we identify groups whose size is already a multiple of batchSize. These groups consume complete batches and leave no leftovers, so they can all be placed at the beginning and will all be happy:

ans = 0
for v in groups:
    i = v % batchSize
    ans += i == 0  # Count groups with remainder 0

Step 2: State Compression for Remaining Groups

For groups not divisible by batchSize, we only care about their remainder when divided by batchSize. We encode the frequency of each remainder into a single integer state:

• Use 5 bits to represent the count of each remainder value
• Remainder i is stored at bit positions [i*5, i*5+4]
• Maximum batchSize is 9, so we need at most 8 * 5 = 40 bits

state = 0
for v in groups:
    i = v % batchSize
    if i:  # If remainder is not 0
        state += 1 << (i * 5)  # Increment count for remainder i

Step 3: Memoized Search with DFS

The core algorithm uses a recursive function dfs(state, mod) where:

• state: Current frequency distribution of remainders
• mod: Current cumulative remainder (sum of served groups mod batchSize)
• Returns: Maximum number of happy groups possible from this state

@cache
def dfs(state, mod):
    res = 0
    x = int(mod == 0)  # 1 if current group will be happy, 0 otherwise
  
    for i in range(1, batchSize):
        if state >> (i * 5) & 31:  # Check if remainder i exists
            # Try using a group with remainder i
            t = dfs(state - (1 << (i * 5)), (mod + i) % batchSize)
            res = max(res, t + x)
  
    return res

How the DFS works:

1. If mod == 0, the next group to be served will be happy (gets fresh donuts), so x = 1
2. Try each possible remainder i from 1 to batchSize-1:
   • Check if we have groups with remainder i: (state >> (i * 5)) & 31 extracts the 5-bit count
   • If available, remove one group with remainder i from state: state - (1 << (i * 5))
   • Update cumulative remainder: (mod + i) % batchSize
   • Recursively solve the subproblem
   • Add x to account for whether the current group is happy
3. Return the maximum happiness achievable

Step 4: Final Answer

The total answer combines:

• Groups with size divisible by batchSize (all happy)
• Optimal arrangement of remaining groups found by dfs(state, 0)

return ans + dfs(state, 0)

The @cache decorator provides memoization, preventing redundant calculations for the same (state, mod) pairs, which significantly improves performance.

Example Walkthrough

Let's walk through an example with batchSize = 4 and groups = [1, 3, 2, 5, 2, 2, 1, 6].

Step 1: Identify and handle groups divisible by batchSize

First, we check which groups have sizes divisible by 4:

• Group of 1: 1 % 4 = 1 (remainder 1)
• Group of 3: 3 % 4 = 3 (remainder 3)
• Group of 2: 2 % 4 = 2 (remainder 2)
• Group of 5: 5 % 4 = 1 (remainder 1)
• Group of 2: 2 % 4 = 2 (remainder 2)
• Group of 2: 2 % 4 = 2 (remainder 2)
• Group of 1: 1 % 4 = 1 (remainder 1)
• Group of 6: 6 % 4 = 2 (remainder 2)

No groups are divisible by 4, so ans = 0.

Step 2: Build state representation

Count the frequency of each remainder:

• Remainder 1: 3 groups (sizes 1, 5, 1)
• Remainder 2: 4 groups (sizes 2, 2, 2, 6)
• Remainder 3: 1 group (size 3)

Encode this into state using 5 bits per remainder:

• Remainder 1 at bits [5-9]: 3 << 5 = 96
• Remainder 2 at bits [10-14]: 4 << 10 = 4096
• Remainder 3 at bits [15-19]: 1 << 15 = 32768
• Total state: 96 + 4096 + 32768 = 36960

Step 3: Execute DFS search

Start with dfs(36960, 0):

Since mod = 0, the first group we place will be happy (x = 1).

We try different first groups:

Option A: Start with remainder 1

• Remove one group with remainder 1: new state = 36960 - 32 = 36928
• New mod = (0 + 1) % 4 = 1
• Recursively solve dfs(36928, 1)
• In this recursive call, mod = 1, so next group won't be happy (x = 0)

Option B: Start with remainder 2

• Remove one group with remainder 2: new state = 36960 - 1024 = 35936
• New mod = (0 + 2) % 4 = 2
• Recursively solve dfs(35936, 2)

Option C: Start with remainder 3

• Remove one group with remainder 3: new state = 36960 - 32768 = 4192
• New mod = (0 + 3) % 4 = 3
• Recursively solve dfs(4192, 3)

Let's trace one promising path: Start with remainder 3, then remainder 1:

1. Place group with remainder 3: mod = 3, this group is happy
2. Place group with remainder 1: mod = (3 + 1) % 4 = 0, this group is happy
3. Place group with remainder 2: mod = (0 + 2) % 4 = 2, this group is happy
4. Place group with remainder 2: mod = (2 + 2) % 4 = 0, this group is not happy
5. Continue placing remaining groups...

The algorithm explores all possible arrangements. One optimal arrangement might be: [3] → [1] → [2, 2] → [1, 1, 2, 2]

Where groups in brackets are served together to reach mod = 0:

• Group with remainder 3: happy (starts fresh)
• Group with remainder 1: happy (cumulative = 4, mod = 0)
• First group with remainder 2: happy (starts fresh after mod = 0)
• Second group with remainder 2: not happy (cumulative = 2)
• Continue analysis...

Step 4: Return final answer

The DFS will find that the maximum number of happy groups is 4, achieved by carefully arranging the groups so that the cumulative remainder becomes 0 as often as possible.

Final answer: 0 + 4 = 4 happy groups.

Solution Implementation

Time and Space Complexity

Time Complexity: O(batchSize * 31^(batchSize-1))

The analysis breaks down as follows:

• The state is represented using 5 bits for each remainder value from 1 to batchSize-1, allowing up to 31 occurrences of each remainder
• The total number of possible states is at most 31^(batchSize-1) since we have batchSize-1 different remainder values (excluding 0)
• For each state, the DFS function iterates through batchSize-1 possible remainders
• Each state is computed only once due to memoization with @cache
• Therefore, the time complexity is O(batchSize * 31^(batchSize-1))

For typical constraints where batchSize ≤ 9, this gives us 31^8 * 9 ≈ 8.5 * 10^11 in the worst case, but in practice, many states are unreachable (since the total number of groups is limited), keeping the actual complexity well below O(10^7).

Space Complexity: O(31^(batchSize-1))

The space analysis:

• The memoization cache stores at most 31^(batchSize-1) different states
• Each cached entry stores the state tuple and the result value
• The recursion depth is at most the total number of groups, which is typically much smaller than the number of states
• For batchSize ≤ 9, this gives us 31^8 ≈ 9.5 * 10^10 theoretical states, but the actual number of valid states encountered is significantly smaller due to the limited number of groups

In practice, with reasonable constraints on the number of groups (typically ≤ 30), the actual space used is much less than the theoretical maximum, keeping it below O(10^6).

Learn more about how to find time and space complexity quickly.

Common Pitfalls

1. Incorrect State Representation - Using Lists or Tuples Instead of Bit Compression

Many developers initially try to use a list or tuple to track remainder frequencies:

# WRONG APPROACH - Will cause Time Limit Exceeded
@cache
def dfs(remainder_counts, current_remainder):
    remainder_counts = list(remainder_counts)  # Convert tuple for modification
    # ... rest of logic

Why this fails:

• Python's @cache requires hashable arguments
• Converting between lists and tuples repeatedly is inefficient
• The state space becomes much larger with tuple representations

Solution: Use bit compression as shown in the correct solution. Each remainder gets 5 bits, allowing efficient state representation in a single integer.

2. Forgetting to Handle Groups Divisible by BatchSize Separately

A common mistake is including groups with remainder 0 in the DFS state:

# WRONG - Including remainder 0 in DFS
for group_size in groups:
    remainder = group_size % batchSize
    compressed_state += 1 << (remainder * 5)  # Includes remainder 0!

Why this fails:

• Groups with remainder 0 always leave the "plate clean" for the next group
• They can all be placed at the beginning and all be happy
• Including them in DFS unnecessarily complicates the state space

Solution: Count groups with remainder 0 separately and add them directly to the answer.

3. Incorrect Bit Manipulation for Extracting/Updating Counts

Developers often make mistakes with bit operations:

# WRONG - Incorrect extraction
count = compressed_state >> (remainder * 5)  # Missing mask!

# WRONG - Incorrect decrement
new_state = compressed_state - remainder  # Should subtract (1 << bit_position)

Solution:

• Extract count: (compressed_state >> (remainder * 5)) & 31
• Decrement count: compressed_state - (1 << (remainder * 5))

4. Not Understanding When a Group is Happy

A critical misunderstanding is when to count a group as happy:

# WRONG - Counting happiness after serving the group
def dfs(state, mod):
    for remainder in range(1, batchSize):
        if has_remainder(state, remainder):
            new_mod = (mod + remainder) % batchSize
            # Wrong: checking if new_mod == 0
            bonus = int(new_mod == 0)  # INCORRECT!

Why this fails:

• A group is happy if they START with fresh donuts (mod == 0 BEFORE serving)
• Not if they FINISH exactly at a batch boundary

Solution: Check mod == 0 at the beginning of each recursive call, before choosing which group to serve next.

5. Off-by-One Error with Remainder Range

# WRONG - Including batchSize as a possible remainder
for remainder in range(1, batchSize + 1):  # WRONG!
    # Process remainder...

Why this fails:

• Remainders range from 0 to batchSize-1
• A remainder of batchSize would actually be 0

Solution: Use range(1, batchSize) since remainder 0 is handled separately.