1815. Maximum Number of Groups Getting Fresh Donuts

HardBit ManipulationMemoizationArrayDynamic ProgrammingBitmask

Problem Description

The goal in this problem is to optimize the happiness of groups visiting a donut shop. A donut shop bakes donuts in batches of a fixed size, batchSize. Groups of customers arrive one after another, and each group size is represented as an element in the groups array. For example, if groups is [3, 5, 1, 2], this means 4 groups of customers will visit the shop in this sequence, with respective group sizes of 3, 5, 1, and 2. Each customer from a group must be served a donut at the same time, and a group is defined as 'happy' if they all receive fresh donuts — no donuts left over from the previous group's batch.

Your task is to rearrange the groups in such a way that maximizes the number of happy groups. You are to return the maximum number of happy groups possible after doing any rearrangements necessary.

Intuition

The solution involves a bit of clever counting and dynamic programming to avoid redundant calculations. The intuition behind the solution is to classify and count the groups based on the remainder they leave when their size is divided by batchSize (groups[i] % batchSize). If the remainder is zero, then that group is automatically happy, as they will exactly finish a batch. For groups that do leave a remainder, we need to track not just how many groups leave the same remainder, but also the state of current leftover donuts when deciding to serve subsequent groups to maximize happiness.

We can represent the state with a bitmask, where for each remainder r, we assign it 5 bits (since 2^5 is more than enough to store count information) and bitwise shift into position r.

To navigate through all possible group arrangements efficiently, we use recursion with memoization (dfs function). At each state, we try serving a group with a remainder that, when added to the current leftover (modeled by mod), will not leave any leftover donuts (i.e., will complete a batch). Each recursive step, we attempt to serve a different remainder group by decreasing the count of that remainder in the state and adjusting the leftover. The result is the number of happy groups following the current group.

The dynamic programming approach, enhanced by memoization with the cache decorator, significantly reduces the number of recalculations by storing intermediate results. The dfs function takes the current state and 'mod' (the current leftover donuts) and returns the maximum number of happy groups that can follow.

Initially, all groups that don't leave a remainder when divided by batchSize are counted as happy (ans variable). The remaining groups are processed to form the initial state, and the recursive dfs function is called to find the maximum number of happy groups from the rearranged groups.

Learn more about Memoization, Dynamic Programming and Bitmask patterns.

Solution Approach

Our solution approach uses both recursion and dynamic programming with memoization, taking advantage of Python's @cache decorator for efficiency. The key is to combine counting, bitmasking, and depth-first search (DFS) to explore possible orderings that maximize the number of happy groups.

Let's break down the solution step by step to better understand the implementation:

Step 1: Initialization

First, we create variables to hold the overall state and the count of already happy groups. The state will be an integer (state), and the initial count of happy groups (ans) is 0.
Step 2: Count Happy Groups with no Remainder

We iterate over the groups array to count the groups that will already be happy without rearrangement, i.e., those groups whose size is a multiple of the batchSize (i % batchSize == 0). For each such group, we increment ans since they are guaranteed to be happy.
Step 3: Construct the Initial State

For the remaining groups that do leave a remainder, we construct a bitmask state. For each remainder r (where r = groups[i] % batchSize), we shift 1 left by r * 5 bits and increment that part of state. This packs the remainders count compactly in a single integer.
Step 4: The DFS Function

The dfs function is at the heart of the solution. Given the current state and the mod (number of donuts that would be left over), this function calculates the maximum number of additional happy groups that can be formed by serving the groups in the best possible order.

Here's how the function works:
- Base Case: If there are no more groups left to rearrange (i.e., state is 0), return 0 because we can't make any more happy groups.
- Recursive Case: Loop through possible remainders from 1 to batchSize - 1 and check if there's a group available with that remainder (using bitwise operations to extract count information from state). If such a group exists, serve it:
  - Recurse with the new state (which has one less group of the current remainder) and the new modulus mod, which accounts for the donuts that will be left over after serving this group.
  - Keep track of whether this group will be happy (i.e., mod == 0) and add this to the total from recursion.
  - Update the result res to be the maximum of itself and the newly calculated total number of happy groups.
The return value of the dfs function is the maximum number of happy groups that can follow after serving a particular arrangement of the groups.

Since computing dfs is expensive and has overlapping subproblems, we apply memoization to cache results and reuse them when the same state recurs.
Step 5: Compute Final Answer

After counting the groups that are already happy and preparing the initial state, we call dfs(state, 0) to compute how many additional groups can be made happy by rearrangement.

We then add the result from the DFS to ans for the final count of maximum happy groups and return it.

This combination of intelligent counting, bitmask to encode states, and memoized recursion builds up the solution to an otherwise intractable problem due to its combinatorial nature.

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Example Walkthrough

Let's walk through a small example to illustrate the solution approach. Assume we have a batchSize of 3 and the groups array is [1,2,3,4]. This means that every batch of donuts baked is made up of 3 donuts and 4 groups are visiting the shop with sizes of 1, 2, 3, and 4 respectively.

Step 1: Initialization

We initialize the state to 0 and ans which counts the already happy groups to 0.
Step 2: Count Happy Groups with no Remainder

The group of size 3 does not leave a remainder when divided by batchSize (3 % 3 == 0). Thus, we know this group will be automatically happy. We set ans = 1.
Step 3: Construct the Initial State

Now for the rest of the groups [1, 2, 4], we consider their remainders when divided by batchSize. Their remainders are [1, 2, 1]. We represent this state using a bitmask. The state integer will be updated as follows: we increment the state by 1 << (1 * 5) for the first group with remainder 1, 1 << (2 * 5) for the second group with remainder 2, and again by 1 << (1 * 5) for the fourth group with remainder 1. The state now encodes the remainders count: state = 0b1000000100001.
Step 4: The DFS Function

The dfs function takes (state, mod) and tries all possible ways of serving the groups to maximize happiness. Starting with mod = 0 (no leftover donuts), we explore states by serving groups with different remainders.

Suppose we serve the group with remainder 1 first. After serving, mod becomes 1, and we update the state to reflect one fewer group with remainder 1. The new state becomes 0b1000001. Since mod was 0 before serving, this group isn't happy.

Next, if we serve the group with remainder 2, mod would become 3 (as 1 from the previous step + 2 from this group), which is a full batch. We update the state again, which becomes 0b1, representing one fewer group with remainder 2.

As we just finished a batch with this serving, we mark this group as happy, increment the happy groups counter from this recursion branch, and continue exploring with the new mod of 0 (since the batch was completed).

We continue in a similar fashion, exploring all combinations, memoizing the results for repeated states, and updating the result res to keep track of the maximum number of happy groups.
Step 5: Compute Final Answer

We call dfs(state, 0) with the state we formed to get the number of additional groups we can make happy. Let's assume our dfs returns 2 for this example.

We then add this to ans where we already had 1 from the group of size 3, yielding a final answer of 3 happy groups.

This small example demonstrates how we use the combination of counting, bitmasking the state, and applying a memoized depth-first search to efficiently calculate the maximum number of happy groups from the visiting groups to the donut shop.

Solution Implementation

1from functools import lru_cache
2from typing import List
3
4class Solution:
5    def max_happy_groups(self, batch_size: int, groups: List[int]) -> int:
6        # Use memoization to store results of subproblems
7        @lru_cache(maxsize=None)
8        def dfs(state, remaining):
9            max_happy = 0
10            is_happy_group = int(remaining == 0)
11          
12            # Iterate through all possible group sizes (from 1 to batch_size-1)
13            for size in range(1, batch_size):
14                count = (state >> (size * 5)) & 31
15                # If there are groups of current size left, try taking one group out and continue DFS
16                if count:
17                    next_state = state - (1 << (size * 5))
18                    next_remaining = (remaining + size) % batch_size
19                    happy_count = dfs(next_state, next_remaining)
20                    # Update max_happy if the resulting configuration gives more happy groups
21                    max_happy = max(max_happy, happy_count + is_happy_group)
22            return max_happy
23
24        # Initialize state to keep track of the number of groups of each size modulo batch_size
25        state = 0
26        # Start with initial max_happy groups where the group size is a multiple of batch_size
27        max_happy_groups = 0
28        for group_size in groups:
29            mod_group_size = group_size % batch_size
30            max_happy_groups += mod_group_size == 0
31            # If the group size is not a multiple of batch_size, encode it into state
32            if mod_group_size:
33                state += 1 << (mod_group_size * 5)
34
35        # Launch DFS to find the rest of the happy groups
36        max_happy_groups += dfs(state, 0)
37        return max_happy_groups
38

1import java.util.Map;
2import java.util.HashMap;
3
4class Solution {
5    // HashMap to memoize solutions of states
6    private Map<Long, Integer> memo = new HashMap<>();
7    private int batchSize;
8
9    public int maxHappyGroups(int batchSize, int[] groups) {
10        this.batchSize = batchSize;
11        int happyGroupsCount = 0; // Count of happy groups
12        long state = 0; // State representation of groups modulo batchSize
13        // Pre-processing: Count groups that are already happy (size % batchSize == 0)
14        for (int group : groups) {
15            int remainder = group % batchSize;
16            if (remainder == 0) {
17                happyGroupsCount++;
18            } else {
19                // Left-shift by 5 places for each remainder slot to represent groups in state bitmask
20                state += 1L << (remainder * 5);
21            }
22        }
23        // Recursive Depth-First Search (DFS) to find max happy groups
24        happyGroupsCount += dfs(state, 0);
25        return happyGroupsCount;
26    }
27
28    private int dfs(long state, int runningMod) {
29        // If the state has already been computed, return the stored result
30        if (memo.containsKey(state)) {
31            return memo.get(state);
32        }
33        int maxGroupsFromState = 0; // Max happy groups from current state
34        for (int i = 1; i < batchSize; ++i) {
35            // Check if groups with current modulo exist
36            if ((state >> (i * 5) & 31) != 0) {
37                // Recurse with the new state, removing one group of current modulo
38                int t = dfs(state - (1L << (i * 5)), (runningMod + i) % batchSize);
39                // If remaining modulo is 0, it can be a new happy group
40                maxGroupsFromState = Math.max(maxGroupsFromState, t + (runningMod == 0 ? 1 : 0));
41            }
42        }
43        memo.put(state, maxGroupsFromState); // Memoize the result
44        return maxGroupsFromState;
45    }
46}
47

1class Solution {
2public:
3    // Function to find the maximum number of happy groups.
4    int maxHappyGroups(int batchSize, vector<int>& groups) {
5        using ll = long long;  // Alias for long long type as 'll'.
6        unordered_map<ll, int> memo;  // Memoization for dynamic programming, stores state and result.
7      
8        ll state = 0;  // Bit representation of the state.
9        int numHappyGroups = 0;  // Initialize number of happy groups to zero.
10      
11        // Pre-process groups to count happy groups divisible by batchSize and to represent the remaining.
12        for (auto& groupSize : groups) {
13            int remainder = groupSize % batchSize;
14            numHappyGroups += remainder == 0;  // Increment count if the group is already happy.
15            if (remainder) {
16                // Increment the state by setting the bits for the remainder count.
17                state += 1ll << (remainder * 5);
18            }
19        }
20
21        // Define the recursive function using dynamic programming to find maximum happy groups.
22        function<int(ll, int)> dfs = [&](ll currentState, int currentModulo) {
23            // Check if the result for the current state has already been computed.
24            if (memo.count(currentState)) {
25                return memo[currentState];
26            }
27            int maxGroups = 0;  // Track the max groups for this state.
28            // Check if the current group can be happy.
29            int isNewGroupHappy = currentModulo == 0;
30
31            // Try different group sizes and update the states accordingly.
32            for (int i = 1; i < batchSize; ++i) {
33                if (currentState >> (i * 5) & 31) {  // Check if this group size can be used.
34                    // Recursively call dfs with the new state and modulo.
35                    int temp = dfs(currentState - (1ll << (i * 5)), (currentModulo + i) % batchSize);
36                    maxGroups = max(maxGroups, temp + isNewGroupHappy);  // Update maxGroups.
37                }
38            }
39            // Save the result in memoization and return.
40            return memo[currentState] = maxGroups;
41        };
42
43        // Calculate the result starting from the state and 0 modulo.
44        numHappyGroups += dfs(state, 0);
45
46        // Return the final count of happy groups.
47        return numHappyGroups;
48    }
49};
50

1// Type alias for large integers.
2type ll = bigint;
3
4// Memoization map to store the dynamic programming states and their results.
5const memo: Map<ll, number> = new Map();
6
7// Variable to hold the number of happy groups.
8let numHappyGroups = 0;
9
10// Pre-process function to count happy groups divisible by batchSize and represent the remaining.
11function preprocessGroups(batchSize: number, groups: number[]): ll {
12    let state: ll = BigInt(0);
13
14    for (let groupSize of groups) {
15        let remainder = groupSize % batchSize;
16        numHappyGroups += remainder === 0 ? 1 : 0; // Increment count if the group is already happy.
17        if (remainder) {
18            // Increment the state by setting the bits for the remainder count.
19            state += BigInt(1) << (BigInt(remainder) * BigInt(5));
20        }
21    }
22    return state;
23}
24
25// Recursive function using dynamic programming to find maximum happy groups.
26function dfs(currentState: ll, currentModulo: number, batchSize: number): number {
27    // Check if the result for the current state has been computed.
28    if (memo.has(currentState)) {
29        return memo.get(currentState)!;
30    }
31
32    let maxGroups = 0; // Track the max groups for this state.
33    // Check if the current group can be happy.
34    let isNewGroupHappy = currentModulo === 0 ? 1 : 0;
35
36    // Iterate through different group sizes to update the states accordingly.
37    for (let i = 1; i < batchSize; ++i) {
38        if ((currentState >> (BigInt(i) * BigInt(5))) & BigInt(31)) {
39            // Recursively call dfs with the new state and modulo.
40            let temp = dfs(currentState - (BigInt(1) << (BigInt(i) * BigInt(5))), (currentModulo + i) % batchSize, batchSize);
41            maxGroups = Math.max(maxGroups, temp + isNewGroupHappy);
42        }
43    }
44    // Save the result in memoization map.
45    memo.set(currentState, maxGroups);
46  
47    return maxGroups;
48}
49
50// Main function to find the maximum number of happy groups.
51function maxHappyGroups(batchSize: number, groups: number[]): number {
52    let state: ll = preprocessGroups(batchSize, groups);
53
54    // Calculate the result starting from the state and 0 modulo.
55    numHappyGroups += dfs(state, 0, batchSize);
56
57    // Return the final count of happy groups.
58    return numHappyGroups;
59}
60

Time and Space Complexity

Time Complexity

The given Python code defines a recursive function that uses memoization for optimizing the solution to the problem of finding the maximum number of happy groups given batchSize and groups. Here's an analysis of the time complexity:

The dfs function uses a memoization decorator @cache, which ensures that each unique state-mod pair is only computed once. The state variable is a bitmask that holds up to batchSize different group sizes, each represented by a 5-bit number. Hence, there are 2^(5*batchSize) possible states.
For each state, dfs iterates over batchSize - 1 possibilities, since it skips the 0 group size.
The recursion depth is at most batchSize.

Combining all the above factors, the worst-case time complexity is O(batchSize * 2^(5*batchSize)).

Space Complexity

The space complexity mainly comes from two parts:

The space for the cache used by dfs. Since it needs to cache each unique state-mod pair, and there are 2^(5*batchSize) possible states and batchSize different mod values, the space complexity for the memoization cache is O(batchSize * 2^(5*batchSize)).
The stack space due to recursive calls which would be O(batchSize), since that's the deepest the recursion can go.

Overall, the space complexity is dominated by the cache used for memoization, resulting in O(batchSize * 2^(5*batchSize)).

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

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