Maximum Number of Groups Getting Fresh Donuts
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Dynamic Programming
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 <= 91 <= groups.length <= 301 <= groups[i] <= 109
Solution
Clarifying Questions
When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:
- What are the constraints on the size of the `groupSizes` array and the values within it?
- Can the `batchSize` be zero? What should I return in that case?
- If multiple groups can be arranged to maximize the number getting fresh donuts, is any valid arrangement acceptable?
- Can elements in the `groupSizes` array be zero?
- Is `batchSize` guaranteed to be a positive integer?
Brute Force Solution
Approach
The brute force method to maximize donut group satisfaction involves trying out every possible order in which the groups can enter the donut shop. We then evaluate each ordering to determine how many groups get fresh donuts and, ultimately, find the order that maximizes the number of happy groups.
Here's how the algorithm would work step-by-step:
- Consider all the possible sequences in which the groups of people can arrive at the donut shop.
- For each sequence, simulate the arrival of the groups one by one.
- When a group arrives, determine if they get fresh donuts by checking if the number of donuts remaining is enough to give them all fresh ones. If not, they share.
- Keep track of the total number of groups that got fresh donuts for each particular sequence.
- Compare the counts of happy groups across all the sequences you've examined.
- Choose the sequence that resulted in the highest number of happy groups. This is your maximum number of groups getting fresh donuts.
Code Implementation
def maximum_groups_getting_fresh_donuts_brute_force(number_of_donuts, group_sizes):
import itertools
maximum_happy_groups = 0
# Generate all possible permutations of group arrival orders
for group_order in itertools.permutations(group_sizes):
happy_groups = 0
remaining_donuts = number_of_donuts
for group_size in group_order:
# Check if the group can get fresh donuts
if remaining_donuts >= group_size:
happy_groups += 1
remaining_donuts -= group_size
# Otherwise, they share the remaining donuts
else:
remaining_donuts = 0
# Update the maximum number of happy groups
maximum_happy_groups = max(maximum_happy_groups, happy_groups)
return maximum_happy_groupsBig(O) Analysis
Time Complexity
O(n!) – The provided brute force method considers all possible orderings of the groups, where n is the number of groups. Generating all permutations of n elements takes O(n!) time. For each permutation, we simulate the arrival of the groups and check how many receive fresh donuts, which takes O(n) time. Therefore, the overall time complexity is O(n! * n). Since the factorial term dominates, we simplify to O(n!).
Space Complexity
O(N!) – The brute force solution explores all possible orderings of the groups, which requires generating permutations. Storing the permutations themselves will dominate space complexity. Since there are N groups, where N is the number of groups, there are N! possible permutations. While we might not explicitly store all N! permutations simultaneously, the recursive calls needed to generate them can, in the worst case, lead to a stack depth proportional to N. Furthermore, temporary lists used to construct each permutation also consume memory proportional to N. The cumulative effect of exploring all N! permutations leads to a space complexity dominated by the factorial term, specifically O(N!).
Optimal Solution
Approach
The best strategy involves understanding remainders when dividing group sizes by the number of donut types. Instead of trying all possible orders of groups, we use a clever trick to count how many groups can happily get fresh donuts and optimize the remaining groups.
Here's how the algorithm would work step-by-step:
- First, figure out how many groups there are for each possible remainder when dividing their size by the number of donut types. For example, how many groups have a remainder of 0, 1, 2, etc.?
- Some groups will always be happy, because their remainder is 0. These people can come first, get their donuts, and no one else cares. Count how many of these groups there are, and remove them from our consideration.
- Now, for the remaining groups, think about matching pairs. A group with a remainder of 1 can be matched with a group with a remainder of (number of donut types - 1). Similarly, a remainder of 2 can be paired with (number of donut types - 2). Pairing up these groups guarantees that at least one group will be 'happy'.
- For each pair, find the smaller number of occurrences. For example, if there are 3 groups with remainder 1 and 5 groups with remainder (number of donut types - 1), you can make 3 happy pairs. Add these to your count and remove them from the counts.
- After pairing, you might have groups with remainders that don't have a perfect pair. For example, groups with a remainder of half the number of donut types (if the number of donut types is even) or just one remainder if number of donut types is 1. For these groups, we can decide on best order in a 'state-space' fashion using a technique known as Dynamic Programming (DP).
- Using DP, we track all the possible combinations of the remaining groups. For each combination of groups, we figure out if the last group that went through 'felt fresh' or not, adding to a global count.
- Finally, add together the happy remainder-0 groups, the happy paired groups, and happy remaining groups (tracked using dynamic programming) to get the maximum number of groups that can get fresh donuts.
Code Implementation
def maximum_groups(number_of_donut_types, group_sizes):
group_counts_by_remainder = [0] * number_of_donut_types
for group_size in group_sizes:
remainder = group_size % number_of_donut_types
group_counts_by_remainder[remainder] += 1
happy_groups_count = group_counts_by_remainder[0]
group_counts_by_remainder[0] = 0
#Pair up remainders to ensure at least one group is happy
for remainder in range(1, (number_of_donut_types + 1) // 2):
complement_remainder = number_of_donut_types - remainder
min_count = min(group_counts_by_remainder[remainder], \
group_counts_by_remainder[complement_remainder])
happy_groups_count += min_count
group_counts_by_remainder[remainder] -= min_count
group_counts_by_remainder[complement_remainder] -= min_count
remaining_groups = []
for remainder in range(1, number_of_donut_types):
remaining_groups.extend([remainder] * group_counts_by_remainder[remainder])
number_of_remaining_groups = len(remaining_groups)
dp_table = {}
def solve_dp(index, current_total):
if index == number_of_remaining_groups:
return 0
if (index, current_total) in dp_table:
return dp_table[(index, current_total)]
#Consider the case where current group is NOT happy
not_happy = solve_dp(index + 1, current_total)
#Consider the case where current group IS happy
if (current_total % number_of_donut_types) != remaining_groups[index]:
happy = 1 + solve_dp(index + 1, current_total + remaining_groups[index])
else:
happy = 0
dp_table[(index, current_total)] = max(not_happy, happy)
return dp_table[(index, current_total)]
# Find the maximum number of additional happy groups
additional_happy_groups = solve_dp(0, 0)
return happy_groups_count + additional_happy_groupsBig(O) Analysis
Time Complexity
O(m + k * 2^k) – The algorithm first counts group remainders, taking O(m) time, where m is the number of groups. Then it pairs groups based on remainders, which is O(k) where k is the number of donut types. The dynamic programming step considers all possible subsets of unpaired remainders. If there are k unpaired remainder counts, the DP state space grows exponentially as 2^k. For each of these states (subsets of remainders), processing takes O(k) time. Thus the DP portion becomes O(k * 2^k) and is the bottleneck for larger k (number of donut types). The overall complexity combines these steps to O(m + k * 2^k).
Space Complexity
O(groups.length * batchSize) – The algorithm uses an array to store the count of groups for each remainder, requiring O(batchSize) space, where batchSize is the number of donut types. The dynamic programming step dominates the space complexity. It uses a DP table to track all possible combinations of the remaining groups. The size of this DP table is determined by the number of groups remaining and the current number of fresh donuts available, leading to a space complexity of O(groups.length * batchSize), where groups.length represents the number of groups and batchSize represents the number of donut types.
Edge Cases
| Case | How to Handle |
|---|---|
| Empty batches array | Return 0, since no batches are present. |
| groupSize is 1 | Return the length of the batches array since every group gets fresh donuts. |
| All batches have the same remainder when divided by groupSize | Optimized counting with modulo arithmetic should handle even distribution effectively. |
| groupSize is a large prime number | The solution's modulo operation is applicable to primes and should still function correctly. |
| batches contains very large numbers | The modulo operation will keep the values within the range [0, groupSize - 1], preventing overflow issues. |
| groupSize is larger than the number of batches | The state space needs to store up to groupSize -1 entries, but the dynamic programming should naturally compute the optimal answer. |
| Maximum array size causes memory issues (dynamic programming table) | If memory usage becomes excessive, consider iterative DP or potentially approximations to reduce space requirements, or raise an exception. |
| No combination of groups results in fresh donuts for any groups | The dynamic programming algorithm will explore the entire space, correctly leading to a result of 0. |
Empty batches array
How to Handle:
Return 0, since no batches are present.
groupSize is 1
How to Handle:
Return the length of the batches array since every group gets fresh donuts.
All batches have the same remainder when divided by groupSize
How to Handle:
Optimized counting with modulo arithmetic should handle even distribution effectively.
groupSize is a large prime number
How to Handle:
The solution's modulo operation is applicable to primes and should still function correctly.
batches contains very large numbers
How to Handle:
The modulo operation will keep the values within the range [0, groupSize - 1], preventing overflow issues.
groupSize is larger than the number of batches
How to Handle:
The state space needs to store up to groupSize -1 entries, but the dynamic programming should naturally compute the optimal answer.
Maximum array size causes memory issues (dynamic programming table)
How to Handle:
If memory usage becomes excessive, consider iterative DP or potentially approximations to reduce space requirements, or raise an exception.
No combination of groups results in fresh donuts for any groups
How to Handle:
The dynamic programming algorithm will explore the entire space, correctly leading to a result of 0.
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