Number of Unique Flavors After Sharing K Candies

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You are given an array of candies, where each candy has a unique flavor. You are also given a positive integer k. You need to divide these candies between two friends such that each friend gets exactly k candies.

After the division, each friend will eat the candies they received. The number of unique flavors a friend experiences is the number of distinct flavors in the candies they received.

Your task is to find the maximum possible sum of unique flavors for both friends after they divide the candies optimally.

Note:

Each candy can only be given to one friend.
If there are not enough candies, return 0.

Example 1:

Input: candies = [1,2,2,3,4,3], k = 3
Output: 5
Explanation:
- Friend 1 can take [1,2,3] with 3 unique flavors.
- Friend 2 can take [2,3,4] with 3 unique flavors.
The total number of unique flavors is 3 + 2 = 5. This is the maximum number that can be obtained.

Example 2:

Input: candies = [2,2,2,2,3,3], k = 2
Output: 3
Explanation:
- Friend 1 can take [2,3] with 2 unique flavors.
- Friend 2 can take [2,3] with 2 unique flavors.
The total number of unique flavors is 2 + 1 = 3. This is the maximum number that can be obtained.

Example 3:

Input: candies = [1,2,3,4,5], k = 1
Output: 2
Explanation:
- Friend 1 can take [1] with 1 unique flavor.
- Friend 2 can take [2] with 1 unique flavor.
The total number of unique flavors is 1 + 1 = 2. This is the maximum number that can be obtained.

Constraints:

2 * k == candies.length
1 <= candies.length <= 10⁵
1 <= candies[i] <= 10⁵
1 <= k <= candies.length / 2

Try coding it in LeetCode

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 `candies` array and the value of `k`?
Can the `candies` array contain negative numbers, zero, or other non-positive integers?
If `k` is larger than the number of candies available, what should I return?
Are there any performance expectations regarding the runtime or memory usage?
If there are multiple subsets of size `k` that maximize the number of unique flavors, is any of those subsets acceptable?

Brute Force Solution

Approach

The brute force approach to this problem involves trying every possible combination of candies that can be shared. We check each combination to see how many unique flavors remain after sharing them.

Here's how the algorithm would work step-by-step:

First, consider all the different ways you can choose candies to give away.
For each way of choosing candies, remove them from the original set of candies.
Count how many different flavors are left after you have shared the chosen candies.
Keep track of the largest number of unique flavors you find after trying all possibilities.
After trying all possible combinations of candies to share, report the maximum number of unique flavors that remained.

Code Implementation

def max_unique_flavors_brute_force(candies, candies_to_share):
    number_of_candies = len(candies)
    max_unique_flavors = 0

    # Iterate through all possible combinations of candies to share
    for i in range(1 << number_of_candies):
        shared_candies = []
        for j in range(number_of_candies):
            if (i >> j) & 1:
                shared_candies.append(candies[j])

        if len(shared_candies) == candies_to_share:
            # Create list of remaining candies after sharing

            remaining_candies = []
            for candy in candies:
                if candy not in shared_candies:
                    remaining_candies.append(candy)
                else:
                    shared_candies.remove(candy)
            # Count unique flavors of remaining candies
            unique_flavors = len(set(remaining_candies))

            # Update max unique flavors
            max_unique_flavors = max(max_unique_flavors, unique_flavors)

    return max_unique_flavors

Big(O) Analysis

Time Complexity

O(2^n * n) – The brute force approach involves considering all possible subsets of candies to share. There are 2^n possible subsets for an array of n candies. For each subset, we need to iterate through the remaining candies (in the worst case, all n candies) to count the number of unique flavors. Therefore, the total number of operations is proportional to 2^n * n, which gives us a time complexity of O(2^n * n).

Space Complexity

O(2^N) – The brute force approach involves exploring all possible combinations of candies to share, where N is the total number of candies. For each combination, we conceptually create a subset of candies to be shared. Since there are 2^N possible subsets of a set of size N, the space complexity for storing these combinations implicitly can be considered O(2^N). Additionally, although not explicitly stated, keeping track of the unique flavors left involves using a set or hashmap to store the remaining flavors, which, in the worst case, could contain all N flavors; however, the dominant space cost comes from the combinations. Thus the algorithm has a space complexity of O(2^N).

Optimal Solution

Approach

The problem asks us to figure out the most unique candy flavors we can have after sharing some of our candies. The key is to recognize that we only care about unique flavors and sharing reduces the number of candies we own, and the best strategy is to keep the most popular flavors.

Here's how the algorithm would work step-by-step:

First, count how many candies we have of each flavor.
Figure out how many candies we have left after sharing K candies.
Keep track of the flavors we have in a collection. We want to figure out the maximum possible number of unique flavors we can keep after sharing.
If the number of unique flavors we have to start is smaller than how many candies are left after the sharing, we can definitely keep all of them and that's our answer.
If not, we want to get rid of candies from the least popular flavors until we have only the candies we are allowed to keep. The number of unique flavors that have candies remaining will be the maximum possible number of flavors that can be kept.

Code Implementation

def max_unique_flavors(candies, k_candies_shared):
    candy_counts = {}
    for candy in candies:
        candy_counts[candy] = candy_counts.get(candy, 0) + 1

    number_of_unique_flavors = len(candy_counts)

    # Check if we can share all duplicate candies
    if k_candies_shared >= len(candies) - number_of_unique_flavors:
        #The other person can have all unique flavors

        return number_of_unique_flavors
    else:
        # We can only share 'k' candies so
        # the other person can only get 'k' unique flavors

        return k_candies_shared

Big(O) Analysis

Time Complexity

O(n) – The algorithm's time complexity is dominated by two main operations. First, counting the frequency of each flavor takes O(n) time, where n is the number of candies. Second, iterating through the frequencies to eliminate the least frequent flavors until the number of remaining candies is within the limit also takes O(n) in the worst case, as we might need to consider all unique flavors. Therefore, the overall time complexity is O(n) + O(n), which simplifies to O(n).

Space Complexity

O(N) – The primary space complexity comes from counting the candies of each flavor, which requires a hash map or dictionary. This data structure stores the counts of each unique flavor. In the worst case, all N candies could be of different flavors, leading to the hash map storing N key-value pairs. Therefore, the auxiliary space used is proportional to the number of unique flavors, which, in the worst case, is N, resulting in a space complexity of O(N).

Edge Cases

Case	How to Handle
candies is null or empty	Return 0 because no candies can be shared.
k is zero	Return 0 because no candies can be shared.
k is greater than the number of candies in the array	Return the number of unique flavors in candies since we can share all of them.
candies array contains only one type of candy (all elements are the same)	Return 1, as only one unique flavor exists.
candies array contains all unique candies (no duplicates)	Return min(number of unique candies, k).
Large input array of candies	Use a HashSet to efficiently count unique flavors.
candies contains negative numbers	The HashSet will handle negative numbers correctly as it treats them as distinct values.
k is a large number close to the maximum integer value	No specific handling is needed as the algorithm focuses on array size and unique flavors, which will not be affected by integer overflow of k.

candies is null or empty

How to Handle:

Return 0 because no candies can be shared.

k is zero

How to Handle:

Return 0 because no candies can be shared.

k is greater than the number of candies in the array

How to Handle:

Return the number of unique flavors in candies since we can share all of them.

candies array contains only one type of candy (all elements are the same)

How to Handle:

Return 1, as only one unique flavor exists.

candies array contains all unique candies (no duplicates)

How to Handle:

Return min(number of unique candies, k).

Large input array of candies

How to Handle:

Use a HashSet to efficiently count unique flavors.

candies contains negative numbers

How to Handle:

The HashSet will handle negative numbers correctly as it treats them as distinct values.

k is a large number close to the maximum integer value

How to Handle:

No specific handling is needed as the algorithm focuses on array size and unique flavors, which will not be affected by integer overflow of k.