Number of Unique Flavors After Sharing K Candies

class Solution {
public:
    int distributeCandies(vector<int>& candies, int windowSize) {
        int uniqueCount = 0;
    
        // Using a hash map to count frequency of each candy flavor
        unordered_map<int, int> flavorCount;
        for (int candy : candies) {
            if (++flavorCount[candy] == 1) {
                uniqueCount++;
            }
        }
    
        // Determine initially how many flavors are completely used
        int flavorsFullyUsed = 0;
        for (int i = 0; i < windowSize; i++) {
            if (--flavorCount[candies[i]] == 0) {
                flavorsFullyUsed++;
            }
        }
    
        // Compute initial max possible unique flavors outside the window
        int maxUniqueFlavors = uniqueCount - flavorsFullyUsed;
            
        // Slide over the remaining candies to update used flavors
        for (int i = windowSize; i < candies.size(); i++) {
            // Restore flavor count of the leaving candy
            if (++flavorCount[candies[i - windowSize]] == 1) {
                flavorsFullyUsed--;
            }
    
            // Decrement flavor count of the newly included candy
            if (--flavorCount[candies[i]] == 0) {
                flavorsFullyUsed++;
            }
            maxUniqueFlavors = max(maxUniqueFlavors, uniqueCount - flavorsFullyUsed);
        }
    
        return maxUniqueFlavors;
    }
};

The given C++ code implements a solution to find the maximum number of unique candy flavors that can be obtained after excluding exactly k candies from an input list of candies, where each candy is associated with a specific flavor.

• The code starts by counting the frequency of each candy flavor using an unordered_map to store the count. This part of the code also tracks the total number of unique flavors.
• After setting up the initial count of flavors, the code then processes the first windowSize (k) candies to simulate them being taken away. This adjustment allows the evaluation of how many flavors are left after these candies are removed.
• Using a sliding window technique, the loop iterates through the candy list while updating the flavor counts as if windowSize candies are continuously being removed from the beginning of the remaining candies list and adjusting the counts accordingly.
• The maximum number of unique flavors is updated at each step of the sliding window to ensure the highest possible number of unique flavors is identified.
• Finally, the function returns the computed maximum unique flavors available after removing k candies.

This implementation efficiently calculates the number of unique flavors after exclusions using the sliding window and hash map, ensuring optimal performance over simply recalculating frequencies for each possible subset of size k. The approach minimizes the time spent recalculating the number of unique flavors for each possible selection of candies to remove.

import java.util.HashMap;
import java.util.Map;
    
class Solution {
    
    public int distributeCandies(int[] sweets, int num) {
        int distinctFlavors = 0;
    
        Map<Integer, Integer> flavorCount = new HashMap<>();
        for (int sweet : sweets) {
            flavorCount.put(sweet, flavorCount.getOrDefault(sweet, 0) + 1);
            if (flavorCount.get(sweet) == 1) {
                distinctFlavors++;
            }
        }
    
        int usedFlavors = 0;
        for (int j = 0; j < num; j++) {
            flavorCount.put(sweets[j], flavorCount.get(sweets[j]) - 1);
            if (flavorCount.get(sweets[j]) == 0) {
                usedFlavors++;
            }
        }
    
        int maxUniqueFlavors = distinctFlavors - usedFlavors;
    
        for (int i = num; i < sweets.length; i++) {
            flavorCount.put(sweets[i - num], flavorCount.get(sweets[i - num]) + 1);
            if (flavorCount.get(sweets[i - num]) == 1) {
                usedFlavors--;
            }
    
            flavorCount.put(sweets[i], flavorCount.get(sweets[i]) - 1);
            if (flavorCount.get(sweets[i]) == 0) {
                usedFlavors++;
            }
    
            maxUniqueFlavors = Math.max(maxUniqueFlavors, distinctFlavors - usedFlavors);
        }
    
        return maxUniqueFlavors;
    }
}

The provided Java solution addresses the problem of determining the maximum number of unique candy flavors that can be obtained after sharing a certain number (k) of candies. Here's how the algorithm works:

1. Initialize distinctFlavors to keep track of the total number of unique flavors.
2. Use a HashMap called flavorCount to count occurrences of each flavor in the sweets array.
3. Count distinct flavors by iterating through the sweets array and updating the flavorCount map. Whenever a flavor is encountered for the first time (count == 1), increment the distinctFlavors.
4. Initialize usedFlavors to zero, which will represent the reduction in distinct flavors due to sharing the first num candies.
5. Decrease the count of each candy flavor in the first num range of the sweets array. If the count of any flavor becomes zero after decrement, increment usedFlavors.
6. Calculate maxUniqueFlavors as distinctFlavors - usedFlavors, representing the initial count of unique flavors after the initial sharing.
7. Slide through the sweets array beyond the initial num candies. For each candy i:
   • Increase the count of the flavor that leaves the window (i - num) and adjust usedFlavors if a flavor count increases from zero to one.
   • Decrease the count of the entering flavor and adjust usedFlavors if its count decreases to zero.
   • Update maxUniqueFlavors to capture the maximum number of unique flavors possible as the window slides.

The distributeCandies function finally returns maxUniqueFlavors, signifying the maximum number of unique candy flavors achievable after sharing k candies. The usage of the sliding window technique allows efficiently updating the count of distinct flavors as each new candy is considered in the range, and each candy exits the window.

class Solution:
    def distributeCandies(self, candies_list, num):
        # Store the frequency of each candy flavor.
        flavor_count = defaultdict(int)
        for candy in candies_list:
            flavor_count[candy] += 1
    
        # Count of distinct candy flavors.
        total_unique_flavors = len(flavor_count)
    
        # Count the flavors completely used in the initial block.
        depleted_flavors_count = 0
        for j in range(num):
            flavor_count[candies_list[j]] -= 1
            if flavor_count[candies_list[j]] == 0:
                depleted_flavors_count += 1
    
        # Calculate maximum distinct flavors possible after shifting the window.
        max_distinct = total_unique_flavors - depleted_flavors_count
    
        # Adjust window over the list of candies.
        for j in range(num, len(candies_list)):
            # Include candy going out of the window.
            flavor_count[candies_list[j - num]] += 1
            if flavor_count[candies_list[j - num]] == 1:
                depleted_flavors_count -= 1
    
            # Include new candy into the window.
            flavor_count[candies_list[j]] -= 1
            if flavor_count[candies_list[j]] == 0:
                depleted_flavors_count += 1
    
            # Recalculate the maximum distinct candy flavors.
            max_distinct = max(max_distinct, total_unique_flavors - depleted_flavors_count)
    
        return max_distinct

In this Python 3 solution, you determine the maximum number of unique candy flavors one can have after sharing exactly k candies. The approach involves using a sliding window combined with a dictionary for counting the occurrences of each flavor. Here's how the code operates:

1. First, a dictionary flavor_count stores the frequency of each flavor present in the candies_list.

2. You calculate the total number of unique flavors available, stored in total_unique_flavors.

3. Then, iterate over the first num candies to determine how many flavors are completely used up in this segment, updating depleted_flavors_count.

4. The potential maximum distinct flavors one can have after this initial distribution is then computed.

5. Next, as you progress through the candies_list, adjust the window by one candy at a time. This involves:

   • Re-including the candy that exits the window and updating the flavor count.
   • Decreasing the count of the new candy entering the window and adjusting the depleted flavors if any flavor count reaches zero.

6. After each adjustment, the maximum number of distinct flavors is recalculated using the current state of flavor depletion.

The function finally returns the maximum distinct flavors possible. This approach ensures efficient management and updating of candy distributions to optimize the variety of flavors.