Problem Description

You have an array of integers named candies where each element represents the flavor of a candy. The position of the flavored candy in the list is its index, starting from 0. You are tasked with sharing a certain number k of consecutive candies with your sister. The goal is to retain the highest number of unique candy flavors for yourself after you share k consecutive candies with her. You must figure out the maximum number of unique flavors you can keep after giving away k candies.

Intuition

The key to solving this problem lies in understanding that when you give away a range of consecutive candies, the unique flavors you have left are the ones not included in the range you gave away. To maximize the unique flavors, you want to give away the segment which has the least number of unique flavors. To find the solution, you can use a sliding window approach that moves across the candies array. This sliding window will represent the candies you will give to your sister.

• Starting after the first k candies, count the unique flavors you have (all the candies except the first k ones).
• Then, for each subsequent candy, adjust the count by removing the flavor going out of the window and adding the flavor coming into the window at the end.
• Note that if after removing a candy from the count its frequency drops to zero, remove it from the count to maintain an accurate representation of unique flavors.
• Update and keep track of the maximum count of unique flavors after each step.

By iterating through the array and adjusting the window of candies given to your sister, we can determine the maximum number of unique flavors that can be kept.

Learn more about Sliding Window patterns.

Solution Approach

The solution uses a sliding window approach alongside Python's Counter class (from collections) to keep track of the frequency of each flavor of candy that we can keep (i.e., candies not included in the k consecutive candies given to the sister).

Here's a step-by-step explanation of the implementation:

1. Begin by counting the unique flavors of the candies that are initially outside the range of the first k candies using Counter(candies[k:]). This gives us the starting set of flavors we have excluding the first k to be given away.

2. Initialize the count of unique flavors (ans) to the length of the initial count, representing the number of unique flavors after giving away the first segment of k candies.

3. Iterate over the candies starting from index k up to the end:

   • For each candy at index i, decrement the count of the flavor encountered since it now becomes part of the segment to give away.
   • Increment the count of the flavor i-k, which is now coming into the window of candies we can keep, as the sliding window moves forward.
   • After adjusting the counts, check if the flavor we just decremented has a count of zero, indicating there are no more of that flavor to keep, and if so, remove it from the count.
   • Update the maximum count of unique flavors (ans) with the larger of the current ans or the current number of unique flavors after the adjustments.

The use of Counter makes it efficient to keep track of the frequency of each flavor and its updates, and the max() function assists in retaining the highest number of unique flavors after each iteration. By the end of the loop, ans holds the maximum number of unique flavors that can be kept.

class Solution:
    def shareCandies(self, candies: List[int], k: int) -> int:
        cnt = Counter(candies[k:])
        ans = len(cnt)
        for i in range(k, len(candies)):
            cnt[candies[i]] -= 1
            cnt[candies[i - k]] += 1
            if cnt[candies[i]] == 0:
                cnt.pop(candies[i])
            ans = max(ans, len(cnt))
        return ans

This code snippet effectively finds the optimal number of unique flavors that can be retained, utilizing the efficiency of Counter for frequency tracking within the sliding window operated by the loop.

Example Walkthrough

Let's consider the candies array as [1, 2, 3, 2, 2, 1, 3, 3] and k equals 3. Now, let's apply the solution step by step:

1. We start with the initial Counter(candies[k:]) which excludes the first k candies: Counter([2, 2, 1, 3, 3]), resulting in a count of {2: 2, 1: 1, 3: 2}. Here, you have two unique flavors of #2, one of #1, and two of #3. Therefore, the initial ans (maximum number of unique flavors) is 3, as there are three unique flavors in this count.

2. We now iterate over candies starting from the k-th element, which is the 3rd index (using zero-based indexing) in the original candies array.

3. At index 3 (candies[3] is 2), we would give away the second flavor candy, which was in the initial segment to be kept. So, we decrement the count of flavor 2 from {2: 2, 1: 1, 3: 2} to {2: 1, 1: 1, 3: 2}.

4. We also include the flavor at index 0 (because the window is sliding forward), so flavor 1 is to be added back into our count. The count remains the same because flavor 1 is already in our count {2: 1, 1: 1, 3: 2}.

5. The current count of unique flavors is still 3, and thus ans remains 3.

6. Continue the iteration:

   • Move to index 4: Remove one occurrence of flavor 2 and add back flavor 1. Count is now {2: 0, 1: 2, 3: 2}. After removing the flavor 2 count which is zero, the count becomes {1: 2, 3: 2}. ans remains 3.

   • Move to index 5: Remove flavor 1 and add back flavor 2. New count is {1: 1, 3: 2} (since 2 was not in the count, it doesn't get added). ans remains 3.

   • Move to index 6: Remove flavor 3 and there's no flavor at i-k (since we're now at the end of the array). Current count is {1: 1, 3: 1}. ans does not change and remains 3.

By the end of the process, we have iterated through the array and adjusted the window properly. The final ans remains 3, which signifies the maximum number of unique flavors we can retain after sharing k consecutive candies with the sister. The sliding window approach made it efficient to find this out.

Solution Implementation

Time and Space Complexity

Time Complexity

The initial setup of the Counter cnt with the slice candies[k:] is O(n-k) where n is the length of candies.

The loop runs for n-k iterations (since it starts from k and goes up to n). Within each iteration, the operations on the Counter object (cnt[candies[i]] -= 1, cnt[candies[i - k]] += 1, and cnt.pop(candies[i])) can be considered O(1) on average, as they involve updating the hash table.

Therefore, the total time complexity of the loop is O(n-k). Since the setup time is also O(n-k) and these are done sequentially, the overall time complexity of the function is O(n-k + n-k), which simplifies to O(n), assuming k is much smaller than n.

Space Complexity

The space complexity is dominated by the space needed to store the Counter object cnt. In the worst case, if all candies elements are unique, the Counter will have up to n-k entries, leading to a space complexity of O(n-k). In the best-case scenario, if all elements are the same, the Counter will have just one entry, but since n is the determining factor (assuming k doesn't scale with n), the space complexity remains O(n).

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