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Maximum Elegance of a K-Length Subsequence

Difficulty: Hard

Problem Description

You are given a 0-indexed 2D integer array items of length n and an integer k. Each item is represented by a pair of values indicating profit and category. The elegance of a subsequence of items is defined as total_profit + distinct_categories^2, where total_profit is the sum of profits in the subsequence and distinct_categories is the number of distinct categories in the selected subsequence. Your task is to find the maximum elegance from all subsequences of size k in items.

Key Insights

  • Elegance combines both the total profit and the square of the number of distinct categories.
  • To maximize elegance, we need to carefully select items that not only provide high profits but also maximize the number of distinct categories.
  • A greedy approach can be effective, where we sort items based on profit and use a priority queue (or a set) to maintain distinct categories.
  • The constraints suggest that an efficient O(n log n) solution is necessary due to the potential size of n (up to 100,000).

Space and Time Complexity

Time Complexity: O(n log n) - due to sorting the items. Space Complexity: O(n) - to store the items and maintain distinct categories.

Solution

To solve this problem, we can follow these steps:

  1. Sort the items based on profit in descending order.
  2. Use a min-heap (or a set) to keep track of selected items and their categories.
  3. Iterate through the sorted items and select up to k items, ensuring we maximize the number of distinct categories.
  4. Calculate the total profit and the number of distinct categories for the selected items to compute the elegance.

Code Solutions

def maxElegance(items, k):
    from collections import defaultdict
    import heapq

    # Sort items based on profit in descending order
    items.sort(reverse=True, key=lambda x: x[0])
    
    total_profit = 0
    distinct_categories = 0
    category_count = defaultdict(int)
    selected_items = []
    
    for profit, category in items:
        if len(selected_items) < k:
            selected_items.append((profit, category))
            total_profit += profit
            if category_count[category] == 0:
                distinct_categories += 1
            category_count[category] += 1
        else:
            # If we already have k items, we need to consider replacing the smallest profit item
            min_profit_item = min(selected_items)
            if profit > min_profit_item[0]:
                # Replace the item with the smallest profit
                total_profit += profit - min_profit_item[0]
                category_count[min_profit_item[1]] -= 1
                if category_count[min_profit_item[1]] == 0:
                    distinct_categories -= 1
                selected_items.remove(min_profit_item)
                selected_items.append((profit, category))
                if category_count[category] == 0:
                    distinct_categories += 1
                category_count[category] += 1

    # Calculate elegance
    elegance = total_profit + distinct_categories ** 2
    return elegance
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