Introduction

In this blog post, we will explore how to solve the LeetCode problem "2813" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

You are given a 0-indexed 2D integer array items of length n and an integer k. items[i] = [profiti, categoryi], where profiti and categoryi denote the profit and category of the ith item respectively. Let's define the elegance of a subsequence of items as total_profit + distinct_categories2, where total_profit is the sum of all profits in the subsequence, and distinct_categories is the number of distinct categories from all the categories in the selected subsequence. Your task is to find the maximum elegance from all subsequences of size k in items. Return an integer denoting the maximum elegance of a subsequence of items with size exactly k. Note: A subsequence of an array is a new array generated from the original array by deleting some elements (possibly none) without changing the remaining elements' relative order. Example 1: Input: items = [[3,2],[5,1],[10,1]], k = 2 Output: 17 Explanation: In this example, we have to select a subsequence of size 2. We can select items[0] = [3,2] and items[2] = [10,1]. The total profit in this subsequence is 3 + 10 = 13, and the subsequence contains 2 distinct categories [2,1]. Hence, the elegance is 13 + 22 = 17, and we can show that it is the maximum achievable elegance. Example 2: Input: items = [[3,1],[3,1],[2,2],[5,3]], k = 3 Output: 19 Explanation: In this example, we have to select a subsequence of size 3. We can select items[0] = [3,1], items[2] = [2,2], and items[3] = [5,3]. The total profit in this subsequence is 3 + 2 + 5 = 10, and the subsequence contains 3 distinct categories [1,2,3]. Hence, the elegance is 10 + 32 = 19, and we can show that it is the maximum achievable elegance. Example 3: Input: items = [[1,1],[2,1],[3,1]], k = 3 Output: 7 Explanation: In this example, we have to select a subsequence of size 3. We should select all the items. The total profit will be 1 + 2 + 3 = 6, and the subsequence contains 1 distinct category [1]. Hence, the maximum elegance is 6 + 12 = 7. Constraints: 1 <= items.length == n <= 105 items[i].length == 2 items[i][0] == profiti items[i][1] == categoryi 1 <= profiti <= 109 1 <= categoryi <= n 1 <= k <= n

Explanation

Here's the breakdown of the solution:

• Greedy Approach: Prioritize items with unique categories to maximize the distinct_categories^2 part of the elegance.
• Balance Profit and Category: If after selecting all unique category items, we still haven't reached size k, greedily add items with the highest profit from the remaining items, even if they duplicate categories. If we exceed k by picking all unique categories, we'll need to strategically remove the lowest-profit items from the unique categories to make room for higher-profit, potentially duplicate category items.

• Optimization: Use appropriate data structures (e.g., sorted lists, sets) to efficiently find maximum profit and manage categories.

• Time Complexity: O(n log n), primarily due to sorting. Space Complexity: O(n), for storing items and categories.

Code

    def maximumElegance(items, k):
    items.sort(key=lambda x: x[0], reverse=True)
    unique_categories = set()
    profit = 0
    duplicates = []
    elegance = 0

    for i in range(len(items)):
        if items[i][1] not in unique_categories:
            unique_categories.add(items[i][1])
            profit += items[i][0]
            k -= 1
        else:
            duplicates.append(items[i][0])

        if k == 0:
            break

    elegance = profit + len(unique_categories)**2

    if k > 0:
        duplicates.sort(reverse=True)
        for i in range(min(k, len(duplicates))):
            profit += duplicates[i]
        elegance = profit + len(unique_categories)**2
    elif k < 0:
        unique_profits = []
        categories_used = set()
        for p, c in items:
            if c in unique_categories and c not in categories_used:
                unique_profits.append(p)
                categories_used.add(c)

        unique_profits.sort()

        for i in range(-k):
          if len(duplicates) > 0 and unique_profits[i] < duplicates[-1]:
            profit -= unique_profits[i]
            profit += duplicates[-1]
            duplicates.pop()
          else:
            profit -= unique_profits[i]
        elegance = profit + len(unique_categories)**2

    return elegance