# Solving Leetcode Interviews in Seconds with AI: Maximum Elegance of a K-Length Subsequence


	# 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](https://www.chatmagic.app), 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
	```python
	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
	```