LeetCode 2813 - Maximum Elegance of a K-Length Subsequence

Difficulty: 🔴 Hard
Topics: Array, Hash Table, Stack, Greedy, Sorting, Heap (Priority Queue)

Solution

Problem Understanding

The problem presents a 0-indexed 2D array items of length n, where each element represents an item with two values: a profit and a category. You are asked to select exactly k items to form a subsequence. A subsequence preserves the original order of items but can skip elements. The goal is to maximize the elegance of the subsequence, defined as:

elegance = total_profit + (distinct_categories)^2

where total_profit is the sum of the profits of the selected items, and distinct_categories is the number of unique categories among the selected items.

The expected output is the maximum possible elegance for any subsequence of size k.

Constraints tell us that n can be as large as 10^5 and profits can be very large (<= 10^9). This implies that a naive brute-force approach, such as generating all subsequences of size k and computing their elegance, is infeasible.

Edge cases include scenarios where all items belong to the same category, k equals n, or the distribution of categories is such that selecting high-profit items may conflict with maximizing distinct categories.

Approaches

Brute-Force Approach

The brute-force method would iterate over all subsequences of size k, compute total_profit and the number of distinct categories for each, then calculate elegance. While this guarantees correctness, its time complexity is exponential, specifically $O(\binom{n}{k})$, which is far too large for n up to 10^5. This approach is impractical due to combinatorial explosion.

Optimal Approach

The key observation is that elegance is influenced by both the total profit and the square of distinct categories. To maximize elegance efficiently:

Sort all items in descending order of profit. This ensures that any greedy selection starts with the most profitable items.
Initially, select the top k items by profit. Track the total profit and the count of items per category using a hashmap.
Identify duplicates: items that share a category already counted. These are candidates to swap out for items from categories not yet in the subsequence.
Maintain a heap (or sorted list) of the lowest profit duplicates. Then iterate through remaining items, attempting to replace duplicates with items from new categories to increase distinct_categories.
Each replacement increases the distinct_categories by 1 while possibly decreasing the total_profit minimally, balancing the square term increase against potential profit loss.

This method leverages greedy selection and replacement optimization. It is efficient because each item is considered at most twice, giving linearithmic performance after sorting.

Approach	Time Complexity	Space Complexity	Notes
Brute Force	O(C(n,k) * k)	O(k)	Generates all k-sized subsequences
Optimal	O(n log n)	O(n)	Sort items by profit and optimize category count with a heap and hashmap

Algorithm Walkthrough

Sort items by profit descending. This ensures we prioritize high-profit items initially.
Select the top k items. Compute total_profit and maintain a hashmap of category counts.
Identify duplicates. For any category appearing more than once in the selection, add the excess items’ profits to a min-heap for potential replacement.
Process remaining items. Iterate over items outside the initial selection. If an item has a new category not yet in the selection, attempt to replace the lowest-profit duplicate with it.
Update elegance. Each time a replacement occurs, increment distinct_categories, adjust total_profit if necessary, and compute the new potential elegance. Track the maximum encountered.
Return the maximum elegance. After considering all possible replacements, return the best value.

Why it works: By initially selecting high-profit items and then strategically replacing duplicates with new categories, we maximize the sum of profits while increasing the square of distinct categories. The greedy choice is valid because higher profit items are always beneficial, and replacements only improve the distinct category term.

Python Solution

from typing import List
from collections import defaultdict
import heapq

class Solution:
    def findMaximumElegance(self, items: List[List[int]], k: int) -> int:
        # Step 1: Sort items by profit descending
        items.sort(key=lambda x: -x[0])
        
        total_profit = 0
        category_count = defaultdict(int)
        duplicate_profits = []
        distinct_categories = 0
        
        # Step 2: Select top k items
        for i in range(k):
            profit, category = items[i]
            total_profit += profit
            category_count[category] += 1
            if category_count[category] == 1:
                distinct_categories += 1
            else:
                heapq.heappush(duplicate_profits, profit)
        
        max_elegance = total_profit + distinct_categories ** 2
        
        # Step 3: Try replacing duplicates with new categories
        for i in range(k, len(items)):
            profit, category = items[i]
            if category_count[category] == 0 and duplicate_profits:
                removed_profit = heapq.heappop(duplicate_profits)
                total_profit += profit - removed_profit
                category_count[category] += 1
                distinct_categories += 1
                max_elegance = max(max_elegance, total_profit + distinct_categories ** 2)
        
        return max_elegance

Explanation: The code first sorts items and computes the total profit for the top k items. It uses a hashmap to track categories and a min-heap to manage duplicates. Replacement of low-profit duplicates with items from new categories increases the distinct categories count and updates the maximum elegance efficiently.

Go Solution

package main

import (
    "container/heap"
    "sort"
)

type Item struct {
    profit, category int
}

type MinHeap []int

func (h MinHeap) Len() int            { return len(h) }
func (h MinHeap) Less(i, j int) bool  { return h[i] < h[j] }
func (h MinHeap) Swap(i, j int)       { h[i], h[j] = h[j], h[i] }
func (h *MinHeap) Push(x interface{}) { *h = append(*h, x.(int)) }
func (h *MinHeap) Pop() interface{} {
    n := len(*h)
    x := (*h)[n-1]
    *h = (*h)[:n-1]
    return x
}

func findMaximumElegance(items [][]int, k int) int64 {
    n := len(items)
    // Sort by profit descending
    sort.Slice(items, func(i, j int) bool {
        return items[i][0] > items[j][0]
    })

    totalProfit := int64(0)
    categoryCount := make(map[int]int)
    duplicateHeap := &MinHeap{}
    heap.Init(duplicateHeap)
    distinctCategories := int64(0)

    for i := 0; i < k; i++ {
        profit, category := int64(items[i][0]), items[i][1]
        totalProfit += profit
        categoryCount[category]++
        if categoryCount[category] == 1 {
            distinctCategories++
        } else {
            heap.Push(duplicateHeap, int(profit))
        }
    }

    maxElegance := totalProfit + distinctCategories*distinctCategories

    for i := k; i < n; i++ {
        profit, category := int64(items[i][0]), items[i][1]
        if categoryCount[category] == 0 && duplicateHeap.Len() > 0 {
            removedProfit := int64(heap.Pop(duplicateHeap).(int))
            totalProfit += profit - removedProfit
            categoryCount[category]++
            distinctCategories++
            newElegance := totalProfit + distinctCategories*distinctCategories
            if newElegance > maxElegance {
                maxElegance = newElegance
            }
        }
    }

    return maxElegance
}

Go-specific notes: Go requires explicit heap interface implementation and type casting. Integer overflow is handled using int64 because profits can be very large. Maps and slices are used similarly to Python dictionaries and lists.

Worked Examples

Example 1

items = [[3,2],[5,1],[10,1]], k = 2

Step 1: Sort by profit → [[10,1],[5,1],[3,2]]

Step 2: Take top 2 → [[10,1],[5,1]]

total_profit = 15, distinct_categories = 1
duplicates = [5]

Step 3: Check remaining items: [3,2] → new category 2

Replace 5 with 3 → total_profit = 13, distinct_categories = 2
elegance = 13 + 2^2 = 17

Maximum elegance = 17

Example 2

items = [[3,1],[3,1],[2,2],[5,3]], k = 3

Sort → [[5,3],[3,1],[3,1],[2,2]]

Top 3 → [[5,3],[3,1],[3,1]]

total_profit = 11, distinct_categories = 2
duplicates = [3]

Remaining [