Problem Description

You are given a 2D integer array items where each element items[i] = [profit_i, category_i] represents an item with its profit and category. You need to select exactly k items to form a subsequence.

The elegance of a subsequence is calculated as: total_profit + distinct_categories²

Where:

• total_profit is the sum of all profits in the selected subsequence
• distinct_categories is the number of unique categories among the selected items

Your task is to find the maximum possible elegance by selecting exactly k items from the array.

For example, if you select items with profits [5, 3, 2] and categories [1, 1, 2], then:

• total_profit = 5 + 3 + 2 = 10
• distinct_categories = 2 (categories 1 and 2)
• elegance = 10 + 2² = 14

The key insight is that there's a trade-off between maximizing total profit and maximizing the number of distinct categories (since the latter is squared in the formula). Sometimes it's better to choose an item with lower profit if it adds a new category, as the squared term can significantly boost the elegance score.

A subsequence maintains the relative order of elements from the original array but doesn't need to be contiguous. You must select exactly k items to form your subsequence and return the maximum elegance achievable.

Intuition

The formula total_profit + distinct_categories² creates an interesting optimization problem. Since we want to maximize this value, we need to balance between high profits and category diversity.

Let's think greedily: we should start by taking the k items with the highest profits. This gives us the maximum possible total_profit initially. However, this might not be optimal because we haven't considered the squared term for distinct categories.

The key observation is that once we have our initial k items, we can potentially improve the elegance by swapping items. When should we make a swap?

Consider replacing an item from our selected k items with an item outside. This swap makes sense only if:

1. The new item introduces a new category (increasing distinct_categories by 1)
2. The item we remove is from a duplicate category (so we don't lose any distinct categories)

When we add a new category, the elegance changes by:

• Profit change: new_profit - old_profit (likely negative since we sorted by profit)
• Category bonus: (distinct_categories + 1)² - distinct_categories² = 2 * distinct_categories + 1

The quadratic term means that adding new categories becomes increasingly valuable as we have more distinct categories. This could outweigh the profit loss from swapping.

To implement this efficiently:

1. Sort items by profit (descending) and take the top k items
2. Track which items in our selection have duplicate categories - these are candidates for removal
3. For remaining items (ranked k+1 onwards), if they bring a new category, consider swapping them with the lowest-profit duplicate-category item
4. Keep track of the maximum elegance throughout this process

We only swap with the minimum profit among duplicates because that minimizes our profit loss while gaining the category diversity bonus. This greedy strategy of always removing the least profitable duplicate ensures we explore the best possible trade-offs between profit and category diversity.

Learn more about Stack, Greedy, Sorting and Heap (Priority Queue) patterns.

Solution Approach

Following our greedy intuition, here's how we implement the solution:

Step 1: Sort and Initial Selection

items.sort(key=lambda x: -x[0])

Sort all items by profit in descending order. This ensures we start with the highest profit items.

Step 2: Select Top k Items

tot = 0
vis = set()
dup = []
for p, c in items[:k]:
    tot += p
    if c not in vis:
        vis.add(c)
    else:
        dup.append(p)

• Take the first k items and calculate their total profit in tot
• Use a set vis to track which categories we've seen
• Use a list dup to store profits of items that have duplicate categories
• When we encounter a duplicate category, we add its profit to dup (these are candidates for replacement)

Step 3: Calculate Initial Elegance

ans = tot + len(vis) ** 2

Calculate the elegance with our initial selection: total profit plus the square of distinct categories.

Step 4: Consider Swapping

for p, c in items[k:]:
    if c in vis or not dup:
        continue
    vis.add(c)
    tot += p - dup.pop()
    ans = max(ans, tot + len(vis) ** 2)

For each remaining item (ranked k+1 onwards):

• Skip if its category is already in vis (no new category to add)
• Skip if dup is empty (no duplicate items to replace)
• Otherwise, this item brings a new category, so:
  • Add the new category to vis
  • Replace the highest-profit duplicate with this item: tot += p - dup.pop()
  • Note: dup.pop() removes the last element, which is the highest profit among duplicates since we added them in order
  • Update ans with the new elegance

Why This Works:

• We always maintain exactly k items in our selection
• We only swap when we can add a new category (increasing distinct_categories)
• We swap out the most expensive duplicate (last in dup), minimizing profit loss
• The algorithm explores all beneficial trades between profit and category diversity

Time Complexity: O(n log n) for sorting, where n is the number of items Space Complexity: O(k) for storing the selected items' information

Example Walkthrough

Let's walk through a concrete example to illustrate the solution approach.

Given: items = [[5,1], [4,1], [3,2], [2,3], [1,1]], k = 3

Step 1: Sort by profit (descending) After sorting: items = [[5,1], [4,1], [3,2], [2,3], [1,1]] (already sorted in this case)

Step 2: Select top k=3 items We take the first 3 items: [[5,1], [4,1], [3,2]]

• Process [5,1]: profit=5, category=1

  • tot = 5
  • vis = {1} (new category)
  • dup = [] (first occurrence of category 1)

• Process [4,1]: profit=4, category=1

  • tot = 5 + 4 = 9
  • vis = {1} (category 1 already exists)
  • dup = [4] (duplicate category, add profit to dup list)

• Process [3,2]: profit=3, category=2

  • tot = 9 + 3 = 12
  • vis = {1, 2} (new category)
  • dup = [4] (not a duplicate)

Step 3: Calculate initial elegance

• elegance = tot + len(vis)² = 12 + 2² = 16
• ans = 16

Step 4: Consider swapping with remaining items

• Consider [2,3] (profit=2, category=3):

  • Category 3 is NOT in vis = {1, 2} ✓
  • dup = [4] is not empty ✓
  • We can swap!
  • Add category 3: vis = {1, 2, 3}
  • Update profit: tot = 12 + 2 - 4 = 10 (add new item's profit, remove duplicate's profit)
  • New elegance: 10 + 3² = 19
  • Update ans = max(16, 19) = 19
  • dup is now empty

• Consider [1,1] (profit=1, category=1):

  • Category 1 is already in vis = {1, 2, 3} ✗
  • Skip this item

Final Answer: Maximum elegance = 19

The key insight: We initially selected [[5,1], [4,1], [3,2]] with elegance 16. By swapping [4,1] (a duplicate category item) with [2,3] (introduces new category 3), we sacrificed 2 profit points but gained 3² - 2² = 5 points from the category bonus, resulting in a net gain of 3 points for a final elegance of 19.

Solution Implementation

Time and Space Complexity

Time Complexity: O(n × log n)

The dominant operation in this algorithm is the sorting of the items list at the beginning with items.sort(key=lambda x: -x[0]), which takes O(n × log n) time where n is the total number of items.

After sorting:

• The first loop iterates through the first k items, taking O(k) time
• The second loop iterates through the remaining n - k items in the worst case, taking O(n - k) time
• All operations inside both loops (set operations like add, in, list operations like append, pop) take O(1) time

Since k ≤ n, the total time after sorting is O(k) + O(n - k) = O(n).

Therefore, the overall time complexity is O(n × log n) + O(n) = O(n × log n).

Space Complexity: O(n)

The algorithm uses the following additional space:

• vis: A set that can contain at most n unique category values in the worst case, requiring O(n) space
• dup: A list that can contain at most k duplicate profit values, requiring O(k) space where k ≤ n
• A few scalar variables (tot, ans, p, c) requiring O(1) space

The total auxiliary space complexity is O(n) + O(k) + O(1) = O(n) since k ≤ n.

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

Common Pitfalls

1. Incorrectly Managing the Duplicate Stack Order

The Pitfall: The most critical mistake is misunderstanding which duplicate item to replace. The code stores duplicate profits in the order they appear (highest to lowest due to initial sorting), but pop() removes the last element. Many developers mistakenly think they're removing the smallest profit duplicate, when they're actually removing the largest.

Why This Happens:

• Items are sorted by profit descending: [10, 8, 6, 4, ...]
• Duplicates are added in this order: dup = [8, 6] (if 8 and 6 are duplicates)
• dup.pop() removes 6, not 8!

The Problem: If you want to minimize profit loss when swapping, you should replace the smallest profit duplicate, not the largest. Removing the wrong item leads to suboptimal elegance scores.

Solution:

# Option 1: Use a min-heap to always pop the smallest
import heapq
duplicate_profits = []  # Will be used as min-heap

# When adding duplicates:
heapq.heappush(duplicate_profits, profit)

# When replacing:
total_profit += profit - heapq.heappop(duplicate_profits)  # Removes smallest

# Option 2: Reverse the duplicate list before popping
duplicate_profits.append(profit)
# Later, before the swapping loop:
duplicate_profits.reverse()  # Now pop() removes the smallest

2. Not Checking if Duplicates List is Empty

The Pitfall: Forgetting to check if not duplicate_profits before attempting to pop can cause an IndexError when trying to pop from an empty list.

Example Scenario:

• All initial k items have distinct categories
• duplicate_profits remains empty
• Attempting duplicate_profits.pop() crashes the program

Solution: Always check both conditions together:

if category in visited_categories or not duplicate_profits:
    continue

3. Modifying the Wrong Total When Swapping

The Pitfall: Some developers create a new variable for tracking profit during swaps but forget to update it consistently, or they update the original total but compare against a stale maximum.

Incorrect Example:

# Wrong: Creating new variable but not maintaining it properly
swap_total = total_profit
for profit, category in items[k:]:
    if category not in visited_categories and duplicate_profits:
        swap_total = total_profit + profit - duplicate_profits.pop()  # Bug: always uses original total_profit
        max_elegance = max(max_elegance, swap_total + len(visited_categories) ** 2)

Solution: Consistently update the same total_profit variable:

total_profit += profit - duplicate_profits.pop()  # Accumulative update