CPMagical CP Library

Problem solution · Python

Maximum Elegance of a K-Length Subsequence

Maximum Elegance of a K-Length Subsequence: a Python solution using sorting and greedy selection. Learn the idea, check the complexity, and read the full code, with credit to walkccc LeetCode Solutions.

Technique
Sorting and greedy selection
Source
walkccc LeetCode Solutions
Length
33 lines
Start with the idea.

Try the problem first. If you get stuck, read the approach below, then write your own solution. The full code is at the bottom.

Approach

Sorting and greedy selection

For Maximum Elegance of a K-Length Subsequence, the implementation first exposes a useful order, then scans that order while making locally justified choices.

  1. Choose the key that reveals the greedy or grouping structure.
  2. Sort the relevant records by that key.
  3. Scan in order, maintaining the invariant that makes each local choice safe.

Code notes

  • 33 lines of Python from the credited upstream file 2813.py.
  • The implementation visibly relies on sequence storage, ordered lookup.
  • No explicit loop blocks detected.

Complexity

Sorting is typically the dominant term unless the subsequent scan uses a more expensive nested operation.

Check the problem constraints before deciding whether this complexity will pass.

Source

Code and credit

This code comes from walkccc LeetCode Solutions by P.-Y. Chen (walkccc) and is used under the MIT licence.

Open problem ↗Original source ↗
Full codeMaximum Elegance of a K-Length Subsequence · PythonPython
Use this to learn the idea, then write your own version.
class Solution:  def findMaximumElegance(self, items: list[list[int]], k: int) -> int:    ans = 0    totalProfit = 0    seenCategories = set()    decreasingDuplicateProfits = []     items.sort(reverse=True)     for i in range(k):      profit, category = items[i]      totalProfit += profit      if category in seenCategories:        decreasingDuplicateProfits.append(profit)      else:        seenCategories.add(category)     ans = totalProfit + len(seenCategories)**2     for i in range(k, len(items)):      profit, category = items[i]      if category not in seenCategories and decreasingDuplicateProfits:        # If this is a new category we haven't seen before, it's worth        # considering taking it and replacing the one with the least profit        # since it will increase the distinct_categories and potentially result        # in a larger total_profit + distinct_categories^2.        totalProfit -= decreasingDuplicateProfits.pop()        totalProfit += profit        seenCategories.add(category)        ans = max(ans, totalProfit + len(seenCategories)**2)     return ans

Keep practising

Did this explanation save you time? I'm a Grade 11 student building this free library to make difficult algorithms easier to understand.

Buy me a coffee ↗